Class 11

Odisha State Board Elements of Mathematics Class 11 Solutions CHSE Odisha Chapter 8 Permutations and Combinations Ex 8(c) Textbook Exercise Questions and Answers.

CHSE Odisha Class 11 Math Solutions Chapter 8 Permutations and Combinations Exercise 8(c)

Question 1.
Compute the following :
(i) ¹²C₃
Solution:
¹²C₃ = $\frac{(12) !}{3 ! 9 !}=\frac{12 \cdot 11 \cdot 10}{3 \cdot 2}$ = 220

(ii) ¹⁵C₁₂
Solution:
¹⁵C₁₂ = $\frac{(15) !}{(12) ! 3 !}=\frac{15 \cdot 14 \cdot 13}{3 \cdot 2}$
= 5.7.13 = 455

(iii) ⁹C₄ + ⁹C₅
Solution:
⁹C₄ + ⁹C₅ = $\frac{9 !}{4 ! 5 !}+\frac{9 !}{5 ! 4 !}$
$=\frac{9 \cdot 8 \cdot 7 \cdot 6}{4 \cdot 3 \cdot 2 \cdot 1}$ × 2 = 252

(iv) ⁷C₃ + ⁶C₄ + ⁶C₃
Solution:
⁷C₃ + ⁶C₄ + ⁶C₃ = ⁷C₃ + ⁶C₄ + ⁶C4-1
= ⁷C₃ + ⁶⁺¹C₄ = ⁷C₃ + ⁷C₄
(∴ ⁿc_r + ⁿC_r-1– = ⁿ⁺¹c_r)
= ⁷C₄ + ⁷C_4-1 = ⁷⁺¹C₄
= ⁸C₄ = $\frac{8 !}{4 !(8-4) !}=\frac{8 \cdot 7 \cdot 6 \cdot 5}{4 \cdot 3 \cdot 2 \cdot 1}$ = 70

(v) ⁸C₀ + ⁸C₁ + …….. + ⁸c₈
Solution:
⁸C₀ + ⁸C₁ + …….. + ⁸c₈ = 2⁸ = 256

Question 2.
Solve :
(i) ⁿC₄ = ⁿC₁₁ ;
Solution:
ⁿC₄ = ⁿC₁₁ ;  (∴ n = 4 + 11 = 15)

(ii) ²ⁿC₃ : ⁿC₃ = 44: 5
Solution:
²ⁿC₃ : ⁿC₃ = $\frac{44}{5}$
⇒ $\frac{2 n !}{2 n-3 !} / \frac{n !}{n-3 !}=\frac{44}{5}$

Question 3.
Find n and r if ⁿP_r = 1680, ⁿC_r = 70.
Solution:
ⁿP_r = 1680, ⁿC_r = 70
∴ $\frac{{ }^n \mathrm{P}_r}{{ }^n \mathrm{C}_r}=\frac{1680}{70}$
or, r ! = 24 = 4!
∴ r = 4
Again, ⁿC_r = 70 or ⁿC₄ = 70
or, $\frac{n !}{4 !(n-4) !}=70$
or, n(n – 1) (n – 2) (n – 3)
= 70 × 4! = 7 × 10 × 4 × 3 × 2
= 8 × 7 × 6 × 5
or, n(n – 1) (n – 2) (n – 3)
or, 8(8 – 1) (8 – 2) (8 – 3)
∴ n = 8

Question 4.
How many diagonals can an n-gon(a polygon with n sides) have?
Solution:
A polygon of n – sides has n vertices.
∴ The number of st. lines joining the n-vertices is ⁿC₂.
∴ The number of diagonals is ⁿC₂ – n

Question 5.
If a set A has n elements and another set B has m elements, what is the number of relations from A to B?
Solution:
If |A| = n, |B| = m
then |A × B| = mn
∴ The number of possible subsets of
A × B = 2^mn
∴ The number of relations from A to B is 2^mn.

Question 6.
From five consonants and four vowels, how many words consist of three consonants and two vowels?
Solution:
Words of consisting of 3 consonants and 2 vowels are to be formed from five consonants and 4 vowels.
∴ The number of ways = ⁵C₃ × ⁴C₂
Again, 5 letters can be arranged among themselves in 5! ways.
∴ The total number of ways
= ⁵C₃ × ⁴C₂ × 5! = 10 × 6 × 120 = 7200.

Question 7.
In how many ways can a committee of four gentlemen and three ladies be formed out of seven gentlemen and six ladies?
Solution:
A committee of 4 gentlemen and 3 ladies is to be formed out of 7 gentlemen and 6 ladies.
∴ The number of ways in which the committee can be formed.
⁷C₄ × ⁶C₃ = $\frac{7 \cdot 6 \cdot 5}{3.2} \times \frac{6 \cdot 5 \cdot 4}{3 \cdot 2}$ = 700

Question 8.
A bag contains 4 black and 5 white balls out of which 6 balls are drawn arbitrarily. In how many ways can this be done? Find also the number of ways such that at least 3 black balls can be drawn.
Solution:
A bag contains 4 black and 5 white balls out of which 6 balls are drawn arbitrarily.
∴ The number of ways in which balls are drawn ${ }^9 \mathrm{C}_6=\frac{9 \cdot 8 \cdot 7}{3 \cdot 2 \cdot 1}$ = 84 as the total number of balls is 9. If at least 3 black balls are drawn, then the drawing can be made as follows.

Black(4)	White(5)
3	3
4	2

The number of ways in which at least 3 black balls are drawn
= (⁴C₃ × ⁵C₃) + (⁴C₄ × ⁵C₂)
= (4 × 10) + (1 × 10) = 50

Question 9.
How many triangles can be drawn by joining the vertices of a decagon?
Solution:
A decagon has 10 vertices and 3 noncollinear points are required to be a triangle.
∴ The number of triangles formed by the joining of the vertices of a decagon is
¹⁰C₃ = $\frac{10 !}{3 ! 7 !}=\frac{10 \cdot 9 \cdot 8}{3 \cdot 2 \cdot 1}$ = 120

Question 9.
How many triangles can be drawn by joining the vertices and the center of a regular hexagon?
Solution:
A regular hexagon has six vertices. Triangles are to be formed by joining the vertices and center of the hexagon. So there is a total of 7 points. So the number of triangles formed.
⁷C₃ = $\frac{7 !}{3 ! 4 !}=\frac{7 \cdot 6 \cdot 5}{3 \cdot 2 \cdot 1}$ = 35
As the hexagon has 3 main diagonals, which pass through the center hence can not form 3 triangles.
∴ The required number of triangles 35 – 3 = 32

Question 11.
Sixty points lie on a plane, out of which no three points are collinear. How many straight lines can be formed by joining pairs of points?
Solution:
Sixty points lie on a plane, out of which no. 3 points are collinear. A straight line required two points. The number of straight lines formed by joining 60 points is
⁶⁰C₂ = $\frac{60 !}{2 \times 58 !}=\frac{60 \times 59}{2}$ = 1770

Question 12.
In how many ways can 10 boys and 10 girls sit in a row so that no two boys sit together?
Solution:
10 boys and 10 girls sit in a row so that no two boys sit together. So a boy is to be seated between two girls or at the two ends of the row. So the boys are to be sitted in 11 positions in ¹¹C₁₀ ways. Again 10 boys and 10 girls can be arranged among themselves in 10! and 10! ways respectively.
∴ The total number of ways = ¹¹C₁₀ × 10! × 10! = (11)! × (10)!

Question 13.
In how many ways can six men and seven girls sit in a row so that the girls always sit together?
Solution:
Six men and seven girls sit in a row so that the girls always sit together. Considering the 7 girls as one person, there are a total of 7 persons who can sit in 7! ways. Again the 7 girls can be arranged among themselves in 7! ways.
∴ The total number of arrangements
= 7! × 7!
= (7!)²

Question 14.
How many factors does 1155 have that are divisible by 3?
Solution:

∴ In order to be a factor of 1155 divisible by 3, we have to choose one or two of 5, 7, and 11 along with 3 or 3 alone.
∴ The number of ways = ³C₁ + ³C₂ + ³C₀ = 2³ – 1 = 7
∴ The number of factors is 7 excluding 1155 itself.

Question 15.
How many factors does 210 have?
Solution:

∴ We can choose at least one 2, 3, 5, or 7  to be a factor of 210.
∴ The number of factors.
= ⁴C₁ + ⁴C₂ + ⁴C₃ + ⁴C₄ = 2⁴ – 1 = 15
∴ The number of factors is 210 is 15. (Including 215 itself and excluding 1).

Question 16.
If n is a product of k distinct primes what is the total number of factors of n?
Solution:
n is a product of k distinct primes.
∴ In order to be a factor, of n, we have chosen at least one of k distinct primes.
∴ The number of ways = ^kC₁ + ^kC₂ + ……… ^kC_k-1 = 2^k  – 1 – 1
∴ The number of factors of n is 2^k – 2.
(Excluding 1 as 1 is not prime. It is also not include n.)

Question 17.
If m has the prime factor decomposition P₁^r1, P₂^r2 ….. P_n^rn, what is the total number of factors of m (excluding 1)?
Solution:
m has the prime factor decomposition P₁^r1, P₂^r2 ….. P_n^rn,
∴ m = P₁^r1, P₂^r2 ….. P_n^rn,
P₁ is a factor of m which occurs r₁ times. Each of the factors P₁^r1 will give rise to (r₁ + 1) factors.
Similarly
P₂^r2 gives (r₂ + 1) factors and so on.
∴ The total number of factors (r₁ + 1) (r₂ + 1) ….. (r_n + 1) – 1 (including m).

Question 18.
If 20! were multiplied out, how many consecutive zeros would it have on the right?
Solution:
If 20! were multiplied out, then the number of consecutive zeros on the right is 4. due to the presence of 4 x 5, 10, 14 x 15,20.

Question 19.
How many factors of 10,000 end with a 5 on the right?
Answer:
We have 1000 = 2⁴ × 5⁴ The factors of 10000 ending with 5 are 5, 5 × 5 = 25, 5 × 5 × 5 = 125
5 × 5 × 5 × 5 = 625
∴ There are 4 factors ending With 5.

Question 20.
A man has 6 friends. In how many ways can he invite two or more to a dinner party?
Solution:
A man has 6 friends. He can invite 2 or more of his friends to a dinner party.
∴ He can invite 2, 3, 4, 5, or 6 of his friends in
⁶C₂ + ⁶C₃ + ⁶C₄ + ⁶C₅ + ⁶C₆ = 2⁶ – ⁶C₀ – ⁶C₁
= 64 – 7 = 57 ways.

Question 21.
In how many ways can a student choose 5 courses out of 9 if 2 courses are compulsory?
Solution:
A student is to choose 5 courses out of 9 in which 2 courses are compulsory. as 2 courses are compulsory, he is to choose 3 courses out of 7 courses in ⁷C₃ = 35 ways.

Question 22.
In how many ways can a student choose five courses out of the courses? C₁, C₂, …………. C₉ if C₁, C₂ are compulsory and C₆, C₈ cannot be taken together?
Solution:
A student chooses five courses out of the courses? C₁, C₂, …………. C₉ if C₁, C₂ are compulsory and C₆, C₈ cannot be taken together.
∴ He is to choose 3 courses out of C₃, C₄, ………… ,C₈, C₉.
Without taking any restrictions 3 courses out of C₃, C₄, …… C₉, i.e. from 7 courses in ⁷C₃ ways. If C₆, C₈ are taken together then one course only to be choosen from C₃, C₄, C₅, C₇, C₉ by ⁵C₁ ways. Hence required number of ways.
= ⁷C₃ – ⁵C₁
= $\frac{7 \times 6 \times 5}{3 \times 2 \times 1}$ – 5
= 35 – 5 = 30 ways.

Question 23.
A cricket team consisting of 11 players is to be chosen from 8 batsmen and 5 bowlers. In how many ways can the team be chosen so as to include at least 3 bowlers?
Solution:
A cricket team consisting of 1 1 player is to be chosen from 8 batsmen and 5 bowlers, as to include at least 3 bowlers.
The selection can be made as follows :

Batsmen(8)	Bowlers(5)
8	3
7	4
6	5

The number of selections is
(⁸C₈ × ⁵C₃) + (⁸C₇ × ⁵C₄) + (⁸C₆ × ⁵C₅)
= (1 × 10) + (8 × 5) + (28 × 1)
= 10 + 40 + 28 = 78

Question 24.
There are n + r points on a plane out of which n points lie on a straight line L and out of the remaining r points that lie outside L, no three points are collinear. What is the number of straight lines that can be formed by joining pairs of their points?
Solution:
There are n + r points on a plane out of which n points are collinear and out of which r points are not collinear.
∴ We can form a straight line by joining any two points.
n-collinear points form one line and r-non-collinear points form ^rC₂ lines.
Again, each of the r non-collinear points when joined to each of the noncollinear points, forms n lines.
∴ The number of such limes is r x n.
∴ The total number of lines

Question 25.
There are 10 books in a shelf with different titles; five of these have red covers and others have green covers. In how many ways can these be arranged so that the red books are placed together?
Solution:
There are 10 books in a shelf with different titles, 5 of these are red covers and others are green covers considering 5 red-covered books as one book, we have a total of 6 books which can be arranged in 6! ways. The five red cover books are arranged among themselves in 5! ways.
∴ The total number of arrangements
= 5! x 6!

Odisha State Board Elements of Mathematics Class 11 Solutions CHSE Odisha Chapter 11 Straight Lines Ex 11(b) Textbook Exercise Questions and Answers.

CHSE Odisha Class 11 Math Solutions Chapter 11 Straight Lines Exercise 11(b)

Question 1.
Fill in the blanks in each of the following, using the answers given against each of them :
(a) The slope and x-intercept of the line 3x – y + k = 0 are equal if k = _________ . (0, -1, 3, -9)
Solution:
-9

(b) The lines 2x – 3y + 1 = 0 and 3x + ky – 1=0 are perpendicular to each other if k = ___________ . (2, 3, -2, -3)
Solution:
2

(c) The lines 3x + ky – 4 = 0 and k – Ay – 3x = 0 are coincident if k = _____________. (1, -4, 4, -1)
Solution:
4

(d) The distance between the lines 3x – 1 = 0 and x + 3 = 0 is _________ units. (4, 2, $\frac{8}{3}$, $\frac{10}{3}$)
Solution:
$\frac{10}{3}$

(e) The angle between the lines x = 2 and x – √3y + 1 = 0 is _________. (30°, 60°, 120°, 150°)
Solution:
60°

Question 2.
State with reasons which of the following are true or false :
(a) The equation x = k represents a line parallel to x – axis for all real values of k.
Solution:
False. As the line x = k is parallel to y- axis for all values of k.

(b) The line, y + x + 1 = 0 makes an angle 45° with y – axis.
Solution:
y + x + 1 = 0
∴ Its slope = -1 = tan 135°
∴ It makes 45° with y – axis, as it makes 135° with x – axis. (True)

(c) The lines represented by 2x – 3y + 1 = 0 and 3x + 2y – k = 0 are perpendicular to each other for positive values of k only.
Solution:
2x – 3y + 1 = 0, 3x + 2y – k = 0
∴ $m_1 m_2=\frac{2}{3} \times \frac{(-3)}{2}=-1$
∴ The lines are perpendicular to each other for + ve values of k only. (False)

(d) The lines represented by px + 2y – 1 = 0 and 3x + py + 1 = 0 are not coincident for any value of ‘p’.
Solution:
px + 2y – 1 = 0, 3x + py + 1=0
∴ $\frac{p}{3}=\frac{2}{p}=\frac{-1}{1} \Rightarrow p^2=6$
and p = -3 or -2
There is no particular value of p for which $\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}$ (True)

(e) The equation of the line whose x and y – intercepts are 1 and -1 respectively is x – y + 1 = 0.
Solution:
Equation of the line whose intercepts 1 and -1 is $\frac{x}{1}+\frac{y}{-1}$ = 1
or, x – y = 1 (False)

(f) The point (-1, 2) lies on the line 2x + 3y – 4 = 0.
Solution:
Putting x = – 1, y = 2
we have 2 (- 1) + 3 × 2 – 4
= -2 + 6 – 4 = 0
∴ The point (-1, 2) lies on the line 2x + 3 – 4 = 0 (True)

(g) The equation of a line through (1, 1) and (-2, -2) is y = – 2x.
Solution:
The equation of the line through (1, 1) and (-2, -2) is y – y₁ = $\frac{y_2-y_1}{x_2-x_1}$ (x – x₁)
or, y – 1 = $\frac{-2-1}{-2-1}$ (x – 1)
or, y – 1 = x – 1
or, x – y = 0 (False)

(h) The line through (1, 2) perpendicular to y = x is y + x – 2 =0.
Solution:
The slope of the line y = x is 1.
∴ The slope of the line perpendicular to the above line is -1.
∴ The equation of the line through (1, 2) having slope – 1 is y – y₁ = m(x – x₁)
or, y – 2 = -1 (x – 1)
or, y – 2= -x + 1
or, x + y = 3 (False)

(i) The lines $\frac{x}{a}+\frac{y}{b}$ = 1 and $\frac{y}{a}-\frac{x}{b}$ = 1 are intersecting but not perpendicular to each other.
Solution:
$\frac{x}{a}+\frac{y}{b}$ = 1 and $\frac{y}{a}-\frac{x}{b}$ = 1
∴ $m_1 m_2=\frac{\left(-\frac{1}{a}\right)}{\frac{1}{b}} \times \frac{\left(-\frac{1}{b}\right)}{\left(-\frac{1}{a}\right)}=-1$
∴ The lines intersect and are perpendicular to each other. (False)

(j) The points (1, 2) and (3, – 2) are on the opposite sides of the line 2x + y = 1.
Solution:
2x + y = 1
Putting x = 1, y = 2,
we have 2 × 1 + 2 = 4 > 1
Putting x – 3, y = -2,
we have 2 × 3 – 2 – 4 > 1
∴ Points (1, 2) and (3, – 2) lie on the same side of the line 2x + y = 1 (False)

Question 3.
A point P (x, y) is such that its distance from the fixed point (α, 0) is equal to its distance from the y – axis. Prove that the equation of the locus is given by, y² = α (2x – α).
Solution:

Question 4.
Find the locus of the point P (x, y) such that the area of the triangle PAB is 5, where A is the point (1, -1) and B is the tie point (5, 2).
Solution:

= $\frac{1}{2}$ (-3x + 4y + 7) = 5
or, – 3x + 4y + 7 = 10
or, 3x – 4y + 3 =0 which is the locus of the point P (x, y).

Question 5.
A point is such that its distance from the point (3, 0) is twice its distance from the point (-3, 0). Find the equation of the locus.

Question 6.
Obtain the equation of straight lines:
(a) Passing through (1, – 1) and making an angle 150°.
Solution:
The slope of the line
= tan 150° = –$\frac{1}{\sqrt{3}}$
∴ The equation of the line is y – y₁ = m(x – x₁)
or, y + 1 = –$\frac{1}{\sqrt{3}}$ (x – 1)
or, y√3 + √3 = -x + 1
or, x + y√3 + √3 – 1 = 0

(b) Passing through (-1, 2) and making intercept 2 on the y-axis.
Solution:
Let the equation of the line be
y – mx + c or, y = mx + 2
∴ As the line passes through (-1, 2)
we have 2 = – m + 2, or, m = 0
∴ Equation of the line is y = 2.

(c) Passing through the points (2, 3) and (-4, 1).
Solution:
The equation of the line is

(d) Passing through (- 2, 3) and a sum of whose intercepts in 2.
Solution:
Let the equation of the line be $\frac{x}{a}+\frac{y}{b}$ = 1 where a + b = 2     …….(1)
Again, as the line passes through the point (-2, 3), we have $\frac{-2}{a}+\frac{3}{b}$ = 1      ………(2)
From (1), we have a= 2 – b
∴ From (2) $\rightarrow \frac{-2}{2-b}+\frac{3}{b}=1$
or, – 2b + 6 – 3b = (2 – b)b
or, 6 – 5b = 2b – b²
or, b² – 7b + 6 = 0
or, (b – 6)(b – 1) = 0
∴ b = 6, 1
∴ a =2 – b = 2 – 6 = -4
or, 2 – 1 = 1
∴ Equation of the lines are $\frac{x}{-4}+\frac{y}{6}$ = 1 or $\frac{x}{1}+\frac{y}{1}$ = 1
i, e. -3x + 2y = 12 or, x + y = 1

(e) Whose perpendicular distance from the origin is 2 such that the perpendicular from the origin has indication 150°.
Solution:
Here p = 2, α = 150°
The equation of the line in normal form is x cos α + y sin α = p
or, x cos 150° + y sin 150° = 2
or, $\frac{-x \sqrt{3}}{2}+y \cdot \frac{1}{2}$ = 2
or, -x √3 + y = 4
or, x√3 – y + 4 = 0

(f) Bisecting the line segment joining (3, – 4) and (1, 2) at right angles.
Solution:
The slope of the line $\overline{\mathrm{AB}}$ is

(g) Bisecting the line segment joining, (a, 0) and (0, b) at right angles.
Solution:
Refer to (f)

(h) Bisecting the line segments joining (a, b), (a’, b’) and (-a, b), (a’, -b’).
Solution:

(i) Passing through the origin and the points of trisection of the portion of the line 3x + y – 12 = 0 intercepted between the coordinate axes.
Solution:

(j) Passing through (-4, 2) and parallel to the line 4x – 3y = 10.
Solution:
Slope of the line 4x – 3y = 10 is $\frac{-4}{-3}=\frac{4}{3}$
∴ The slope of the line parallel to the above line is $\frac{4}{3}$.
∴ Equation of the line through (- 4, 2) and having slope $\frac{4}{3}$ is y – y₁ = m(x – x₁)
or, y – 2 = $\frac{4}{3}$ (x + 4)
or, 3y – 6 = 4x + 16
or, 4x – 3y + 22 = 0

(k) Passing through the point (a cos³ θ, a sin³ θ) and perpendicular to the straight line x sec θ + y cosec θ = α.
Solution:
The slope of the line x sec θ + y cosec θ = a is $\frac{-\sec \theta}{{cosec} \theta}$ = -tanθ
∴ Slope of the required line  = cot θ
∴ Equation of the line through (a cos³ θ, a sin³ θ) is y – y₁ = m(x – x₁)
or, y – a sin³ θ = cot θ(x – a cos³ θ)
or, y – a sin³ θ = $\frac{\cos \theta}{\sin \theta}$ (x – a cos³ θ)
or y sin θ – a sin⁴ θ = x cos  θ – a cos⁴ θ
or (x cos θ – y sin θ) + a(sin⁴ θ – cos⁴ θ) = 0

(l) Which passes through the point (3, -4) and is such that its portion between the axes is divided at this point internally in the ratio 2: 3.
Solution:

(m) which passes through the point (α, β) and is such that the given point bisects its portion between the coordinate axis.
Solution:

x = 2α , y = 2β
∴ Equation of the line $\overleftrightarrow{\mathrm{AB}}$ is $\frac{x}{2 \alpha}+\frac{y}{2 \beta}$ = 1(Intercept form)

Question 7.
(a) Find the equation of the lines that is parallel to the line 3x + 4y + 7 = 0 and is at a distance 2 from it.
Solution:
3x + 4y + 7 = 0
or, $\frac{3 x}{5}+\frac{4 y}{5}+\frac{7}{5}$ = 0(Normal form)
∴ Equation of the lines parallel to the above line and 2 units away from it are $\frac{3 x}{5}+\frac{4 y}{5}+\frac{7}{5}$ ± 2 = 0
or, 3x + 4y + 7 ± 10 = 0
∴ 3x + 4y + 17 = 0 and 3x + 4y – 3 = 0

(b) Find the equations of diagonals of the parallelogram formed by the lines ax + by = 0, ax + by + c = 0, lx + my = 0, and lx + my + n = 0. What is the condition that this will be a rhombus?
Solution:

(c) Find the equation of the line passing through the intersection of 2x – y – 1 = 0 and 3x – 4y + 6 = 0 and parallel to the line x + y – 2 = 0.
Solution:
Let the equation of the required line be (2x – y – 1) + λ(3x – 4y + 6) = 0
or, x(2 + 3λ) + λ(-1 – Aλ) + 6λ – 1 = 0
As this line is parallel to the line x + y – 2 = 0
we have their slopes are equal.
∴ $-\left(\frac{2+3 \lambda}{-1-4 \lambda}\right)=\frac{-1}{1}$
or, 2 + 3λ = -1 – 4λ
or, 7λ = -3 or, λ = $\frac{-3}{7}$
∴ Equation of the line is (2x – y – 1) – $\frac{3}{7}$ (3x – 4y + 6) = 0
or, 14x – 7y – 1 – 9x + 12y – 18 = 0
or, 5x + 5y – 25= 0
or, x + y = 5

(d) Find the equation of the line passing through the point of intersection of lines x + 3y + 2 = 0 and x – 2y – 4 = 0 and perpendicular to the line 2y + 5x – 9 = 0.
Solution:
Let the equation of the line be (x + 3y + 2) + λ(x – 2y – 4) = 0
or, x(1 + λ) + y(3 – 2λ) + 2 – 4λ = 0
As this line is perpendicular to the line 2y + 5x – 9 = 0.
We have the product of their slopes is -1.
∴ $\frac{1+\lambda}{3-2 \lambda} \times \frac{5}{2}$ = -1
or, 5 + 5λ = – 6 + 4λ
or, λ = -6 – 5 = -11.
∴ Equation of the required line is (x + 3y + 2) – 11(x – 2y – 4) = 0
or, x + 3y + 2- 11x + 22y + 44 = 0
or, – 10x + 25y + 46 = 0
or, 10x – 25y – 46 = 0

(e) Find the equation of the line passing through the intersection of the lines x + 3y – 1 = 0 and 3x – y + 1 = 0 and the centroid of the triangle whose vertices are the points (3, -1) (1, 3) and (2, 4).
Solution:
Let the equation of the required line (x + 3y – 1) + λ(3x – y + 1) = 0   … (1)
Again, the centroid of the triangle with vertices (3, – 1), (1, 3), and (2, 4) is $\left(\frac{3+1+2}{3}, \frac{-1+3+4}{3}\right)$ = (2, 2)
As line (1) passes through (2, 2), we have (2 + 6 – 1) +1(6 – 2 + 1) = 0
or, 7 + 5λ = 0 or, λ = $\frac{-7}{5}$
∴ Equation of the line (x + 3y – 1) – $\frac{7}{5}$ (3x – y + 1) = 0
or, 5x + 15y – 5 – 21x + 7y – 7 = 0
or, 22y – 16x – 12 = 0
or, 11y – 8x – 6 = 0
or, 8x – 11y + 6 = 0

Question 8.
If lx + my + 3 = 0 and 3x – 2y – 1 = 0 represent the same line, find the values of l and m.
Solution:
lx + my + 3 = 0 and 3x – 2y – 1 = 0 represents the same line
∴ $\frac{l}{3}=\frac{m}{-2}=\frac{3}{-1}$
∴ l = -9, m = 6

Question 9.
Find the equation of sides of a triangle whose vertices are at (1, 2), (2, 3), and (-3, -5).
Solution:
Equation of $\overline{\mathrm{AB}}$ is $y-y_1=\frac{y_2-y_1}{x_2-x_1}\left(x-x_1\right)$
$y-2=\frac{3-2}{2-1}(x-1)$
or, y – 2 = x – 1
or, x – y + 1 = 0

Question 10.
Show that origin is within the triangle whose sides are given by equations, 3x – 2y = 1, 5x + 3y + 11 = 0, and x – 7y + 25 = 0.
Solution:



∴ The origin lies within the triangle ABC.

Question 11.
(a) Find the equations of straight lines passing through the point (3, -2) and making an angle 45° with the line 6x + 5y = 1.

or, 11x – y = 35, x + 11y + 19 = 0

(b) Two straight lines are drawn through the point (3, 4) inclined at an angle 45° to the line x – y – 2 = 0. Find their equations and obtain area included by the above three lines.
Solution:

Slope of L₂ = 0
Then as L₂ ⊥ L₃
Slope of L₃ = ∞
∴ Equation of L₂ is
y – y₁ =m(x – x₁)
or, y – 4 = 0(x – 3) = 0
or, y = 4
∴ Equation of L₃ is y – 4 = ∞ (x – 3)
or, x – 3 = 0 or, x = 3
∴ Sloving L₁ and L₂, we have
x – y – 2 = 0, y = 4
or, x = 6
The coordinates of A are (6, 4).
Again solving L1 and L3, we have
x – y – 2 = 0, x = 3
or, y = x – 2 = 3 – 2 = 1
∴ The coordinates of B are (3, 1).
Area of the triangle PAB is

(c) Show that the area of the triangle formed by the lines given by the equations y = m₁x + c₁,y = m₂x + c_2, and x = 0 is $\frac{1}{2} \frac{\left(c_1-c_2\right)^2}{\left[m_2-m_1\right]}$
Solution:

Question 12.
Find the equation of lines passing through the origin and perpendicular to the lines 3x + 2y – 5 = 0 and 4x + 3y = 7. Obtain the coordinates of the points where these perpendiculars meet the given lines. Prove that the equation of a line passing through these two points is 23x + 11y – 35 = 0.
Solution:
The slopes of the line 3x + 2y – 5 = 0 and 4x + 3y = 7 are $\frac{-3}{2}$ and $\frac{-4}{3}$
∴ Slopes of the lines perpendicular to the above lines are $\frac{2}{3}$ and $\frac{3}{4}$
∴ Equation of the lines through the origin and having slopes $\frac{2}{3}$ and $\frac{3}{4}$
y = $\frac{2x}{3}$ and y = $\frac{3x}{4}$
Now solving 3x + 2y – 5 = 0 and y = $\frac{2x}{3}$
we have 3x + $\frac{4x}{3}$ – 5 = 0
or, 9x + 4x – 15 = 0
or, x = $\frac{15}{13}$
∴ y = $\frac{2 x}{3}=\frac{2}{3} \times \frac{15}{13}=\frac{10}{13}$
∴ The perpendicular y = $\frac{2 x}{3}$ meets the line 3x + 2y – 5 = 0 at $\left(\frac{15}{13}, \frac{10}{13}\right)$
Again, solving 4x + 3y = 7 and y = $\frac{3 x}{4}$
we have 4x + 3 × $\frac{3 x}{4}$ = 7
or, 16x + 9x = 28 or, x = $\frac{28}{25}$

Question 13.
(a) Find the length of a perpendicular drawn from the point (-3, -4) to the straight line whose equation is 12x – 5y + 65 = 0.
Solution:
The length of the perpendicular drawn from the point (- 3, -4) to the straight line 12x – 5y + 65 = 0 is

(b) Find the perpendicular distances of the point (2, 1) from the parallel lines 3x – 4y + 4 = 0 and 4y – 3x + 5 = 0. Hence find the distance between them.
Solution:
The distance of the point (2, 1) from the line 3x – 4y + 4 = 0 is $\left|\frac{3 \times 2-4 \times 1+4}{\sqrt{9+16}}\right|=\frac{6}{5}$
Again distance of the point (2, 1) from the line 4y – 3x + 5 = 0 is

(c) Find the distance of the point (3, 2) from, the line x + 3y – 1 = 0, measured parallel to the line 3x – 4y + 1 = 0
Solution:

Let the coordinates of M be (h, k).
As $\overline{\mathrm{PM}} \| \mathrm{L}_1$, We have their slopes are equal.

(d) Find the distance of the point (-1, -2) from the line x + 3y – 7 = 0, measured parallel to the line 3x + 2y – 5 = 0
Solution:
Slope of the line 3x + 2y – 5 = 0 is $\left(-\frac{3}{2}\right)$
Equation of the line through (-1, -2) and parallel to this line is y + 2 = – $\frac{3}{2}$ (x + 1)
⇒ 2y + 4 = -3x – 3
⇒ 3x + 2y + 7 = 0 …(1)
Given line is : x + 3y – 7 = 0    ….(2)
from (1) and (2) we get 7y – 28 = 0 Py = 4 and x = $-\frac{35}{7}$ = -5
Thus the required distance is $\sqrt{(-1+5)^2+(-2-4)^2} \quad=\sqrt{16+36}$
= √52 = 2√3 units.

(e) Fine the distance of the line passing through the points (a cos α, a sin α) and (a cos β, a sin β) from the origin.
Solution:
The equation of the line passing through the points (a cos α, a sin α) and (a cos β, a sin β) is y – y₁ = $\frac{y_2-y_1}{x_2-x_1}$ x – x₁
or, y – a sin α

Question 14.
Find the length of perpendiculars drawn from the origin on the sides of the triangle whose vertices are A( 2, 1), B (3, 2), and C (- 1, -1).
Solution:

Question 15.
Show that the product of perpendicular from the points $\left(\pm \sqrt{a^2-b^2}, 0\right)$ upon the straight line $\frac{x}{a}$ cos θ + $\frac{y}{b}$ sin θ = 1 is b²
Solution:

Question 16.
Show that the lengths of perpendiculars drawn from any point of the straight line 2x + 11y – 5 = 0 on the lines 24x + 7y – 20 = 0 and 4x – 3y – 2 = 0 are equal to each other.
Solution:
Let P(h, k) is any point on the line
2k + 11y – 5 = 0
∴ 2h + 11k – 5 = 0
Now the length of the perpendicular from P on the line 24x + 7y – 20 = 0 is

Clearly d₁ = d₂

Question 17.
If p and p’ are the length of perpendiculars drawn from the origin upon the lines x sec α + y cosec α = 0 and x cos α – y sin α – a cos 2α = 0
Prove that 4p² + p’² = a²
Solution:

Question 18.
Obtain the equation of the lines passing through the foot of the perpendicular from (h, k) on the line Ax + By + C = 0 and bisect the angle between the perpendicular and the given line.
Solution:

Slope of the line L is $\frac{-\mathrm{A}}{\mathrm{B}}$
∴ Slope of the line $\overline{\mathrm{PM}} \text { is } \frac{\mathrm{B}}{\mathrm{A}}$
∴ Equation of the line $\overline{\mathrm{PM}}$ is y – y₁ = m(x – x₁)
or, y – k = $\frac{\mathrm{B}}{\mathrm{A}}$ (x – h)
or, Ay – Ak =Bx – Bh
or, Bx – Ay + Ak – Bh = 0
∴ Equation of the bisectors of the angles between the lines L and $\overline{\mathrm{PM}}$ is
$\frac{\mathrm{A} x+\mathrm{B} y+\mathrm{C}}{\sqrt{\mathrm{A}^2+\mathrm{B}^2}}=\pm \frac{\mathrm{B} x-\mathrm{A} y+\mathrm{A} k-\mathrm{B} h}{\sqrt{\mathrm{B}^2+\mathrm{A}^2}}$
or, Ax + By + C = ± (Bx – Ay +Ak – Bh)

Question 19.
Find the direction in which a straight line must be drawn through the point(1, 2) such that its point of intersection with the line x + y – 4 = 0 is at a distant $\frac{1}{3} \sqrt{6}$ from this point.
Solution:

Question 20.
A triangle has its three sides formed by the lines x + y = 3, x + 3y = 3, and 3x + 2y = 6. Without solving for the vertices, find the equation of its altitudes and also calculate the angles of the triangle.
Solution:

or 1 + 3λ = – 3 – 6λ
or 9λ = – 4 or , λ = $\frac{-4}{9}$
∴ Equation of $\overline{\mathrm{AD}}$ is (x + y – 3) – $\frac{4}{9}$ (3x + 2y – 6) = 0
or, 9x + 9y – 27 – 12x – 8y + 24 = 0
or, -3x + y – 3 = 0
or, 3x – y + 3 = 0
Let the equation of $\overline{\mathrm{BE}}$ be (x + y – 3) + 1(x + 3y – 3) = 0
or, x(1 + λ) + y( 1 + 3λ) – 3 – 3λ = 0
As $\overline{\mathrm{BE}} \perp \overline{\mathrm{AC}}$
we have $\frac{1+\lambda}{1+3 \lambda} \times \frac{3}{2}$ = -1
or, 3 + 3λ = -2 – 6λ
or, 9λ = – 5 or λ = $\frac{-5}{9}$
∴ Equation of $\overline{\mathrm{BE}}$ is (x + y – 3) – $\frac{5}{9}$ (x + 3y – 3) = 0
or 9x + 9y – 27 – 5x – 15y + 15 = 0
or, 4x – 6y – 12 = 0
or, 2x – 3y – 6 = 0
Let the equation of $\overline{\mathrm{CE}}$ be (3x + 2y – 6) + λ (x + 3y – 3) = 0
x(3 + λ) + y (x + 3λ) – 6 – 3λ = 0
As $\overline{\mathrm{CF}} \perp \overline{\mathrm{AB}} .$
we have $\frac{3+\lambda}{2+3 \lambda}$ × 1 = -1
or, 3 + λ = -2 – 3λ
or, 4λ = -2 – 3 = -5
or, λ = $\frac{-5}{4}$
∴ Equation of is $\overline{\mathrm{CF}}$ (3x + 2y – 6) $\frac{-5}{4}$ (x + 3y – 3) = 0
or, 12x + 8y – 24 – 5x – 15y + 15 =0
or, 7x – 7y – 9 = 0

Question 21.
A triangle has its vertices at P(1, -1), Q(3, 4) and R(2, 5). Find the equation of altitudes through P and Q and obtain the coordinates of their point of intersection. (This point is called the ortho-center of the triangle.)
Solution:

Question 22.
(a) Show that the line passing through (6, 0) and (-2, -4) is concurrent with the lines
2x – 3y – 11 = 0 and 3x – 4y = 16
Solution:
The equation of the line through (6,0) and (-2, -4) is

(b) Show that the lines lx + my + n = 0 mx + ny + 1 = 0 and nx + ly + m = 0 are concurrent, l + m + n = 0
Solution:
As the lines
lx + my + n = 0
mx + ny + 1 = 0
and nx + ly + m = 0 are concurrent.

Question 23.
Obtain the equation of the bisector of the acute angle between the pair of lines.
(a) x + 2y = 1, 2x + y + 3 = 0
Solution:
Equation ofthe bisectors ofthe angles between the lines x + 2y – 1 = 0 and 2x + y + 3 = 0 are $\frac{x+2 y-1}{\sqrt{1^2+2^2}}=\pm \frac{2 x+y+3}{\sqrt{2^2+1^2}}$
or, x + 2y – 1 = ± (2x + y + 3)
∴ x + 2y – 1 = 2x + y + 3 and
x + 2y – 1= -2x – y – 3
∴ x – y + 4 = 0 and 3x + 3y + 2 = 0
Let θ be the angle between x + 2y – 1 = 0 and x – y + 4 = 0
∴ tan θ = $\frac{a_1 b_2-a_2 b_1}{a_1 a_2+b_1 b_2}$
$=\frac{1 \cdot(-1)-(+1) \cdot 2}{1 \cdot 1+2(-1)}$
$=\frac{-1-2}{1-2}=\frac{-3}{-1}=3$
sec² θ = 1 + tan² θ = 1 + 9 = 10
cos² θ = 1/10
cos θ = $\frac{1}{\sqrt{10}}<\frac{1}{\sqrt{2}}$ ⇒ θ > 45°
∴ x – y + 4 = 0 is the obtuse angle bisector.
⇒ 3x + 3y + 2 = 0 is acute angle bisector.

(b) 3x – 4y = 5, 12y – 5x = 2
Solution:
Given equation of lines are
3x – 4y – 5 = 0    …..(1) and  5x – 12y + 2 = 0     ……(2)
Equation of bisectors of angles between these Unes are:

Question 24.
(a) Find the coordinates of the center of the inscribed circle of the triangle formed by the line x cos α + y sin α = p with the coordinate axes.
Solution:


$\left(\frac{p}{\sin \alpha+\cos \alpha+1}, \frac{p}{\sin \alpha+\cos \alpha+1}\right)$

(b) Find the coordinates of the circumcentre and incentre of the triangle formed by the lines 3x – y = 5, x + 2y = 4, and 5x + 3y + 1 = 0.
Solution:

Question 25.
The vertices B, and C of a triangle ABC lie on the lines 3y = 4x and y = 0 respectively, and the side $\overline{\mathbf{B C}}$ passes through the point (2/3, 2/3). If ABOC is a rhombus, where O is the origin, find the equation of $\overline{\mathbf{B C}}$ and also the coordinates of A.
Answer:
Let the coordinates of C be (a, 0) so that the length of the side of the rhombus is ‘a’

Question 26.
Find the equation of the lines represented by the following equations.
(a) 4x² – y² = 0
Solution:
4x² – y² = 0
or, (2x + y)(2x – y) = 0
∴ 2x + y = 0 and 2x – y = 0 are the two separate lines.

(b) 2x² – 5xy – 3y²
Solution:
2x² – 5xy – 3y²
or, 2x² – 6xy + xy – 3y² = 0
or, 2x(x – 3y) + y(x – 3y) = 0
or, (x – 3y)(2x + y) = 0
∴ x – 3y = 0 and 2x + y = 0 are the two separate lines.

(c) x² + 2xy sec θ + y² = 0
Solution:
x² + 2xy sec θ + y² = 0
∴ a = 1, b = 2y sec θ, c = y²
x = $\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}$
= $\frac{2 y \sec \theta \pm \sqrt{4 y^2 \sec ^2 \theta-4 y^2}}{2}$
= $\frac{-2 y \sec \theta \pm 2 y \tan \theta}{2}$
= y (- sec θ ± tan θ)
∴ x = y (- sec θ + tan θ) and x – y (- sec θ – tan θ) are the two separate lines.

(d) 3x² + 4xy = 0
Solution:
3x² + 4xy = 0
or x (3x + 4y) = 0
∴ x = 0 and 3x + 4y = 0 are the two separate lines.

Question 27.
From the equations which represent the following pair of lines.
(a) y = mx; y = nx
Solution:
y – mx = 0, y – nx = 0
or (y – mx) (y – nx) = 0
or, y² – nxy – mxy + mnx² = 0
or, y² – xy (m + n) + mnc² = 0 which is the equation of a pair of lines.

(b) y – 3x = 0 ; y + 3x = 0
Solution:
y – 3x = 0, y + 3x = 0
∴ (y – 3x) (y + 3x) = 0
or, y² – 9x² = 0 which is the equation of a pair of lines.

(c) 2x – 3y + 1 = 0 ; 2x + 3y + 1 = 0
Solution:
2x – 3y + 1 = 0; 2x + 3y + 1 =0
or, (2x – 3y + 1)(2x + 3y + 1) = 0
or, (2x + 1)² – 9y² = 0
or, 4x² + 1 + 4x – 9y² = 0
or, 4x² – 9y² + 4x + 1= 0 which represents a pair of lines.

(d) x = y. x + 2y + 5 = 0
Solution:
x = y, x + 2y + 5 = 0
∴ (x – y) (x + 2y + 5) = 0
or, x² + 2xy + 5x – xy – 2y² – 5y =0
or, x² – 2y² + xy + 5x – 5y = 0 which represents a pair of lines.

Question 28.
Which of the following equations represents a pair of lines?
(a) 2x² – 6y² + 3x +  y + 1 = 0
Solution:
a = 2, b = -6, 2g = 3
2f = 1, c = 1
∴ g = $\frac{3}{2}$, f = $\frac{1}{2}$, h = 0
∴ abc + 2fgh – ah² – bg² – ch²
= 2(-6). 1 + 2 × $\frac{1}{2}$ × $\frac{3}{2}$ – 0 – (-6) × $\frac{9}{4}$ – 1 × 0
= -12 + $\frac{3}{2}$ + $\frac{27}{2}$ = $\frac{6}{2}$ = 3 ≠ 0
∴ The given equation does not represent a pair or lines.

(b) 10x² – xy – 6y² – x + 5y – 1 = 0
Solution:
a = 10. 2h = 1
B = -6, 2g = -1
2f = 5. C= -1
∴ h = –$\frac{1}{2}$, g = –$\frac{1}{2}$ , f = $\frac{5}{2}$
∴ abc + 2fgh – ah² – bg² – ch²
= 10(-6)(-1) + 2 × $\frac{5}{2}$ × (-$\frac{1}{2}$) × (-$\frac{5}{2}$) – 10 × $\frac{2.5}{4}$ – (-6)$\frac{1}{4}$ – (-1)$\frac{1}{4}$
= 60 + $\frac{5}{4}$ – $\frac{250}{4}$ + $\frac{6}{4}$ + $\frac{1}{4}$
= $\frac{240+5+6-250+1}{4}=\frac{2}{4}$
∴ The given equation does not represent a pair of lines.

(c) xy + x + y + 1 = 0
Solution:
xy + x + y + 1= 0
or, x(y + 1) + 1(y + 1 ) = 0
or (y + 1 )(x + 1) =0
∴ x + 1 = 0
and y + 1 = 0 are the two separate lines,
∴ The given equation represents a pair of lines.

Question 29.
For what value of λ do the following equations represent pair of straight lines?
(a) λx² + 5xy – 2y² – 8x + 5y – λ = 0
Solution:
λx² + 5xy – 2y² – 8x + 5y – λ = 0
∴ a = λ, 2h = 5, b = -2, 2g = -8
2f = 5, c = -1
∴ h = $\frac{5}{2}$, g = -4, f = $\frac{5}{2}$
As the given equation represent a pair of lines, we have abc + 2fgh – ah² – bg² – ch² = 0
or, λ(-2)(-λ) + 2. $\frac{5}{2}$ (-4). $\frac{5}{2}$ -λ × $\frac{25}{4}$ – (-2) (-4)² – (-λ) × $\frac{25}{4}$ = 0
or, 2λ² – 50 – $\frac{25 λ }{4}$ + 32 + $\frac{25 λ }{4}$ = 0
or, 2λ² = 18 or, λ² = 9
λ = ±3

(b) x² – 4xy – y² +6x + 8y + λ = 0
Solution:
Here a = 1, 2h = -1, b = -1, 2g = 6, 2f = 8, c = τ
As the given equation represent a pair of lines, we have
abc + 2fgh – af² – bg² – ch² = 0
⇒ (-1) τ + 2.4.3 (-2) – 1. 4² – (-1). ³2 – τ(-2)² = 0
⇒ -τ – 48 – 16 + 9 – 4τ = 0
⇒ -5τ – 55 = 0 ⇒ τ = -11

Question 30.
(a) Obtain the value of λ for which the pair of straight lines represented by 3x² – 8xy + λy² = 0 are perpendicular to each other.
Solution:
3x² – 8xy + λy² = 0
∴ a = 3. 2h = -8, b = λ
As the pair of lines are perpendicular to each other, we have a + b = 0.
or, 3 + λ = 0 – or, λ = -3

(b) Prove that a pair of lines through the origin perpendicular to the pair of lines represented by px² – 2qxy + ry² = 0 is given by rx² – 2qxy + py² = 0
Solution:
px² – 2qxy + ry² = 0
∴ a = p, b = 2qy, c = ry²

(c) Obtain the condition that a line of the pair of lines ax² + 2hxy + by² = 0,
(i) Coincides with
Solution:

(ii) is perpendicular to, a line of the pair of lines px² + 2qxy + ry² = 0
Solution:

Question 31.
Find the acute angle between the pair of lines given by :
(a) x² + 2xy – 4y² = 0
Solution:
x² + 2xy – 4y² = 0
∴ a=1, 2h = 2, b = -4
∴ tan θ = $\frac{\pm 2 \sqrt{h^2-a b}}{a+b}=\frac{\pm 2 \sqrt{1+4}}{1-4}$
$=\pm \frac{2 \sqrt{5}}{-3}=\mp \frac{2 \sqrt{5}}{3}$
∴ The acute angle between the pair of lines is tan^-1 $\frac{2 \sqrt{5}}{3}$

(b) 2x² + xy – 3y² + 3x + 2y + 1 = 0
Solution:
2x² + xy – 3y² + 3x + 2y + 1 = 0
∴ a = 2, 2h = 1, b = -3, 2g = 3 2f= 2, c = 1.
tan θ = $\frac{\pm 2 \sqrt{h^2-a b}}{a+b}$
$=\pm \frac{2 \sqrt{\frac{1}{4}+6}}{2-3}=\pm \frac{2 \times 5}{2(-1)}$ = ± 5
∴ The acute angle is tan^-1 5

(c) x² + xy – 6y² – x – 8y – 2 = 0
Solution:
Given Equation is x² + xy – 6y² – x – 8y – 2 = 0
here a = 1, 2h = 1, b = -6 thus if 0 is the acute angle between two lines then
tan θ = $=\left|\frac{2 \sqrt{h^2-a b}}{a+b}\right|=\left|\frac{2 \sqrt{\frac{1}{4}+6}}{-5}\right|$
= $\left|\frac{2 \times 5}{-10}\right|$ = 1
∴ θ = 45°

Question: 32.
Write down the equation of the pair of bisectors of the following pair of lines :
(a) x² – y² = 0 ;
Solution:
x² – y² = 0
∴ a = 1, b = -1, h = 0
∴ The equation of the bisectors of the angles between the pair of lines are $\frac{x^2-y^2}{a-b}=\frac{x y}{h}$
or, $\frac{x^2-y^2}{1+1}=\frac{x y}{0}$
or, xy = 0

(b) 4x² – xy – 3y² = 0
Solution:
4x² – xy – 3y² = 0
∴ a = 4, 2h = -1, b = -3
∴ Equation of the pair of bisectors are $\frac{x^2-y^2}{(a-b)}=\frac{x y}{h}$
or, $\frac{x^2-y^2}{7}=\frac{x y}{\left(-\frac{1}{2}\right)}$
or, x² – y² = -14xy
or, x² + 14xy – y² = 0

(c) x² cos θ + 2xy – y² sin θ = 0
Solution:
x² cos θ + 2xy – y² sin θ = 0
∴ a = cos θ, 2h = 2, b = – sin θ
∴ Equation of the pair of bisectors are $\frac{x^2-y^2}{a-b}=\frac{x y}{h}$
or, $\frac{x^2-y^2}{\cos \theta+\sin \theta}=\frac{x y}{1}$
or, x² – y² = xy(cos θ + sin θ)

(d) x² – 2xy tan θ – y² = 0
Solution:
x² – 2xy tan θ – y² = 0
∴ a = 1, 2h = -2 tan θ, b = -1
∴ Equation of the pair of bisectors are $\frac{x^2-y^2}{2}=\frac{x y}{-\tan \theta}$
or, x² – y² = 2xy cot θ
or, x² + 2xy cot θ – y² = 0

Question 33.
If the pair of lines represented by x² – 2pxy – y² = 0 and x² – 2qxy – y² = 0 be such that each pair bisects the angle between the other pair, then prove that pq = -1.
Solution:

Question 34.
Transform the equation: x² + y² – 2x – 4y + 1 = 0 by shifting the origin to (1, 2) and keeping the axes parallel.
Solution:
x² + y² – 2x – 4y + 1 = 0     ……(1)
Let h = 1, k = 2
Taking x’ + h and y = y’ + k we have
(x’ + h)² + (y’ + k)² – 2(x’ + j) -4(y’ + k) + 1=0
or, (x + 1)² + (y’ + 2)² – 2(x’ + 1) – 4(y’+ 2) + 1=0
or, x^‘2 + 1 + 2x’ + y^‘2 + 4 – 4y’ – 2x’ – 2 – 4y’- 8 + 1 = 0
or, x^‘2 + y’² – 4 = 0
∴ The transformed equation is x² + y² = 4

Question 35.
Transform the equation: 2x² + 3y² + 4xy – 12x – 14y + 20 = 0. When referred to parallel axes through(2, 1).
Solution:
2x² + 3y² + 4xy – 12x – 14y + 20 = 0
Let h = 2, k = 1
Taking x = x’ + 1 and y = y’ + 1
we have
2(x’ + 2)² + 3(y’ + 1)² + 4(x’ + k)(y’ + 1) – 12 (x’ + 2)- 14 (y’ + 1) + 20 = 0
or, 2x^‘2 + 8 + 8x’ + 3 + 6y’ + 3y^’2 + 4x’y’ + 4x’ + 8y’ + 8 – 12x’ – 14y’ – 18 = 0
or, 2x^‘2 + 3y^’2 + 4x’y’ + 1=0
The transformed equation is
2x² + 3y² + 4xy + 1 = 0

Question 36.
Find the measure of rotation so that the equation x² – xy + y² = 5 when transformed does not contain xy- term.
Solution:
x² – xy + y² = 5
Taking x = x’ cos α – y’ sin α
y = x’ sin α – y’ cos α
We get (x’ cos α – y’ sin α)² – (x’ cos α – y’ cos α) (x’ sin α + y’ cos α) + (x’ sin α + y’ cos α)² = 5
⇒ x^‘2 cos² α + y^‘2 sin² α – 2x’y sin α.
cos α – x^‘2 sin α. cos α – x’y cos² α + x’y’ sin² α + y^‘2 sin α. cos α + x^‘2 sin² α + y^‘2 cos² α + 2x’y’ sin α cos α = 5
Given that the transformed equation does not xy term.
Hence the co-efficient of x’y’ is zero.
That is sin² α – cos² α = 0
⇒ sin² α = cos² α
⇒ tan² α = 1 ⇒ tan α = 1 ⇒ α= 45°

Question 37.
What does the equation x + 2y – 10 =0 become when the origin is changed to (4, 3)?
Solution:
x + 2y – 10 = 0
Let h = 4, k = 3
Taking x = x’ + 4, y = y’ + 3
we have x’ + 4 + 2 (y’ + 3) – 10 = 0
or, x + 2y’ = 0
∴ The transformed equation is x + 2y = 0.

Odisha State Board Elements of Mathematics Class 11 Solutions CHSE Odisha Chapter 14 Limit and Differentiation Ex 14(a) Textbook Exercise Questions and Answers.

CHSE Odisha Class 11 Math Solutions Chapter 14 Limit and Differentiation Exercise 14(a)

Question 1.
$\lim _{x \rightarrow 3}$(x + 4)
Solution:
Clearly, if we take x very close to 3, x + 4 will go very close to 7.
Now let us use ε – δ technique to confirm the result.
Given ε > 0, we seek for δ > 0 depending on ε such that
|x – 3| < δ ⇒ |(x + 4) – 7|< ε
Now |(x + 4) – 7| < ε
if |x – 3| < ε
∴ We can choose ε = 8
Hence for given ε > 0, there exist 8 = ε > 0
such that |x – 3| < δ ⇒ |(x + 4) – 7| < ε
∴ $\lim _{x \rightarrow 3}$(x + 4) = 7

Question 2.
$\lim _{x \rightarrow 1}$(4x – 1)
Solution:
By taking very close to 1 we have 4x- 1 tends to 3.
Let us use ε – δ technique to confirm the result.
Given ε > 0. We shall find δ > 0 depending on ε such that
|x – 1| < 5 ⇒ |(4x – 1) – 3| < ε
Now |(4x – 1 ) – 3| < ε
if |4x – 1| < ε i.e.|x – 1| < $\frac{\varepsilon}{4}$
Let us choose δ = $\frac{\varepsilon}{4}$
∴ For given ε > 0 there exists δ = $\frac{\varepsilon}{4}$ > 0
such that |x – 1| < δ
⇒ |(4x – 1) – 3| < ε
∴ $\lim _{x \rightarrow 1}$(4x – 1) = 3

Question 3.
$\lim _{x \rightarrow 1}$(√x + 3)
Solution:
As x → 1 we see √x + 3 → 4
We will confirm the result using ε – δ technique
Let ε > 0, we will choose δ > 4
such that |x – 1| < 8 ⇒ |√x + 3 – 4| < ε
Now |√x + 3 – 4| = |√x – 1|
$=\frac{|x-1|}{|\sqrt{x}+1|}$
But |√x + 1| > 1
⇒ $\frac{1}{|\sqrt{x}+1|}$ < 1
⇒ $\frac{|x-1|}{|\sqrt{x}+1|}<\frac{\delta}{1}$
∴ (√x + 3) – 4 < $\frac{\delta}{1}$
We can take δ < ε i.e. δ = min {1, ε}
∴ |x – 1| < δ ⇒ |(√x + 3) – 4| < ε
for given ε > 0 and (δ = ε)
⇒ $\lim _{x \rightarrow 1}$(√x + 3) = 4

Question 4.
$\lim _{x \rightarrow 0}$ (x² + 3)
Solution:
As x → 0 we observe that x³ + 3 → 3
Let us use ε – δ technique to confirm the result.
Let ε > 0, we seek for a δ > 0 such that
|x – 0| < ε ⇒ |x² + 3 – 3| < ε
Let |x| < 8
Now |x² + 3 – 3| < ε
We have |x|² < ε ⇒| x| < √ε
(∴ |x| and ε are positive.)
∴ we can choose δ = √ε
∴ We have for given δ > 0 there exists
δ = √ε > 0 such that |x| < δ ⇒ |x² + 3 – 3| < ε
∴ $\lim _{x \rightarrow 0}$ (x² + 3) = 3

Question 5.
$\lim _{x \rightarrow 0}$ 7
Solution:
If x → 0 we observe that 7 → 7.
Let us use e- 8 technique to confirm the limit.
Let f(x) = 7
Given ε > 0, we will choose a δ > 0
such that |x – 0| < δ ⇒ |f(x) – 7| < ε
Now |f(x) – 7| < ε
If f(x) ∈ (7 – ε . 7 + ε)
But for every x, f(x) = 7
⇒ for|x| < δ also f(x) = 7 ∈ (7 – ε . 7 + ε)
∴ Choosing ε = δ we have
|x| < δ ⇒ |f(x) – 7| < ε
∴ $\lim _{x \rightarrow 0}$ (7) = 7

Question 6.
$\lim _{x \rightarrow 1} \frac{(x-1)^3}{(x-1)^3}$
Solution:
We guess the limit is 1.
Let us confirm using ε – δ technique.
Let ε > 0, f(x) = $\frac{(x-1)^3}{(x-1)^3}$
We will choose a δ > 0 such that
|x – 1| < δ ⇒ |f(x) – 1)| < ε
Now |f(x) – 1| < ε
if 1 – ε < f(x) < 1 + ε
∴ We will choose a δ > 0 such that
x ∈ (1 – δ, 1 + δ) – { 1 }
⇒ f(x) ∈ ( 1 – ε, 1 + ε)
As f(x) = for x ≠ 1
We have f(x) ∈ (1 – ε. 1 + ε) for all x ∈ (1 – δ, 1 + δ) – [1]
∴ We can choose δ = ε
for given ε > 0, there exists δ = ε
s.t. |x – 1| < δ ⇒ |f(x) – 1| < ε
∴ $\lim _{x \rightarrow 1}$ f(x) = 1

Question 7.
$\lim _{x \rightarrow 3} \frac{x^3-9}{x-3}$
Solution:
If we take x very close to 3 (≠ 3)
we have $\frac{x^3-9}{x-3}$
= $\frac{(x-3)\left(x^2+3 x+3^2\right)}{2}$ → 27
Let ε > 0 and x ≠ 3
Now |$\frac{x^3-9}{x-3}$ – 27| = |x² + 3x +9 – 27|
=|x² – 9 + 3(x – 3)| ≤ |x² – 9| + 3|x – 3|
= |x – 3| [|x + 3| + 3] ≤ |x – 3| [|x + 6| < |x – 3| [|x – 3 + 9|]]
If |x – 3| < δ and δ < 1 then |x – 3| [x – 3 + 9| < δ {1 + 9} = 10 δ
Let δ = min {1, $\frac{\varepsilon}{10}$}
∴ For given ε > 0 we have a δ = min {1, $\frac{\varepsilon}{10}$} >0 such that
|x – 3| < δ ⇒ |$\frac{x^3-9}{x-3}$ – 27|
∴ $\lim _{x \rightarrow 3} \frac{x^3-9}{x-3}$ = 27

Question 8.
$\lim _{x \rightarrow 1} \frac{3 x+2}{2 x+3}$
Solution:
we observe that as x → 1, $\frac{x+2}{2 x+3}$ → 1
To establish this let ε > 0,
we seek a δ > 0,

Question 9.
$\lim _{x \rightarrow 0}|x|$
Solution:
We see that when x → 0,|x| → 0
Let us establish this using ε – δ technique.
Let ε > 0 we seek a δ > 0 depending on
ε s.t.|x – 0| < ε ⇒ ||x| – 0| < ε
Now ||xl – 0| = ||x|| = |x| < δ
By choosing ε = δ we have |x| < ε ⇒ ||x| – 0| < ε
∴ $\lim _{x \rightarrow 0}|x|$ = 0

Question 10.
$\lim _{x \rightarrow 2}(|x|+3)$
Solution:
We see that as x → 2, |x| + 3 → 5
Let ε > 0 we were searching for a, δ > 0
such that |x – 2| < δ ⇒ ||x| + 3 – 5| < ε
Now ||x|| + 3 – 5| = ||x| – 2| < |x – 2| < δ
∴ Choosing ε = δ
We have |x – 2| < δ ⇒ ||x| + 3 – 5| < ε
∴ Choosing ε = δ
We have |x – 2| < δ ⇒ ||x| + 3 – 5| < ε
∴ $\lim _{x \rightarrow 2}(|x|+3)$ = 5

Odisha State Board CHSE Odisha Class 11 Math Solutions Chapter 12 Conic Sections Ex 12(a) Textbook Exercise Questions and Answers.

CHSE Odisha Class 11 Math Solutions Chapter 12 Conic Sections Exercise 12(a)

Question 1.
Fill in the blanks by choosing the correct answer from the given alternatives :
(a) The center of the circle x² + y² + 2xy – 6y + 1 = 0 is _____________. [(2, -6), (-2, 6), (-1, 3), (1, -3)]
Solution:
(-1, 3)

(b) The equation 2x² – ky² – 6x + 4y – 1 = 0 represents a circle if k = ____________. [2, -2, 0, 1]
Solution:
-2

(c) The point (-3, 4) lies ______________ the circle x² + y² = 16 [outside, inside, on]
Solution:
Outside

(d) The line y = x + k touches the circle x² + y² = 16 if k = _______________. [±2√2, ±4√2, ±8√2, ±16√2]
Solution:
±4√2

(e) The radius of the circle x² + y² – 2x + 4y + 1 = 0 is _______________. [1, 2, 4, √19]
Solution:
2

Question 2.
State (with reasons), which of the following is true or false :
(a) Every second-degree equation in x and y represents a circle.
Solution:
Every 2nd-degree equation in x and y represents a circle if the coefficients of x and y are equal and the equation does not contain xy term (False)

(b) The circle (x – 1)² + (y – 1)² = 1 passes through origin.
Solution:
(0 – 1)² + (0 – 1)² = 1 + 1 = 2 ≠ 1.
So the circle does not pass through the origin. (False)

(c) The line y = 0 is a tangent to the circle (x + 1)² + (y – 2)²  = 1.
Solution:
The line y = 0 is a tangent to the circle centre at (-1, 2) and the radius is 1. (True)
∴ The distance of the centre from the line y = 0 is 1 which is equal to its radius.

(d) The radical axis of two circles always passes through the centre of one of the circles,
Solution:
As radical axis is the common chord of the circles, which should not pass through the centre of one of the circles. (False)

(e) The circle x² + (y – 3)² = 4 and (x – 4)² + y² = 9 touch each other.
Solution:
The distance between the centres is $\sqrt{(0-4)^2+(3-0)^2}$ = 5 which is equal to the sum of the radii. (True)

Question 3.
Find the equation of circles determined by the following conditions.
(a) The centre at (1, 4) and passing through (-2, 1).
Solution:

(b) The centre at (-2, 3) and passing through origin.
Solution:
Centre at (-2, 3) and circle passes through origin.
∴ Radius of the circle = $\sqrt{(-2)^2+3^2}=\sqrt{13}$
∴ Equation of the circle is (x – h)² + (y – k)² = a²
or, (x + 2)² + (y – 3)² = 13

(c) The centre at (3, 2) and a circle is tangent to x – axis.
Solution:

(d) The centre at (-1, 4) and circle is tangent to y – axis.
Solution:

(e) The ends of diameter are (-5, 3) and (7, 5).
Solution:
The endpoints of the diameter of the circle are (-5, 3) and (7, 5).
∴ Equ. of the circle is
(x – h)² + (y – k)² = a²
(x- x₁)(x – x₂) + (y – y₁)(y – y₂) = 0
or, (x + 5)(x – 7) + (y – 3)(y – 5) = 0
or, x² – 7x + 5x – 35 + y² – 5y – 3y + 15 = 0
or, x² + y² – 2x – 8y – 20 = 0

(f) The radius is 5 and circle is tangent to both axes.
Solution:
As the circle is tangent to both axes, we have its centre at (5, 5).
∴ Equation of the circle is
or, (x ± 5)² + (y ± 5)² = 25

(g) The centre is on the x-axis and the circle passes through the origin and the point (4, 2).
Solution:

∴ $\sqrt{(4-a)^2+4}$ = a
or, (4 – a)² + 4 = a²
or, 16 + a² – 8a + 4 = a²
or, 8a = 20 or, a = $\frac{20}{8}=\frac{5}{2}$
∴ Equation of the circle is
(x – h)² + (y – k)² = a²
or, (x – $\frac{5}{2}$)² + (y – 0)² = ($\frac{5}{2}$)²
or, x² + $\frac{25}{4}$ – 5x + y² = $\frac{25}{4}$
or, x² + y² – 5x = 0

(h) The centre is on the line 8x + 5y = 0 and the circle passes through the points (2, 1) and (3, 5).
Solution:
Let the equation of the circle be x² + 2gx + y² + 2fy + c = 0
∴ Its centre at (- g, -f). As the centre lies on the line 8x + 5y = 0
We have -8g – 5f = 0      …..(1)
Again, as the circle passes through points (2, 1) and (3, 5)
We have
4 + 4g + 1 + 2f + c = 0  …..(2)
and 9 + 6g + 25 + 10f + c = 0   …..(3)
Now from (1), we have g = $\frac{-5 f}{8}$
From equation (2), 4g + 2f + c + 5 = 0
or, 4 $\frac{-5 f}{8}$ + 2f + c + 5 = 0
or, -5f + 4f + 2c + 10 = 0
or, f = 2c + 10    …..(4)
(2) 6g + 10f + c + 34 = 0
or, 6$\frac{-5 f}{8}$ + 10f + c + 34 = 0
or, -15f + 40f + 4c + 136 = 0
or, 25f = -4c – 136
or, f = $\frac{-4 c-136}{25}$
∴ 2c + 10 = $\frac{-4 c-136}{25}$
or, 25 (c + 5) = -2c – 68
or, 25c + 2c = -68 – 125
or, 27c = -193 or, c = $\frac{-193}{27}$
∴ f = 2C + 10 = 2($\frac{-193}{27}$) + 10
= $\frac{-386+270}{27}=\frac{-116}{27}$
∴ g = $\frac{-5 f}{8}=\frac{-5}{8} \times\left(\frac{-116}{27}\right)=\frac{145}{54}$
Eqn. of the circle is x² + y² + 2 × $\frac{145}{54}$ x + 2 $\frac{-116}{27}$ y + $\frac{-193}{27}$ = 0
or, 27x² + 27y² + 145x – 232y – 193 = 0

(i) The centre is on the line 2x + y – 3 = 0 and the circle passes through the points (5, 1) and (2, -3).
Solution:
Let the eqn. of the circle be x² + y² + 2gx + 2fy + c = 0
As it passes through (5, 1) and (2, -3),
we have 25 + 1 + 10g + 2f + c = 0 …(1)
and 4 + 9 + 4g – 6f + c = 0    …..(2)
Again as the centre lies on the line 2x + y – 3 = 0,
we have- 2g – f – 3 = 0 or, f= -2g – 3
∴ From equation (1)
10g + 2 (-2g – 3) + c + 26 = 0
or, 10g – 4g – 6 = -c – 26
or, 6g = -c – 20
or, g = $\frac{-c-20}{6}$
∴ From equation (2)
4g – 6 (-2g – 3) + c + 13 = 0
or, 4g + 12g + 18 + c + 13 = 0
or, 16g = -c – 31
or, g = $\frac{-c-31}{16}$
∴ $\frac{-c-20}{6}=\frac{-c-31}{16}$
or, -8c – 160 = -3c – 93
or, 5c = -160 + 93 = -67
or, c = –$\frac{67}{5}$
∴ g = $\frac{-c-20}{6}=\frac{\frac{67}{5}-20}{6}=\frac{67-100}{5 \times 6}$
= $\frac{-33}{5 \times 6}=\frac{-11}{10}$
∴ f = -2g – 3 = (-2)$\left(\frac{-11}{10}\right)$
= $\frac{11-15}{5}=\frac{-4}{5}$
∴ Eqn. of the circle is x² + y² + 2 ($\frac{-11}{10}$)x + 2($\frac{-4}{5}$)y – $\frac{67}{5}$ = 0
or, 5x² + 5y² – 11x – 8y – 67 = 0

(j) The circle is tangent to the line x + 2y – 9 = 0 at (5, 2) and also tangent to the line 2x – 3y – 7 = 0 at (2, -1).
Solution:

(k) The circle touches the axis of x at (3, 0) and also touches the line 3y – 4x = 12.
Solution:
Let the centre be at (3, k)
Radius = k

or, 3k – 24 = ±5k or, 2k = -24
or, k = -12
Also k = 3
∴ Equation of the Circle is
(x – 3)² + (y – k)² = k²
or, x² + 9 – 6x + y² + k² – 2ky = k²
or, x² + y2 – 6x – 2 (-12)y = 0
or, x² + y² – 6x + 24y + 9 = 0
and x² + y² – 6x – 6y + 9 = 0

(l) Circle is tangent to x – axis and passes through (1, -2) and (3, -4).
Solution:
Let the centre be at (h, k).
So the radius is k.

∴ Equation of the circle is (x – h)² + (y – k)² = k²
or, (x + 5)² + (y + 10)² = 100 and (x – 3)² + (y + 2)² = 4

(m) Circle passes through origin and cuts of intercepts a and b from the axes.
Solution:
Let the equation of the circle be x² + y² + 2gx + 2fy + c = 0

(n) Circle touches the axis of x at a distance of 3 from the origin and intercepts a distance of 6 on the y-axis.
Solution:
Let the centre be at (3, k).
So the radius is k.
∴ Equation of the circle is (x – 3)² + (y – k)² = k²
or, x² + 9 – 6x + y² + k² – 2ky = k²
or, x² + y² – 6x – 2xy + 9 = 0

∴ |y₂ – y₁| = 2$\sqrt{k^2-9}$ = 6
or, $\sqrt{k^2-9}$ = 3
or, k² = 18, or, k = ±3√2
∴ Equation of the circle is x² – y² – 6x ± 6y√2 + 9 = 0

Question 4.
Find the centre and radius of the following circles:
(a) x² + y² + 6xy – 4y – 12 = 0
Solution:
x² + y² + 6xy – 4y – 12 = 0
∴ 2g = 6, 2f = – 4, c = -12
∴ 8 = 3, f = -2
Centre of (-g, -f) = (-3, 2) and radius = $\sqrt{g^2+f^2-c}$
= $\sqrt{9+4+12}$ = 5

(b) ax² + ay² + 2gx + 2fy + k = 0
Solution:
ax² + ay² + 2gx + 2fy + k = 0
or, x² + y² + $\frac{2 g}{a}$x + $\frac{2 f}{a}$y + $\frac{k}{a}$ = 0
∴ Centre of $\left(\frac{-g}{a}, \frac{-f}{a}\right)$
and radius = $\sqrt{\frac{g^2}{a^2}+\frac{f^2}{a^2}-\frac{k}{a}}=\sqrt{\frac{g^2+f^2-a k}{a}}$

(c) 4x² + 4y² – 4x + 12y – 15 = 0
Solution:
4x² + 4y² – 4x + 12y – 15 = 0
or, x² + y² – 4 + 3y  – $\frac{15}{4}$ = 0
∴ 2g = -1, 2f = 3, c = $\frac{15}{4}$
∴ g = – $\frac{1}{2}$, f = $\frac{3}{2}$
∴ Centre at (-g, -f) = ($\frac{1}{2}$, $\frac{-3}{2}$) and radius $\sqrt{g^2+f^2-c}$
= $\sqrt{\frac{1}{4}+\frac{9}{4}+\frac{15}{4}}=\frac{5}{2}$

(d) a(x² + y²) – bx – cy = 0
Solution:
a(x² + y²) – bx – cy = 0
or, x² + y² – $\frac{b x}{a}$ – $\frac{c y}{a}$ = 0

Question 5.
Obtain the equation of circles passing through the following points and determine the coordinates of the centre and radius of the circle in each case:
(a) the points (3, 4) (4, -3) and (-3, 4).
Solution:
Let the centre be at (h, k)

(b) the points (2, 3), (6, 1) and (4, -6).
Solution:
Let the centre be at (h, k).

we have $|\overline{\mathrm{PC}}|=|\overline{\mathrm{QC}}|=|\overline{\mathrm{RC}}|$
∴ (h – 4)² + (k + 6)² = (h – 6)² + (k – 1)² and (h – 4)² + (k + 6)² = (h – 2)² + (k – 3)²
∴ h² + 16 – 8h + k² + 36 + 12k
= h² + 36 – 12h + k² + 1 – 2k
and h² + 16 – 8h + k² + 36 + 12k
= h² + 4 – 4h + k² + 9 – 6k
or, 14k = -4h – 15 and 18k = 4h – 39
or, k = $\frac{-4 h-15}{14}$ and k = $\frac{4 h-39}{18}$

(c) the points (a, 0), (-a, 0) and (0, b).
Solution:
Let the equation of the circle be x² + y² + 2gx + 2fy + c = 0. As it passes through the points (a, 0), (-a, 0) and (0, b). We have
a² – 2ga + c = 0   …..(1)
a² + 2ga + c = 0   …..(2)

(d) the points (-3, 1), (5, -3) and (-3, 4).
Solution:
Let the equation of the circle be x² + y² + 2gx + 2fy + c = 0
As it passes through the points. we have (-3, 1), (5, -3) and (3, 4).
We have 9 + 1 – 6g + 2f + c = 0    …..(1)
25 + 9 + 10g – 6df + c = 0      …(2)
9 + 16 – 6g + 8f + c = 0     …..(3)

Question 6.
Find the equation of the circles circumscribing the triangles formed by the lines given below :
(a) the lines x = 0, y = x, 2x + 3y = 10
Solution:

∴ The coordinates. C are (0, 0) of
Lastly, solving $\overline{\mathrm{AB}}$ and $\overline{\mathrm{AC}}$
we have y = x, 2x + 3y = 1 0
we have 5x = 10
or, x = 2 and y = 2.
∴ The coordinates of A are (2, 2).
∴ The circle passes through the points (2, 2), (0, $\frac{10}{3}$) and (0, 0)
Let the equation of the circle be x² + y² + 2gx + 2fy + c = 0
As it passes through the points A, B, C we have c = 0, 4 + 4 + 4g + 4f = 0,
$\frac{100}{9}$ + 2. f. $\frac{10}{3}$ + 0 = 0
∴ f = $\frac{-100}{9} / \frac{20}{3}=\frac{-5}{3}$
and g = $\frac{-4 f-8}{4}=\frac{-4\left(\frac{-5}{3}\right)-8}{4}$
= $\frac{20-24}{3 \times 4}=\frac{-1}{3}$
∴ Equation of the circle is  x² + y² + 2($\frac{-1}{3}$)x + 2 $\frac{-5}{3}$y + 0 = 0
or, 3(x² +  y²) – 2x – 10y = 0

(b) The lines x = 0, 4x + 5y = 35, 4y = 3x + 25
Solution:



or, 4x² + 4y² – 24x – 53y + 175 = 0

(c) The lines x = 0, y = 0, 3x + 4y – 12 = 0
Solution:
The coordinates of A, B and C are (4, 0), (0, 3) and (0, 0).
Let the equation of the circle be x² + y² + 2gx + 2fy + c = 0

(d) The lines y = x, y =2 and y = 3x + 2
Solution:

(e) the lines x + y = 6, 2x + y = 4 and x + 2y = 5
Solution:

Question 7.
Find the coordinates of the points where the circle x² + y² – 7x – 8y + 12 = 0 meets the coordinates axes and hence find the intercepts on the axes. [Hint: If a circle intersects a line at points A and B, then the length AB is its intercepts on line L]
Solution:
x² + y² – 7x – 8y + 12 = 0
Putting x = 0, we have y² – 8y + 12 = 0 or, (y – 6) (y – 2) = 0, or, y = 6, 2.
∴ The circle meets the Y-axis at (0, 6) and (0, 2) and its Y-intercept is 6 – 2 = 4.
Again putting y = 0,
we have x² – 7x + 12 = 0
or, (x – 4)(x – 3) = 0 or, x = 4, x = 3.
∴ The circle meets the X-axis at (4, 0) and (3, 0) and its x-intercept is 4 – 3 = 1.

Question 8.
Find the equation of the circle passing through the point (1, -2) and having its centre at the point of intersection of lines 2x – y + 3 = 0 and x + 2y – 1 =0
Solution:

Question 9.
Find the equation of the circle whose ends of a diameter are the points of intersections of the lines and x + y – 1 = 0, 4x + 3y + 1 = 0 and 4x +y + 3 = 0, x – 2y +3 = 0.
Solution:
Solving x + y – 1 = 0, 4x + y + 3 = 0

∴ The endpoints of the diameter are (-4, 5) and (-1, 1).
∴ Equation of the circle is
(x – x₁) (x – x₂) + (y – y₁) (y – y₂) = 0
or, (x + 4) (x + 1) + (y – 5) (y – 1) = 0
or, x² + x + 4x + 4 + y² – y – 5y + 5 = 0
or, x² + y² + 5x – 6y + 9 = 0.

Question 10.
Find the equation of the circle inscribed inside the triangle formed by the line $\frac{x}{4}+\frac{y}{3}$ = 1 and the coordinate axes.
Solution:
The circle is inscribed in the triangle formed by x = 0, y = 0 and $\frac{x}{4}+\frac{y}{3}$ = 1
∴ If (h, k) is the centre and r is the radius of the circle then h = k = r.
The perpendicular distance of the centre (h, h) from the line 3x + 4y = 12 is the radius.
⇒ $\left|\frac{3 h+4 h-12}{5}\right|$ = h
⇒ 7h – 12 = ±5h
⇒ 2h = 12 or 2h = 12
⇒ h = 6 or h = 1
But h can not be 6 thus the circle has equation (x – 1)² + (y – 1)² = 1
⇒ x² + y² – 2x – 2y + 1 =0

Question 11.
(a) Find the equation of the circle with its centre at (3, 2) and which touches to the line x + 2y – 4 = 0.
Solution:

(b) The line 3x + 4y + 30 = 0 is a tangent to the circle whose centre is at ($-\frac{12}{5},-\frac{16}{5}$). Find the equation of the circle.
Solution:

(c) Prove that the points (9, 7), and (11, 3) lie on a circle with centre at origin. Find the equation of the circle.
Solution:

(d) Find the equation of the circle which touches the line x = 0, x = a and 3x + 4y + 5a = 0.
Solution:

(e) If a circle touches the co-ordinate axes and also touches the straight line $\frac{x}{a}+\frac{y}{b}$ = 1 and has its centre in the 1st quadrant, And its equation.
Solution:
Let the centre be at (k, k) and the radius is k.

Question 12.
ABCD is a square of side ‘a’ If AB and AD are taken as co-ordinate axes, prove that the equation of the circle circumscribing the square is
x² + y² = a(x + y)
Solution:

or, x² + $\frac{a^2}{4}$ – ax + y² + $\frac{a^2}{4}$ – ay = $\frac{a^2}{2}$
or, x² + y² – ax – ay = 0
or, x² + y² = a(x + y)

Question 13.
(a) Find the equation of the tangent and normal to the circle x² + y² = 25 at the point (3, -4).
Solution:
Equation of the tangent to the circle x² + y² = 25 at the point (3, -4) is
xx₁ + yy₁ = a²
3x – 4y = 25
Equation of the normal is x₁y = xy₁
or, 3y = -4x or, 4x + 3y = 0

(b) Find the equation of the tangent and normal, to the circle, x² + y² – 3x + 4y – 31 = 0 at the point (-2, 3).
Solution:
Equation of the tangent of the circle x² + y² – 3x + 4y – 31 = 0 at the point (-2, 3) is
xx₁ + yy₁ + g(x + x₁) + f(y + y₁) + c = 0
or, -2x + 3y – $\frac{3}{2}$ (x – 2) + 2(y + 3) – 31 = 0
or, – 4x + 6y – 3x + 6 + 4y + 12 – 62 = 0
or, -7x + 10y – 44 = 0
or, 7x – 10y + 44 = 0
Equation of the normal is x(f + y₁) – y(g + x₁) fx₁ + gy₁ = 0
or, x(2 + 3) – y($-\frac{3}{2}$ – 2) – 2(-2) – $\frac{3}{2}$ × 3 = 0
or, 5x + $\frac{7y}{2}$ + 4 – $\frac{9}{2}$ = 0
or, 10x + 7y – 1 = 0

(c) Find the equation of the tangents to the circle x² + y² + 4x – 6y – 16 = 0 at the point where it meets the y – axis.
Solution:
Putting x = 0 in the circle equation, we have
y² – 6y – 16 = 0
or, y² – 8y + 2y – 16 = 0
or, y(y – 8) + 2(y – 8) = 0
or, (y – 8)(y + 2) = 0
y = 8 or, -2
The circle meets y – axis at (0, 8) and (0, -2).
Eqn. of the tangents are
xx₁ + yy₁ + g(x + x₁) + f(y + y₁) + c = 0
or, 0 + 8y + 2 (x + 0) – 3(y + 8) – 16 = 0
or, 8y + 2x – 3y – 24 – 16 = 0
or, 2x + 5y = 40 and
x × 0 – 2y + 2 (x + 0) – 3 (y – 2) – 16 = 0
or, -2y + 2x – 3y + 6 – 16 = 0
or, 2x – 5y – 10 = 0

(d) Find the condition under which the tangents at (x₁, y₁) and (x₂, y₂) to the circle x² + y² + 2gx + 2fy + c = 0 are perpendicular.
Solution:
Equation of tangent to the circle
x² + y² + 2gx + 2fy + c = 0 at the point (x₁, y₁) is xx₁ + yy₁ + g(x + x₁) + f(y + y₁) + c = 0
or, (g + x₁)x + y(f + y₁) + gx₁ + fy₁ + c = 0
Again equation of the tangent to the circle at (x₂, y₂) is
x(g + x₂) + y(f + y₂) + gx₂ + fy₂ + c = 0
As the tangent (1) and (2) are perpendicular, we have the product of their slopes is -1.
∴ $\frac{g+x_1}{f+y_1} \times \frac{g+x_2}{f+y_1}$ = -1
or, (g + x₁)(g + x₂) = -(f + y₁)(f + y₂)
or, (g + x₁)(g + x₂) + (f + y₁)(f + y₂) = 0

(e) Calculate the radii and distance between the centres of the circles, whose equations are, x² + y² – 16x – 10y + 8 = 0; x² + y² + 6x – 4y – 36 = 0. Hence or otherwise prove that the tangents drawn to the circles at their points of intersection are perpendicular.
Solution:
x² + y² – 16x – 10y + 8 = 0;
x² + y² + 6x – 4y – 36 = 0.
g₁ = -8, f₁ = -5, c₁ = 8,
g₂ = 3, f₂ = -2, c₂ = -36
The centres are (-g₁, -f₁) and (-g₂, -f₂)

Question 14.
(a) Find the equation of the tangents to the circle x² + y² = 9 perpendiculars to the line x – y – 1 = 0
Solution:

(b) Find the equation of the tangent to the circle x² + y² – 2x – 4y = 40, parallel to the line 3x – 4y = 1.
Solution:


(c) Show that the line x – 7y + 5 = 0 a tangent to the circle x² + y² – 5x + 5y = 0. Find the point of contact. Find also the equation of tangent parallel to the given line.

Solution:
we have the line is x – 7y + 5 = 0
or, y = $\frac{x+5}{7}$
Now putting the value of y in the circle, we have x² + y² – 5x + 5y = 0
or, x² + ($\frac{x+5}{7}$)² – 5x + 5 $\frac{x+5}{7}$ = 0
or, 49x² + x² + 25 + 10x – 245x + 35x + 175 = 0
or, 50x² – 200x + 200 = 0
or, x² – 4x + 4 = 0
∴ a = 1, b = -4, c = 4
∴ b² – 4ac = (-4)² – 4 × 1 × 4
= 16 – 16 = 0
∴ The line x – 7y + 5 = 0

x – 7y – 45 and x – 7y + 5 = 0

(d) Prove that the line ax + by + c = 0 will be the tangent to the circle x² + y² = r² if r²(a² + b²) = c².
Solution:
We know that a line is a tangent to the circle if the distance of the line from the centre is equal to the radius.
Now the circle is x² + y² = r²
⇒ Centre is at (0, 0) and radius r. The distance of (0, 0) from ax + by + c = 0 is

(e) Prove that the line 2x + y = 1 tangent to the circle x² + y² + 6x – 4y + 8 = 0.
Solution:

(f) If the line 4y – 3x = k is a tangent to the circle x² + y² + 10x – 6y + 9 = 0 find ‘k’. Also, find the coordinates of the point of contact.
Solution:
Center of the circle is (-5, 3) and the radius is $\sqrt{25+9-9}$ = 5
Distance of the centre from the line 4y – 3x – k = 0

Question 15.
(a) Find the length of the tangent, drawn to the circle x² + y² + 10x – 6y + 8 = 0 from the centre of the circle x² + y² + 4x = 0.
Solution:
Center of the circle x² + y² + 4x = 0 is (2, 0)
∴ Length of the tangent drawn from the point (2, 0) to the circle x² + y² + 10x – 6y + 8 = 0
$\sqrt{\mathrm{x}_1{ }^2+\mathrm{y}_1{ }^2+2 g \mathrm{~g}_1+2 \mathrm{fy}_1+\mathrm{c}}$
= $\sqrt{4+0+10 \times 2+0+8}=\sqrt{32}=4 \sqrt{2}$

(b) Find the length of the tangent drawn from the point (2, -1) to the circle x² + y² + 6x + 10y + 18 = 0
Solution:
Length of the tangent drawn from the point (2, -1) to the circle x² + y² + 6x + 10y + 18 = 0 is
$\sqrt{\mathrm{x}_1{ }^2+\mathrm{y}_1{ }^2+2 \mathrm{gx}_1+2 \mathrm{fy}_1+\mathrm{c}}$
= $\sqrt{4+1+(-6) \times 2+10(-1)+18}$
= $\sqrt{5-12-10+18}$ = 1

(c) Find the length of the tangent drawn from the point (4, 7) to the circle x² + y² = 15.
Solution:
Length of the tangent drawn from the point (4, 7) to the circle x² + y² = 15 is $\sqrt{16+49-15}$ = √50 = 5√2

Question 16.
(a) Prove that the circle given by the equations x² + y² + 2x – 8y + 8 = 0 and x² + y² + 10x – 2y + 22 = 0 touches each other externally. Find also the point of contact
Solution:
x² + y² + 2x – 8y + 8 = 0
g₁ = 1, f₁ = -4, c₁ = 8
Hence centre = c₁(-g₁, -f₁) = c₁(-1, 4)
Radius = r₁ = $\sqrt{1+16-8}$ = 3
Again x² + y² + 10x – 2y + 22 = 0
g₂ = 5, f₂ = -1, c₂ = 22
Centre c₂(-g₂, -f₂) = c₂(-5, 1)
Radius r₂ = $\sqrt{25+1-22}$ = 2
Now

(b) Prove that the circle is given by the equations x² + y² = 4 and x² + y² + 6x + 8y – 24 = 0, touch each other and find the equation of the common tangent.
Solution:
x² + y² = 4,
x² + y² + 6x + 8y – 24 = 0
Their centres are (0, 0) and (-3, -4) and radii are 2 and $\sqrt{9+16+24}$ = 7
∴ Distance between the centres is $\sqrt{(-3)^2+(-4)^2}$ = 5, which is equal to the difference between the radii.
∴ The circles touch each other internally.
∴ Equation of the common tangent is S₁ – S₂ = 0
or, (x² + y² + 6x + 8y – 24) – (x² + y² – 4) = 0
or, 6x + 8y – 20 = 0
or, 3x + 4y = 10

(c) Prove that the two circle x² + y² + 2by + c² = 0 and x² + y² + 2ax + c² = 0,  will touch each other $\frac{1}{a^2}+\frac{1}{b^2}=\frac{1}{c^2}$.
Solution:
x² + y² + 2by + c² = 0,
x² + y² + 2ax + c² = 0
g₁ = 0, f₁ = b, c₁ = c².
g₂ = a, f₂ = 0, c₂ = c².
The centres of the circle are (0, -b) and (-a, 0) and radii are $\sqrt{b^2-c^2}$ and $\sqrt{a^2-c^2}$. As the circles touch each other, we have the distance between the centres is equal to the sum of the radii.

(d) Prove that the circles given by x² + y² + 2ax + 2by + c = 0, and x² + y² + 2bx + 2ay + 2c = 0, touch each other, if (a + b) = 2c.
Solution:
x² + y² + 2ax + 2by + c = 0
x² + y² + 2bx + 2ay + 2c = 0,
The centre of the circle is (-a, -b) and (-b, -a). the radii of the circle are $\sqrt{a^2+b^2-c}$ and $\sqrt{b^2+a^2-c}$. As the circles touch each other we have, the distance between the centres is equal to the sum of the radii.

Question 17.
Find the equation of the circle through the point of intersection of circles x² + y² – 6x = 0 and x² + y² + 4y – 1 = 0 and the point (-1, 1).
Solution:
Let the equation of the circle be (x² + y² – 6x) + λ(x² + y² + 4y – 1) = 0
As it passes through the point (-1, 1),
we have (1 + 1 + 6) + λ(1 + 1 + 4 – 1) = 0
or, 8 + 5λ = 0 or, λ = $\frac{-8}{5}$
∴ Equation of the circle is (x² + y² – 6x) – $\frac{8}{5}$ (x² + y² + 4y – 1) = 0
or, 5x² + 5y² – 30x – 8x² – 8y² – 32y + 8 = 0
or, 3x² + 3y² + 30x + 32y – 8 = 0

Question 18.
Find the equation of the circle passing through the intersection of the circles, x² + y² – 2ax = 0 and x² + y² – 2by = 0 and having the centre of the line $\frac{x}{a}-\frac{y}{b}$ = 2
Solution:

Question 19.
Find the radical axis of the circles x² + y² – 6x – 8y – 3 = 0 and 2x² + 2y² + 4x – 8y = 0
Solution:
x² + y² – 6x – 8y – 3 = 0
2x² + 2y² + 4x – 8y = 0
x² + y² – 6x – 8y – 3 = 0
x² + y² + 2x – 4y = 0
∴ The equation of the radical axis is S₁ – S₂ = 0
or, (x² – y²– 6x – 8y – 3) – (x² + y² + 2x – 4y) = 0
or, -6x – 8y- 3 – 2x + 4y = 0
or, -8x – 4y – 3 = 0
or, 8x + 4y + 3 = 0

Question 20.
Find the radical axes of the circle x² + y² – 6x + 8y – 12 = 0  and x² + y² + 6x – 8y + 12 = 0 Prove that the radical axis is perpendicular to the line joining the centres of the two circles.
Solution:
Equation of the radical axes of the circle x² + y² – 6x + 8y – 12 = 0  and x² + y² + 6x – 8y + 12 = 0
(x² + y² – 6x + 8y – 12) – (x² + y² + 6x – 8y + 12) = 0
or, -12x + 16y – 24 = 0
or, 3x – 4y + 6 = 0
Again, slope of the radical axis is $\frac{3}{4}$ = m₁ (say)
Centres of the circles are (3, -4) and (-3, 4).
Slope of the line joining the centres is $\frac{4+4}{-3-3}=\frac{8}{-6}=-\frac{4}{3}$ = m₂ (say)
m₁. m₂ = $\frac{3}{4}\left(-\frac{4}{3}\right)$ = -1
∴ The radical axis is perpendicular to the line joining centres of the circles. (Proved)

Question 21.
If the centre of one circle lies on or inside another, prove that the circles cannot be orthogonal.
Solution:
The orthogonality condition for two circles.
x² + y² + 2g₁x + 2f₁y + C₁ = 0   …..(1)
and x² + y² + 2g₂x + 2f₂y + C₂ = 0    …..(2)
is 2(g₁g₂ + f₁f₂) – C₁ – C₂ = 0
Let us consider two circles
Case-1. Let the centre of (2) which is C (-g₂, -f₂) lies on the circle (1). Hence it satisfies the equation (i)
i.e., g₂² + f₂² – 2g₁g₂ – 2f₁f₂ + C₂ = 0
⇒ 2g₁g₂ + 2f₁f₂ – C₁ – C₂ = g₂² + f₂² – C₂
Its right-hand side is the square of the radius of 2nd circle which can not be equal
to zero i.e., 2(g₁g₂ + f₁f₂) – C₁ – C₂ ≠ 0
Hence circles are not orthogonal.
Case-2. Let the centre of (2) which is (-g2, -f2) lies inside the circle (1).
Distance between their centres < radius of the first circle.
i,e. $\sqrt{\left(g_1-g_2\right)^2+\left(f_1-f_2\right)^2}<\sqrt{g_1^2+f_1^2-C_1}$
⇒ g₁² – 2g₁g₂ + f₁² + f₂² + 2f₁f₂ < g₁² + f₁² – C₁
⇒ 2g₁g₂ – 2f₁f₂ – C₁ – C₂ > g₂² + f₂² – C₂
= square of the radius of 2^nd circle. Hence greater than 0.
⇒ 2(g₁g₂ + f₁f₂) – C₁ – C₂ > 0
So two circles are not orthogonal. By case -1 and case -2 we conclude that if the centres of one circle lie on or inside another, then circles cannot be orthogonal.

Question 22.
If a circle S intersects circles S₁ and S₂ orthogonally. Prove that the centre of S lies on the radical axis of S₁ and S₂. [Hints: Take the line of centres of S₁ and S₂ as x – axis and the radical axis as y – axis. Use conditions for the orthogonal intersection of S, S₁ and S, S₂ simultaneously and prove that S is centred on the y – axis.]
Solution:
Let the equation of the circle S, S₁ and S₂ are
x² + y² + 2gx + 2fy + C = 0      …(1)
x² + y² + 2g₁x + 2f₁y + C = 0      …(2)
and x² + y² + 2g₂x + 2f₂y + C = 0      …(3)
According to the question, the circle S intersects circles S₁ and S₂ orthogonally.
Hence 2 (g₁g + f₁f) – C₁ – C = 0 …(4)
and 2 (g₂g + f₂f) – C₂ – C = 0  ….(5)
Subtracting (4) from (3) we get
2g(g₁ – g₂) + 2f (f₁ – f₂) – C₁ + C₂ = 0 …(6)
Now radical axis of circles S₁ and S₂ is S₁ – S₂ = 0
i, e. 2x (g₁ – g₂) + 2y (f1 – f2)+ C₁ – C₂ = 0 ….(7)
The centre of the circle S is (-g, -f).
If it lies in the radical axis then equation (7) will be satisfied by the centre.
i,.e, 2g (g₁ – g₂) + 2f (f₁ – f₂) – C₁ + C₂ = 0
which is nothing but equation (5). Hence centres of S lie on the radical axis of S₁ and S₂.

Question 23.
R is the radical centre of circles S₁, S₂ and S₃. Prove that if R is on/inside/outside one of the circles then it is similarly situated with respect to the other two.
Solution:
Given R is the radical centre of S₁, S₂ and S₃
The radical centre is the intersection point of three radical axes whose equations are
S₁ – S₂ = 0
S₂ – S₃ = 0    …..(1)
S₃ – S₁ = 0
Let S₁ : x² +y² + 2g₁x + 2f₁y + C₁ =0
S₂ : x² + y² + 2g₂x + 2f₂y + C₂ =0
S₃ : x² + y² + 2g₃x + 2f₃y + C₃ =0
Now equations of radical axes by set of equation (1) are
2x(g₁ – g₂) + 2y(f₁ – f₂) + C₁ – C₂ =0 …(2)
2x(g₂ – g₃) + 2y(f₂ – f₃) + C₂ – C₃ =0 …(3)
and 2x(g₃ – g₁) + 2y(f₃ – f₁) + C₃ – C₁ = 0 …(4)
Let the co-ordinate of R be (x1, y1) the
point R must satisfy (2), (3) and (4).
i.e., 2x₁(g₁ – g₂) + 2y₁(f₁ – f₂) + C₁ – C₂ = 0 …(5)
2x₁(g₂ – g₃) + 2y₁(f₂ – f₃) + C₂ – C₃ =0 …(6)
2x₁(g₃ – g₁) + 2y₁(f₃ – f₁) + C₃ – C₁ =0 …(7)
Subtracting (6) for (5) we get
2x₁(g₁ – g₃) + 2y₁(f₁ – f₃) + C₁ – C₃ =0
⇒ 2g₁x₁ + 2f₁y₁ + C₁ =2g₃x₁ + 2f₃y₁ + C₃
Similarly subtracting (7) from (6) we get
2g₂y₁ + 2f₂y₁ + C₂ = 2g₁x₁ + 2f₁y₁ + C₁
Combining the above two equations we get
2g₂x₁ + 2f₂y₁ + C₂ = 2g₁x₁ + 2f₁y₁ + C₁ = 2g₃x₁ + 2f₃y₁ + C₃
If R x₁ y₁ lies on / inside / outside of S₁ …(8) then x₁² + y₁² + 2g₁x₁ + 2f1y1 + C₂ (= / < / >)0 respectively.
⇒ x₁² + y₁² + 2g₂x₂ + 2f₂y₂ + C₂(=/</>) 0
⇒ x₁² + y₁² + 2g₃x₃ + 2f₃y₃ + C₃(=/</>) 0
respectively by Eqn (8).
This concludes that if R is on /inside/outside. One of the circles then it is similarly situated with respect to the other two.

Question 24.
Determine a circle which cuts orthogonally to each of the circles.
S₁: x² + y² + 4x – 6y + 12 = 0
S₂: x² + y² + 4x + 6y + 12 = 0
S₃: x² + y² – 4x + 6y + 12 = 0
[Hints: The centre of the required circle S must be the radical centre R (why?), which lies outside all the circles. Then show that the radius of S must be the length of the tangent from R to any circle of the system.
Solution:
Let the equation of the required circle is x² + y² + 2gx + 2fy + C = 0    …..(1)
We know if two circles
x² + y² + 2g₁x + 2f₁y + C₂ = 0 and
x² + y² + 2g₂x + 2f₂y + C₂ = 0
are orthogonal then
2(g₁g₂ + f₁f₂) – C₁ – C₂ =0   ….(2)
According to the question circle (1) is orthogonal to the circles
S₁: x² + y² – 4x – 6y + 12 = 0     ….(3)
S₂: x² + y² + 4x + 6y + 12 = 0   ….(4)
S₃: x² + y² – 4x + 6y + 12 = 0     ….(5)
For these circles equation (2) will be
2(-2g – 3f) – C – 12 = 0     ….(6)
2(2g + 3f) – C – 12 = 0     …..(7)
2(-2g + 3f) – C – 12 = 0     …..(8) respectively.
Now subtract eqn. (7) from (6) and (8) from (7) we get
2(- 4g – 6f) = 0
⇒ 2(4g) = 0 ⇒ g = 0 , and f = 0
Using the value of g and f in eq. (6) we get
C = -12
Using g = 0, f= 0 , C = -12 in (1) we get
x² + y² – 12 = 0 is the required equation of the circle.

Question 25.
Prove that no pair of concentric circle can have radical axes.
Solution:
Let the centre of pair of concentric circles is C (h, k) and radii are r₁ and r₂.
So equation of the circles are
S₁: (x – h)² + (y – k)² = r₁²
S₂: (x – h)² + (y – k)² = r₂²
Equation of the radical axis is S₁ – S₂ = 0
⇒ r₁² – r₂² = 0
which is not a straight line as r₁ and r₂ are constants.
Hence it concludes that no pair of concentric circles have a radical axis.

Odisha State Board CHSE Odisha Class 11 Invitation to English 3 Solutions Writing Description Textbook Activity Questions and Answers.

CHSE Odisha 11th Class English Writing Description

Description

Descriptions are of two types: factual (or realistic) and impressionistic. For a factual or realistic description, you will have to go looking for descriptions of objects or processes. An impressionistic description, on the other hand, maybe factual in places but it is chiefly concerned with recording the impression produced by the describer of a person, place or object. Above all, there is the description of a person.
Some common words used to describe a person
Words that go with hair: long, wavy, curly, brown, dark
eyes : pale, blue, black, flashing
nose : long, high, fleshy
lips : full, thin
shoulder : broad, drooping
age : mid-thirties
voice : commanding
Important points in connection with the description of the person :
A person’s height (tall/short), lean or fat, age, physical appearance, the shape of the face with its prominent features* eyes (bright/dull/dark), hair, forehead, dress and nature.

Activity 5

Read passage C again. Note that there are some points of description which are not very favourable to the person being described. Can you replace them with more favourable descriptions? Reorganise the passage, starting with the sentence: “She was fair and her face was round.”
Answer:
She was fair and her face was round. The prominent feature of her face was its sparkle. She was tall and full of youth. She was lean. She had thick hair and her forehead was not that broad. Her nose was long and her cheeks chubby. She was elegantly dressed. She was brimming with confidence.

Activity 6

You met the people in the picture at a party. Describe them to your friend in a letter. (Have a close look at the picture first.)

Answer:

Satyanagar, Plot No. 121
1st January 20

Dear Sarika,
Yesterday evening I had been to a great party. There was plenty to eat, and numerous games to play but the highlight of it all was the music and dance with which the party closed. Besides the music, we had a dance competition in which the best couple was chosen. The dance went on for an hour. Mr and Mrs Das were adjudged the best couple. They were a perfect match for each other. Mr Das was dressed in a black suit, white shirt and a black bow. He was tall, dark and handsome and he held himself elegantly as he held his wife.

His jet black hair, thick moustache and long sharp nose showed forth a man of character. Mrs Das just a shade shorter than her husband was a cute-looking lady. Her hair was tied in a chignon, her eyebrows were shaped like thin orange pieces. She too had a sharp nose, and a face as clear as glass. She was fair, slim and tall. She sported a short sleeveless gown and a plain slip-on. She also held a dupatta in her hand. She looked really chick and beautiful. They deserved the prize. Well, the party was over at around 4.00 a.m. in the morning. I really enjoyed the party even though it was exhausting.

Please write back.
Yours sincerely,
Rita.

Activity 7

Describe the person in the picture below as interestingly as you can.
Answer:
He wore a long full-sleeved magenta gown and long loose magenta trousers to go along with it. The gown opened onto a large star-shaped neck which was shining black with confetti pasted onto it. His face he had put on a mask with his face farded with cosmetic paint. He had a huge bulbous nose painted in red, large lips painted again in red. He wore no shoes but grey socks which looked extremely funny. In his hand, he held a conical hat which had feathers on its top.

Activity 8

Here is a short dialogue between two friends. They are talking about a mutual friend whose name is Prabhakar. Read the dialogue carefully and write a short description of Prabhakar.
Asaf : You remember Prabhakar? He has become a doctor. He is doing very well, in fact.
Krishna : Prabhakar? The name is familiar, but I don’t remember who you are talking about. What did he look like?
Asaf : He was that short chap with a shining pink face. Always dressed in white. He had long hair, like a girl’s. We used to call him Prabhavati, and how he used to blush then!
Answer:
Prabhakar is a short chap with a shining pink face. He sported long hair like a girl and was always clad in white.

Activity 9

(a) A stranger visited your house during your father’s absence. You received him and talked to him. When your father returned, he wanted to know if any visitor had come while he was away. Describe the visitor to your father so that he can know who you are referring to.
(b) Your mother is looking for a bride for your elder brother. You have seen a girl who, in your opinion, will be ideal for your brother. Describe her to your mother.
Answer:
(a) Dad, this man was around six feet high and darkly complexioned. He had peculiar hairstyle which was parted right in the middle like a girl’s. He also wore ear¬rings in both ears and four stone rings of different colours on his right hand. Two of his toes on the right also had rings in them. He was carrying a shoulder bag and putting on a white dhoti as well as white kameez. There was a long tilak on his forehead. He spoke in the Sambalpuri dialect and he is perhaps in his thirties. I hope you recognise him.

(b) She is very fair and has eyes shaped like petals of a rose, eyes-lashes thin and long like leaves of the touch-me-not. Her hair is thick and black, falling down even below the knees. Her face is spotlessly clean and has a soft look. She is twenty-four but looks like she is in her teens. There is a black birthmark on her chin which adds to the beauty of her face. Her sharp nose and jaw looked as if it was sculpted like a Grecian statue. Her height is about five-feet and she is slim. On the whole, she looked like a model.

Activity 10

Here is a picture of a rhinoceros. Write a short description of the animal for a friend who has not seen a rhinoceros.
Answer:
The very look of it is ferocious. It is a huge mammal, 4 feet high and perhaps 5 feet long, that is bigger than a cow and smaller than an adult elephant. Just like a cow, it has a long face with the snout protruding forward, underneath which is a large mouth. The lower jaw looks like a concave plateau. Its most remarkable features are two 1 horns protruding right out from the snout. Moreover, it has three toes on each of its legs. Its skin is so thick that they look like shielding pad. Besides this, it has a short tail. It is found mostly in Asia and Africa and it is a herbivore.

Activity 11

Here is the description of a particular dog. All the details are present, but not in order. Rewrite the description. Begin with the general appearance and size, then describe the features of the animal, which you find most striking.
(a) He has huge paws, with joined fingers and retractable claws.
(b) Achilles isn’t an ordinary dog.
(c) But the most incredible characteristic is his face, which looks sad and solemn.
(d) Firstly, he is larger than any dog I have ever seen, and he is more like a wolf.
(e) It seems as if he can almost speak if he is given the chance.
Answer:
Achilles isn’t an ordinary dog. He is larger than any normal dog and looks more like a wolf. He has huge paws, with jointed fingers and retractable claws. But the most incredible characteristic is his face, which looks solemn and sad. It seems as if he can speak, if he is given a chance.

Activity 12

Write short descriptions of the following animals. A few questions are given to help you organise your descriptions.

(a) A giraffe How tall is it? What makes it look so tall? Where is it found? What does it eat and how? How does it fight other giraffes and enemies? What kind of sound does it make?
(b) A tiger Where is it found? How tall/long/heavy is it? What is its colour? What is its food? What are man-eaters? How long does a tiger live (life span)? How do tiger cubs look?

Answer:
(a) A giraffe :
Excepting the now extinct Dinosaurs, the giraffe is the tallest mammal found originally on the African continent. Its most characteristic feature is its long neck that protrudes out angularly from its body and is usually about 4 feet to 6 feet long. Besides, just like the leopard, it has black spots spread all over its body which is white-skinned. It is due to this that it was formerly called a came lopard. The giraffe is herbivorous and lives on the leaves of plants which it can easily reach due to its height. It has a very long stride and therefore it is difficult for any preying animal to catch it.

(b) A tiger :
The tiger is a savage and cruel animal. We say, “As cruel as a tiger”. It is really a great cat. It is a large, strong, and fine-looking animal. Its hair is yellow, marked with black stripes. It is shaped like a cat, with a long tail, round head, and thickly padded feet. It has sharp claws and strong teeth. The tiger is an Indian animal. There are many in the jungles of Bengal. Like cats, tigers hunt at night. They kill big animals, like deer, cows, sheep and goats. They attack men, too. Some become “man-eaters”. They like men’s flesh best to eat. Tigers are feared by farmers. They come at night to steal their sheep and cows. Its average lifespan is 17 years and its cubs look like domestic cats with shining black eyes.

Activity 13

Rewrite passage (c), using your own words as far as possible. Divide your description into two paragraphs.

A telephone comes with a bell which can ring, a microphone which converts human speech into electrical signals, an earphone which converts incoming electrical signals back into speech, and a dial which is used to send electrical pulses along the line to an automatic exchange.
Answer:
When we want to make a call, we must lift the handset and then dial the number we want to call. Immediately after we dial the number an automatic selector connects us to an outgoing junction cable that is linked to the exchange we want. A ringing bell indicates the fact that a telephone has come. Then the operation starts. The exchange first connects our phone to automatic selector equipment which in turn connects us to an outgoing cable linked to the exchange that is connected to the number we rang. Finally, this exchange connects us to the phone we are interested in.

Activity 14

Add a short paragraph to passage (d). The hints below will help you in writing the paragraph.

Butter is a rich food made from the cream of milk. It is usually eaten as a spread on bread, but cooks may use butter for frying and making cakes and pastries. Butter contains about 80 per cent fat, the remainder being water, salt and protein. Butter is made from cream, by churning the cream so that the fat [ is separated out. For many centuries, farmers have made butter from cream by churning it by hand in a wooden vat. Nowadays, however, butter is made by machines. First, the milk is whirled in a centrifuge to separate the cream. The l cream is then pasteurised by heating it and then cooling it quickly. This action kills germs in the cream and prevents the butter from going rancid quickly. The pasteurised cream is then churned in huge revolving drums, which separate fat from the liquid in the cream. When the liquid, called the buttermilk, is ‘ drained away, the resulting mass of butter is then cut into pieces and packed.
Hints : How does butter feel when you touch it? Is it tasty to eat? / Is it expensive? j How is butter used in India? etc.
Answer:
Butter is hard but smooth to the touch because it is kept mostly in the refrigerator. Of course, it is tasty but expensive. A mere hundred grams cost thirteen rupees. It is mostly used in cakes and as a spread on bread sandwiches. Sometimes it is used to fry almonds and cashew nuts. However, its use is limited and common people seldom buy it.

Activity 15

Here is a conversation in which an uncle describes a saw to his nephew. Read through the conversation and write a paragraph describing a saw.

Boy : What is a saw, uncle?
Uncle: It’s something we use to cut a piece of wood into two.
Boy : You mean it’s a sort of axe, uncle?
Uncle: No, not an axe, This one has a thinner blade and a short ring-like handle of wood.
Boy : Oh, I know what it is. It’s like a sword.
Uncle: Not really. A saw has one edge sharp. The other edge does not cut.
Boy : Like a big knife?
Uncle: Partly, but the sharp edge does not cut like a knife. There are teeth on the sharp edge. When you press the blade against the wood and move it forward and backwards, like the bow of a violin, the wood gets cut along that line. There are big saws, too, which two people hold at either end to make cuts along the whole length of a log.
Answer:
A saw is an instrument made of either iron or steel with curved teeth on one side or both. The blade of a saw is thin but strong. There is a ring-like wooden handle fitted to one end of the instrument held during sawing wood or wood planks. When it is pressed against the wood, the wood gets cut along that line. Nowadays, the saw has developed a lot from its crude form to a sophisticated one.

Activity 16

Write a short description of the following objects :
(a) A football
Hints:
(i) Size, shape and colour
(ii) Is it smooth, rough or soft to the touch? Is it hard? Is it light or heavy?
(iii) How does it smell?
(iv) Does it bounce? How high?

(b) A ripe mango
Hints:
(i) Size, shape, colour and smell –
(ii) How does it feel to the touch?
(iii) What happens when you press it hard?
(iv) How does it taste?

(c) A pressure cooker
Hints:
(i) What is it? (Definition)
(ii) What does it look like? (parts, size, make, etc.)

Answer:
(a) A football :
A football is a spherical object made of hexagonal leather pieces of alternately white and black colour or plain grey. It has a bladder inside which is inflated by air to give it a round shape. The leather is smooth and soft in the evening but gradually becomes rough because of wear and tear as a result of frequent use. It is light when inflated and bounces up to a height of 15 feet to 20 feet depending on how much it has inflated and how hard it is hit.

(b) A ripe mango :
It is a tropical fruit which consists of a hard kernel, a central core around which is a fleshy pulp. It is yellowish-red in colour and in ovalish in shape. The mango smells sweet and is soft and smooth to the touch; when pressed hard the outer pulp along with the juice comes out. It has a very sweet taste to it.

(c) A pressure cooker :
A pressure cooker is a vessel in which food is cooked in steam under pressure. It consists of a very strong vessel, made of aluminium alloy with a lid that fits tightly on the top. The lid can be sealed onto the vessel by means of a rubber ring. At the centre of the lid, there is a vent or hole through which steam can escape. The food to be cooked is placed in a smaller vessel inside the cooker and a little water is poured into the outer vessel.

Water boils in the vessel and steam begins to escape through the vent. Then we stop the steam by placing a weight on the vent. Steam pressure inside increases and the temperature rises. So the food gets cooked at a higher temperature. This takes only one-third of the time taken by the ordinary method.

Activity 17

Describe the following objects :
(a) A bicycle
(b) A teapot
(c) A fountain pen
(d) A gold necklace

Answer:
(a) A bicycle :
A bicycle is the cheapest and simplest form of transport on wheels. It consists of the main frame and a secondary frame both joined together and triangular in shape. The main frame has a head tube in its front. The handle of the bicycle protrudes out from the upper end of the head tube while the fork protrudes out from the lower end of the head tube. An inner bolt holds both the handle and the fork in place. At the lower end of the fork is the front axle which holds the wheel.

At the opposite end of the upper end of the head, the tube is a tube that protrudes out of the hollow of the main frame. This has a nut-bolt arrangement to hold the seat. Similarly, the peak of the triangular main frame has a hole and axle arrangement to which the crank is connected. The secondary frame has a seat stay which serves as a support for the weight on the seat. The upper end of the seat stay is joined to the main frame while the lower end forks out into two legs which hold the rear wheel in place.

The wheels consist of a central spoke holder from which spokes radiate out into the rim of the wheel where it is screwed. The spokes keep the rim in shape and support it. Besides this, the wheel has an inflatable tube and an outer tyre. The tube has a valve through which air is pumped into it. This valve emerges on to the outer side of the rim through a hole in it. The crank is held by the main frame while the rear frame holds a sprocket wheel. A chain extends from the crank and is wound around the sprocket wheel. The chain is fitted onto them and locked.

The crank further has two pedals joined to it. When force is applied to the pedals, the crank turns and this chain transmits this force applied to the sprocket wheel which is attached to the rear wheel, thereby moving it. Consequently, the cycle moves. To facilitate proper control of the bike, there are brakes. Brake levers are attached to the handle and have brake brackets with rubber on them which are fitted close to the rear and front wheels. Besides this mudguards are provided for both wheels. Finally, a bell and a stand complete the bike. The stand serves as a prop to keep the bicycle standing.

(b) A teapot:
A teapot is a vessel used to brew tea. It can be of various shapes and sizes but most often it is cylindrical in shape with a hollow inside. It is open at one end and this top end has a lid which can be closed or opened as required. The lid is attached to the body of the teapot. The teapot also has a snout with an opening in it from which brewed tea is poured. In teapots of other kinds, instead of a snout, there is a long neck with a mouth at one from which the tea is to be poured. The teapot can be made of various materials like clay, bone china, wood, steel or copper.

(c) A fountain pen:
A fountain pen has two parts. A hollow cylindrical lower part two inches in diameter and a nib holder that is screwed onto it. Besides this, it has a cap with which to cover the nib and protect the ink from spreading. The lower cylindrical half of the pen is filled with ink. This ink passes to the nib which has a sharp pointed end to which it drips. It is with this pointed end of the nib that one writes.

(d) A gold necklace :
Gold is the most precious of all metals. Its bright yellow colour is very pretty. It takes fine polish. Gold is used to make many ornaments out of which a necklace is one. The goldsmith artistically makes it. It is of different sizes and designs. Each gold necklace has a beauty of its own. It is studded with rare stones and diamonds. This necklace is made by hand as well as by machine.

Activity 18

Your father has bought the item in the picture for you. Write a letter to your sister describing what it is, how it looks and what you are going to do with it.

Answer:

Shanti Vihar
7 August 20

Dear sister,
Dad presented me with a very useful gift. I really needed it. Well, don’t hazard any guesses because you might think of the wrong thing. It is a wall clock. The clock is a fairly big one. It has a huge round dial which is fixed onto a round plastic case. This in turn is encrusted into a squarish plastic body. The glass on the dial case is spotlessly clean. One significant feature of the watch is its radium-coated hands.

These shine in the darkness and allow me to know the time even though the lights are off. Besides this, the clock has an alarm system. This serves the purpose of waking up a lazy boy like me. Nowadays I use the alarm to wake up at 5.00 a.m. to go out jogging. That is a thoughtful present from dad, isn’t it? I am writing to him separately to thank him but do tell him how useful it is to me. Thanks for sending cakes and biscuits through your classmate Suneeta.

Your loving brother,
Sushil.

Activity 19

Read through the following paragraph and answer the following questions.

Coal mining-digging coal out of the earth – is a very big industry. Some coal is mined on the surface, but most of it has to be mined deep underground. Both forms of mining are now highly mechanised. On or near the surface, coal is mined by the open-cast method. Huge power shovels first strip off the earth’s overburden above the coal seam. Then the coal is broken up by explosives and shovelled into trucks.

Underground mining is more complicated, more expensive, and more dangerous. Shafts are sunk down into the earth and tunnels are struck outwards from the shafts to the coal seams. Then a machine, called a continuous miner, rips coal from the mine face and loads it onto a conveyor belt, which carries the coal up.

Now answer the following questions :

(i) What is the paragraph about?
(iii) The sentences below give us a simple description of the process of surface ( mining, but they are not in order. Rewrite them in the proper order and use the connectives “first,” “second”, “third” and “finally”. “The coal is thus broken up. Explosives are detonated. The earth above the coal seam is removed. It is loaded into the trucks”.
(iii) There are certain steps involved in underground mining. Write down the steps in proper order. The first one is given to you as an example.
a. Shafts are sunk down in the ground.
b. _______________________
c. Coal is __________________
d. The coal is _______________
e. Then it is ________________

Answer:
(i) The paragraph defines coal mining and enumerates the two kinds of mining. The paragraph also describes the open-cast method of mining.
(ii) Firstly the earth above the coal seam is removed. Secondly, explosives are detonated. Thirdly, the coal is thus broken up. Finally, it is loaded into trucks.
(iii) (b) Tunnels are struck outward from the shafts to the coal seam.
(c) Coal is then ripped from the mine face by a machine called a continuous miner.
(d) The coal is broken up by explosives and shovelled into trucks.
(e) Then it is loaded onto a conveyor belt which carries the coal up.

Activity 20

Here is a description of an experiment demonstrating the process of photosynthesis in green leaves. Read it carefully and note the steps in the experiment.

Two leaves are removed from a de-starched plant. The upper side of one and the lower side of the other are greased with vaseline. The stalk of ‘ each leaf is dipped in water and the leaves are left in the light for four hours so that photosynthesis takes place. Most of the vaseline is wiped off and the leaves are placed in a solution of potassium iodide. The leaf greased on the upper side develops a blue colour, showing that starch has formed by photosynthesis from carbon dioxide, which entered through the leaf pores which are mainly on the underside. No colour develops in the other leaf in which vaseline blocked the pores.

Have you understood the steps involved in the experiment? Can you now help your younger sister conduct this experiment? For this, you may have to give her instructions and let her do the experiment. Give her instructions step by step. You may proceed like this :

1. Take two leaves from a de-starched plant.
2. Grease one leaf on the upper side.
3. _______________________
Continue the instructions till the experiment is over.
Answer:
1. Take two leaves from a de-starched plant
2. Grease one leaf on the upper-side
3. Grease the other on the lower side
4. Dip both leaves in water
5. Then leave it under light for four hours so that photosynthesis takes place.
6. After this wipe off vaseline from the leaves and place it in a solution of potassium iodide.
7. You will now notice that the leaf greased on the upper side develops a blue colour.
8. This shows that starch has formed photosynthesis from carbon dioxide which entered through the leaf pores on the underside of the leaf.
9. No colour develops on the leaf in which vaseline blocked the pores. Structural items used in the passages. We use technical writing while describing the processes etc. These technical writings are commonly impersonal and formal. In this type, the action referred to is more important than the doer of that action. Hence, we express this importance using active voice. Try to fill in the blanks.

1. Coal is mined ___________.
2. Both forms ________ are now highly mechanised.
3. Then the coal is broken up ________ and shovelled
4. Shafts are sunk ________ and tunnels are stuck ________.
5. Two leaves are removed ___________.
6. ________ are greased with vaseline
7. Stalks are dipped __________.
8. Leaves are left ____________.

Answer:
1. Coal is mined on the surface. but most of it has to be mined deep underground.
2. Both forms of mining are now highly mechanised.
3. Then the coal is broken up by explosives and shovelled into trucks.
4. Shafts are sunk down into the earth and tunnels are stuck outwards from the shafts to the coal seams.
5. Two leaves are removed from a de-starched plant.
6. The upper side of one and the lower side of the other are greased with vaseline.
7. Stalks are dipped in water.
8. Leaves are left in light for four hours.

Activity 21

Here is a set of instructions for an experiment on transpiration in plants. Rewrite the description in the passive voice. Select a potted plant and water it sufficiently before the experiment. Cover the soil surface by means of oil paper and check the ordinary evaporation of water. Put the pot on the workbench of the laboratory and cover it with a bell jar. Allow the experimental set-up to continue for one hour. Observe that drops of water stick to the inner wall of the bell jar.
Hint: A plotted plant is selected and it is watered sufficiently before the experiment. (Continue)
Answer:
A potted plant is selected and watered sufficiently before the experiment. Then its soil surface is covered by an oil paper to check ordinary evaporation of water. After this, the pot is put on the workbench of the laboratory and covered with a bell- jar. This experimental setup is allowed to continue for an hour. It is observed that drops of water stick to the inner wall of the bell jar.

Activity 22

The following sentences are from a passage, which tells us about the ideal temperature necessary for the growth of plants. But the sentences are not in order. Put them in order.
1. At lower temperatures the activity of enzymes is reduced; therefore, the growth is also retarded.
2. Most plants grow well between 20-30 degrees centigrade, which may be called the optimum temperature range.
3. The effect of temperature on growth may be indirectly related to the activity of enzymes.
4. But some plants grow well at temperatures lower than 20° C, while other plants grow best at temperatures higher than 30° C.
5. At higher temperatures, the activity of the enzymes in the plant is considerably increased, leading to a kind of ‘exhaustion’ of the plant. Beyond 40° C, the enzymes themselves are destroyed.
Answer:
Most plants grow well between 20°-30° centigrade, which may be called the optimum temperature range. But some plants grow well at temperatures lower than 20° C while other plants grow best at temperatures higher than 30° C. At higher temperatures, the activity of enzymes in the plant is considerably increased, leading to a kind of ‘exhaustion’ of the plant. Beyond 40° C, the enzymes themselves are destroyed. At lower temperatures the activity of enzymes is reduced; therefore, growth is also retarded. Thus the effect of temperature on growth may be indirectly related to the activity of enzymes.

Activity 23

Given below is a diagram which describes how water mixed with solid substances or impurities is distilled. Write a description of the process of distillation.

Answer:
The water mixed with solid substances or impurities is first put into a round-bottomed flask. This flask is then placed on a tripod stand. A rectangular glass tube is then put into the flask through the hole in the cork covering the flask. The tube must reach down to the depth of the water level in the flask. The other hand of the tube must be kept under an empty glass beaker. After this, the flask is to be heated by a glass flame. As the flask is heated, it gradually reaches boiling point and water starts turning into water vapour.

This steam passes through the rectangular tube. As it passes through the tube, the water vapour condenses and droplets of water start falling into the glass beaker. They quicken the process of condensation and the rectangular tube can be attached to a condenser tube through which cold water passes. As all the water in the flask evaporates, the impurities or the solid substance will remain behind in the flask and pure water will be deposited in the beaker.

(a) In ironing a shirt, you first press the cuffs and the sleeves. You then press the collar, inside and outside. After that you ……………….
Answer:
While ironing a shirt, first we press the cuffs and then the sleeves. We then press the collar on both sides with the iron. Then we can press the front part and then | the back ……………….

Activity 24

Describe the following simple processes.
(a) how to make a glass of lassi
(b) how to make a booking for a berth in a reserved compartment (on a train)
(c) how to clean and polish your shoes
(d) how to cook rice
(e) how to send a letter by registered post

Answer:
(a) how to make a glass of lassi:
1. First, take the required amount of curd and milk.
2. Pour them into the jar of the mixer.
3. Add ice cubes and sugar (to your own taste) to it.
4. Then churn and grind it in the mixer using the whipper till the mixture of curd, milk sugar and ice, foam is.
5. Pour it back into a glass.
6. Top ft with garnished coconut, cream, ground cashew nut, bournvita powder or cocoa powder and dried grapes.
7. The lassi is now ready.

(b) how to make a booking for a berth in a reserved compartment (on a train):
1. Procure a reservation slip/form from the reservation counter at the railway station or city booking centres.
2. Fill in the form giving details of the train you want to travel, the class you want to travel to, the date of your journey, your name, age, sex and preference for lower, middle or upper berth.
3. Then give this reservation slip/form to the reservation clerk.
4. The clerk will then check on his computer to find out whether a berth is available on the train and on the particular date you asked for.
5. After finding the availability if it is available, he will print the details on the ticket and pass it to you asking you the fee for it.
6. If the clerk finds that no berth is available, he will tell you what other options are available and you can fill out a new reservation slip with the options available and thereby start the whole process again.

(c) how to clean and polish your shoes:
1. First bring a cherry blossom or a polishing cream (white or black).
2. Use a soft polishing brush for cleaning the dust and dirt.
3. Apply the cherry cream on the brush.
4. Polish the shoes slowly and continuously for some time so as to give them a shining colour.
5. Then apply the cream for a better glaze on the shoes.

(d) how to cook rice:
1. Clean the rice off stones, chaff and burnt rice.
2. Then clean it with water.
3. After this take water that is twice the volume of rice you have taken and set it to boil on the stove in a pot or vessel.
4. When you notice the water boiling, pour the rice into it.
5. Keep it over the fire till the grains of rice become soft.
6. Then drain the water from the pot, so that the cooked rice is left behind ready for consumption.

(e) how to send a letter by registered post:
1. Procure an envelope of the size required by you from the stationery shop.
2. Put the letter inside it and seal it with gum or cello tape.
3. Then write the name and address of the person you want to send it to on the right-hand side of the envelope. Add your name and address to the envelope in the left bottom corner.
4. Take it to the post office and hand it to the registration clerk. He will weigh it and tell you how much stamp it requires. Buy the required amount of stamp from him and paste it on the envelope.
5. Then hand it back to him. He will enter it in a registration journal, write the registration number and date on the envelope and put his Initials on it. He will then hand you a receipt for the letter he received from you.
6. The registration work is done.

Read the following description of a hill station.

(a) Ootacamund, or Ooty (as it is popularly known), which nestles in the Nilgiri Hills, lies on the borders of Kerala, Tamil Nadu and Karnataka. Tourists from both home and abroad flock to this beautiful little hill station for a holiday. The most prominent attraction for them is the Botanical Garden, which was established in 1847. A variety of exotic and ornamental plants adorn this garden. The chief attraction of the garden is a fossil tree trunk which is 20 million years old. A small lake runs through the garden. The government organises in this garden a flower festival in May every year.

(i) What is Ootacamund’s other name? Where is it situated?
(ii) What is its main attraction?
(iii) Where is the lake?
Answer:
(i) Ootacamund’s other name is Ooty. It is situated in the Nilgiri Hills which . “ lies on the borders of Kerala, Tamilnadu and Karnataka.
(ii) The most prominent attraction is the Botanical Garden, which was established in 1847.
(iii) The lake runs through the Botanical Garden of Ooty.

Can you draw up an outline of the passage above and how the description progresses?
Answer:
Paragraph 1: Popular name and location of Ooty – a tourist spot.
Paragraph 2: The Botanical Garden – the most prominent tourist spot.
Paragraph 3: The lake and the flower festival
Now read another description of a place of tourist interest in India.

(b) Junagadh is an ancient city in Gujarat. It is situated among the shadows of Mount Gimar. The name “Junagadh”- Juna (old) and Gadh (fort)- literally means “old fort”. On the outskirts of the city, there is a dark basalt rock. It stands on the way to Mount Gimar. The rock holds the inscriptions of three mighty dynasties. They include the Maurya and Gupta dynasties. The inscriptions are in Sanskrit.

Notice some keywords and phrases used in the descriptions.
existence: Ooty lies on the borders of Kerala, Tamil Nadu and Karnataka,
location: on the outskirts of the city there is a dark basalt rock.

Activity 25

Describe the following places, highlighting their size, location and type. Also, mention the interesting or outstanding features of each place.
(a) Your home town or village
(b) An important place you have visited
(c) Your college

Answer:
(a) My Village :
My village, Mahendragiri is situated in the Gajapati district. It is one-hundred and eighty km away from the silk city, of Berhampur. To reach my village one has to take a bus from Berhampur and after the Tapta Pani Ghat take the route leading to Ramgiri Udayagiri. This place was in the news recently because of communal clashes. Mahendragiri is just 60 km away from R. Udaygiri. The village is situated at the foot of the Mahendra Hills and hence it is called Mahendragiri.

The whole village consists of a cluster of huts, asbestos roof houses and a few concrete buildings. It has only two sahis namely the Nuasahi and the Puranasahi. These sahis are situated one after the other. When one approaches Mahendragiri from R. Udayagiri, one will first see the Nuasahi and after that the Puranasahi. Each ship has rows of houses facing each other. Thus in Nuasahi, we have two rows of houses facing each other and in Puranasahi too there are two rows of houses facing each other.

There are only 200 families living in the whole village. The village has only one main road, the state highway which comes from Berhampur goes past R. Udayagiri to our village Mahendragiri and then continues upto Parlakhemundi, the district headquarters of Gajapati district, which is just 20 km from our village. Nuasahi which is in the south of Berhampur is surrounded by a huge mango grove and tamarind trees.

Puranasahi which is on the north of Parlakhemundi is bordered by cashew-nut plantations. Beyond the mango grove and the cashew, plantation lie the hills. On the top of a hill is a Shiva temple. It can be reached after climbing 480 steps. The temple is a very ancient one. It is now almost in ruins because of a lack of maintenance. Nevertheless, one can see the crude Shiva Lingam in the inner sanctuary always covered with fresh flowers.

The view from the temple courtyard is thrilling. One can see the streams flowing down on the rear of the hill. The sahis looked like tiny rows of toy houses. T.V. antennas look like minute clothes hangers and the mango grove and casuarina trees look like beds of cauliflowers. 3 km away from the village, on the road to Parlakhemundi is our marketplace. It does not have any permanent shops but only rows of rectangular concrete platforms on which businessmen put up their shops.

The market meets on three days of the week, Monday, Wednesday and Friday. It is a very colourful market where one can get everything necessary. Beyond the market is a thick jungle. On the hills, one can also see patches of barren land. This is because the farmers of our village practice shifting cultivation. Some of the plants on the hill are seen as half burnt. Some are already yellow with the flowers of the mustard.

(b) An important place I have visited :
One of the most important and unforgettable places that I have ever visited is the Taj Mahal at Agra. I have not seen anything else that surpasses the beauty of this marble mausoleum. Built by Shah Jahan, as a tomb for his wife and as an enduring symbol of his love, the Taj Mahal is true “an elegy in stone.” It has a gateway of red stone with verses from the Quran inscribed on it.

The gateway leads to a garden with three pathways. Besides that, there are fountains and pillars that lead to the marble platform at whose four comers are four towers or minarets. In the middle is the main dome with two smaller domes flanking it on either side. The red and white marble walls are decorated with stones of various colours encrusted in them. Their insides too are covered by flowers wrought in stone and lace work of green foliage.

The hall of death has a verse inscribed on it. Words cannot describe the splendour of the Tajmahal in the moonlight. It glitters and appears radiant as a bride. Moreover, the large rooms, cool ambience and solitariness about it, give it a sober air whereby one becomes reverential and meditative.

(c) My college :
My college is situated on the National Highway No.5 between Cuttack and Bhubaneswar on the outskirts of our village Kamalpur when proceeding towards Cuttack. It is a red-brick, single-storeyed straight-line building that faces the East. It is constructed only on an acre of land which is marked by a similar red-brick boundary wall that has only one gate which opens onto the highway. The gate has a gravel path which leads to the portico of the college building.

Immediately after the portico is steps which lead to the principal’s office. On the right of the principal’s office is the staff common-room, while to its left is the Administrative Office of the college. All three are housed in single rooms. Beyond the staff common- room and the administrative office on both sides are the stores, the right one storing sports equipment while the left one has stationary. Following the sports store on the right are the classrooms. There are three classrooms in all for 1st Year Arts students.

Similarly on the left beyond the stationeries store are three classrooms for 2nd year Arts students. Beyond the boundary wall, on every side are paddy fields. It is interesting that the college itself has been constructed on an erstwhile paddy field donated by a farmer whose son is one of the teachers in the college. Thus in the rainy season the earth there does not absorb water and as a result, we often have 2-3 ft. of standing water in the College compound. One can even fish inside the college during rainy reason.

Activity 26

Describe the place shown in the picture below.

Answer:
It was a wonderful and picturesque sight just like a picture postcard. A waterfall nestling among the mountains peopled by the evergreen pines standing tall on the mountains. The water cascaded down the fall spraying itself into the air looking like tiny globules of diamond and then crashing into the rock below where it turns into white foam and then cascades down the mountain forming divergent streams that end up in a rivulet.

Activity 27

Describe the following people of our country and the places they live in.
(a) The Kashmiris
(b) The Sikhs
(c) The Santals

Here are some helpful points for (a)
(i) Live in the valley of Kashmir, fair complexion, tall, long noses, about 5 million people
(ii) Very cold winters – snow, frozen lakes, poorly heated mud houses, individual firepots. Woollen clothes, long gowns and rubber shoes
(iii) Food: meat, fish and rice; fruit (apples, pears, peaches, cherries, etc.); Drink: a lot of tea with or without milk
(iv) handicrafts: carpets, silk, wood carvings, etc.
(v) well-known tourist spots: Shalimar Gardens, Gulmarg, Dal lake, etc.

Answer:
(a) The Kashmiris:
The Kashmiris are a fair complexioned people most of who live in the valley of Kashmir which lies in the north of India. They usually have tall and long noses. They experience very cold winters because of snow, frozen lakes and badly heated mud houses. To get rid of the cold, they wear woollen clothes, long gowns and rubber shoes. However, most of the population is poor and hence they live in poorly heated mud houses but each of them has separate firepots. The majority of the Kashmiris are Muslims. As a result, most men wear caps while women are veiled. They wear purdah. This is of course more common among the orthodox folk.

The Kashmiris are mostly non-vegetarian people eating meat, fish and rice. They also consume fruits like apples, pears, peaches, cherries etc. and drink a lot of tea. Their main occupation is handicrafts. Whole families including young children are engaged in weaving carpets, silk clothes, wood carvings etc. all of which are exquisitely done. The main revenue, however, comes from tourism. Kashmir which is considered earth’s paradise has many famous tourist spots like the Dal Lake, Shalimar Gardens and the Gulmarg. Terrorism has however decreased tourist trade in recent times.

(b) The Sikhs :
The Sikhs are the residents of Punjab but over the years they have migrated to almost all the states of India and to many countries abroad. Sikhism began as a socio-religious movement which was more interested in fighting evils but in its process of evolution, it was forced by circumstances to become a militant sect. It was Guru Gobind Singh who transformed the Sikhs into a militant sect and created Khalsa. The Sikh people are easily distinguishable by the turban they put on. Every Sikh is bound by the laws of his religion to never have his hair cut.

Men, therefore, tie their hair in a plait, bind it on the head and wear a turban upon it. Besides this, all Sikhs who are part of the Khalsa are armed with a Kirpan and put on a steel bangle called Kada. Most men are dressed in long Kurtas that reach down to the knees and pyjamas. Women are dressed in salwar kameez. The Sikhs are very hard-working and industrious people. They mainly cultivate wheat, rice, maize, gram and pulses. They produce the largest amount of wheat in India.

Unlike other States, the Sikhs use all mechanised equipment for agriculture and adopt the latest techniques and methods of production. Besides this, they are engaged in several industries like bicycle parts, auto parts, sports and leather goods, hosiery, knitwear, footwear, nuts and bolts, textiles etc. Most Sikhs eat roti or parathas along with Makhan, and dal and drink large glasses of milk. They celebrate the birthdays of their Guru by offering prayer and distributing sweets. The important tourist centres and places of worship in Punjab are the Golden Temple at Amritsar, the Durgiana Mandir, the Anandpur Sahib and the Jalianawala Bagh.

(c) The Santhals :
The Santhals or Santals are an indigenous aboriginal tribe inhabiting Bihar and some parts of Orissa. They lend their name to the Santhal Pargana district of Bihar, which is known after them. They are short dark-skinned people having broad noses, thick lips, coarse and curly hair and very prominent cheekbones. Their main occupation is cultivating and cattle breeding. Most of them are uneducated and illiterate and rarely mix with mainstream society. They are also good hunters.

They are animistic in their beliefs and enjoy sexual liberty practising polyandry and polygamy. Their dialect is called Santali. They live in mud houses short in height. But their villages are extremely clean. The Santals have an elaborate tribal structure, with 12 exogamous clans. More over, each village has its cadre of village officials the head of whom is the chief.

Activity 28

Describe how the Money Order which you send reaches the addressee.
Answer:
The money order form is first filled up and the money to be sent as well as the amount of exchange is given to the postal clerk who gives a receipt in return. It is understood that a customer has commissioned money to be sent to a customer in another place for paying a sum for the service. The postal clerk thus sends this form to the post office where the addressee is to be found. This is taken as a direction by that post office to pay the sum to the addressee, which is done by a postal peon.

Activity 29

Write about 150 words on each of the following :
(a) A peacock
(b) A cat
(c) An elephant

Answer:
(a) A Peacock :
The male peacock is a beautiful bird. Its neck is covered with lovely blue feathers. Its body is green and blue. Its glory is its long tail. It can open its tail like a great fan. The colours are blue and green and gold. The bird is very proud of its tail. It opens it, and struts about to be admired. The peacock stands for pride. We say, “as proud as a peacock”. The lady peacock is a plain brown bird. She has no tail like her husband. He has all the beauty.

In India, peacocks are sacred birds. The peacock is called the mount of Saraswati, the goddess of learning. So it is very wrong to kill a peacock. But peacocks are great thieves. They do great damage to growing crops. Some people say that peacocks kill snakes. Some say they can smell the coming rain. Then they give harsh screams.

(b) A Cat :
People keep cats as pets. Cats are pretty animals, covered with soft fur. They are of different colours. Some are black, some white, some grey, and some brown. Kittens, or young cats, are very playful. They will play for hours with little balls, fallen leaves, or bits of string. The chief use of cats is to catch mice and rats. Like their big cousins, lions and tigers, cats can see in the dark. They hunt for mice at night. Mice are a great pest in a house. A cat will soon kil them, or drive them away.

Cats have been tamed for thousands of years. They were kept as pets in ancient Egypt. Cats are very different from dogs. Dogs love persons, but cats love places. A dog will follow his master anywhere. But a cat loves the comfort of the house and stays at home. Their love is what we call “cupboard love”.

(c) An elephant :
The elephant is the largest of all animals. It is a strange animal to look at. It has thick legs, huge sides and back, large ears, small eyes, a short tail, and great white tusks. Its long nose, or trunk, is the strangest thing about it. It uses its trunk like a hand. It picks things up with its trunk and puts them into its mouth. It sucks up water with its trunk and squirts it into its mouth for drinking.

Elephants are very strong. And they are very clever. So tame elephants are very useful. They are trained to draw heavy loads. They are taught to carry logs of wood on their tusks and pile them up in perfect order. They are used, too, in hunting tigers in the jungle. In old days they were used in battle. And Indian Rajas ride on elephants in state processions.

Activity 30

Write short paragraphs on :
(a) A refrigerator
(b) A screwdriver
(c) A motorcycle

Answer:
(a) A refrigerator :
A refrigerator is a common household electrically powered equipment/appliance that is used to chill or freeze food items for preservation. It consists of an outer metal cabinet or box, rectangular in shape and an inner polyurethane foam lining (pdf) to ensure zero gaps, insulation and provide no space for insects to breed. Its size ranges from 165 litres to 310 litres. Its inside can be cooled to temperatures as low as – 16°C.

The cooling is affected by a thermostat which controls the temperature inside the refrigerator as well as of the freezer compartment. The cabinet contains a separate freezer compartment in which ice can be formed and food kept frozen. The freezer also has ice trays with which ice cubes are made. Below it is the chill tray which is used to store soft drinks as well as milk jackets for quick cooling. Besides, the cabinet may have adjustable shelves, which are found mostly in domestic refrigerators.

In which vessels of different sizes can be accommodated to store cooked food, jellies, pies etc. Right below is the San crisper which is a compartment for storing leafy vegetables and fresh fruits. The door inside also has a dairy compartment for cheese and butter, removable egg racks and adjustable bottle racks. Today frost-free refrigerators are available.

(b) A screwdriver:
A screwdriver is a common tool used for turning screws. It consists of a metal rod that is fixed in a wooden, plastic or rubber groove that has been moulded into a handle grip. The rod is chiselled in the front to facilitate its getting into the groove of the screw. The rod varies in length and diameter.

(c) A motorcycle :
Motor cycle is one of the most popular means of conveyance. Now different brands are manufactured by different companies. Each of them has a distinctive feature. The motorcycle consists of various parts, such as a handle, brake, fuel tank, silencer pipe, engine (two-stroke/four-stroke) carburettor, clutch lever, speedometer, two tyres, indicator (front and rear), and battery compartment. These parts are systematically set in the bike’s comfortable seat, headlight, shock absorber and so on.

Four-stroke motor-bike is superior to a two-stroke one because the former has smooth pick-up. Besides, it doesn’t produce defeaning sound. On the other hand, the motorbike has a two-stroke engine that doesn’t have that smooth pick-up. It produces sound. The fuel consumption of a four-stroke engine is better than that of a two-stroke engine. The former is economical. Replacement of engine oil at the scheduled time is of great importance. Now wherever we notice, we see varieties of wonderful bikes playing on the road.

Odisha State Board CHSE Odisha Class 11 Invitation to English 3 Solutions Writing Narration Textbook Activity Questions and Answers.

CHSE Odisha 11th Class English Writing Narration

There are different ways of developing a paragraph depending on the topic and the purpose of writing. We shall study some of those ways now and in a subsequent chapter, we will develop our ideas into a longer piece of writing in the same light. We will also learn to mix different ways of developing our ideas into one or more paragraphs.

The common ways of developing paragraphs are :
(i) Narration – To tell an event, incident., or experience in chronological order.
(ii) Description -To describe a person, animal, object, place or process.
(iii)Exposition – To explain an idea, instrument, or problem.
(iv)Argumentation or persuasion -To argue for or against a view, in order to influence the reader’s opinion.

Narration

Read the following paragraph and see how the sentences have been arranged.

On Sunday morning, I get up at six in the morning. After a quick wash, I get into my jogging rig and go for a run. By 6.30 I am on the road. I run half an hour. I return home and have a leisurely bath, a luxury. I cannot afford it on weekdays. The bath is over, and I get ready quickly. What have you done? You have described your activities on a Sunday morning, in the order in which they take place.

You start with what you do first, then go on to what you do next, and so on and you come to your last activity. This brings you to the most important thing about a narrative paragraph. In a narrative paragraph, the events or happenings are arranged in chronological order, that is, in order of time.

A. Narrative Passage

A. Account of Events with Sequence of Action:
It was late in the evening. The bride’s place was richly illuminated and decorated. There was music and dancing all around. Around half past 10, the bridegroom’s procession arrived and there was a flurry of excitement. The bridegroom was sitting erect on a young horse. The bride’s parents came out to receive the bridegroom and the bride was brought out too, her face covered with a pink and gold veil. She was barely 18. But, suddenly, without any provocation, the horse broke away from his master’s grip and ran away taking the bridegroom with him.

The bridegroom shouted for help and clung on to the horse for dear life. Soon his turban fell off, revealing his sparse white hair flying in the breeze. “He was too old to marry”, everyone concluded. He couldn’t be less than 70. The young bride stood aghast. Turning to her parents she cried aloud, “He’s old ! He’s a doddering old man ! I won’t marry him”. Tearing off her finery she stormed back into the house. And the old man was taken to hospital with multiple fractures.

B. A Different point of view:
At last, we reached the bride’s place. It was well lighted and there was music and dancing all around. The atmosphere was exciting and I liked it. The bride’s parents came out to receive the bridegroom and then the bride was brought too, her face covered with a pink and gold veil. I stood on the tips of my hooves to get a better view of her. Suddenly the veil on her face blew up in the breeze and I had a glimpse of her face. I was shocked.

She was only a child! She couldn’t be more than 18. And she was about to marry a man old enough to be her grandfather! “Something is seriously wrong”, I thought. “This marriage must be stopped! I then looked sly at my master. He was looking the other way. Without losing any time I broke away from his hold and ran, taking the bridegroom with me. When I reached the street I threw him on the ground and bolted away.

Activity 1

Read the accounts in passages A and B again. Then narrate the incident from the point of view of :
(i) the bridegroom
(ii) the bride’s father
(iii) the bride
Answer:
(i) The bridegroom:
It was late in the evening when I reached the bride’s place which was richly decorated and illuminated. It was a treat to watch. The bride’s parents then came out to receive me and my friends. She was there too, with her face covered with a pink and gold veil. I wanted to take a good look at her as I sat on my horse. Fortunately, a breeze blew her veil away and I got a full glimpse of her face. she was young and beautiful! My heart beat fast in excitement. I couldn’t wait to get married.

Suddenly without warning my horse broke away from my grip and galloped away as if frightened. The sudden jerk loosened my turban and it fell off my head. Alas ! my white hairs were now visible! I could hear shouts of amazement from people gathered for the marriage. They were calling me an old man, fit to be a grandad. All my hopes of marriage were dashed. Suddenly, I was thrown off the horse and consciousness. When I regained it, I found myself in the hospital. I was told that the girl refused to marry me.

(ii) The bride’s father:
It was the happiest day of my life. My one and only daughter were getting married. I spent money lavishly on decorating my house. The bridegroom had not arrived till now. It was almost ten o’clock. Fortunately, however, they arrived half an hour later. I went out to receive him along with my wife and daughter. He sat astride on a horse. He had put on a coffee-colored sherwani suit. A gold brocaded yellow silk turban wrapped his head.

He looked handsome. I was proud of him – a man of many means. We now faced to face and I was asking him to get down from the horse so that the rituals to welcome him could be started, but he was inattentive. I followed his eyes. He was looking at my daughter. “Natural”, I thought. Suddenly, the horse raised its forelegs high up in the air, broke its master’s grip, and galloped away toward the street. I thought I could hold its reins but it was too fast for me.

My would-be son-in-law was now crying out loudly for help but in vain. Being afraid, I ran behind the horse. At that moment my would-be son-in-law’s turban fell off and I was aghast by what I saw. There was only a little white hair clinging to his bald head! O God, he could not.be less than seventy! “Marry my daughter”, would he ?” I said to myself. I then looked back to see my daughter rushing into the house with tears rolling down her cheeks.

(iii) The bride:
The moment of my marriage had arrived. My parents took me out with them to receive the bridegroom. As usual, this was an arranged marriage and I had not even had so much as a glimpse of my fiance before. So when we neared him sitting astride on a house, I tried to look at him through my veil but was not satisfied with the hazy figure that I perceived. Luckily, a breeze lifted off my veil and I had a fair look at him. He looked like a chivalrous knight sitting on a horse. He looked smart and handsome.

Suddenly, however, his horse neighed loudly, raised its forelegs high up into the air, and bolted away. This movement disrobed his head and his turban fell off. What I saw turned me speechless for a moment. He was completely bald, except for a few strands of impeccably white hair. I was shocked. I couldn’t believe my eyes. I had been deceived! Screaming, I ran into the house. “I couldn’t marry this old man, no not at all”. I thought to myself. Tears filled my eyes and I was disconsolate. It was indeed providential that the horse had bolted and the turban had fallen off.

Activity 2

One day you saw a woman faint at a bus stop. Narrate what happened; describe the sequence of events and the reaction of the people at the bus stop.
Answer:
It was a blazing summer afternoon. I could feel the heat almost scalding my cheeks. I was perspiring like a faucet as I reached the bus stop from the college. It was mid-noon then. The streets were deserted and desolate. The tar on the road was melting. There was no one at the bus stop except for a lady in a synthetic saree. “How the hell can she put on a synthetic saree on this hot noon ?” I thought she was holding a bag of vegetables – “A housewife, returning home after marketing44, I surmised.

My lips were dry and so I crossed the road to have sugarcane juice. There were four to five people at the crushing trolley. As I ordered a glass of juice and started sipping it all the while facing the bus stop to see if the bus is arriving. Suddenly I saw the woman standing across the road, at the stop, fell down unconscious. For seconds I did not realize that she had fainted.

We all did but none moved to help – they were men and she was a woman. She was old enough to be my mother. I ran to her; saw that she was faint and her lips were trembling, her mouth was completely dry. I ran back, took an ice block from the juice seller and a glass of water, and rushed back. Meanwhile, all my fellow drinkers had gathered around her. They asked me to sprinkle water on her face which I did. But then I also poured water into her mouth and massaged her head with the ice block.

Within a few minutes, she regained consciousness and blushed with confusion as she saw the crowd around her. Holding her hand I lifted her up and handed her another glass of water. She drank that and her cheeks regained color. I helped her pick up the vegetables that had fallen all around her, hailed an auto at her request, and saw her off.

Activity 3

Write short accounts of the following imaginary incidents :
(i) You saw a man trying to steal someone’s wallet (purse) and caught hold of him.
(ii) A road accident you witnessed.
Answer:
(i) It was at the Bhubaneswar railway station that this happened. I was standing in the queue, nonchalantly like the rest of the others, waiting for my turn at the booking counter. It was around 10 o’clock in the morning. The heat and sweat were making everyone restless. To add to this, persons in the queue were having unnecessary arguments with those trying to bypass the queue and go to the counter.

I was fidgeting with my watch, noting how much time each person’s booking was taking and calculating roughly how much time it would take for my turn to come. Just then a man, handsomely dressed arrived and went straight to the counter. He then asked something to the clerk and stood there. Watching his demeanor, I did not like to ask him to move from the I place and take his place in the queue. Instead, I was admiring his mustache jeans. It looked smart on him.

Suddenly I saw him reach for the wallet in the back pocket of the person, first in the queue. He picked it up and turned back. Spontaneously I shouted out, [ “you thief’ and embraced him in a hug. He threw it on the floor and feigned ignorance about it. However, everyone had seen him throw it and so he was caught red-handed. Meanwhile, hearing the commotion the railway police arrived and arrested him.

(ii) It was a Sunday morning, 7.00 a.m. I was off on my bicycle to IRC village to buy vegetables from the Gandhi Market. I was on the road that runs parallel to the National Highway in Jaydev Vihar. All of a sudden a mini truck sped past me, raising a lot of dust. It immediately turned left to go onto the highway taking the mud track road. Just before it reached the highway, there was a loud thud. The four wheels of the truck had fallen into the ditch that had been recently dug to make a drain along the highway.

The driver did not know this and had taken that track to reach the highway. He was unfortunate. The four wheels were deep in the ditch and the truck’s back had been thrown up into the air with the rear tires hanging and wheeling. I rushed to the spot. The driver had escaped unhurt but the cleaner was lying unconscious in the driver’s cabin. We quickly brought him out through the open door of the truck. Pieces of glass had pierced his cheeks and forehead. He was bleeding profusely. The driver hailed a taxi and took him immediately to the hospital.

Activity 4

The Prime Minister is to visit Bhubaneswar next week and the following is the tour program. Write a short account of the planned tour, using the points below.
10.0 am : arrival by a special Air Force plane.
10.10 a.m.: reception by the Chief Minister at the airport.
10.30 a.m.: meeting at the State Secretariat; discussion with the Chief Minister
11.30 a.m.: meeting with party workers.
12.30 p.m.: lays the foundation of the Software Technology Park.
1.0 p.m. : meeting the press.
2.0 p.m. : lunch at Raj Bhavan.
3.0 p.m. : return to Delhi.
Answer:
The Prime Minister arrives at the Biju Pattanaik Airport at 10.00 a.m. sharp by the special Air Force plane. He is to be received there by the Chief Minister and other Cabinet Colleagues. After this, he heads straight for the Secretariat where he discusses relief measures granted by the World Bank and other funding agencies for the cyclone-affected area. After this, at 11.30 p.m.

he reaches the B.J.B. party office to meet party workers. There he discusses organizational elections. At 12.30 p.m. he reaches the site for the Software Technology Park and lays its foundation stone. Following this, he attends a press conference organized by the BBSR Press Club. At 2.00 p.m. he has lunch with the Governor at the Raj Bhavan. At 3.00 he once again boards the plane to leave for Delhi.

Odisha State Board CHSE Odisha Class 11 Invitation to English 3 Solutions Writing a Paragraph Textbook Activity Questions and Answers.

CHSE Odisha 11th Class English Writing a Paragraph

A. Form And Function Of A Paragraph

A paragraph is a carefully and deliberately ordered arrangement of sentences built around a central theme or topic. Every paragraph should have a central theme either started in general terms in one sentence and reinforced by other sentences or implied when all the sentences are taken together. In short, a good paragraph comprises unity, order and organization.

Example
The octopus is one of the strangest creatures in the world. The octopus has eight long arms or tentacles round its head. Each tentacle has two rows of suckers along its length. The tentacles of the common octopus are less than one metre long. The tentacles of the largest octopuses are as long as 4.5 metres. The five sentences in the paragraph above contain five pieces of information. These five pieces are related on some way; otherwise, they would not be in the same paragraph.

Secondly, they are put together for a particular purpose: to describe the strange creature called the ‘octopus’. The opening sentence states the central theme of the paragraph in general terms. This is followed by five particular statements, all of them in support of the general statement. We should notice, again, the order in which the sentences are arranged. If we try to rearrange the sentences in any other order, we shall find that they do not add up to a satisfactory paragraph. In short, we must remember the most important principle, “One point, one paragraph”.

Activity 1

Each group of sentences below belong to a paragraph, but they are not in the proper order. Rewrite them in their correct order to form a coherent paragraph.

Question 1.
(a) The day is not far off when there will be regular space-flights to these planets.
(b) He is now planning to travel to Venus and back.
(c) Man has already been to the moon.
Answer:
Man has already been to the moon. He is now planning to travel to Venus and back. The day is not far off when there wil be regular space-flights to these planets.

Question 2.
(a) They can also act in indirect ways – by refusing to buy products made from rare animals, and by compelling law-makers to pass sound conservation legislation.
(b) People can help protect wildlife in simple, direct ways.
(c) They can join forces with other citizens in supporting organisations that are devoted to the cause of preserving nature and wildlife.
(d) They can maintain a bird feeding station, confine family dogs at night, or put a bell on the cat.
Answer:
People can help protect wildlife in simple, direct ways. They can maintain bird feeding station, confine family dogs at night, or put a bell on the cat. They can also act in indirect ways – by refusing to buy products made from rare animals, and by compelling laW-makers to pass sound conservation legislation. They can join forces with other citizens in supporting organisations that are devoted to the cause of preserving nature and wild life.

Question 3.
(a) The third runs across the south of Europe and North Africa into the middle of Asia.
(b) There are several other much smaller fault lines, but these three are the main ones.
(c) The first runs along the east coast of the Asian continent down to the west coast of North America.
(d) There are three large regions in the world where earthquakes are most likely to happen.
(e) The second runs down the west coast of South America.
Answer:
There are three large regions in the world where earthquakes are most likely to happen. The first runs along the east coast of the Asian continent down to the west coast of North America. The second runs down the west coast of South America. The third runs across the south of Europe and North Africa into the middle of Asia. There are several other much smaller fault lines, but these three are the main ones.

Question 4.
(a) But then it was discovered how paper could be prepared from wood-pulp.
(b) Paper was first made in AD 105 by a China man, Tsai Lun.
(c) Until the mid 1805 most paper was hand-made from rags or from grasses.
(d) He discovered that certain plant materials could be broken down into fibres and pressed into a sheet which made a good writing material.
Answer:
Paper was first made in AD 105 by a China man, Tsai Lun. He discovered that certain plant materials could be broken down into fibres and pressed into a sheet which made a good writing material. Until the mid 1805 most paper was hand-made from rags or from grasses. But then it was discovered how paper could be prepared from wood-pulp.

Activity 2

You will find below, in (a), (b) and (c), some advertisements. The sentences in these advertisements have been jumbled up. Re-arrange them in the proper order.

Question (a)
(i) Silver sterilisation inhibits the growth of bacteria with its oligodynamic effect.
(ii) Yes, AMP is 100% safe, pure drinking water.
(iii) Acqua Minerale Pura (AMP) is purified water, passed through multiple filters and treated with the special Italian process of silver sterilisation.
Answer:
Yes, AMP is 100% safe, pure drinking water. Acqua Minerale Pura (AMP) is purified water, passed through multiple filters and treated with the special Italian process of silver sterilisation. Yes, AMp is 100% safe, pure drinking water. Silver sterilisation inhibits the growth of bacteria with its oligodynamic effect.

Question (b)
(i) Your hair will stay healthy and beautiful.
(ii) Use, Neo-Karpin Hair Vitaliser regularly each morning one hour before your bath.
(iii) Use it also before going to bed.
Answer:
Use, Neo-Karpin Hair Vitaliser regularly each morning one hour before your bath. Use it also before going to bed. Your hair will stay healthy and beautiful.

Question (c)
(i) It gently smooths away premature wrinkles and firms your skin.
(ii) Qadil Cucumber Face Pack gives your skin a youthful and firm look.
(iii) It also nourishes your skin with proteins and vitamins.
(iv) Finally, it leaves the skin soft to the touch.
Answer:
Qadir Cucumber Face Pack gives your skin a youthful and firm look. It gently smooths away premature wrinkles and firms your skin. It also nourishes your skin with proteins and vitamins. Finally, it leaves the skin soft to the touch.

1.2 Topic Sentence

The sentence starting with the central theme of the paragraph is called a topic sentence.
Example :
The life cycle of a virus consists of two phases. The first is extracellular. This means that the virus lives outside the cell. The second is intracellular and this means that the virus lives inside the cell where it reproduces. Here the first sentence, “The life cycle of a virus consists of two phases” is the central theme of the paragraph and hence it is called the topic sentence. Within the paragraph, the topic sentence can be
(a) in the beginning
(b) at the end
(c) in the middle
(d) split and placed in two places
(e) implicit where the main idea is distributed over a number of sentences. Let us now see some paragraphs containing topic sentences at different places within them. When it comes towards the end, it usually serves as a climax.

Activity 3

Now, pick out the topic sentences in paragraphs II and III, and write their outlines.

Occasionally, the topic sentence can come in the middle of the paragraph. In such a case, the paragraph shows a contrast between two ideas. Read the following paragraph and its outlines.
¹Traditionally we have tended to think of meat, dairy produce and eggs, as the normal protein supplements to our diet. ²Peas, beans, lentils and other grains and vegetables are regarded as cheaper alternatives. ³It may come as a surprise to many to learn that in fact, the humble beans and lentils in our diet are richer in protein content than the products in the first category (meat and dairy). ⁴For example, cheese contains about 25 per cent protein, beef and lamb about 14 per cent, and pork, chicken and eggs about 11 per cent. ⁵In comparison, grains contain 20 to 40 per cent, depending on the variety.

Sentence  1	:	traditional belief	meat, dairy products and eggs considered normal sources of protein.
Sentence  2	:	traditional belief	peas, beans, lentils, etc. considered cheaper alternatives (sources of protein)
Sentence  3	:	a surprising fact	beans, etc. are richer in proteins than meat etc. (Topic sentences)
Sentence  4	:	supporting examples	cheese has 25% (etc.)
Sentence  5	:	further examples	beans have 40%

Answer:
Topic sentence :
Traditionally, we have tended to think of meat, dairy produce and eggs as the normal protein supplements to our diet.

¹Primarily democracy is the conviction that there are extraordinary possibilities in ordinary people, and that if we throw wide the doors of opportunity so that all boys and girls can bring out the best that is in them, we will get amazing results from unlikely sources. ²Shakespeare was the son of a bankrupt butcher and a woman who could not write her name. ³Beethoven was the son of a consumptive mother, herself daughter of a cook and a drunken father. ⁴Schubert was the son of a peasant father and mother who had been in domestic service. ⁵Faraday, one of the greatest scientific experimenters of all time, was bom in a stable, his father an invalid blacksmith and his mother a common drudge. ⁶Such facts as these underlie democracy. ⁷That is why, with all its discouraging blunders, we must everlastingly believe in it.

Sentence 1	:	Democracy defined + its primary benefits	Topic sentence (first half)
Sentence 2	:	Example	Shakespeare
Sentence 3	:	Example	Beethoven
Sentence 4	:	Example	Schubert
Sentence 5	:	Example	Faraday
Sentence 6	:	A transitional sentence preparing us	for the conclusion
Sentence 7	:	Writer’s conclusion	Topic sentence (second half)

Answer:
Topic sentence :
(i) Primarily democracy is the conviction that there are extraordinary possibilities in ordinary people, and that if we throw wide the doors of opportunity so that all boys and girls can bring out the best that is in them, we will get amazing results from unlikely sources.
(ii) That is why, with all its discouraging blunders, we must everlastingly believe in it.

Activity 4

Read the paragraph below and write its central idea.
It arrived at 5.00 a.m. after a sleepless night, spent thinking about questions I wanted to ask Guruji. He told me, to sit in the lotus posture and to look into his eyes without blinking. He put his hand on my chest and told me to concentrate there. As soon as he touched me, I felt as if an electrical current was passing through me. I lost consciousness. When I regained consciousness, I found that I had my head on the lap of Guruji and that he had kept his hand on my chest. Guruji told me to remember this experience but not to tell these things publicly anybody.
The central idea :
The disciple’s interaction with his Guruji, his extraordinary experience in response to the Guruji’s touch on his body and Guruji’s instruction to him to keep the fact a secret.

Activity 5

Find the topic sentences of the following paragraphs. Underline them. Then write on the margin the supporting details in single words or short phrases.

Question 1
¹Friction creates many problems for the designers of automobile engines. ²First, friction wastes energy and reduces the power produced by the engine. ³Secondly, friction creates heat, which can lead to over-heating of the engine and jam up the moving parts. ⁴Thirdly, frictions lead to excessive wear and tear of the engine.
Answer:
Topic sentence :
Friction creates many problems for the designers of automobile engines. It leads to wear and tear of the engines.

Question 2
¹In Europe, in pre-historic times, people used to tame and keep wolves as hunting companions. ²The Ancient Babylonians tamed hyenas for the same purpose. ³The Egyptians considered cats to be sacred animals but used them to kill mice and rats in their granaries. ⁴People in the Middle Ages tamed falcons and other birds of prey and used them to hunt. ⁵The keeping of animals as pets has gone on since the earliest times.
Answer:
Topic sentence :
The keeping of animals as pets has gone on since the earliest times.
This paragraph has the following points :

Sentence 1	:	Pre-historic Europe	Wolves used as hunting companions.
Sentence 2	:	Ancient Babylon	Hyenas used as hunting companions.
Sentence 3	:	Egyptians	Considered rats sacred but used them to kill mice and rats in granaries.
Sentence 4	:	Middle Ages	Used Falcons and other preying birds for hunting.
Sentence 5	:	Animals kept as pets from earliest times.

Question 3
¹Most people think of science merely as a servant which can make their lives easier. ²Science adds to the wealth of nations and creates better living conditions. ³This may be true. ⁴But science is more than that. ⁵The history of science shows that it does not simply better the old; it sometimes upsets the old. ⁶It does not merely add new truths to the old ones, but sometimes the new truths it discovers destroy some part of the old truths and thereby upsets the way of men’s thinking and living.
Answer:
Main idea :
The historv of science shows that it does not simplv better the old : but it sometimes upsets the old.

Sentence 1	:	Common idea about science	It is a servant which makes people’s lives easier.
Sentence 2	:	Supporting detail	Science adds to the wealth of nations and creates better living conditions.
Sentence 3	:	Confirming propositions of sentence 1 and 2	This may be true
Sentence 4	:	Introducing another proposition	But science is more than that.
Sentence 5	:	Main idea introduced	Science simply does not better the old; it sometimes upsets the old.
Sentence 6	:	Elaboration of main idea	Not only adds new truths but such new truths destroy parts of the old truths which upset men’s way of thinking and lilving.

Question 4.
The Youth Hostel movement has, during the last half-century, contributed to international understanding and racial equality. One of the principles embodied in the constitution of the International Youth Hostel Federation (IIHF) is that in the hostels ‘there shall be no distinctions of race, nationality, colour, religion, class or political opinion.’ In the 4363 Youth Hostels of the world, the young men and women of the world meet and make friends. Making no distinction between the rich and the poor, the white and the coloured, the conservative and the radical, young people from all parts of the world share their experiences in an atmosphere of informality and friendliness. In these hostels, equality and brotherhood are practised in a real sense, and mutual understanding and trust are fostered.
Answer:
Topic Sentence :
The Youth Hostel movement has, during the last half-century, contributed to international understanding and racial equality.

Sentence 1	:	One of the principles enshrined in its constitution	No discriminations in the hostels, irrespective of race or nationality.
Sentence 2	:	Fact        Meeting of the young men and women of the world and	meeting friends in the 4364 youth hostels of the world takes place.
Sentence 3	:	Details	Young people from all over the world share their experiences in a coordial atmosphere. No distinction between the rich and the poor, the white and coloured, the conservative and radical is marked.
Sentence 4	:	Hallmark of these hostels	Practising equality and brotherhood in word and spirit and fostering mutual understanding and trust.

Question 5
Just after sunset, when all our work was over, I wanted to eat an apple. I went on deck, I got into the apple barrel, but all these apples had been finished. I sat down there in the dark and the motion of the Hispaniola (name of a ship) made me feel sleepy. I was closing my eyes in sleep when a heavy man sat down close by the apple barrel. I was about to jump up when the man began to speak. It was John Silver. Before I had heard a dozen words, I was trembling with fear. I prayed they would not discover me because Silver’s dozen words told me that all our lives were in danger. Only I knew it, the only I could warm them before it was too late.
Answer:
Topic Sentence :
Just after sunset, when all our work was over. I wanted to eat an apple.

Sentence 2	:	The narrator’s interest to fulfil his wish	Entered the apple barel, but no apple.
Sentence 3	:	Result	Sitting there in the dark, felt sleepy, the presence of a heavy man by the apple barrel
Sentence 4	:	Result	The narrator’s readiness to jump; he heard the man speak
Sentence 5	:	Identify that person	John Silver.
Sentence 6	:	John Silver’s utterance and impact on the narrator	Became fear-stricken.
Sentence 7	:	The narrator’s prayer and its reason	Failure to discover me, Silver’s dozen words were indicative of their danger
Sentence 8	:	Reason	He alone knew it to warn them soon.

Question 6
There are many reasons why the Korowai people build their huts high up among the trees. If a rival clan attacks, the women, children and the old men can climb to safety while the warriors fight it out on the ground. The mosquitoes cannot fly so high and the air up there is cooler. Lastly from the verandahs of their huts the Korawais can shoot arrows at birds flying by and also keep watch on their plantations, where they grow many types of bananas and sweet potatoes.
Answer:
Topic Sentence :
There are many reasons why the Korowai people build their huts high up among the trees.
1. Security to all in the face of rival attack
2. Free from mosquito-bite
3. Close watch on their plantations

Question 7
On my way home from work yesterday evening, I saw a small boy standing outside a door struggling to reach the door bell. “Here let me help you”, I said pressing the door bell for him. “Thanks, mister” said the small boy, starting to run away. “But do not stand there, you will get caught, “he added as he disappeared down the road.
Answer:
Topic Sentence :
On my way home from work yesterday evening, I saw a small boy standing outside a door struggling to reach the door bell.
1. The narrator’s help
2. The boy’s delight
3. The latter’s caution to the former
4. The boy’s disappearance

Question 8
When the motor car was first introduced, it was a clumsy, noisy machine that laboured along the street at a pace no faster than that of a trotting horse. People looked at it with suspicion and fear. Since then, the motor car has come a long way and is today a combination of beauty, luxury and efficiency. The modem automobile, sleek and streamlined with its brilliant colours, is almost a work of art. You can travel in it in absolute comfort at an incredible speed, and yet feel no fatigue at the end of the journey. The engine is a masterpiece of mechanical ingenuity and seldom lets you down if you maintain it in good condition.
Answer:
Topic Sentence :
The motor car is today a combination of beauty, luxury and efficience.

S.1	:	when introduced it was a clumsy noisy machine and was only as fast as a horse.	–
S.2	:	Its original image:	viewed with suspicion and fear.
S.3	:	Its modern image:	a combination of beauty, luxury and efficiency.
S.4	:	Its beauty (detail to support S: 3)	considered a work of art.
S.5	:	The luxury it provides: (detail to support S: 3)	can travel in absolute comfort, incredible speed without feeling tired over long distances.
S.6	:	Its efficiency: (detail to support s: 3)	a masterpiece of mechanical ingenuity which never lets anyone down if properly maintained.

Question 9
Air pollution affects our health in many ways. Large quantities of smoke and other particles in the air can cause lung-diseases including cancer. Sulphur dioxide, in particular, causes serious damage to the respiratory system. Besides, other pollutants present in the air can cause skin infections as well as infections of the eyes.
Answer:
Topic Sentence :
Air pollution affects our health in many ways.
Sentence 2 : Example 1 : Smoke and other particles cause lung diseases including cancer.
Sentence 3 : Example 2 : Sulphur dioxide causes serious respiratory problems.
Sentence 4 : Example 3 : Other pollutants cause skin as well as eye infections.

Question 10
‘In their natural habitat animals face different hazards, such as diseases, storms, fires and floods. 2They may also be hunted by men or attacked by predators. 3This results in their natural life span being cut short prematurely. 4If we want to find out about the longevity of animals, the best way is to study them in captivity, in zoos, protected animal parks, etc. ‘
Answer:
Topic Sentence :
If we want to find out about the longevity of animals, the best way is to study them in captivity, in zoos, protected animal parts, etc.
Sentence 1 : Fact 1 : Animals face hazards like diseases, storms, fires and floods in their natural habitat.
Sentence 2 : Fact 2 : They may also be hunted or attacked by predators in their natural habitat.
Sentence 3 : Consequence : Their natural life span is thus cut short prematurely.

Activity 6

Use the following sentences as topic sentences in paragraphs of your own.
(a) We wear clothes for various reasons.
(b) Trees are our best friends.
(c) Smoking should be banned.
(d) Most of the traffic accidents can be prevented.
(e) English should be made optional in schools.
Answer:
(a) People wear clothes for many reasons. The basic reason for wearing clothes arose from the need to cover up one’s nakedness. This first happened when God killed an animal and used its skin to cover up the nakedness of Adam and Eve. Since then man has used animal skin as clothes. Another fundamental reason arose from the need to keep oneself warm, to preserve one’s body-heat. Today, however clothes have become a necessity for one’s personality. More and more people are changing clothes in pace with the changing fashions, entirely forgetting the original intentions with which clothes were worn by people.

(b) Trees are our best friends. They provide us with shade, food and with wood for lighting fires and building houses. They also help in causing rain and prevent the top soil from being eroded by the wind. Through them we produce papers on which we write; chairs and tables which make and use as our furniture and even coffins to bury our dead in. Above all, they are the reason for which we continue to breathe oxygen and enjoy the beauty of the environment. Thus from birth to death trees stand by us thereby proving to be our best friends.

(c) Smoking is injurious to health. This warning is given on every cigarette pack and tells us how great a hazard it is. Yet numerous people smoke tobacco and endanger their lives. Even the fact that smoking causes such dangerous disease as cancer does not deter people. Victims of smoking do not only hurt themselves, they effect a chain reaction whereby others too are affected by it. Hospitals have to nurse them back to health and the expense towards this is borne by the government and the institution which the person works for.

Besides this it affects the smoker’s family. Sometimes it leaves them without a bread earner causing untold misery. Besides this, smokers are a greater danger to non-smokers. When a smoker breathes out smoke, it is a more potent poison since carbon-dioxide is added to it. Thus non-smokers are more often than not forced to become passive smokers and they too can fall into the pit that the smoker will eventually fall. Hence smoking should not only be banned in public, it should be totally banned because the social and economic price to be paid for allowing smoking is too high.

(d) The rapid rise of the number of vehicles on the roads are the cause of many accidents which can be prevented through several steps. First of all lane driving must be introduced so that heavy vehicles and light vehicles are driven in separate lanes. Secondly, traffic signals must be constructed at important points and crossroads in the city and these must be manned and monitored by traffic policemen. Thirdly, public awareness of traffic rules and regulation must be raised. Finally these rules and regulations must be strictly enforced by the police with exemplary punishment meted out to the offenders. Such steps can only prevent traffic accidents.

(e) When a student passes out from school and joins college, he seldom understands what his teacher is teaching because he is not proficient in English. In almost all the states of India, the medium of instruction in Higher Education is English. This is especially true of technical and professional courses. Moreover, most of the books, whether they belong to Science or Humanities, Medical or Engineering, Veterinary Science or Agriculture and Forestry or vocational course, are all written in English.

This makes it necessary for student to know English. In addition, as both Gandhi and Nehru point out, English is the language used in international commerce and diplomacy and that it contains some of the richest treasures of literature. All these are very good reasons why English should be made compulsory in schools so that when entering the portals of Higher Education they are competent in skills.

Activity 7

Question (a)
Imagine that you are on the beach at Puri, enjoying a view of the sea. Note down the things you see and write a descriptive paragraph, to form a part of a letter to be written to a friend who has never seen the sea. You can start either with distant things or the things nearest to you.
Answer:

Bhubaneswar
10 October, 20

Dear Prafulla,
I received your letter yesterday. It was really surprising to hear from you after such a long time. Your description of Ooty and the Nilgiris was fascinating. From your letter it is obvious that you greatly enjoyed your vacations there. We too had been on a holiday to Puri. Though Puri is not such an exotic place as Ooty, I had a good time there. All of us had put up at Panth Nivas right on the beach.

Early every morning, I and Reena used to run off to the beach to watch the sunrise. The beach is not crowded then and it is wonderful to smell and feel the fresh morning air. The beach is extremely clean now and for the most part untrodden by people. One can see everywhere waste matter brought onto the beach by the night tide. These all form a meandering line on the sand as left by the wave.

There are pieces of wood, rubber, plastic, a slipper, and all kinds of small shells. I even noticed a dead Jelly fish lying with all its entrails uncovered. Besides there are plenty of small crabs butting in an out of one-inch holes. They quickly disappear into the safety of their holes as one approaches them. The sea itself seems calm and quiet, with waves splashing on to the sand, murmuring softly as it were and spraying its foam into the air majestically.

It is really a beautiful sight. Moreover, just before sunrise, one can see fishermen, naked except for loin- clothes, skilfully pushing their catamaran across the waves and then boarding it to go out fishing in the deep. Far away there are boats with their sails up and the whole scene looks greatly picturesque. Further the sea seems to extend up to as far as the eye can see.

The further one tries to see the more does one’s eyes become misty and the sky seems to touch the water. Right at one corner, a ball of fire rises suddenly, resplendent in its beauty and casting a red shadow on the waters. It showly rises from the waters and rests in the sky, now shining, brightly and freshly.
I think I should stop here. Please write back.
With Love,
Sushil.

Question (b)
Write a recipe, in the form of step-by-step instructions, on how to cook a pot of rice.
Answer:
Take the required amount of rice and clean it with water. Meanwhile take water which is double the volume of rice and boil it. When this water boils, add the cleaned rice to it. Cook over fire till the rice is cooked soft. After this drain the water from the pot till no water is left. One can also place the rice over a metal sieve so that the water is drained off and only dry rice is left behind. The rice is now ready for serving.

Question (c)
Write a paragraph to form a part of a letter to a friend telling him about a study tour of Orissa that you undertook last week with your classmates under the supervision of a teacher.
Steps
(i) Prepare a date-wise account of the places you visited and what you saw and did at each place.
(ii) Based on these notes, write a short paragraph to form a part of the letter.
(iii) You may begin your paragraph with the following sentence : Last week I took part in a study tour to
Answer:
20.6.20 :
Arrived at Bhubaneswar by train. Lodged at Yatri Nivas
21.6.20 :
Started at 6.00 am for Konark temple. Saw temple. Left at 10.30 for Puri. Had lunch at Panth Nivas, Puri at 12.00 noon. Went to Jagannath Temple. Participated in puja. Brought sovenirs from shops in front of Temple. Went to the beach. Had coffee and snacks there. Bought conches. Returned to BBSR.
22.6.20 :
Started out for Khandagiri / Udayagiri caves by 9.00 a.m. Had a hill top view of BBSR from Udayagiri. Studied the habits of Buddhist monks. Left for Nandankanan at 11.00 a.m. Reached at 12.00 noon. Had lunch at OTDC restaurant. Went to the zoo. Went boating. Returned at 5.00 p.m. Boarded East Cost at 7.30 p.m.

Hyderabad
30 June, 20

Dear Mitali,
Thanks a lot for writing to me. I am glad that you enjoyed your trip to Ooty. I knew the change of place would do you good and improve your spirits. Anyway, you must be eager to know what I did during the holidays. Well, you will be surprised. Last week I took part in a study tour to Orissa. Of course we did not visit all of Orissa but we visited what is called the “Golden Triangle” Puri, Konark and Bhubaneswar.

It was a hectic tour no doubt, but I thoroughly enjoyed it. On 19th we left for BBSR by Falaknamma arrived there on 20th and lodged at Yatri Nivas. Since the train was rather late; we did not go out anywhere that night. Instead we took the much needed rest. The following day, however we went straight to the Konark Temple.

What a beauty it was ! statues of dancers, devadasis depicting intercourse decorated the temple walls. But the wheels were marvellous. They had unimaginably skilled intricate work. We took a number of snaps and then travelled to Puri where we had lunch at the Panth Nivas and then visited the Jagannath Temple. There I and others of our group participated in Puja, ate prasad and sought the blessing of the Lord. After this, I bought some brass sovenirs in shops in front of the temple. Then we went off to the beach, had coffee and snacks at a restaurant and sat on the sand, building castles and occasionally running off to wet our legs in the sea.

We returned to Bhubaneswar at 10.00 The following day on 22.06.20 , we visited the Khandagiri and Udayagiri caves. We tried to delve back into history thinking of the past when Buddhist monks sat in meditation here. After this we visited Nandankanan. Everything seemed in disarray there in the aftermath of the cyclone. All animals looked diseased. However, we enjoyed our boating. By 5.00 we came back to Yatri Nivas where we washed and refreshed ourselves. That night we boarded the East Coast for Hyderabad once more. The tour was greatly enjoyed by everyone. I hope you had accompanied us.
Do write back.
With Love,
Snigdha.

Activity 8

Study the following paragraph and replace the appropriate noun phrases by pronouns so as to make the paragraph a coherent one. The first one has been answered as an example. There are seven more. Human beings are destroying their environment. It is ironic how people pollute people’s (their) surroundings and then go to great pains to clean up the messes that people have made. A cheap method of getting rid of industrial wastes is to dump the wastes into rivers or lakes. The result of such intrusion into nature is the death of many forms of life in and around the waters. Only when the pollution reaches the people’s reservoirs, however, do people figure that the time has come to clean up the people’s environment. Then people create a series of expensive projects to restore the water to the water’s original purity. Of course, the cheapest and most effective way to get clean water is not to pollute the water in the first place.
Answer:
Human beings are destroying their environment. It is ironic how people pollute people’s (their) surroundings and then go to great pains to clean up the messes that they have made. A cheap method of getting rid of industrial wastes is to dump them into rivers or lakes. The result of such intrusion into nature is the death of many forms of life in and around the waters. Only when the pollution reaches the peopled reservoirs, however, do they figure that the time has come to clean up the their environment. Then they create a series of expensive projects to restore the water to its original purity. Of course, the cheapest and most effective way to get clean water is not to pollute it in the first place.

Activity 9

Complete the following paragraph by filling in each blank with the right connective from the list.
(and, but, because, on the other hand, which)
(1)     (2)      (3)           (4)                         (5)
In the recent craze about the seven wonders of the world, we looked at buildings made of stone and clay. __________ there are seven other wonders, those of the inner world, _____________ are present in all human beings. They are the seven vices : lust, anger, greed, attachment, ego, jealousy and laziness. These are wonders _____________ they take us away from our.original virtuous self. They lead us into an unreal world of short living pleasure _____________ take us away from Truth. ___________ , if these are conquered, the earth becomes a wonderful place to live in. Life becomes a wonderful gift.
Answer:
In the recent craze about the seven wonders of the world, we looked at buildings made of stone and clay. But there are seven other wonders, those of the inner world, which are present in all human beings. They are the seven vices : lust, anger, greed, attachment, ego, jealousy and laziness. These are wonders because they take us away from our original virtuous self. They lead us into an unreal world of short living pleasure and take us away from Truth. On the other hand, if these are conquered, the earth becomes a wonderful place to live in. Life becomes a wonderful gift.

Activity 10

Read the following text carefully. In the first paragraph, back reference has been indicated by means of a box and an arrow. Other linking devices in the same paragraph have been boxed. Mark the rest of the text in the same way. SmokingTlwhich 1 may be a pleasure for some people, is a source of serious discomfort for their fellow. [Further,] medical authorities express [their j concern about the effect of smoking on the health not only of those who] smoke but also those who must involuntarily inhale the contribution of smokers to the atmosphere. As you are doubtlessly aware, a considerable number of our students have joined together in an effort to persuade the University to ban smoking in the classrooms. I believe they are entirely right in their aim. However, I would hope that it is possible to achieve this by an appeal to reason and to concern for others rather than by regulation. Smoking is prohibited by city by-laws in theatres and halls used for showing films as well as laboratories where there may be a fire hazard. Elsewhere, it is upto your own good sense. I am therefore asking you to maintain “No Smoking” in the auditorium, classrooms and seminar rooms where you teach. This proof of your interest for their health and well-being is very important to a large number of students.
Answer:
Smoking which may be a pleasure for some people, is a source of serious discomfort for their fellows. Further, medical authorities express their concern about the effect of smoking on the health not only of those who smoke but also those who must involuntarily inhale the contribution of smokers to the atmosphere. As you are doubtlessly aware a considerable number of our students have joined together in an effort to persuade the University to ban smoking in the classrooms.

I believe they are entirely right in their aim. However, I would hope that it is possible to achieve this by an appeal to reason and to concern for others rather than by regulation. Smoking is prohibited by city by-laws in theatres and in halls used for showing films as well as laboratories where there may be a fire hazard. Else where, it is upto your own good sense. I am therefore asking you to maintain “No Smoking” in the auditorium, classrooms and seminar rooms where you teach. This proof of your interest for their health and well-being is very important to a large number of students.

Activity 11

Dear Harish,
Remember that I told you I was trying to get a job at ICTL?
(1) __________, I finally managed to get one! Of course, I haven’t been working there long, (2) ____________ I can already tell that it’s a wonderful place to
work. All the staff, (3) ___________ the directors, are very friendly with everybody, and (4) ___________, they have marvellous facilities for the employees. (5) _________, there’s a bar and a gym, and lots of other things. I’m called the Safety Equipment Officer. It may sound like an impressive title, but it’s not a very accurate description of what I do. My main job is to provide protective clothing, (6) ___________ overalls, helmets, and so on. I estimate what the different departments will need, and (7) ____________ I order it from the suppliers. (8) __________ I make sure that the various departments have everything they want. (9) ____________, stationery is also my responsibility. (10) ____________, I have to supply all the offices with paper, envelopes, and so on. I find the job very interesting (11) _____________ I get a chance to go all over the factory and to meet everybody. (12) ___________ the pay is a lot better than in my old job.
(13) _____________ that’s my news. What about yours? Drop me a line when you have time. Regards to your family, and best wishes to you.
Yours truly,
Tarun

	(a)	(b)	(c)
1.	Then	Well	And
2.	but	because	so
3.	until	and	even
4.	so	what’s more	on the other hand
5.	For instance	However	Even
6.	however	in fact	such as
7.	then	after	so
8.	By the way	Anyway	In this way
9.	Indeed	Moreover	But
10.	Secondly	In other words	Also
11.	why	because	then
12.	Besides	But	On the other hand
13.	At the end	Anyway	After all

Answer:
(1) Well, (2) but, (3) even, (4) what’s more, (5) For instance, (6) such as, (7) then, (8) In this way, (9) Moreover, (10) In other words, (11) because, (12) Besides, (13) Anyway.

Activity 12

Each of the following sentences has a blank where there should be a linking word or phrase. Put an appropriate linking device in each blank so that the relation between the two statements is made clear.
(i) The pay and conditions are very good. ____________ , it’s only five minutes’ walk from where I live.
(ii) I didn’t apply for the job ____________ I didn’t think I had much chance of getting it.
(iii) A lot of professional groups, ________________ doctors and lawyers, have strong associations that protect their members’ rights.
(iv) The hours are short, the pay’s excellent, and the people I work with are very nice. ____________, it’s a great job.
(v) You may think it’s boring, ___________ it’s really very interesting.
(vi) All my relatives were at the wedding, ___________ my cousins from Australia.
(vii) At first I didn’t feel happy with so much responsibility. ____________ , now I feel quite confident that I can manage.
(viii) There are several things that make it a nice place to live in. ____________ , there’s a park right across the road.
Answer:
(i) The pay and conditions are very good. Besides, it’s only five minutes’ walk from where I live.
(ii) I didn’t apply for the job anyway I didn’t think I had much chance of getting it.
(iii) A lot of professional groups, such as. doctors and lawyers, have strong associations that protect their members’ rights.
(iv) The hours are short, the pay’s excellent, and the people I work with are very nice. In other words, it’s a great job.
(v) You may think it’s boring, but it’s really very interesting.
(vi) All my relatives were at the wedding, even my cousins from Australia.
(vii) At first I didn’t feel happy with so much responsibility. But, now I feel quite confident that I can manage.
(viii) There are several things that make it a nice place to live in. What’s more, there’s a park right across the road.

Activity 13

What is wrong in the following paragraphs ?
Machines have turned human society from an agrarian one into an I industrial one. Today in countries like the USA, the UK and Japan, only a small section of the population is engaged in agriculture, industry forms the basis of the life and progress of these nations. Machines have also changed the life of the individual in many ways. Life at home has been made more comfortable, and the drudgery of household work has been removed. People travel to schools and offices in buses or trains, and spend their evenings in – amusements made possible by machine civilization.
Answer:
The first paragraph is actually two paragraphs combined into one.
(i) Machines have turned human society from an agrarian one into an industrial one. Today in countries like the USA, the UK and Japan, only a small section of the population is engaged in agriculture. Industry forms the basis of the life and progress of these nations.
(ii) Machines have also changed the life of the individual in many ways. Life at home has been made more comfortable, and the drudgery of household work has been removed. People travel to schools and offices in buses or trains, and spend their evenings in amusements made possible by machine civilization. The twentieth century is the age of machines. From the time the . Industrial Revolution began in Europe, man’s life has been changing in many ways. At first the change was slow. But in the second half of the nineteenth century there was an increase in the rate of mechanization and as a result, life began to change more quickly. During the last fifty years, machines of all kinds have become part of our daily life and have transformed it in the most incredible manner.
Answer:
This passage is also two passages combined into one. There is no justification to have two paragraphs because it speaks of the same thing of machines.

Activity 14

Find out the reasons for the lack of unity in the following paragraphs, and rewrite each paragraph so that it has unity.

Question 1
In their eating habits goats are often very destructive. The famous Swiss scientist, Karl Vogt, says that the goat does more harm to the forests than any other animal. Goat’s milk is highly valued in many places and is often used to make cheese. It is inadvisable, therefore, to let goats run at large where valuable green things are growing. They will eat even the barks of trees.
Answer:
In their eating habits goats are often very destructive. They will eat even the barks of trees. The famous Swiss scientist, Karl Vogt, says that the goat does more harm to the forests than any other animal. It is inadvisable, therefore, to let goats run at large where valuable green things are growing. Here the sentence, “Goat’s milk is highly valued in many places and is often used to make cheese.” is irrelevant in the context of the topic. Hence this passage lack unity.

Question 2
In most vertebrates, limbs that have been lost will not grow again. If large masses of tissues are destroyed, new tissues will not take their place. Some new tissues will, of course, appear on the site of the injury, but they serve rather as a protection to the remaining tissues than as a replacement. But plants are superior to animals in this regard : they can repair their damaged tissues very efficiently and can often replace lost parts.
Answer:
(i) In most vertebrates, limbs that have been lost will not grow again. But plants are superior to animals in this regard : they can repair their damaged tissues very efficiently and can often replace lost parts.
(ii) If large masses of tissues are destroyed, new tissues will not take their place. Some new tissues will, of course, appear on the site of the injury, but they serve rather as a protection to the remaining tissues than as a replacement. Thus the paragraph is actually two paragraphs combined into one.

Question 3
The most memorable day in my life is the day the President of India shook my hand and handed me the Young Scientist of 1987 Award. There was loud applause from the packed hall. Many cameras including the TV cameras clicked and I was in a pool of light. I felt that the long hours I spent in my little laboratory in the midst of grumblings from my family who wanted me to help them with housework have been rewarded. I could not withhold my tears when I saw my father, husband and sisters, sitting in the front row, wiping their tears of joy. I am a scientist.
Answer:
The most memorable day in my life is the day the President of India shook my hand and handed me the Young Scientist of 1987 Award. There was loud applause from the packed hall. Many cameras including the TV cameras clicked and I was in a pool of light. I felt that the long hours I spent in my little laboratory in the midst of grumblings from my family who wanted me to help them with housework have been rewarded. I could not withhold my tears when I saw my father, husband and sisters, sitting in the front row, wiping their tears of joy. If the paragraph begins with the sentence No. 6 “I am a scientist.” it has definitely a unity.

Question 4
‘It is important to keep our forests from being destroyed for cultivation and wood. 2Forests may be of different kinds. 3Forests are essential for maintaining the rate of rainfall and temperature. 4Forests prevent soil erosion and the growth of deserts. 5Forests house a variety of wild animals and birds which will all perish if forests are destroyed. 6The losses – immediate as well long-term-following the destruction of forests are far greater than the paltry financial gain from cultivation and wood.
Answer:
It is important to keep our forests from being destroyed for cultivation and wood. Forests are essential for maintaining the rate of rainfall and temperature. Forests prevent soil erosion and the growth of deserts. Forests house a variety of wild animals and birds which will all perish if forests are destroyed. The losses – immediate as well long-term-following the destruction of forests are far greater than the paltry financial gain from cultivation and wood. Here the passage loses its unity, because the sentence “Forests may be of different kinds.” does not fit into its main idea.

Question 5
What makes a job perfect at one time and undesirable at another is simply change. The change may be in yourself, in the position or in the job market. A job could be perfect for you at this time of your life. But you may some day become totally dissatisfied with the very same job for the simple reason that you have changed. Some jobs are rather easy. You may not consider a high salary as important now as you did a few years ago. You may not be able to travel as much as you used to. A job which you were keen on getting a few years ago may not interest you now simply because you have become older and your system of values has changed. Some young people like jobs that involves a lot of travelling.
Answer:
This paragraph is actually two paragraphs combined into one.
(i) What makes a job perfect at one time and undesirable at another is simply change. The change may be in yourself, in the position or in the job market. A job could be perfect for you at this time of your life. But you may some day become totally dissatisfied with the very same job for the simple reason that you have changed. Some jobs are rather easy.
(ii) You may not consider a high salary as important now as you did a few years ago. You may not be able to travel as much as you used to. A job which you were keen on getting a few years ago may not interest you now simply because you have become older and your system of values has changed. Some young people like jobs that involves a lot of travelling. The passage mentions the same idea – A man’s changing attitude towards a job.

Question 6
Baba, Khaire’s first leopard cub, was brought to his foster father when he was only 15 days old: He had been found in a cattle-shed, deserted by his mother. After Baba had refused milk, prawns, minced meat and fish for two days, Khaire started his ward on Farex and the orphaned cat grew up to be a bonny baby. Leopards are affectionate animals with a remarkable memory and so far they have not been observed to have in them the instinct to kill. They never attack until provoked. Within a year, Baba measured seven feet from nose to tail, and when he attained a body weight of 70 kilos, he had to be tearfully sent off to the zoo. Baba had a passion for cars. He used to love being taken for a drive in a Maruti car.
Question
There are two sentences in this paragraph which are not related to the main idea and distract our attention away from the main idea. Can you spot them ?
Answer:
The two distinct sentences are-
(i) “Leopards are affectionate animals with a remarkable memory and so far they have not been observed to have in them the instinct to kill.”
(ii) They never attack until provoked.

Question 7
Bears are generally good-natured animals, yet there are times when they attack human beings. My aunt hates bears; she says they look ugly. Getting between a mother bear and her cubs is certain to provoke an attack. Some people wear bear skin caps. “They look cute. Acts of cruelty and ill treatment by unthinking persons have often provoked bears in captivity to attack. There are half a dozen brown bears in our zoo. Bears in the wild state are very unpredictable; they may suddenly lose their usual good nature and attack human beings for no apparent reason.
Answer:
In this paragraph there are few irrelevant details. Let us, first, look at the main idea – that bears, through good natured, may attack human beings. Let us now see what each sentence says about this idea. Sentence 2 is about ‘my aunt’ who thinks bears are ugly; it says nothing about the main point. Sentence 3 does give an example of what provokes an attack by a bear, but sentence 4 is about people who wear bear-skin to look pretty – nothing to contribute to the main idea.

Sentence 5 provides an example of what bears in captivity (e.g. in a zoo) may be forced to do, and the last sentence makes a point about wild bears – they are unpredictable. Thus only sentences 3, 5 and 6 belong to the paragraph : the other two add nothing to it and distract the readers’ attention from the main point.

Hence the correct paragraph is :
Bears are generally good-natured animals, yet there are times when they attack human beings. Getting between a mother bear and her cubs is certain to provoke an attack. They look cute. Acts of cruelty and ill treatment by unthinking persons have often provoked bears in captivity to attack.

Question 8
China assured itself of a place in the final six of the men’s basketball tournament today with a 104-48 win over Iraq. After leading 57-36 at half-time it completely shut out Iraquis for most of the second half. In a period of six minutes, China scored 20 points while conceding only two. At no stage did China relax the pressure, though the tallest player, Mu Tich-Chu, played no part in the match. Outstanding for China was Chang Weipink, who was top scorer with 45 points. The Iraquis, though talented, lacked the discipline of their opponents. Their game crumbled in frustration.
Answer:
The paragraph is actually three paragraphs combined into one.
Para – 1:
China assured itself of a place in the final six of the men’s basketball tournament today with a 104-48 win over Iraq.
Para – 2:
After leading 57-36 at half time it completely shut out Iraquis for most of the second half. In one period of six minutes China scored 20 points while conceding only two. At no stage did China relax the pressure, though the tallest player, Mu Tich-Chu, played no part in the match. Outstanding for China was Chang Wei-pink who was top scorer with 45 points.
Para – 3:
The Iraquis though talented, lacked the discipline of their opponents. Their game crumbled in frustration. 9 Once upon a time there was a tree in the forest which had thin, pointed, leaves. It felt sad when it saw the large, green leaves of the other trees swaying and rustling in the wind. The forest was big and there were more than a million trees in it. “How unlucky lam!” it moaned. “If only God gave me leaves of gold, I could stand proud among my neighbours and shine with dazzling brilliance in the Sun.” The next morning the tree stood transformed. It had leaves of gold. It looked around at the other trees and saw how envious they were. But when night fell, a greedy man crept close to the tree and plucked all the golden leaves, put them into a sack and stole away. “Oh, how unlucky I am,” the tree lamented again. “But perhaps I can have leaves made of glass. They will shine brighter than gold and no one will steal them.” ‘°The next morning the tree was again transformed. “When the rays of the sun fell on the tree, they were reflected in all directions. The tree felt proud of its leaves, but that night there was thunder and lightning, and the wind blew violently and shook the tree. A11 its glass leaves were broken. ,4“Bad luck, again,” sighed the tree. “Let me have beautiful green leaves like the other trees, but let me perfumed.” This wish was granted, too, and the next day the whole forest was filled with the sweet smell of its leaves. God was kind to all other trees. But within a few hours, goats from all around the forest gathered, drawn by the pleasant smell, nibbled at the leaves until not a trace of green was left on the tree. How foolish I have been,” said the tree sadly. “My own leaves are best for me, not any other, I’ve learnt my lesson.” The next day the tree stood covered once again with thin, pointed leaves. 20It felt as proud of its own leaves as the other trees felt of theirs.
Answer:
This paragraph is in reality five paragraphs combined into one.
Para – 1:
Once upon a time there was a tree in the forest which had thin, pointed leaves. It felt sad when it saw the large green leaves of the other trees swaying and rustling in the blind. ‘‘How unlucky I am !” it moaned. ‘‘If only God gave me leaves of gold, I could stand proud among my neighbours and shine with dazzling brilliance in the sun.”
Para – 2:
The next morning the tree stood transformed. It had leaves of gold. It looked around at other trees and saw how envious they were. But when night fell, a greedy man kept close to the tree and plucked all the golden leaves, put them into a sack and stole away. “Oh, how unlucky I am,” the tree lamented again. “But perhaps I can have leaves made of glass. They will shine brighter than gold and no one will steal them.”
Para – 3:
The next morning the tree was again transformed. When the rays of the sun fell on the tree, they were reflected in all directions. The tree felt proud of its leaves. But that night there was thunder and lightning, and the wind blew violently and shook the tree. All its glass leaves were broken. “Bad luck, again” sighed the tree. “Let me have beautiful green leaves like other trees, but let them be perfumed.”
Para – 4:
This wish was granted, too, and the next day the whole forest was filled with the sweet smell of its leaves. But within a few hours, goats from all round the forest gathered, drawn by the pleasant smell, and nibbled at the leaves until not a trace of green was left on the tree. “How foolish I have been,” said the tree sadly, “My own leaves are best for me, not any other, I’ve learnt my lesson.”
Para – 5:
The next day the tree stood covered once again with thin, pointed leaves. It felt as proud of its own leaves as the other trees felt of theirs.

Activity 15

Rearrange the sentences in each of the following paragrahs so as to make it meaningful. Indicate the sequence of sentence by their numbers.

Question 1
Then they left the dead body indoors and went out and wandered through the city, with their breasts bare and beating themselves as they walked. A11 the female relatives would join them and do the same. When these ceremonies were over, the body would be carried away to be embalmed. The men, too, would plaster themselves with mud and beat their breasts. As soon as an important personality died, the women of the family plastered their heads with mud. The following is the way in which ancient Egyptians conducted their mourning.
Answer:
6. The following is the way in which ancient Egyptians conducted their mourning.
5. As soon as an important personality died, the women of the family plastered their heads with mud.
4. The men, too, would plaster themselves with mud and beat their breasts.
3. When these ceremonies were over, the body would be carried away to be embalmed.
1. Then they left the dead body indoors and went out and wandered through the city, with their breasts bare and beating themselves as they walked.
2. All the female relatives would join them and do the same.

Question 2
‘Even if we increase the speed of our spacecraft to 200 miles a second (12000 miles per minute) we will need more than 4000 years. 2If we travel at the speed of 20 miles a second (1200 miles a minute) after getting out of the gravitational pull of the earth and the sun, it will take us 4000 years. 3How long will it take us to get to our nearest star?
Answer:
3. How long will it take us to get to our nearest star?
2. If we travel at the speed of 20 miles a second (1200 miles a minute) after getting out of the gravitational pull of the earth and the sun, it will take us 4000 years.
1. Even if we increase the speed of our spacecraft to 200 miles a second (12000 miles per minute) we will need more than 4000 years.

Question 3
When it was found, a French General claimed it as his personal property. Champollion’s work on the Rosetta stone meant that the writing on many ancient Egyptian manuscripts and monuments could be read, and our knowledge of the old civilisation of Egypt was greatly increased. This was done, and the Rosetta stone was taken to England and put in the British Museum, where it still is. The history of the stone is also interesting because it shows the attitude of people and governments to ancient objects in those days. But the British were at war with the French at that time, and when they beat them, they demanded the Rosetta stone and other valuable antiquities should be handed over to them.
Answer:
2. Champollion’s work on the Rosetta stone meant that the writing on many ancient Egyptian manuscripts and monuments could be read, and our knowledge of the old civilisation of Egypt was greatly increased.
4. The history of the stone is also interesting because it shows the attitude of people and governments to ancient objects in those days.
1. When it was found, a French General claimed it as his personal property.
5. But the British were at war with the French at that time, and when they beat them, they demanded the Rosetta stone and other valuable antiquities should be handed over to them.
3. This was done, arid the Rosetta stone was taken to England and put in the British Museum, where it still is.

Study the following paragraph. What type of arrangement does it follow ? Can you rewrite the paragrah by reversing the arrangement ?
The earlitest calendars invented by man were lunar ones,based on the motion of the moon. The Babylonians, for example, had a year of 12 lunar months, which were alternately 29 and 30 days long. Their year was about 11 days short, so they added on extra month every 3 years. This calendar was not accurate over long periods. The Ancient Egyptians established a solar calendar, based on the sun. They split their year of 365 days into 12 months of 30 days, adding 5 days at the end of the year. Next came the early Roman calendar, which was based on the Ancient Egyptian pattern. It altered the number of days in the months to eliminate the extra 5 days.
Answer:
The paragraph follows a time arrangement. Paragraph in reverse order. The earliest calendars were invented by the Romans. Egyptians and the Babylonians. Each of their calendars had a distinct feature. For example, the early Roman calendar changed the number of days in the months to wipe out the extra 5 days; the Egyptians who established solar calendar added 5 days at the end of the year and the Babylonians’ Calendars had 12 lunar months and their year was short of 11 days, and as a result, they added an extra month every 3 years.

Activity 16

Some of these patterns of arrangement can be found in the following paragraphs. Study each of them, underline the topic sentences, and write the principle of arrangement of ideas on the margin.

Question 1
Insecticides are substances toxic* to insects and are used to control them but in some situations they can cause harm to men, domestic animals, or crops. There are three main kinds of insecticides : stomach insecticides which are eaten by the insects along with food, contact insecticides which get into blood through the skin, and fumigant insecticides which the insects breathe in. While all insecticides are harmful, the stomach insecticides are the most harmful, and should be used very carefully. [*toxic = poisonous]
Answer:
Topic Sentence :
Insecticides are substances toxic to insects and are used to control them but in some situations they can cause harm to men, domestic animals, or crops.
Sentence 2 :
classification and Exemplification – stomach insecticide eaten by insects, contact insecticides that penetrate blood through skin and fumigant insecticides insects breathe in
Sentence 3 :
Consequence and precaution – stomach insecticides most harmful and need careful use

Question 2.
All the great orators have had a gift for remembering words like that of a musician for remembering music. When Daniel Webster, a famous American orator and dictionary maker was a boy, his teacher held up, one Saturday morning, a shiny new jack-knife and promised it to the boy who would commit the most Bible verses to memory by Monday. Daniel came back on Monday and rattled off seventy verses before the astonished teacher handed him the jack-knife – must to Daniel’s disgust, for he still had several chapters to go.
Answer:
Topic Sentence :
All the great orators have had a gift for remembering words like that of a musician for remembering music.
Sentence 2 :
Time and process – the teacher asked the boy to commit the most Bible verses to memory by Monday and promised him the jack-knife
Sentence 3 :
Cause and effect – Daniel rattled off seventy verses before the teacher gave him the jack-knife, but the former was disgusted for not committing several chapters to memory.

Question 3.
If you drink plenty of sea water you will soon be as thirsty and dehydrated as a man in a desert who has nothing to drink at all. This is because sea water cotains about 3.5 per cent minerals, about 1 per cent more than the kidneys can excrete normally with its own resources. Thus, if you drink one litre sea water, the kidneys will need 1.75 litres of water to dilute the minerals in the sea water before they can be excreted. This additional 0.75 litre water is taken from the body cells, and this results in dehydration.
Answer:
Topic Sentence :
1f you drink plenty of seawater you will soon be as thirsty and dehydrated as a man in a desert who has nothing to drink at all.
Sentence 2 :
Effect – seawater contains 3.5% minerals, 1% more than kidneys can excrete normally of their own
Sentence 3 :
Exemplification – drinking one-litre sea water kidney needs 1.75 litres of water to dilute the minerals in the sea water before their excretion
Sentence 4 :
Cause and effect – extracting an additional 0.75 litre from body cells resulting in dehydration

Question 4.
Tapioca tubers which form part of the breakfast of a larger number of farmers in Kerala, are like sweet potatoes. The topioca tubers are peeled, boiled and eaten like sweet potatoes. There are, of course, many important differences between them. While sweet potatoes rarely grow more than ten centimetres, long tapioca tubers can grow thirty centimetres. Sweet potatoes have a thin, pink or white skin whereas tapioca tubers have a thin brown skin and under it a thicker, whitish skin.
Answer:
Topic Sentence :
Tapioca tubers which form part of the breakfast of a larger number of farmers in Kerala, are like sweet potatoes.
Sentence 2 :
Process – tapioca tubers – peeled, boiled and eaten like sweet potatoes
Sentences 3 & 4 :
Important differences between the two – sweet potatoes – short tapioca – long
Sentence :
Comparison – sweet potatoes – a thin, pink or white skin tapioca tubers thin brown skin, a thicker, whitish skin under it

Question 5.
Economics is the social science that studies how limited resources are distributed for unlimited and competing uses. It tries to find out what men and societies do to satisfy their material needs and desires, when the means they have are not enough to fulfil all their material desires.
Answer:
Topic Sentence and Definition :
Economics is the social science that studies how limited resources are distributed for unlimited and competing uses.
Sentence 2 :
Exemplication – a social science that deals with unlimited wants and limited resources of man It studies human behaviour in this respect

Question 6.
Building a good compfire involves a routine which the serious camper learns very early. Before trying to start a fire, the camper prepares a site. He clears an area with a radius of about ten feet to ensure that the fire will not spread. He then gathers the following materials : dry twigs, and some pine and spruce wood. In laying the fire, the camper first makes a small pile of grass in the centre of the fire site. He then stacks twigs in a pyramidal or I tepee shape around the grass. He usually starts the fire with matches, though the experienced camper can also start it by rubbing two sticks together if necessary. As the fire progresses, he adds small sticks of dry pine wood and then larger pieces as the fire spreads out and becomes hotter. When the fire is very hot and is thoroughly established, he adds spruce wood or another ^ long-burning wood if such is available. He takes care to add new wood to the fire stick by stick, for too many pieces assed at one time may put the I fire out or cause a lot of smoke. By following these steps, almost anyone can build a campfire successfully.
Answer:
Topic Sentence :
Building a good compfire involves a routine which the serious camper learns very early. Before trying to start a fire, the camper prepares a site.
Sentences 2-7:
He clears … smoke clearing the area, gathering I dry tags, and so on
Sentence 8 :
Result Following these steps makes anyone build a successfl campfire

Question 7 .
Man has existed for about a million years. He has possessed writing for about 6000 years, agriculture some what longer, but perhaps not much longer. Science, as a dominant factor in determining the beliefs of educated men, has existed for about 300 years; as source of economic technique, for about 150 years. In this brief period it has proved itself an incredibly powerful revolutionary force. When we consider how recently it has risen to power, we find ourselves forced to believe that we are at the very beginning of its work in transforming human life. What its future effects will be is a matter of conjecture, but possibly a study of its effects hitherto may make the conjecture a little less hazardous.
Answer:
Topic Sentence :
Science, as a dominant factor in determining the beliefs of educated men, has existed for about 300 years; as source of economic technique, for about 150 years.
Now let us look at the order in which the facts are presented in the paragraphs.
(a) when man appeared on earth : 10,00,000 years ago
(b) when he learnt to write : 600 years ago
(c) when modem science began : 300 years ago
(d) when science began to provide technique : 150 years ago
(e) the future of science
The statements are placed in order of time. This arrangement can be used in explaining a process or in narrative writing.

Question 8
She led me into a cold dark room, rough and very gloomy, although with two candles burning. I took little heed of the things in it, though I marked that the window was open. That which I heeded was an old man, very stem, with death upon his countenance; yet not lying in his bed, but set upright in a chair, with .a loose red cloak thrown over him. Upon this his white hair fell and his pale fingers lay in a ghastly fashion, without a sign of life or movement, or of the power that kept him up; all rigid, calm, relentless. Only in his great black eyes, fixed upon me solemnly, all the poser of his body dwelt, all the life of his soul was burning.
Answer:
Topic Sentence :
She led me into a cold dark room, rough and very gloomy, although with two candles burning.
Let us look at the order in which the facts are presented in the paragraph.
(a) The author’s eye falls upon the ‘cold, dark room’, ‘the candles’, ‘the windows’.
(b) Next he notices the man, his posture (‘stem’, ‘solemnly’, ‘death upon his countenance’), and his dress (loose and cloak)
(c) Then he notices his features : his hair (white) and his fingers (pale).
(d) Finally his most striking eyes in which ‘his soul was burning’. Here the writer follows a spatial arrangement.

Activity 17

Now, using each of the above paragraphs as a model, write a similar paragraph, (the question for each paragraph is meant go guide you in writing a similar paragraph.)

Question 1.
How many types of insecticides does the writer mention ?
Answer:
The writer mentions three types of insecticides.
Task : Now write a similar paragraph based on these hints.
Mineral rock that bums – five kinds
(i) anthracite – hardest and blackest – cleanest – greatest heat
(ii) cannel – less hard, dull black, clean bright flame
(iii) bituminous – less hard, bums easily but lot of dust and smoke
(iv) lignite – brown and moist, not difficult to bum and smoke
(v) peat – brown, : good heat once dried and burned
Answer:
There are five kinds of mineral rocks that bum. The first is anthracite which is the hardest, blackest and cleanest of all. The heat that it gives is the greatest. The second of its kind is cannel which is less hard and its colour is dull black. This burning mineral rock gives clean bright flame. Then there is bituminous which is equally less hard. It bums without any problem, but gives off lot of dust and smoke. The fourth kind of burning mineral rock is ignite which is brown and moist. It is easy to bum and smoke. The last of all is peat which is brown in colour. This mineral once dried and burned gives us good heat.

Question 2.
Why has the writer mentioned the incident from Daniel Webster’s life ? Now write a similar paragraph.
Answer:
The writer has mentioned the incident from Daniel Webster’s life to bring home his talent as an orator. Rames is a musician, but he is endowed with a gift of remembering words like great orators. When he was a boy, his teacher held up, one Monday afternoon a beautiful guitar and promised it to him who would give a lilting tune to his song within three hours. He rose to the occasion and his teacher’s astonishment knew no bounds. He handed the boy the guitar, but the latter was not vexed, for he would have given still a sweeter music to his teacher’s song.

Question 3.
What is the effect of drinking sea water ?
Answer:
Drinking sea water will make a person thirsty and dehydrated at once.
Task : Your cricket team lost the first match of a tournament. As the captain you have to give a report on your failure to the manager. Your paragraph should include lack of discipline, disobedience, lack of cooperation, lack of fitness, injury, lack of good food and proper accommodation as reasons for your losing the match.
Answer:
Unfortunately, Eleven Gun team has lost its first match of the tournament against Cuttack Cricket Club. The captain of the team apprises his manager of the cause of his team’s dismal failure. He attributes lack of discipline, disobedience, lack of co¬operation among players. Besides other factors, such as, lack of players’ fitness, lack of good food and proper accommodation contribute to the team’s debacle.

Question 4.
How is tapioca similar to sweet potatoes ?
Answer:
Tapioca is similar to sweet potatoes in terms of peeling, boiling and eating.
Task : Khaitan fans look similar to Orient fans – 3 blades – instant pick-up – but differences – angle of blades – pushes more air out – uses less power – double ball bearings – smooth and noiseless.
Answer:
Khaitan fans look similar to Orient fans with three blades and instant pick¬up. But there are glancing differences between the two. They have dissimilar angle of blades. Khaitan fan pushes more air out, uses less power. It has double ball bearings. Above all, it is smoother and natural. People understandably prefer Khaitan fans to Orient ones.

Question 5.
What is Economics ?
Answer:
Economics, a social science, deals with how limited resources are distributed for unlimited and competing uses.
Task : Lie detector doesn’t detect lies – detects emotional changes, catches in breath,blood pressure, pulse rate etc. – suspect’s connection with wife – answer questions – some innocent questions – some important questions at unexpected moments.
Answer:
Lie detector doesn’t detect lies. Instead, it detects a person’s emotional changes. It catches in his breath. Not only does it know his blood pressure and pulse rate, but also it studies his body language to the core. It finds the person in a very unpleasant mood by asking him some unexpected and unnatural questions, such as his relation with other women, etc. At times, a lie detector asks him some innocent questions such as the rationale behind his childish simplicity and so on. It also wants to detect his dreams, hopes and aspirations at a time when he is not ready to answer them because of some unavoidable circumstances.

Question 6.
What should a camper do to make a camp-fire ?
Answer:
A camper should prepare a site before starting a camp fire and then follows several steps beginning from gathering twigs to adding new wood to the fire stick by stick for the purpose.
Task : Write a paragraph on how to make a paper plane or a paper boat.
Answer:
Making a paper boat involves some steps. At first, we take a piece of paper. It should be of a standard size. Then we fold it in a manner that resembles a boat. We never cut the piece of paper. While folding the paper we see that its bottom becomes hollow, the middle portion is elevated a little and two sides look like those of a boat.

Question 7.
What are the first and last sentences of the paragraph ? What do the middle sentences say ?
Answer:
First sentence : Man has existed for about a million years.
Last sentence : What its future effects will be is a matter of conjecture, but possibly a study of its effects hitherto may take the conjecture a little less hazards. The middle sentences throw light on the dominance of science, though it has existed for about 300 years.
Task : A handful of important inventions have changed the course of the world – fire – copper – iron – steam – petrol – electricity. Write a paragraph on these important inventions.
Answer:
A handful of important inventions such as, fire, copper, iron, steam, petrol and electricity have changed the course of the world. Before the advent of fire, man didn’t how to cook food. With its emergence, he cooked his food. Other inventions have led to the industrial revolution. It resulted in machine civilisation. Electricity was the most important invention. Thanks to it, the age of machine started. Iron is instrumental in the construction of houses. Numerous necessary materials are made from it. These inventions made the world small and enhanced the people’s standard of life.

Question 8.
What does the writer see first ? What does he observe next ?
Answer:
The writer first sees ‘the cold, dark room’, ‘the candles’, ‘the windows’. Next he observes an old man, his posture and dress.
Task : Describe your classroom.
Answer:
My classroom is 35′ x 32′ in size. It has two doors on the same side. One door is at the beginning of the room and the other one is at the end of the room. Just opposite to the wall of doors there are four windows, each at a distance of 8′. Our classroom is on the first floor. We have double desks in the classroom. There is one rolling blackboard quite big and black.

There is a huge table called the teacher’s table and well-cushioned chair for the teacher. The classroom is decorated with charts and maps. It is kept clean and nothing is left on the floor. There are three electric fans each at a distance of eleven feet from the other. There are three shelves in our classroom. Last though not the least, it always looks beautiful.

Additional Question On Paragraph With Answers

Question 1.
Each group of sentences below belong to a paragraph, but they are not in proper order. Rewrite them in their correct order to form a coherent paragraph. On fullmoon day people flock to Agra to feast their eyes on this exquisite building which presents to their eyes a blend of beauty and poetry. They remember Shah Jahan, wo erected this monument to love and dedicated it to the memory of his queen, Mumtaz. Tourists who return from India carry indelible memories of their visit to Agra, the city of the Taj Mahal. They remember the glory of the Moghul Empire. Taj Mahal by moonlight is one of the most beautiful sights in the world.
Answer:
Tourists who return from India carry indelible memories of their visit to Agra, the city of the Taj Mahal. Taj Mahal by moonlight is one of the most beautiful sights in the world. On fullmoon day people flock to Agra to feast their eyes on this exquisite building which presents to their eyes a blend of beauty and poetry. They remember the glory of the Moghul Empire. They remember Shah Jahan, wo erected this monument to love and dedicated it to the memory of his queen, Mumtaz.

Question 2.
Write the following paragraph and point out its topic sentence. Making ropes is one of the oldest trades in the world we know that people made ropes more than 5,000 years ago because we have found pieces of rope in very old Egyptian tombs. They made some of these from the hair of camels. They have made others from twisted glass. People use them for tying animals for getting water from deep wells and for pulling large stones which they used in buildings. We have found too, ropes were made of thin copper wire in the city of Pompeii, which a volcano destroyed a little less than 2,000 years ago.
Answer:
Topic sentence:
Making ropes is one of the oldest trades in the world.

Question 3.
Complete the following paragraph by filling in each blank with the right connectives from the list. In addition, for instance, probably, or, because ___________ the most terrible example of superstitions is the belief in witchcraft. In Western Europe, during the sixteenth and seventeenth centuries, three-quarters of a million people were killed, mostly after being tortured, ___________ they were found guilty of witchcraft something for which today we can find no scientific evidence. When people give reasons for persecuting others, we ought to be very sure that their reasons are not merely superstitions, ___________ based on false principles __________ even the civilised nations today, many actions take place and laws are made on the basis of principles which are just as much unproved assumptions as we many of these of the philosophies of the middle ages. ____________ by nature, it is often held as a principle that white people are superior to people of other colours.
Answer:
probably, because, or, in addition, for instance.

Question 4.
What is wrong in the following paragraph?
It is a matter of surprise, a very great surprise indeed, that when we are about to enter the twenty-first century, some of us still believe in superstitions and in the superiority of one set of people over another. For example, some of the whites of South Africa firmly believe that as a race, they are infinite to the block or the Broncos. Mahatma Gandhi fought all through his life for the unity of Hindus and Muslims.
Answer:
The above paragraph is wrong because the last sentence does not match its context.

Question 5.
Re-arrange the sentences in the following paragraph so as to make them meaningful. Indicate the sequences of sentences by their numbers.
1. The rocky material carried by a river is called its load.
2. A river carries its load by rolling the rocks and large stones along its bed, while the finer rock pieces are carried by the water.
3. The river wears away the surface of the land over which it flows.
4. As it is pulled along, the load is slowly broken up into smaller and smaller pieces.
5. The load which a river carries rubs against the sides and the bed or the floor of the river and wears them away.
Answer:
3, 1, 5, 4, 2.

Odisha State Board Elements of Mathematics Class 11 Solutions CHSE Odisha Chapter 14 Limit and Differentiation Ex 14(c) Textbook Exercise Questions and Answers.

CHSE Odisha Class 11 Math Solutions Chapter 14 Limit and Differentiation Exercise 14(c)

Question 1.

Evaluate the following limits :
(i) $\lim _{x \rightarrow 0} \frac{x}{\sin 2 x}$
Solution:

(ii) $\lim _{x \rightarrow 0} \frac{\sin 3 x}{\sin 5 x}$
Solution:

(iii) $\lim _{x \rightarrow 0} \frac{\sin m x}{\sin n x}$
Solution:

(iv) $\lim _{x \rightarrow 0} \frac{\tan \alpha x}{x}$
Solution:

(v) $\lim _{x \rightarrow 0} \frac{1-\cos x}{x^2}$
Solution:

(vi) $\lim _{x \rightarrow 0} \frac{\sin x^{\circ}}{x}$
Solution:

(vii) $\lim _{x \rightarrow \pi} \frac{\sin x}{\pi-x}$
Solution:

(viii) $\lim _{x \rightarrow \frac{\pi}{2}} \frac{1-\sin x}{\left(\frac{\pi}{2}-x\right)^2}$
Solution:

(ix) $\lim _{x \rightarrow 0} \frac{1-\cos ^3 x}{x \sin 2 x}$
Solution:

(x) $\lim _{x \rightarrow 0} \frac{1+\sin x-\cos x}{1-\sin x-\cos x}$
Solution:

(xi) $\lim _{x \rightarrow 0} \frac{\tan x-\sin x}{x^3}$
Solution:

(xii) $\lim _{x \rightarrow 0} \frac{(1-\cos x)^2}{\tan ^3 x-\sin ^3 x}$
Solution:

(xiii) $\lim _{x \rightarrow \frac{\pi}{2}}\left(\frac{\pi}{2}-x\right) \tan x$
Solution:

(xiv) $\lim _{x \rightarrow \frac{\pi}{4}} \frac{\cos x-\sin x}{\cos 2 x}$
Solution:

(xv) $\lim _{x \rightarrow 0} \frac{x-x \cos 2 x}{\sin ^3 2 x}$
Solution:

(xvi) $\lim _{x \rightarrow 0} \frac{\sqrt{1+\sin x}-\sqrt{1-\sin x}}{\tan x}$
Solution:

(xvii) $\lim _{x \rightarrow 0} \frac{2 \sin x-\sin 2 x}{x^3}$
Solution:

(xviii) $\lim _{x \rightarrow 0} \frac{\cos x-\cos 5 x}{\cos 2 x-\cos 6 x}$
Solution:

(xix) $\lim _{x \rightarrow 0} \frac{\sin ^{-1} x}{x}$
Solution:

Question 2.
Evaluate
(i) $\lim _{x \rightarrow \alpha} \frac{x \sin \alpha-\alpha \sin x}{x-\alpha}$
Solution:

(ii) $\lim _{x \rightarrow 0} x \sin \frac{1}{x}$
Solution:

Question 3.
Evaluate the following limits :
(i) $\lim _{h \rightarrow 0} \frac{\sin (x+h)-\sin x}{h}$
Solution:

(ii) $\lim _{h \rightarrow 0} \frac{\cos (x+h)-\cos x}{h}$
Solution:

(iii) $\lim _{h \rightarrow 0} \frac{\tan (x+h)-\tan x}{h}$
Solution:

(iv) $\lim _{h \rightarrow 0} \frac{{cosec}(x+h)-{cosec} x}{h}$
Solution:

(v) $\lim _{h \rightarrow 0} \frac{\sec (x+h)-\sec x}{h}$
Solution:

(vi) $\lim _{h \rightarrow 0} \frac{\cot (x+h)-\cot x}{h}$
Solution:

(vii) $\lim _{h \rightarrow 0} \frac{\sqrt{x+h}-\sqrt{x}}{h}$
Solution:

(viii) $\lim _{h \rightarrow 0} \frac{\log _{\mathrm{a}}(x+h)-\log _a x}{h}$
Solution:

(ix) $\lim _{h \rightarrow 0} \frac{\ln (x+h)-\ln x}{h}$
Solution:

(x) $\lim _{h \rightarrow 0} \frac{a^{x+h}-e^x}{h}$
Solution:

(xi) $\lim _{h \rightarrow 0} \frac{e^{x+h}-e^x}{h}$
Solution:

(xii) $\lim _{h \rightarrow 0}\left\{\frac{1}{(x+h)^3}-\frac{1}{x^3}\right\}$
Solution:

(xiii) $\lim _{h \rightarrow 0} \frac{\sin (x+h)-\sin (x-h)}{h}$
Solution:

(xiv) $\lim _{h \rightarrow 0} \frac{1}{h}\left\{\frac{1}{\sqrt{x+h}-\frac{1}{\sqrt{x}}}\right\}$
Solution:

Question 4.
Evaluate the following :
(i) $\lim _{x \rightarrow 0} \frac{\log _e\left(1+\frac{x}{2}\right)}{x}$
Solution:

(ii) $\lim _{x \rightarrow 1} \frac{x-1}{\log _e x}$
Solution:

(iii) $\lim _{x \rightarrow 1} \frac{\log _e(2 x-1)}{x-1}$
Solution:

(iv) $\lim _{x \rightarrow 0} \frac{\log _e(x+1)}{\sqrt{x+1}-1}$
Solution:

(v) $\lim _{x \rightarrow 2} \frac{\log _e(x-1)}{x^2-3 x+2}$
Solution:

(vi) $\lim _{x \rightarrow 0} \frac{e^{a x}-1}{x}$
Solution:

(vii) $\lim _{x \rightarrow 0} \frac{e^{a x}-e^{-a x}}{x}$
Solution:

(viii) $\lim _{x \rightarrow 0} \frac{e^{3 x}-e^{2 x}}{e^{4 x}-e^{3 x}}$
Solution:

(ix) $\lim _{x \rightarrow 0} \frac{a^{2 x}-1}{x}$
Solution:

(x) $\lim _{x \rightarrow 0} \frac{a^x-b^x}{x}$
Solution:

(xi) $\lim _{x \rightarrow 1} \frac{2^{x-1}-1}{x-1}$
Solution:

(xii) $\lim _{x \rightarrow 0} \frac{a^x-a^{-x}}{x}$
Solution:

(xiii) $\lim _{x \rightarrow 1} \frac{3^x-3}{x-1}$
Solution:

(xiv) $\lim _{x \rightarrow 0} \frac{3^x-2^x}{4^x-3^x}$
Solution:

(xv) $\lim _{x \rightarrow 1} \frac{2^{x-1}-1}{\sqrt{x}-1}$
Solution:

Question 5.
Evaluate the following :
(i) $\lim _{x \rightarrow 0} \frac{\sqrt{x+1}-1}{x}$
Solution:

(ii) $\lim _{x \rightarrow 0} \frac{\sqrt{x+2}-\sqrt{2}}{x}$
Solution:

(iii) $\lim _{x \rightarrow 0} \frac{\sqrt{x}-\sqrt{5}}{x-5}$
Solution:

(iv) $\lim _{x \rightarrow 0} \frac{\sqrt{3-2 x}-\sqrt{3}}{x}$
Solution:

(v) $\lim _{x \rightarrow 5} \frac{\sqrt{x-1}-2}{x-5}$
Solution:

(vi) $\lim _{x \rightarrow 1} \frac{x^2-\sqrt{x}}{\sqrt{x}-1}$
Solution:

(vii) $\lim _{x \rightarrow a} \frac{\sqrt{x-b}-\sqrt{a-b}}{x^2-a^2}$, (a > b)
Solution:

(viii) $\lim _{x \rightarrow 1} \frac{x^{\frac{1}{m}}-1}{x^{\frac{1}{n}}-1}$ (m, n are integers)
Solution:

(ix) $\lim _{x \rightarrow 0} \frac{\sqrt{x^2+1}-1}{\sqrt{x^2+4}-2}$
Solution:

= $\frac{2+2}{1+1}=\frac{4}{2}$ = 2

(x) $\lim _{x \rightarrow \infty}(\sqrt{x+1}-\sqrt{x})$
Solution:

(xi) $\lim _{x \rightarrow \infty}\left(\sqrt{x^2+1}-\sqrt{x^2-1}\right)$
Solution:

(xii) $\lim _{x \rightarrow 0} \frac{\sqrt[3]{1+x}-\sqrt[3]{1-x}}{x}$
Solution:

(xiii) $\lim _{x \rightarrow 0} \frac{(x+9)^{\frac{3}{2}}-27}{x}$
Solution:

(xiv) $\lim _{x \rightarrow 0} \frac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt[3]{1+x}-\sqrt[3]{1-x}}$
Solution:

(xv) $\lim _{x \rightarrow \infty} \frac{a_0+a_1 x+a_2 x^2+\ldots+a_m x^m}{b_0+b_1 x+b_2 x^2+\ldots+b_n x^n}$
Solution:
$\lim _{x \rightarrow \infty} \frac{a_0+a_1 x+a_2 x^2+\ldots+a_m x^m}{b_0+b_1 x+b_2 x^2+\ldots+b_n x^n}$
= $\left\{\begin{array}{lll} \infty & \text { if } & m>n \\ 0 & \text { if } & m<n \\ \frac{a_m}{b_n} & \text { if } & m=n \end{array}\right.$

Question 6.
Evaluate the following :
(i) $\lim _{x \rightarrow \infty} \frac{\sin x}{x}$
Solution:

(ii) $\lim _{x \rightarrow \infty} x\left(a^{\frac{1}{x}}-1\right)$, a > 0
Solution:

(iii) $\lim _{x \rightarrow 0} \frac{x^{\frac{1}{2}}+2 x+3 x^{\frac{3}{2}}}{2 x^{\frac{1}{2}}-2 x^{\frac{5}{2}}+4 x^{\frac{7}{2}}}$
Solution:
$\lim _{x \rightarrow 0} \frac{x^{\frac{1}{2}}+2 x+3 x^{\frac{3}{2}}}{2 x^{\frac{1}{2}}-2 x^{\frac{5}{2}}+4 x^{\frac{7}{2}}}$
= $\lim _{x \rightarrow 0} \frac{1+2 \sqrt{x}+3 x}{2-2 x^2+4 x^3}=\frac{1}{2}$

(iv) $\lim _{x \rightarrow \infty} \sqrt{x}\{\sqrt{x+1}-\sqrt{x}\}$
Solution:

(v) $\lim _{x \rightarrow \infty} x^2\left\{\sqrt{x^4+a^2}-\sqrt{x^4-a^2}\right\}$
Solution:

(vi) $\lim _{x \rightarrow 0} \cos (\sin x)$
Solution:
$\lim _{x \rightarrow 0} \cos (\sin x)$
= cos (sin 0) = cos 0 = 1

(vii) $\lim _{x \rightarrow 0} \log _e \frac{\sqrt{1+x}-1}{x}$
Solution:

(viii) $\lim _{x \rightarrow 2} \log _e \frac{x^2-4}{\sqrt{3 x-2}-\sqrt{x+2}}$
Solution:

(ix) $\lim _{x \rightarrow \infty} \log _e\left(1+\frac{a}{x}\right)^x$
Solution:

(x) $\lim _{x \rightarrow 0} \log _e(1+b x)^{\frac{1}{x}}$
Solution:

(xi) $\lim _{x \rightarrow \frac{\pi}{4}} \frac{\sin \left(\frac{1-\tan x}{1+\tan x}\right)}{\frac{\pi}{4}-x}$
Solution:

(xii) $\lim _{x \rightarrow \frac{\pi}{2}} \log \frac{1-\sin ^3 x}{\cos ^2 x}$
Solution:

(xiii) $\lim _{x \rightarrow \infty} e^x\left(a^{\frac{1}{x}}-1\right)$
Solution:

(xiv) $\lim _{x \rightarrow 0} \frac{x\left(e^{\frac{\sqrt{1+x^2+x^4-1}}{x}-1}\right)}{\sqrt{1+x^2+x^4}-1}$
Solution:

(xv) $\lim _{x \rightarrow 0+} \frac{b \tan x\left(e^{\sin \frac{a x}{b x}-\frac{a}{b}}\right)}{b \sin a x-a \tan b x}$
Solution:

Question 7.
Examine the existence of the following limits :
(i) $\lim _{x \rightarrow 0+} \log _a x$
Solution:
$\lim _{x \rightarrow 0+} \log _a x$
= $\lim _{h \rightarrow 0} \log _a h=-\infty$
∴ The limit exists

(ii) $\lim _{x \rightarrow \frac{\pi}{2}} \tan x$
Solution:

(iii) $\lim _{x \rightarrow 0}{cosec} x$
Solution:

(iv) $\lim _{x \rightarrow 0-} \frac{1}{e^x}$
Solution:
$\lim _{x \rightarrow 0-} \frac{1}{e^x}$ = 0 because as
x → 0, $\frac{1}{x}$ → ∞
So $e^{\frac{1}{x}}$ → 0
∴ The limit exists.

(v) $\lim _{x \rightarrow 0+} \frac{1}{e^x}$
Solution:
$\lim _{x \rightarrow 0+} \frac{1}{e^x}$ = $\lim _{h \rightarrow 0} e^{\frac{1}{h}}=e^{\infty}$ = ∞
The limit exists.

(vi) $\lim _{x \rightarrow 0} \frac{1}{e^{\frac{1}{x}}-1}$
Solution:

Question 8.
(i) $\lim _{x \rightarrow \alpha} \frac{\tan a(x-\alpha)}{x-\alpha}=\frac{1}{2}$
Solution:

(ii) $\lim _{x \rightarrow \alpha} \frac{\tan a x}{\sin 2 x}=1$
Solution:

(iii) $\lim _{x \rightarrow 0} \frac{e^{a x}-e^x}{x}$ = 2
Solution:

(iv) $\lim _{x \rightarrow 1} \frac{5^x-5}{(x-1) \log _e a}$ = 5
Solution:

(v) $\lim _{x \rightarrow 2} \frac{\log _e(2 x-3)}{a(x-2)}$ = 1
Solution:

Odisha State Board CHSE Odisha Class 11 Sanskrit Solutions Grammar पत्रलिखन प्रकरणम् Textbook Exercise Questions and Answers.

CHSE Odisha Class 11 Sanskrit Grammar पत्रलिखन प्रकरणम्

1. कुशलवार्त्ता ज्ञातुं पितु: पुत्रं प्रति पत्रम् ।
Answer:

ଅପର୍ଣ୍ଣାନଗରମ୍
ଦିନାଙ୍କ-୧୫,୦୭,୧୬

ଚିରଞ୍ଜୀବିନେ ଆଶୁତୋଷାୟ ଶୁଭାଶିଷୀ,
ବୟଂ ସର୍ବେ ଅତ୍ର କୁଶଳିନଃ, ତ୍ଵଦୀୟଂ କୁଶଳସମାଚାରଃ ଅପି ପତ୍ରମୁଖେନ ଜ୍ଞାତଃ । ତବ ଅନୁଜଃ ରବିନ୍ଦ୍ର ସମ୍ୟକ୍ ପଠନ୍ ଅସ୍ତି । ଗତପରୀକ୍ଷାୟାଂ ୟଃ ପ୍ରଥମ ଶ୍ରେଣ୍ୟାମ୍ ଉତ୍ତୀର୍ଶୀ । ଭବତଃ ବିଷୟ ସ୍ୱ ତଦା ତଦା ପୃଚ୍ଛତି । ତବ ଅନୁଜା ଅପି ସମ୍ୟକ୍ ପଠନ୍ତୀ ଅସ୍ତି ।

ବତ୍ସ ! ପଠନୀୟବିଷୟଂ ଶ୍ରଦ୍ଧୟା ପଠତୁ । ଆହାରବିଷୟେ ଜାଗରୂକତା ଭବତୁ । ପ୍ରତିଦିନଂ ପ୍ରାତଃ ସାୟଂ ବା କିଞ୍ଚିତ୍କାଳଂ ବ୍ୟାୟାମଂ କୁରୁ । ତବ କୁଶଳବାର୍ତ୍ତା ପ୍ରତିସପ୍ତାହଂ ପତ୍ରଦ୍ୱାରା ସୂଚୟ ।
ତବ ମାତା ଅପି ସସ୍ନେହମ୍ ଆଶୀର୍ବାଦମ୍ ଉକ୍ତବତୀ । ଅନ୍ୟ ବିଶେଷ ନାସ୍ତି ।

ଇତି ତବ ଶୁଭାକାଙ୍‌କ୍ଷୀ
ପିତା

ପ୍ରେଷକଃ	ପ୍ରାପକଃ
ବିଶ୍ଵମୋହନ ସାହୁଃ	ଆଶୁତୋଷ ସାହୁଃ
ଗୃହକ୍ତମାଙ୍କଃ – ୧୮୩୭	ଗୃହକ୍ରମାଙ୍କ – ୭୬୮
ଅପର୍ଣ୍ଣାନଗରମ୍	ବାନ୍ଧୁଲି
କଟକମ୍	ବାଲେଶ୍ଵରମ୍

2. शैक्षणिकप्रवासार्थं पुत्य्रा: मातरं प्रति पत्रम् ।
Answer:

ବାସୁଦେବପୁରମ୍
ଦିନାଙ୍କ-୧୭.୦୭.୧୬

ମାତୃଶ୍ରୀଚରଣସନ୍ନିଧୌ
ସାଦରଂ ବନ୍ଦନାନି,
ଭବତ୍ୟାଃ ଆଶୀର୍ବାଦପତ୍ର ପ୍ରାପ୍ତମ୍ । ପତ୍ର ଦୃଷ୍ଟି ଆନନ୍ଦ ଅଭବତ୍ । ଭବିତ୍ୟା ସୂଚନାନୁସାରଂ ମମ ସ୍ଵାସ୍ଥ୍ୟସ୍ୟ ଯଥାବିଧ ଯତ୍ନ କୁର୍ବତୀ ଅସ୍ଥି ।

ଆଗାମିମାସସ୍ୟ ଦ୍ବିତୀୟସପ୍ତାହେ ଅସ୍ମାକଂ ବାର୍ଷିକୀ ପରୀକ୍ଷା ଭବିଷ୍ୟତି । ପରୀକ୍ଷାମନ୍ତରମେବ ଦିନତ୍ରୟସ୍ୟ ଶୈକ୍ଷଣିକପ୍ରବାଦଃ ଅସ୍ତି । ପ୍ରାଧ୍ୟାପକୈ ସହ ବୟଂ ସର୍ବେ ଗଚ୍ଛାମଃ ପ୍ରବାସଂ ସମାପ୍ୟ ଯାବଚ୍ଛୀଘ୍ର ଗୃହଂ ପ୍ରତି ଆଗମିଷ୍ୟାମି ।

ଏବଂ ବିଷୟଂ ପୂଜ୍ୟ ପିତରଂ ସୂଚତୁ ମମ ନମସ୍କାରଂ ଚ ସୂଚୟତୁ । ପ୍ରିୟାୟ ଅନୁଜାୟ ରାଜାୟ ଆଶିଷଃ

ଇତି ପ୍ରିୟପୁତ୍ରୀ
ଅଙ୍କିତା

ପ୍ରେଷକଃ	ପ୍ରାପକଃ
ଅଙ୍କିତା ମିଶ୍ରଃ	ସୌଦାମିନି ମିଶ୍ରଃ
ଶୈଳବାଳା ମହିଳା ମହାବିଦ୍ୟାଳୟଃ	ବ୍ରଜସୁନ୍ଦରପୁରମ୍
ଚଣ୍ଡିଛକମ୍	ରଣପୁରମ୍
କଟକମ୍	ନୟାଗଡ଼ମ୍

3. संस्कृतप्रबन्धलिखन प्रतियोगितायां सफलता प्रसङ्गे मित्रं प्रति पत्रम् ।
अथवा, स्वमहाविद्यालयविषये वर्णयन्तं स्वमित्रं प्रति पत्रम् ।
Answer:

ଉଦିତନଗରମ୍
ଦିନାଙ୍କ-୧୬.୦୭.୧୬

ପ୍ରିୟ ମିତ୍ର !
ବିଜୟାଦଶମୀପର୍ବଦିନସ୍ୟ ଶୁଭାଶୟଃ ।
ଅତ୍ର ବୟଂ ସର୍ବେ କୁଶଳିନଃ । ଭବାନ୍ ଅପି କୁଶଳୀ ଇତି ମନ୍ୟୁ । ଭବତଃ ପତ୍ର ପ୍ରାପ୍ତମ୍ । ପତ୍ରଲିଖନେ ବିଳମ୍ବୀ ଅଭବତ୍ । କୃପୟା କ୍ଷମ୍ୟତାମ୍ ।

ଭବାନ୍ ସଂସ୍କୃତପ୍ରବନ୍ଧଲିଖନ ପ୍ରତିଯୋଗିତାୟାଂ ପ୍ରଥମପୁରସ୍କାରଂ ପ୍ରାପ୍ତିବାନ୍ ଇତି ଜ୍ଞାତ୍ମା ମମ ଅତୀବ ସନ୍ତୋଷ ଅଭିବତ୍ । ସଫଳତାୟୈ ଅଭିନନ୍ଦନାନି । ଅହମପି ସଂସ୍କୃତକଥାକଥନ ପ୍ରତିଯୋଗିତାୟାଂ ଦ୍ୱିତୀୟ ସ୍ଥାନଂ ପ୍ରାପ୍ତବାନ୍ ଅସ୍ଥି ।

ଭବିଷ୍ଯତି । ପରୀକ୍ଷାୟଃ ଅନନ୍ତରଂ ଡିସେମ୍ବରମାସେ ମହାବିଦ୍ୟାଳୟସ୍ୟ କ୍ରୀଡ଼ୋତ୍ସବେ ତଥା ଜାନୁଆରୀମାସେ ବାର୍ଷିକୋତ୍ସବ ଚ ଅସ୍ତି । ବାର୍ଷିକୋତ୍ସବେ ଏବଂ ସଂସ୍କୃତନାଟକମପି ପ୍ରଦର୍ଶିତଂ ଭବିଷ୍ୟତି । ଦିନାଙ୍କ ସୂଚୟିଷ୍ୟାମି, ଅବଶ୍ୟମ୍ ଆଗଚ୍ଛତୁ । ବାର୍ଷିକୋତ୍ସବସ୍ୟ ଅନନ୍ତରଂ ମାର୍ଚ୍ଚମାସେ ବାର୍ଷିକପରୀକ୍ଷା ଭବିଷ୍ୟତି । ଅତଃ ମମ ଅଧ୍ୟୟନଂ ସମ୍ୟକ୍ ପ୍ରଚଳତି ।

ନାନ୍ୟଃ ବିଶେଷ । ପତ୍ର ଲିଖନ୍ତୁ । ଗୃହେ ସର୍ବାନ୍ ମମ ବନ୍ଦନାନି ସୂଚୟତୁ ।

ଇତି ଭବଦୀୟଂ ମିତ୍ର
ଆକାଶଃ

ପ୍ରେଷକଃ	ପ୍ରାପକଃ
ଶ୍ରୀଯୁକ୍ତ ବିଶାଳ ଶ୍ରୀବାସ୍ତବଃ	ଶ୍ରୀମାନ୍ଆକାଶଃଶ୍ରୀବ।ସ୍ତ୍ରବଃ
ଉଦିତ୍ନଗରମ	ବଳଭଦ୍ରପୁରମ୍
ରାଗରକେଲା	କେନ୍ଦ୍ରାପଡ଼ା

4. स्वविद्याप्रगतिः विषये सखीं प्रति सख्या: पत्रम् ।
Answer:

ଅପର୍ଣ୍ଣାନଖରମ୍
ଦିନାଙ୍କ-୧୮.୦୮.୧୬

ଅୟି ପ୍ରିୟ ସଖ୍ !
ସଂକ୍ରାନ୍ତିପର୍ବଦିନସ୍ୟ ଶୁଭାଶୟା ।
ଅତ୍ର ଅହଂ କୁଶଳିନୀ, ଭବତ୍ୟା ବନ୍ଧୁନାଂ ଚ କୁଶଳୀ ସମ୍ଭାବୟାମି । ବହୁକାଳତଃ ଭବତ୍ୟା ପତ୍ରମେବ ନାସ୍ତି । ଅତଃ ଅହମେବ ଲିଖନ୍ତୀ ଅସ୍ଥି । ଇଦାନୀ ବା ଭବତୀ ଅବଶ୍ୟ ପତ୍ର ଲିଖେତ୍ ଇତି ଚିନ୍ତୟାମି ।

ମହାବିଦ୍ୟାଳୟେ ମମ ବିଦ୍ୟାପ୍ରଗତିଃ ସାଧାରଣୀ ଅସ୍ଥି । ଭବତ୍ୟା ଅଭ୍ୟାସ ସମ୍ୟକ୍ ପ୍ରଚଳତି ଇତି ମନ୍ୟ । ମମ ମହାବିଦ୍ୟାଳୟେ ଆଗାମି ଅଗଷ୍ଟମାସେ ବିବିଧଃ ସଂସ୍କୃତସ୍ପର୍ଷା ଭବିଷ୍ୟନ୍ତ । ତତ୍ର ଅହଂ ସଂସ୍କୃତ ଭାଷଣସ୍ପର୍ଧୟାଂ ଭାଗଂ ନେଷ୍ୟାମି ।

ସମ୍ପ୍ରତି ଖ୍ରୀଷ୍ଟାବକାଶଃ ପ୍ରଚଳତି । ଆଗାମିନି ଏକବିଂଶତିତମେ ଦିନାଙ୍କ ମହାବିଦ୍ୟାଳୟ ଉନ୍ମୋଚିତଃ ଭବିଷ୍ୟତି । ଭବତୀ ଦିବସଦ୍ଵୟନିମିତ୍ତ ମମ ଗୃହମ୍ ଆଗଚ୍ଛତୁ । ଅମ୍ଳାକଂ ଗୃହେ ସର୍ବେ ଆନନ୍ଦିତଃ ଭବିଷ୍ୟନ୍ତ ।

ଭବତ୍ୟା ପତ୍ର ପ୍ରତୀକ୍ଷାମାଣା….
ଇତି ଭବଦୀୟା ସଖା
ଅଭୀପ୍‌ସା

ପ୍ରେଷକଃ	ପ୍ରାପକଃ
ଅଭୀପ୍ସା ପାତ୍ରଃ	ପ୍ରିୟଙ୍କା ଦାସଃ
ଅପର୍ଣ୍ଣାନଗରମ୍	ରବୀନ୍ଦ୍ରନଗରମ୍
କଟକମ୍	ଭୁବନେଶ୍ବରମ୍

5. शैक्षिक भ्रमणाय धनप्रेषणार्थं पितरं प्रति पत्रम् ।
Answer:

ମଧୁସୂତନ ଛାତ୍ର।ବାସଃ
ଭୁବନେଶ୍ବରମ୍
ଦିନାଙ୍କ-୨୫.୦୮.୧୬

ପରମଶ୍ରଦ୍ଧେୟାଃ ପିତୃମହାଭ।ଗାଃ
ସାଦରଂ ପ୍ରଣାମାଃ ।
ଅତ୍ର କୁଶଳଂ ତତ୍ରାସ୍ତୁ । ମମ ବାର୍ଷିକୀ ପରୀକ୍ଷା ସମାପ୍ତା । ମମ ଉତ୍ତରପତ୍ରାଣି ଅପି ଶୋଭନାନି ଅଭବନ୍ । ଅଧୁନା ଅବକାଶେ ଅହଂ ଗୃହଂ ନ ଆଗମିଷ୍ୟାମି ଯତଃ ମହାବିଦ୍ୟାଳୟେନ ଏକସ୍ୟା ଶୈକ୍ଷିକଯାତ୍ରାୟା ଆୟୋଜନଂ କୃତମ୍ । ବସ୍ତୁତଃ ଅରୁଣାଚଳସ୍ୟ ବିଷୟ ପଠିତ୍ବା ବୟଂ ସର୍ବେ ତତ୍ ପ୍ରଦେଶଂ ଦୃଷ୍ଣୁ ସମୁସୁକା ଅଭିବନ୍ ପରୀକ୍ଷାପରାନ୍ତ ଚ ଅନ୍ୟ ପ୍ରଦେଶସ୍ୟ ଭ୍ରମଣାର୍ଥମ୍ ଅଧ୍ଯକ୍ଷାୟ ନିବେଦିତବନ୍ତଃ । ଅଧୁନା ସର୍ବମ୍ ଆୟୋଜିତଂ କୃତମ୍ । ଅସ୍ମାକମ୍ ଅଧ୍ୟାପକ ତଥା ଚ ସର୍ବେ ଛାତ୍ରୀ ଗ୍ରୀଷ୍ମବକାଶସ୍ୟ ପ୍ରଥମେ ଏବ ଦିବସେ ଅରୁଣାଚଳ ପ୍ରତି ବାୟୁମାର୍ଗେଣ ଗମିଷ୍ୟାମଃ । ଯାତ୍ରାବ୍ୟୟାର୍ଥୀ ପଞ୍ଚଦଶସହସ୍ର ରୂପ୍ୟକାଣି ପ୍ରେଷୟନ୍ତୁ ଭବନ୍ତଃ । ସପ୍ତାହୋପରାନ୍ତଃ ତସ୍ମାତ୍ ପ୍ରଦେଶାତ୍ ପ୍ରତିନିବୃତ୍ୟ ଏବଂ ଗୃହମ୍ ଆଗମିଷ୍ୟାମି । ମାତୃଚରଣେଷୁ ମମ ପ୍ରଣାମା ସନ୍ତୁ । ଅନୁଜାୟ ସଲିଳାୟ ସ୍ନେହରାଶିଃ ।

ତ୍ତବତାଂ ପ୍ରିୟପୁତ୍ରଃ
ଅଂଶୁମାନ୍

ପ୍ରେଷକଃ	ପ୍ରାପକଃ
ଶ୍ରୀମାନ୍ ଅଂଶୁମାନ୍ ମହାପାତ୍ର	ଶ୍ରୀଯୁକ୍ତ ସୂର୍ଯ୍ୟପ୍ରକାଶ ମହାପାତ୍ରଃ
ମଧୁସୂଦନ ଛାତ୍ରାବାସୀଃ	ବିକାଶଭବନମ୍
ଭୁବନେଶ୍ୱରମ୍	ରାଉରକେଲା

6. स्वमहाविद्यालये रक्तदान शिविरं वर्णयन्तः मित्रं प्रति पत्रम् ।
Answer:

ଫକାରଣେହିନ ଛ।ତ୍ରାବାସଃ
ବାଲେଶ୍ବରମ୍
ଦିନାଙ୍କ -୨୫.୦୯.୧୬

ପ୍ରିୟ ମିତ୍ର ସୌରଭ !
ସାଦରଂ ବନ୍ଦନାନା
ହ୍ୟୁଃ ଏବ ତବ ପତ୍ର ପ୍ରାପ୍ତମ୍ । ମହାତ୍ମାଗାନ୍ଧିନଃ ଜନ୍ମଦିବସସ୍ୟ ସମାରୋହ ଭବତାଂ ମହାବିଦ୍ୟାଳୟେ ଯଥା ସମ୍ପାଦିତମ୍ ଇତି ଜ୍ଞାତ୍ମା ଅତୀବ ପ୍ରସନ୍ତଃ । ଅହମ୍ ଅପି ସ୍ୱମହାବିଦ୍ୟାଳୟସ୍ୟ ଏତାଦୃଶସ୍ୟ ସମାରୋହସ୍ୟ ବିଷୟେ ଲିଖାମି ଯସ୍ମିନ୍ ମହାପୁରୁଷାକାଂ ଭବତି । ତସ୍ୟ ଇଦମ୍ ଉଦ୍‌ଘୋଷମ୍- ‘ପୂୟଂ ମତ୍ସ୍ୟ ରକ୍ତ ଯଚ୍ଛତ ଅହଂ ଯୁଗ୍ମଭାଂ ସ୍ବାଧୀନତାଂ ପ୍ରଦାସ୍ୟ’’ ଇତି ଆଶ୍ରିତ୍ୟ ଏବ ଅସ୍ମାକଂ ମହାବିଦ୍ୟାଳୟେ ରକ୍ତଦାନଶିବରସ୍ୟ ଆୟୋଜନଂ କୃତମ୍ । ମହାବିଦ୍ୟାଳୟସ୍ୟ କ୍ରୀଡ଼ାପ୍ରମୁଖ ଅସ୍ୟ ଶିବରସ୍ୟ ଆୟୋଜନଃ ଆସୀତ୍ । ରକ୍ତଦାନାର୍ଥମ୍ ‘ଅହଂ ପ୍ରଥମ ପ୍ରଥମୋହମ୍’ ଇତି ଛାତ୍ରାଣାମ୍ ଅଧ୍ୟାପକାନାଂ ତଥା ଚ ପ୍ରଶାସକବର୍ଗସ୍ୟ ଉତ୍ସାହଃ ଦର୍ଶନୀୟ ଆସୀତ୍ । ଅସ୍ୟ କାର୍ଯ୍ୟକ୍ରମସ୍ୟ ସାଫଲ୍ୟମ୍ ଏବ ସୁଭାଷମହୋଦୟଂ ପ୍ରତି ଉଚିତା ଶ୍ରଦ୍ଧାଞ୍ଜଳି । ସୈନିକଚିକିତ୍ସାଳୟସ୍ୟ ସର୍ବୋଽପି ଚିକିତ୍ସକା କାର୍ଯ୍ୟକ୍ରମସ୍ୟ ଅଦ୍ଭୁତ ସାଫଲ୍ୟ ଭୂରି-ଭୂରି ପ୍ରଶଂସିତବନ୍ତଃ ।

ଗୃହେ ମାତୃପିତୃଚରଣୟୋ ମମ ପ୍ରଣାମା । ଶେଷ ମିଳନାନନ୍ତରମ୍ ଏବଂ କଥୟିଷ୍ୟାମି ।

ଭବଦୀୟମ୍ଅଭନ୍ନଂମିତ୍ରମ
ରବିନାରାୟଣାଃ

ପ୍ରେଷକଃ	ପ୍ରାପକଃ
ରବିନାରାୟଣ ପାଣିଃ	ସୌରଭ ପାଟ୍ଟଯୋଶା
ବାଲେଶ୍ଵରମ୍	ସମ୍ବଲପୁରମ୍

7. पितामहः तवजन्मदिवसे उपहारं प्रदत्तवान्, तं प्रति कृतज्ञता पत्रम् ।
Answer:

ବିକ। ଶସଦନମ୍
ବ୍ରହ୍ମପୁରମ୍
ଦିନାଙ୍କଃ -୧୭.୦୭.୧୬

ପରମଶ୍ରଦ୍ଧେୟ ପିତାମହ ଚରଣୟୋଃ
ସାଦର ପ୍ରଣତୟଃ
ଭବତା ପ୍ରେଷିତମ୍ ଉପହାରଂ ଜନ୍ମଦିବସସ୍ୟ ସହଭୋଜନକାଳେ ଏବ ପ୍ରାପ୍ତମ୍ । ତସ୍ମିନ୍ ଉଚିତେ ଅବସରେ ଭବତଃ ଆଶୀର୍ବାଦରୂପେଣ ଯା ଘଟିକା ମୟା ପ୍ରାପ୍ତା ତେନାହଂ ଭବଦର୍ଥେ ଅତୀବ କୃତଜ୍ଞତା ଜ୍ଞାପୟାମି । ଉପହାରପ୍ରେଷଣେନ ସହ ଯଃ ସମୟନିୟୋଜନସ୍ୟ ଭାବନା ଭବତାଂ ହୃଦୟେ ଆସନ୍ ଅହଂ ତାସାଂ ମନସା, ବାଚା, କର୍ମଣା ଆଚରଣଂ କରିଷ୍ୟାମି । ସମୟଂ ନଷ୍ଟ ନ କରିଷ୍ୟାମି । ସର୍ବେଷୁ କାର୍ଯ୍ୟଷୁ ନିୟମିତଃ ଭୂତ୍ଵା ଭାବିଜୀବନସ୍ୟ ନିର୍ମାଣେ ସନ୍ନଦ୍ଧ ଭବିଷ୍ୟାମି ।

ଗ୍ରୀଷ୍ମବକାଶେ ଅହଂ ପିତୃଭ୍ୟା ସହ ଅବଶ୍ୟମେବ ପୁରୀ ନଗରମ୍ ଆଗମିଷ୍ୟାମି । ପିତାମହୀମହାଭାଗାୟଃ ଚରଣେୟୋଃ ନମୋନମଃ।

ଭବତଃ ପ୍ରିୟଃ ପୌନଃ
ଆଦିତ୍ୟଃ

ପ୍ରେଷକଃ	ପ୍ରାପକଃ
ଶ୍ରୀମାନ୍ ଆଦିତ୍ୟ ବର୍ମା	ଶ୍ରୀଯୁକ୍ତ ଭୋଳାଶଙ୍କର ବର୍ମା
ବ୍ରହ୍ମପୁରମ୍	ଶ୍ରୀକ୍ଷେତ୍ରମ୍ ପୁରୀ

8. पर्वतीयस्थलस्य वर्णनं कुर्वन् स्वसखी प्रति पत्रम् ।
Answer:

ଆଚାର୍ଯ୍ୟବିହାରମ୍
ଭୁବନେଶ୍ୱରମ୍
ଦିନାଙ୍କ-୧୮.୦୭,୧୬

ପ୍ରିୟ ସଖ୍ ହରପ୍ରିୟେ !
ସସ୍ନେହଂ ନମସ୍କାରଃ ।
ଅତ୍ର କୁଶଳଂ ତତ୍ରାସ୍ତୁ । ବିଜୟଦଶମ୍ୟାଃ ଅବକାଶେଽହଂ ସ୍ବାଧ୍ୟାପକୌଃ ସହପାଠିଭିଶ୍ଚ ସହ ଶୈକ୍ଷିକ ଭ୍ରମଣାୟ ‘ରାନୀଖେତ’ କିଞ୍ଚିଦ୍ ଦୂର ସ୍ଥିତମ୍ ଇଦଂ ସ୍ଥାନଂ ପ୍ରକୃତଃ ରମ୍ୟା ବଉଁତେ । ବିଶାଲିଂ ଦେବଦାର-ବୃକ୍ଷେ ସୁସଜ୍ଜିତା ଇତି ପର୍ବତୀୟସ୍ଥଳଂ ଗତବତୀ । ‘ନୈନୀତାଲ’ ଇତି ପର୍ବତୀୟ ପ୍ରଦେଶତଃ ଲୀଳାସ୍ଥଳୀ ବର୍ଷତେ । ଅସ୍ୟ ସ୍ଥାନସ୍ୟ ନୈସର୍ଗିକଂ ସୌନ୍ଦର୍ଯ୍ୟ ବସ୍ତୁତଃ ଅଦ୍ଭୁତଂ ଇୟଂ ଦେବଭୂମି ମନମୋହିନୀ ଅସ୍ଥି । ଅତ୍ରତଃ ସୁଦୂରଂ ହିମାଚ୍ଛାଦିତାଃ ପର୍ବତଃ ଅପି ଦ୍ରସ୍ତୁ ଶକ୍ୟନ୍ତେ, ଯେ ରଜତେନ ଆଚ୍ଛାଦିତାଃ ଇବ ପ୍ରତୀୟନ୍ତେ । ଏକସ୍ମିନ୍ ପର୍ବତେ ତୁ ହିମେନ ନିର୍ମିତା ‘ତ୍ରିଶୂଳସ୍ୟ’ ଏବଂ ଆକୃତିଃ ଅପି ପରିଲକ୍ଷ୍ୟତେ । ସଖ୍, ବସ୍ତୁତଃ ମି ସହ ପର୍ବତଭ୍ରମଣ ଅତ୍ୟାନନ୍ଦକରଂ ବର୍ଭତେ, ପୁନଃ ପର୍ବତୀରୋହଣସ୍ୟ ତୁ କା କଥା । ଅହଂ ତ୍ରୟା ସହ ଅପି ଏକବାରଂ ତତ୍ର ଗନ୍ତୁମ୍ ଇଚ୍ଛାମି । ଆଶାସେ ଆବାମ୍ ଅଗ୍ରିମେ ଅବକାଶେ ଶୀଘ୍ର ଗମିଷ୍ୟତଃ । ଅଧୁନା ବିରମ୍ୟତେ ମୟା । ଗୃହେ ସର୍ବେଭ୍ୟଃ ନମୋନମଃ ।

ଭବତ୍ୟାଃ ସଖୀ
ଅଭିନ୍ନହୃଦୟା
ରୂପଶ୍ରୀ

ପ୍ରେଷକଃ	ପ୍ରାପକଃ
ରୂପଶ୍ରୀ ଦାସଃ	ହରପ୍ରିୟା ସାମଲଃ
ଜଳେଶ୍ଵରମ୍	ବାରିପଦା

9. स्वदिनचर्याया: वर्णनं कुर्वन् स्वमातरं प्रति पत्रम् ।
Answer:

ଗୋପବନ୍ଧୁଛାତ୍ରାବାସଃ
ମୟୂରଭଞ୍ଜମ୍
ଦିନାଙ୍କ-୨୦.୧୨.୧୬

ଆଦରଣୀୟ ମାତୃଚରଣଃ
ସାଦରଂ ପଣାମଃ ।
ଅହମତ୍ର କୁଶଳୀ । ଭବତୀନାଂ ପତ୍ର ପ୍ରାପ୍ୟ ସନ୍ତୋଷୀ ଜାତଃ ଯତ୍ ପିତୃମହୋଦୟଃ ଇଦାନୀ ପୂର୍ଣ୍ଣରୂପେଣ ସ୍ଵସ୍ଥ୍ୟଽସ୍ତି । ଧଃ ମମ ସ୍ଵାସ୍ଥ୍ୟସ୍ୟ ଅଧ୍ୟୟନସ୍ୟ ଚ ବିଷୟ ଚିନ୍ତତଃ ଆସୀତ୍ । ପରଂ ଚିନ୍ତାୟା ନ କୋଽପି ବିଷୟ ।

ଅହଂ ପ୍ରତିଦିନଂ ପ୍ରାତଃ ଚତୁର୍ବାଦନେ ଉତ୍ଥାୟ ବ୍ୟାୟାମଂ ଯୋଗାସନାଦିକଂ ଚ କୃତ୍ୱା ଘଣ୍ଟାଦ୍ଵୟଂ ପଠାମି ତତଃ ସ୍ନାତ୍ବା ଦୁଗ୍ଧାଦିକଂ ଚ ପୀତ୍ବା ପାଦୋନସପ୍ତବାଦନେ ମହାବିଦ୍ୟାଳୟଂ ଗଚ୍ଛାମି । ଦ୍ବିବାଦନେ ମହାବିଦ୍ୟାଳୟାତ୍ ଆଗତ୍ୟ ଭୋଜନଂ କୃତ୍ୱା ବିଶ୍ରାମିଂ କରୋମି । ସାର୍ଧଚତୁର୍ବାଦନେ ଉତ୍ଥାୟ ଗୃହକାର୍ଯ୍ୟ କରୋମି । ସାୟଂକାଳେ କ୍ରୀଡ଼ନାୟ ଗଚ୍ଛାମି, ରାକ୍ଷ୍ନୌ ଅହଂ କିଞ୍ଚିତ୍ ପଠାମି । ଗଣିତବିଷୟ ସଂସ୍କୃତବିଷୟ ଚ ଅହଂ ବିଶେଷତୟା ପରିଶ୍ରମଂ କରୋମି ।

ପରିଶ୍ରମସ୍ୟ ଫଳମ୍ ଅପି ମଧୁରଂ ଭବିଷ୍ୟତି ଇତି ଆଶା ଅସ୍ଥି । ପିତୃମହାଭାଗାନାଂ ଚରଣ ପ୍ରଣାମଃ କଥନୀୟଃ ପତ୍ରୋତ୍ତରଂ ପ୍ରତୀକ୍ଷମାଣଃ ।

ଭବତୀନାଂ ବିନୀତଃ ବାଜଃ
ରମେଶଃ

ପ୍ରେଷକଃ	ପ୍ରାପକଃ
ରମେଶଚନ୍ଦ୍ର ଦାଶଃ	ଶ୍ରୀମତୀ ବିଜୟିନୀ ଦାଶଃ
ମୟୂରଭଞ୍ଜମ୍	ଭଦ୍ରକମ୍

10. स्वास्थ्यरक्षणविषये अनुजं प्रति पत्रम् ।

ସାଇସଦନମ୍
ଯାଜପୁରମ୍
ଦିନାଙ୍କ-୨୫.୧୦.୧୬

ପ୍ରିୟ ସୁଧାକର,
ସସ୍ନେହାଶୀଷଃ।
ପୂଜ୍ୟାୟା ମାତୁଃ ପତ୍ରେଣ ଜ୍ଞାତଂ ଯତ୍ ଗତ ସପ୍ତାହେ ଜଂ ରୁଗ୍‌ ଆସୀ। ଜ୍ଞାୟତେ ଯତ୍ ତଂ ସ୍ଵାସ୍ଥ୍ୟରକ୍ଷା ପ୍ରତି ଜାଗୃତଃ ନାସି ।
ଅନୁଜ ! ସ୍ଵାସ୍ଥ୍ୟମ୍ ଅମୂଲ୍ୟ ରତ୍ନମ୍ ଅସ୍ଥି । ତଦ୍ ବିନା ଜଗତି ସର୍ବଂ ଧନଂ ବ୍ୟର୍ଥମ୍ ଏବ । ଅତଏବ ସ୍ଵାସ୍ଥ୍ୟସ୍ୟ ରକ୍ଷାୟେ ଅଧୋଲିଖ୍ ନିୟମା ଅବଶ୍ୟମେବ ପାଳନୀୟଃ!

• ପ୍ରାତଃ ସୂର୍ଯ୍ୟସ୍ୟ ଉଦୟାତ୍ ପୂର୍ବମ୍ ଏବ ଉଦ୍ଘାତବ୍ୟମ୍ । ଏବମ୍ ଆଳସ୍ୟ ଶରୀରଂ ନ ଆକ୍ରାମତି ।
• ଶୁଦ୍ଧ ପବନେ ବ୍ୟାୟାମଂ ଯୋଗସନାଦିକଂ ଚ କର୍ତ୍ତବ୍ୟମ୍ । ଏତେନ ଶରୀର ସବଳଂ ସ୍ବୟଂ ଚ ଭବତି ।
• ତସ୍ମାଦନନ୍ତରଂ କୋଚିଂ ଦୁଗ୍ଧ ପାତବ୍ୟମ୍ ।
• ନିତ୍ୟ ଭୋଜନଂ କୃତ୍ରା ଦନ୍ତାଃ ଅବଶ୍ୟମେବ ମାର୍ଜନୀୟା । ଯତଃ ଯସ୍ୟ ଦନ୍ତଃ ନିର୍ମଳା ଭବନ୍ତ ତଥ୍ୟ ଭୋଜନମ୍ ଉଦରେ ସମ୍ୟକ୍ ପଚତି ।
• ଭୋଜନଂ କୃତ୍ୱ କଦାପି ନ ଧାବିତବ୍ୟମ୍ ।
• ଗରିଷ୍ଠ ଭୋଜନଂ ନୈବ ଗ୍ରହଣୀୟମ୍ । ଫଳାନି ଶାକାନି ଚ ଖାଦିତବ୍ୟାନି ଯତଃ ଫଳାନି ଶରୀରଂ ପୁଙ୍ଖ କୁର୍ବନ୍ତି ।
• ଖେଳନାନନ୍ତରଂ ଜଳସ୍ୟ ପାନଂ ହାନିକରମ୍ ।
• ନିଶାୟାଂ ପ୍ରଥମେ ଏବ ପ୍ରହରେ ଶୟନାୟ ପର୍ଯ୍ୟଙ୍କ ଗନ୍ତବ୍ୟମ୍ ।

ଆଶା ଅସ୍ତା ଯତ୍ ତ୍ୱମ୍ ଅବଶ୍ୟମେବ ଏତାନ୍ ନିୟମାନ୍ ପାଳୟଷ୍ୟସି ଯତଃ ‘ଶରୀରମାଦ୍ୟ ଖଳୁ ଧର୍ମସାଧନମ୍’ । ଗୃହେ ସର୍ବେଭ୍ୟ ଯଥାଯୋଗ୍ୟମ୍ । ପତ୍ରୋତ୍ତରସ୍ୟ ପ୍ରତୀକ୍ଷାୟାମ୍ ।

ତବ ଶୁଭଚିନ୍ତକଃ
ଦିବ୍ୟଜ୍ୟୋତିଃ

ପ୍ରେଷକଃ	ପ୍ରାପକଃ
ଦିବ୍ୟଜ୍ୟୋତି ମହାପାତ୍ରଃ	ସୁଧାକର ପଣ୍ଡା
ଯାଜପୁରମ୍	ରଣପୁରମ୍

 

Odisha State Board CHSE Odisha Class 11 Sanskrit Solutions Grammar ବିଷୟବହିର୍ଭୂତ ବ୍ୟାକରଣ Textbook Exercise Questions and Answers.

CHSE Odisha Class 11 Sanskrit Grammar ବିଷୟବହିର୍ଭୂତ ବ୍ୟାକରଣ

ସମାସ

(୧) ଅବ୍ୟୟୀଭାବ –

ଅଧୁରି = ହରୌ ଇତି
ଉପକୃଷ୍ଣମ୍ = କୃଷ୍ଣସ୍ୟ ସମୀପମ୍
ସୁକଳିଙ୍ଗମ୍ = କଳିଙ୍ଗାନାଂ ସମୃଦ୍ଧିଃ
ଦୁର୍ଭାଗ୍ୟମ୍ = ଭାଗ୍ୟାନାଂ ବୃଦ୍ଧିଃ
ନିର୍ଜନମ୍ = ଜନାନାମ୍ ଅଭାବ
ଅତିହିମମ୍ = ହିମସ୍ୟ ଅତ୍ୟୟଃ
ଅନୁବିଷ୍ଣୁ = ବିଷୋ ପଶ୍ଚାତ୍
ଅନୁରୂପମ୍ = ରୂପସ୍ୟ ଯୋଗ୍ୟମ୍
ଯଥାଶକ୍ତି = ଶକ୍ତିମ୍ ଅନତିକ୍ରମ୍ୟ
ସହରି = ହରଃ ସାଦୃଶ୍ୟମ୍
ସଚକ୍ରମ = ଚକ୍ରଣ ଯୁଗପତ୍
ସକ୍ଷତ୍ରମ୍ = କ୍ଷତ୍ରାମାଂ ସମ୍ପରିଂ
ସାଗ୍ନି = ଅଗ୍ନି ପର୍ଯ୍ୟନ୍ତମ୍ ଅଧୀତେ
ଆମୁକ୍ତି = ଆ ମୁକ୍ତେଃ
ପାରେଗଙ୍ଗମ୍ = ଗଙ୍ଗାୟା ପାରମ୍

(୨) ତତ୍‌ପୁରୁଷ – ଦ୍ଵିତୀୟା – ତତ୍‌ –

କୃଷ୍ଣଶ୍ରିତଃ = କୃଷ୍ଣ ଶ୍ରିତଃ
ଉପକୂଳମ୍ = କୂଳସ୍ୟ ସମୀପମ୍
ସୁମଦ୍ରମ୍ = ମଦ୍ରାମାଂ ସମୃଦ୍ଧିଃ
ଦୁର୍ଯବନମ୍ = ଯବନାନାଂ ବୃଦ୍ଧିଃ
ନିର୍ମକ୍ଷିକମ୍ = ମକ୍ଷିକାଣାମ୍ ଅଭାରଃ
ଦୁର୍ଭିକ୍ଷମ୍ = ଭିକ୍ଷାୟାଃ ଅଭାବ
ଅତିଗ୍ରୀଷ୍ମମ୍ = ଗ୍ରୀଷ୍ମସ୍ୟ ଅତ୍ୟୟଃ
ଅନୁରଥମ୍ = ରଥସ୍ୟ ପଶ୍ଚାତ୍
ପ୍ରତିଦିନମ୍ = ଦିନଂ ଦିନମ୍
ଯଥାବିଧ୍ = ବିଧୂମ୍ ଅନତିକ୍ରମ୍ୟ
ଅନୁଜ୍ୟେଷ୍ଠମ୍ = ଜ୍ୟେଷ୍ଠସ୍ୟ ଆନୁପୂର୍ବେଣ
ସସଖ୍ = ସଖ୍ୟା ସଦୃଶଃ
ସତୃଣମ୍ = ତୃଣମ୍ ଅପି ଅପରିତ୍ୟଜ୍ୟ
ଯାବଚ୍ଛୋକମ୍ = ଯାବନ୍ତଃ ଶ୍ଳୋକା
ଆବାଳମ୍ = ଆ ବାଳେଭ୍ୟ
ମଧ୍ୟଗଙ୍ଗମ୍ = ଗଙ୍ଗାଯା ମଧ୍ୟମ୍ ।
ଦୁଃଖାତୀତଃ = ଦୁଃଖମ୍ ଅତୀତଃ
ଗ୍ରାମଗତଃ = ଗ୍ରାମଂ ଗତଃ
ପ୍ରାପ୍ତଜୀବନଃ = ଜୀବନଂ ପ୍ରାପ୍ତଃ
କୂପପତିତଃ = କୂଫ ପତିତଃ
ମୁହ୍ୱର୍ଭସୁଖମ୍ = ମୁହୂର୍ଭିଂ ସୁଗମ୍

ତୃତୀୟା ତତ୍ –

ଦ୍ରିରିତ୍ରାତଃ = ହରିଶା ତ୍ରାତଃ
ମାସପୂର୍ବଃ = ମାସେନ ପୂର୍ବଃ
ମାତୃସତୃଣଃ = ମାତ୍ରା ସଦୃଶଃ
ଆଚାରନିପୁନଃ = ଆଚାରେଣ ନିପୁନଃ
ମାସାବରଃ = ମାସେନ ଅବରଃ
ନଖଭିନଃ = ନଖିଃ ଭିନଃ
ପିତୃସମଃ = ପିତ୍ରାସମଃ
ବାକ୍କଳହଃ = ବାଚା କଲହଃ
ଗୁଡ଼ମିଶ୍ର = ଗୁଡ଼େନ ମିଶ୍ର
ଗୁଡ଼ଧାନଃ = ଗୁଡ଼େନ ଧାନଃ

ଚତୁର୍ଥା ତତ୍‌ –

ଯୂପଦାରୁ = ପୂପାୟ ଦାରୁ
ଭୂତବହିଃ = ଭୂତେଭ୍ୟଃ ବଳି
ଗୋସୁଖମ୍ = ଗୋଭ୍ୟ ସୁଖମ୍
ବାଳକାର୍ଥିଃ = ବାଳକାୟ ଅୟମ୍
ଗୋହିତମ୍ = ଗୋଭ୍ୟ ହିତମ୍
ଗୋରଷିତମ୍ = ଗୋଭ୍ୟଃ ରଷିତମ୍

ପଞ୍ଚମୀ ତତ୍ –

ଚୋରଭୟମ୍ = ଚୋରାଦ୍ ଭୟମ୍
ଗ୍ରାମନିର୍ଗତଃ = ଗ୍ରାମାତ୍‌ ନିର୍ଗତଃ
ସ୍ଵର୍ଗପଡିତଃ = ସ୍ୱର୍ଗାତ୍ ପତିତଃ
ଦୂରାଦାଗତଃ = ଦୂରାତ୍‌ ଆଗତଃ
ବ୍ୟାଘ୍ରଭୀତଃ = ବ୍ୟାଘ୍ରାତ୍ ଭୀତଃ
ସୁଖାପେତଃ = ସୁଖାତ୍ ଅପେତଃ
ଚନ୍ଦ୍ରମୁନଃ = ଚକ୍ରାତ୍‌ ମୁନଃ
ପରଷ୍ପତଃ = ଶତାତ୍ ପରେ

ଷଷ୍ଠୀ ତତ୍ –

ଦେଶସେବକଃ = ଦେଶସ୍ୟ ସେବକଃ
ବ୍ରାହ୍ମଣଯାଜକଃ = ବ୍ରାହ୍ମଣସ୍ୟ ଯାଜକଃ
ସର୍ବମହାନ୍ = ସର୍ବେକ୍ଷାଂ ମହତ୍ତରଃ
ଅର୍ଥଗୌରବମ୍ = ଅର୍ଥସ୍ୟ ଗୌରବମ୍
ବୃକ୍ଷୋପରି = ବୃକ୍ଷସ୍ୟ ଉପରି
ଦେବପୂଜନମ = ଦେବସ୍ୟ ପୂଜକଃ
ଦେବପୂଜନମ୍ = ଦେବସ୍ୟ ପୂଜନମ୍
ସ୍ୱକର୍ତ୍ତବ୍ଯମ୍ = ସ୍ବସ୍ଯ କର୍ତ୍ତବ୍ଯମ୍
ବୁଦ୍ଧିମାନ୍ଦ୍ୟମ୍ = ବୁଦ୍ଧେଃ ମାନ୍ଦ୍ୟାମ୍
କଳିଙ୍ଗରାଜଃ = କଳିଙ୍ଗାନାଂ ରାଜା

ସପ୍ତମୀ ତତ୍ –

ଅକ୍ଷଶୌନଃ = ଅକ୍ଷେଷୁ ଶୌଣ୍ଡୀ
ବେଦପଣ୍ଡିତଃ = ବେଦେ ପଣ୍ଡିତଃ
କାର୍ଯ୍ୟଚପଳୀ = କାର୍ଯ୍ୟ ଚପଳଃ
ଶାସ୍ତ୍ରପ୍ରବୀଣୀ = ଶାସ୍ତ୍ରପ୍ରବୀଣୀ
କାବ୍ୟକୁ ଶଳଃ = କାବ୍ୟେ କୁଶଳଃ
କାଶୀସିଦ୍ଧଃ = କାଶ୍ୟା ସିଦ୍ଧଃ
ଆତପଶୁଷ୍କଃ = ଆତପେ ଶୁନଃ
ତୀର୍ଥକାକଃ = ତୀର୍ଥେ କାକଃ
ଚକ୍ରବନ୍ଧ = ଚକ୍ରେ ବନ୍ଧଃ
କଟାହପକ୍ଵଃ = କଟାହେ ପକ୍ଵଃ

ନଞ ତତ୍ –

ଅବିଘ୍ନଃ = ନ ବିଘ୍ନଃ
ଅନକଃ = ନ ଅଶ୍ଳଃ
ଅସନ୍ଦେହଃ = ନ ସନ୍ଦେହଃ
ଅପନ୍ଥାଃ = ନ ପନ୍ଥାଃ

ପ୍ରାଦି ତତ୍ –

ସୁପୁରୁଷଃ ଶୋଭନଃ ପୁରୁଷଃ
ପ୍ରାଚାର୍ଯ୍ୟ = ପ୍ରଗତଃ ଆଚାର୍ଯ୍ୟଃ

ଉପପଦ ତତ୍ –

କୁମ୍ଭକାରଃ = କୁମ୍ଭୀ କରୋତି ଇତି
ଜଳଦଃ = ଜଳଂ ଦଦାତି ଇତି
ନିଶାକରଃ = ନିଶା କରୋତି ଇତି
ପାଦପଃ = ପାଦେ ପିବତି ଇତି ।

(୩) କର୍ମଧାରୟ –

ନୀଳୋତ୍ପଳମ୍ = ନୀଳମ୍ ଉତ୍ପଳମ୍
ଭୀମକାନ୍ତଃ = ଭୀମ କାନ୍ତଃ
ଘନଶ୍ୟାମଃ = ଘନଃ ଇବ ଶ୍ୟାମଃ
ବିଦ୍ୟାସାଗରଃ = ବିଦ୍ୟା ଏବ ସାଗରଃ
ଶାକପାର୍ଥ୍ବଃ = ଶାକପ୍ରିୟଃ ପାର୍ଥିବଃ
କୃତାକୃତମ୍ = କୃତମ୍ ଅକୃତମ୍
ତପୋଧନମ୍ = ତପଃ ଏବ ଧନମ୍
ବୈୟାକରଣଖପୂଚିଃ = ବୈୟାକରଣଃ ଖସୂଚିଃ
ରାମକୃଷ୍ଣଃ = ରାମଃ କୃଷ୍ଣଃ
ପୁରୁଷବ୍ୟାଘ୍ର = ପୁରୁଷ ବ୍ୟାଘ୍ର ଇବ
ବଟବୃକ୍ଷଃ = ବରଂ ଇତି ବୃକ୍ଷଃ
ଗ୍ରାମାନ୍ତରମ୍ = ଅନ୍ୟଃ ଗ୍ରାମଃ
ପୂର୍ବପୁରୁଷ = ପୂର୍ବଃ ପୁରୁଷଃ
ମହାପୁରୁଷ = ମହାନ୍ ପୁରୁଷଃ
ମୁଖପଦ୍ମମ୍ = ମୁଖ୍ୟ ପଦ୍ମମ୍ ଇବ
ଅଧ୍ୟାନଗରୀ = ଅଯୋଧ୍ୟା ଇତି ନଗରୀ

(୪) ଦ୍ବିଗୁ –

ଷାଣ୍କ।ତୁରଃ = ଷଣ୍ଣା ମାତୃ ଣାମ୍ ଅପତ୍ୟ ପୁମାନ୍
ପଞ୍ଚବଟୀ = ପଞ୍ଚାନା ବଟାନାଂ ସମାହାରଃ
ତ୍ରିଭୁବନମ୍ = ତ୍ରୟାନାଂ ଭୁବନାନାଂ ସମାହାରଃ
ସପ୍ତାହଃ = ସପ୍ତାନାମ୍ ଅହ୍ନା ସମାହାରଃ
ପଞ୍ଚରବଧନଃ = ପଞ୍ଚଗାବଃ ଧନଂ ଯସ୍ୟ ସଃ
ତ୍ରିଲୋକୀ = ତ୍ରୟାନାଂ ଲୋକାନାଂ ସମାହାରଃ
ଚତୁର୍ୟୁଗମ୍ = ଚତୁର୍ବିଂ ଯୁଗାନାଂ ସମାହାରଃ
ତ୍ରିଫଳା = ତ୍ରୟାଣାଂ ଫଳାନାଂ ସମାହାରଃ

(୫) ଦ୍ବନ୍ଦ୍ବ –

ହରିହରୌ = ହରିଶ୍ଚ ହରଶ୍ଚ
ମାତାପିତଗୌ = ମାତା ଚ ପିତା ଚ
ପାଣିପାଦମ୍ = ପାଣି ଚ ପାଦୌ ଚ
ଘଟପଟମ୍ = ଘଟଣ୍ଟ ପଟଣ୍ଟ ଅନୟୋ ସମାହାରଃ
ଅହିନକୁଳମ୍ = ଅହିଷ୍ଟ ନକୁଳଶ୍ଚ

(୬) ବହୁବ୍ରୀହି –

ପ୍ରାୟୋଦକଃ = ପ୍ରାପ୍ତମ୍ ଉଦକଂ ଯଃ ଡଃ
ଅପୁନଃ = ଅବିଦ୍ୟମାନଃ ପୁନଃ ଯସ୍ୟ ଡଃ
ବୀଣାପାଣିଃ = ବୀଣା ପାଣି ଯସ୍ୟା ସା
କଣ୍ଠେକାଳୀ = କଣ୍ଠେ କାଳେ ଯସ୍ୟ ସ୍ୱ
ଦ୍ଵିତ୍ରାଃ = ଦୌ ବା ତ୍ରୟୋ ବା
ସପୁତ୍ରଃ = ପୁତ୍ରେଣ ସହ ବର୍ତ୍ତମାନଃ
ଅପ୍ରଜାଃ = ଅବିଦ୍ୟମାନା ପ୍ରଜା ଯସ୍ୟ ଡଃ
ସୁଗନ୍ଧି = ଶୋଭନଃ ଗନ୍ଧ ଯସ୍ୟ ସ୍ଃ
ପୀତାମ୍ବରଃ = ପୀତମ୍ ଅମ୍ବରଂ ଯସ୍ୟ ଡଃ
ଦୃଢ଼ାଭକ୍ତି = ଦୃଢ଼ା ଭକ୍ତି ଯସ୍ୟ ଡଃ
ଚନ୍ଦ୍ରମୌଳି = ଚନ୍ଦ୍ର ମୌନୌ ଯସ୍ୟ ଡଃ
ଉପଦଶା = ଦଶାନାଂ ସମୀପେ ଯେ ସନ୍ତ ତେ
ସପୁନଃ = ପୁତ୍ରେଣ ସହ ବର୍ତ୍ତମାନଃ
କେଶାକେଶି = କେଶେଷୁ କେଶେଷୁ ଗୃହୀତ୍ମା ଇଦଂ ଯୁଦ୍ଧଂ ପ୍ରବୃତ୍ତମ୍
ନିର୍ଭୟ = ନିର୍ଗତା ଦୟା ଯସ୍ମାତ୍
ଗଜାନନଃ = ଗଜାନାମ୍ ଇବ ଆନନଂ ଯସ୍ୟ ଡଃ

ଏକପଦୀକରଣ

ଅସ୍କିନ୍‌ ଦିନେ = ଅଦ୍ୟ
ସମୀପମ୍ = ସମକ୍ଷମ୍
ଅହଞ୍ଚ ତ୍ଵଞ୍ଚ = ଆବାମ୍
ଅକ୍ଷଣୋ ପଶ୍ଚାତ୍ = ଅନ୍ନକ୍ଷମ୍
ଅନ୍ତେ ବସତି ଇତି = ଅନ୍ତେବାସୀ
ଆଗମନଂ କୃତ୍ୱା = ଆଗମ୍ୟ|ଆଗତ୍ୟ
ଯୁନଃ ଭାକଃ = ଯୌବନମ୍
ନରସ୍ୟ ଭାକଃ = ନରତ୍ୱ, ନରଡା
ନରତା ବିକାରଃ = ମୃଣ୍ମୟଃ
ବିଦୁଷୀ ପତ୍ନୀ = ବିଦୁଷୀ
ବିଳେ ସ୍ଥିତ୍ରା = ବିଳାତ୍
ବଚାରଂ କୃତ୍ୱା = ବିଚାର୍ଯ୍ୟ
ବିଦ୍ୟା ଅସ୍ୟ ଅସ୍ତୀତି = ବିଦ୍ୱାନ୍
ରାଜ୍ଞଃ ପୁରୁଷଃ = ରାଜପୁରୁଷଃ
ଶୟନଂ କର୍ଭୁମ୍ = ଶୟିତୁମ୍
ଜ୍ଞାତୁ ପ୍ରେରୟତୁ = ଜ୍ଞାପୟତୁ
କୁତ୍ସିତଂ ଅନ୍ତଃ = କଦଂଶଃ
ଆଦାନଂ କୃତ୍ୱା = ଆଦାୟ
ନନ୍ଦୟତି ଯଃ ସ = ନନ୍ଦନଃ
କୃଷ୍ଣସ୍ୟ ସମୀପମ୍ = ଉପକୃଷ୍ଣମ୍
ବସ୍ତୁ ମନଃ ଯସ୍ୟ ଡଃ = ବକ୍ତୁମନାଃ
ମନୋ ସ୍ତ୍ରୀ = ମନବୀ, ମନାୟୀ
ଶାସ୍ତି ଯା = ଶାସତୀ
ଜଳମାନେତୁମ୍ = ଜଳାୟ
କାଲ୍ୟା ଦାସ = କାଳିଦାସ୍ଯ
ଶିବୋ ଦେବତା ଅସ୍ୟ = ଶୈବାଃ
ବୃକ୍ଷମ୍ ଆରୁହ୍ୟ = ବୃକ୍ଷାତ୍
ବିଭେତି ଯା = ବିଦ୍ୟୁତୀ
ବିଷୋ ପୁରମ୍ = ବିଷ୍ଣୁପୁରମ୍
ଦାତ୍ମ୍ଯ ପ୍ରେରୟତୁ = ଦାପୟତୁ
ଶୟନଂ ପ୍ରେରୟତି = ଶାୟୟତି
ଗଚ୍ଛତି ଯା = ଗଚ୍ଛନ୍ତୀ
ଗୃହେ ଇତି = ଅଧ୍ଗୃହମ୍
ବନେଚରତି ଯା ସା = ବନେଚରୀ
ଶ୍ରବଣାଦନନ୍ତରମ୍ = ଶୁଦ୍ଧା
ସୂର୍ଯ୍ୟସ୍ୟ ସ୍ତ୍ରୀ = ସୂରୀ/ସୂର୍ଯ୍ୟା
କୃତ୍ରିମା ଭୂମି = ସ୍ଥଳା
ଈଷତ୍ ଉଷୁମ୍ = କବୋଷମ୍
ଶ୍ରୀ ଅସ୍ୟା ଅସ୍ତୀତି = ଶ୍ରୀମତୀ
ଯୁବତିଃ ଜାୟା ଯସ୍ୟ ଡଃ = ଯୁବଜାନିଃ
ଶୂରସ୍ୟ ଭାବଃ = ଶୂରତା ଣୌର୍ଯ୍ୟମ୍
ସୂରଭେ ଅପତ୍ୟ ସ୍ତ୍ରୀ = ସୌରଭେୟୀ
କନ୍ୟାୟ ଅପତ୍ୟ ପୁମାନ୍ = କାନୀନଃ
ଶୋଭନଂ ହୃଦୟଂ ଯସ୍ୟ କଃ = ସୁହୃତ୍
ତ୍ରୟାନାଂ ଭୁବନାନାଂ ସମାହାର = ତ୍ରିଭୁବନମ୍
ଶୋଭନା ମତି ଯସ୍ୟ ଡଃ = ସୁମତିଃ ଭୋଜନଂ କୁର୍ବନ୍ = ଭୁଜ୍ୟମାନଃ
ଭୋଜନାଦନନ୍ତରମ୍ = ଭୁକ୍କା
ଚଣ୍ଡୀ ଏବ ରୂପଃ ଯସ୍ୟ ଡଃ = ଚଣ୍ଡରବ
ଅନେନି କୁର୍ବନ୍ = ଇତ୍ଥମ୍
ଅପରମ୍ ଅନ୍ତଃ = ଅପରାହ୍ଣ
ତ୍ରୟାନାଂ ଲୋକାନାଂ ସମାହାରଃ = ତ୍ରିଲୋକୀ
ପଞ୍ଚାନାଂ ନଳାନାଂ ସମାହାରଃ = ପଞ୍ଚନଳମ୍
ସକୃତ୍ ପ୍ରସବା ଗୌଃ = ଗୃଷିଃ
ଶୟନଂ କୃତବାନ୍ = ଶୟିତବାନ୍
କ୍ଷୁଧାୟା ହେତୋ = କ୍ଷୁଧୟା
ବାରୀମାଂ ବାହକଃ = ବଳାହକଃ
ନ ଇଷ୍ଟମ୍ = ଅନିଷ୍ଟମ୍
ସାକ୍ଷାତ୍ ଦୃଷ୍ଟବାନ୍ = ସାକ୍ଷୀ
ମନୁନା ପ୍ରୋକ୍ତମ୍ = ମନୁପ୍ରୋକ୍ତମ୍
ଭିକ୍ଷାର୍ଥେ ଇଦମ୍ = ଭିକ୍ଷାର୍ଥମ୍
ହରୌ ଇତି = ଅଧୂରି
ଈଶ୍ବରସ୍ୟ ସ୍ତ୍ରୀ = ଈଶ୍ବରୀ
ମହତ୍ ହିମମ୍ = ହିମାନୀ
ଦୟା ଅସ୍ତି ଅସ୍ୟ = ଦୟାନୁଃ
ପିନାକଃ ଧନୁଃ ଯସ୍ୟ ଡଃ = ପିନାକଧନ୍ବା
ନ ଶଃ = ଅଗ|ନମଃ
କୁଶଣ୍ଢ ଲବଷ୍ଟ = କୁଶୀଲର୍ବୋ
ଦାତ୍ମ୍ଯ ଯୋଗ୍ୟମ୍ = ଦାତବ୍ୟମ୍
ମତିଃ ଅନ୍ୟ ଅସ୍ତୀତି = ମତିମାନ୍
ରାଜ୍ଞ ଧୁରମ୍ = ରାଜଧୁରମ୍
ସମ୍ୟକ୍ ଦାନମ୍ = ସମ୍ପ୍ରଦାନମ୍
ବର୍ଦ୍ଧନଂ କୁର୍ବାଶଃ = ବର୍ଷମାନଃ
ନିଶାୟାଂ ଚରତି ଯା ସା = ନିଶାଚରୀ
ପୁନଃ ପୁନଃ ଗଚ୍ଛତି = ଜଙ୍ଗମ୍ୟତେ
ବୃତ୍ତୟେ ଇଦମ୍ = ବୃତ୍ତୀର୍ଥମ୍
ଦାନଂ କୃତମ୍ = ଦତ୍ତମ୍
ପାନଂ କୃତମ୍ = ପୀତମ୍
ଆସନେ ଉପବିଶ୍ୟ = ଆସନାତ୍
ଉନ୍ନତଂ ମନଃ ଯସ୍ୟ କଃ = ଉନ୍ନତମନାଃ
ଦର୍ଶନଂ କୁର୍ବନ୍ = ପଶ୍ୟନ୍‌
ବର୍ଣ୍ଣ ଏବ = ବର୍ଣ୍ଣମାତ୍ରମ୍
ପାନସ୍ୟ ନିମିତ୍ତମ୍ = ପାନୀୟ, ପାତୁମ୍
ବିବାହିତା ସ୍ତ୍ରୀ = ପାଣିଗୃହିତୀ
ସିଂହାସନେ ଉପବିଶ୍ୟ = ସିଂହାସନାତ୍
ପ୍ରିୟା ସହ ବର୍ତ୍ତମାନଃ = ସସ୍ତ୍ରୀକଃ
ଅସ୍ଥିନ୍ ବର୍ଷେ = ଐଷମ
ପୃଥାୟା ଅପତ୍ୟ ପୁମାନ୍ = ରାଧେୟଃ
ଅଧ୍ୟୟନାତ୍ ପରମ୍ = ଅଧୀତ୍ୟ
ସମାନଃ ପତିଃ ଯସ୍ୟା ସା =
ମଗଧସ୍ୟ (ମଗଧାନାଂ) ରାଜା = ମଗଧରାଜଃ
ମକ୍ଷିକାଣାମ୍ ଭାବଃ = ନିର୍ମକ୍ଷିକମ୍
ତୃଣମପି ଅପରିତ୍ୟଜ୍ୟ = ସତୃଣମ୍
ତେଜଃ ବିଦ୍ୟତେ ଯସ୍ୟାଃ = ତେଜସ୍ଵୀ
ହିରଣସ୍ୟ ବିକାରଃ = ହିରଣ୍ମୟମ୍
କୁତ୍ସିତମ୍ ଅକ୍ଷମ୍ = କାକ୍ଷମ୍
ତରୁଣା ଛାୟା = ତରୁଛାୟମ୍ ଦଦାତି ଯା = ଦଦତୀ
ହରଃ ସାଦୃଶ୍ୟମ୍‌ = ସହରି = ସତୃଣମ୍
ତେଜଃ ବିଦ୍ୟତେ ଯସ୍ୟା = ତେଜସ୍ବୀ
ହିରଣସ୍ୟ ବିକାରଃ = ହିରଣ୍ମୟମ୍
ମୃଗସ୍ୟ ଇବ ନୟନେ ଯସ୍ୟା ସା = ମୃଗନୟନା
କୁସିତମ୍ ଅଷ୍ଟମ୍ = କାନ୍ଧମ୍
ରାଜ୍ଞ ଗୌ = ରାଜଗବ
ତତ୍ ଏବ = ତନ୍ମାତ୍ରମ୍
ଦଦାତି ଯା = ଦଦତୀ
ହରଃ ସାଦୃଶ୍ୟମ୍‌ = ସହରି
କସ୍ମାତ୍ କାରଣାତ୍ = କୃତଃ
ନୀୟତେ ଅନେନ ଇତି = ନୟନମ୍
ଗ୍ରାମେ ବସତି ଇତି = ଗ୍ରାମବାସୀ
ଗ୍ରାମେ ଭରଃ = ଗ୍ରାମ୍ୟ
ପଣ୍ଡା ଅସ୍ତି ଅସ୍ୟ ଇତି = ପଣ୍ଡିତଃ
ଗସ୍ତ୍ୟ ଯୋଗ୍ୟମ୍ = ଗନ୍ତବ୍ୟମ୍
ଗାନଂ କୃତଃ = ଗୀତଃ
ନୟନଂ ପ୍ରେରୟତି = ନାୟୟତି

ନ ଆଗତଃ = ଅନାଗତଃ
ପଥସ୍ୟ ସ୍ତ୍ରୀ = ପଥୁକୀ
ଲିଖନଂ କୃତବତୀ = ଲିଖ୍ବତବତୀ
ନ ମିତ୍ରମ୍ = ଅମିତ୍ରଃ
ଆଚାର୍ଯ୍ୟସ୍ଯ ପତ୍ନୀ = ଆଚାର୍ଯ୍ୟାଣୀ
ବୃହତାଂ ପତିଃ = ବୃହସ୍ପତିଃ
ବ୍ରାହ୍ମଣସ୍ୟ ପତ୍ନୀ = ବ୍ରାହ୍ମଣୀ
ମନୋ ଅପତ୍ୟ ପୁମାନ୍ = ମନୁଜଃ/ମାନମଃ
ପଞ୍ଚାନାଂ ରାତ୍ରାମିଂ ସମାହାରଃ = ପଞ୍ଚର।ତ୍ରମ୍
ମହତ୍ ଅରଣ୍ୟମ୍ = ଅରଶ୍ୟାନୀ
ଏକାବାରମ୍ = ସକୃତ୍
ନାସ୍ତି କିଞ୍ଚନ ଯସ୍ୟ ଡଃ = ଅକିଞ୍ଚନଃ
ବସୂନି ଧରତି ଯା ସା = ବସୁନ୍ଧରା
ବନସ୍ୟ ସମୀପମ୍ = ଉପବନମ୍
କୁତ୍ସିତଂ ଜଳମ୍ = କାଜଳମ୍
ଶୂଦ୍ରଜାତୀୟା ସ୍ତ୍ରୀ = ଶୂଦ୍ରା
ବାଚା କଳହଃ = ବାକ୍‌କଳହଃ
ଯା ଗଚ୍ଛତି = ଗଚ୍ଛନ୍ତି
ଶରଦଃ ସମୀପମ୍ = ଉପଶରଦମ୍
ଅକ୍ଷ୍ଶୋ ପଶ୍ଚାତ୍ = ଅନ୍ବକ୍ଷମ୍

ସ୍ତ୍ରୀ ପ୍ରତ୍ୟୟ

Odisha State Board CHSE Odisha Class 11 Sanskrit Solutions Grammar निमन्त्रणपत्रम् Textbook Exercise Questions and Answers.

CHSE Odisha Class 11 Sanskrit Grammar निमन्त्रणपत्रम्

1. स्वपुत्रस्य विवाहनिमन्त्रणपत्रम् ।
Answer:
ଓଁ ଶ୍ରୀ ଶ୍ରୀ ପ୍ରଜାପତୟେଃ ନମଃ
ଅୟି ମାସ୍ୟା !
ସାଦରଂ ବନ୍ଦନାନି ।

ପ୍ରଚଳିତଃ ଆଷାଢ଼ମାସସ୍ୟ ଶୁକ୍ଳପକ୍ଷସ୍ୟ ପଞ୍ଚମାଂ ତିର୍ଥେ ବୃହସ୍ପତି ବାସରେ ୩-୭-୧୬ ମିତେ ଦିନାଙ୍କ ମମ.ତୃତୀୟ ପୁତ୍ରସ୍ୟ ଆୟୁଷ୍ମତଃ ଦୁଷ୍ୟନ୍ତସ୍ୟ ପୁରୀନିବାସିନଃ ଶ୍ରୀମତଃ ପଦ୍ମନାଭ ମିଶ୍ର ଶର୍ମଶଃ ଦ୍ୱିତୀୟପୁତ୍ର ଆୟୁଷ୍ମତ୍ୟା ଶକୁନ୍ତଳୟା ସହ ବୈଦିକବିଧ୍ୟାନୁସାରଂ ଶୁଭପରିଣୟଃ କଟକ ବାରବାଟୀ ଅତିଥିଭବନେ ଅନୁଷୀୟତେ ।

ଅସ୍ମିନ୍ ଶୁଭବିବାହୋତ୍ସବେ ଭବନ୍ତଃ ସପରିବାରଂ ସମୁପସ୍ଥିତ ସନ୍ତଃ ବଧୂବର୍ରେ ଶୁଭାଶୀଭଃ ଅନୁଗ୍ରହଣନ୍ତୁ ଇତି ପ୍ରାର୍ଥୀତେ ।

କାର୍ଯ୍ୟାସୂଚୀ

୨-୭-୧୬ (ପଣ୍ଡିତବାସରଃ) – ମଙ୍ଗଳକୃତ୍ୟମ୍
୩-୭-୧୬ (ବୃହସ୍ପତିବାସରଃ) – ବରାନୁଗମନଂ ପାଣିଗ୍ରହଣଂ ଚ
୬-୭-୧୬(ରବିବାସରଃ) – ପ୍ରୀତିଭୋଜନମ୍ – ସାୟଂ ସପ୍ତବାଦନତଃ

2. महाविद्यालये वार्षिकोत्सवस्य निमत्रणपत्रम् ।
Answer:
ସାଲେପୁର ମହ।ବଦ୍ୟାଳୟଃ
ସାଲେପୁରମ୍, କଟକମ୍
ମହାଶୟଃ !
ସାଦରଂ ପ୍ରଣାମା ।
ବିଷୟମିତଂ ହର୍ଷୀ ହରଂ ନିବେଦୟାମଃ ଯଦକଂ ମହାବିଦ୍ୟାଳୟସ୍ୟ ବାର୍ଷିକୋତ୍ସବ ଆଗାମିନି ଜାନୁୟାରୀମାସସ୍ୟ ଦ୍ଵାଦଶେଦିନାଙ୍କ ସମ୍ପତ୍ସ୍ୟତେ । ଉତ୍ସବେଽସ୍ମିନ୍ ନିମ୍ନାଙ୍କିତା ଅତିଥୟ ଯୋଗଦାନାୟ ସଦୟଂ ସ୍ବୀକୃତଂ ପ୍ରଦତ୍ତବନ୍ତଃ ।
ମୁଖ୍ୟାତିଥୟ – ମାନ୍ୟବରଃ ମୁଖ୍ୟମନ୍ତ୍ରୀତଃ
ମୁଖ୍ୟବକ୍ତାରଃ – ମୁଖ୍ୟସମ୍ପାଦକା, ଦୈନିକ ପତ୍ରିକା ‘ସମାଜ’ ସଭାକାର୍ଯ୍ୟ ପାର୍ଶ୍ୱସ୍ଥସୂଚ୍ୟନୁସାରଂ ପ୍ରଚଳିଷ୍ୟତି | ଅନ୍ତେ ଛାତ୍ରେ ମନୋରଞ୍ଜନକାର୍ଯ୍ୟକ୍ରମା ସମାୟୋଜିତ ଭବିଷ୍ୟନ୍ତି । ସଦବସରେଽସ୍ମିନ୍ ଭବନ୍ତଃ ଉପସ୍ଥିତ ସନ୍ତଃ ପ୍ରୋତ୍ସାହୟନ୍ତୁ ଇତି ସବିନୟଂ ନିବେଦ୍ୟତେ ।

ଇତି
ଛାତ୍ରସଂସଦଃ
ସଭାପତି

ସମ୍ପାଦକଃ

ଅଧ୍ୟକ୍ଷ
କାର୍ଯ୍ୟସୂଚୀ-
୧। ଅତିଥୀନାମ୍ ଆସନଗ୍ରହଣମ୍
୨। ଦୀପପ୍ରଜ୍ବାଳନମ୍
୩। ସ୍ଵାଗତଗୀତମ୍
୪। ପରିଚୟପ୍ରଦାନଂ ମାଲ୍ୟାର୍ପଣ ଚ
୫। ବିବରଣୀପାଠ୍ୟଃ
୬। ମୁଖ୍ୟାତିଥ୍ୟ ଭାଷଣମ୍
୭। ମୁଖ୍ୟବତଃ ଭାଷଣମ୍
୮। ସଭ।ପତେଃ ଅଭିଭାଷଣମ୍
୯ । ପୁରସ୍କାର ବିତରଣମ୍
୧୦ । ଧନ୍ୟବାଦାର୍ପଣମ୍
୧୧। ମନୋରଞ୍ଜନକାର୍ଯ୍ୟକ୍ରମ

3. पूजा निमन्त्रणपत्रम् ।
Answer:
॥ ଓଁ ଶ୍ରୀ ଶ୍ରୀ ସରସ୍ଵତ୍ୟେ ନମଃ ॥
ସରସ୍ଵତୀ ମହାମାୟେ ବଢ୍ୟେ କମଳଲୋଚନୋ
ବିଦ୍ୟାରୂପେ ବିଶାଳାକ୍ଷି ବିଦ୍ୟା ଦେହି ନମୋଽସ୍ତୁତେ ॥
ମାନ୍ୟ ମହୋଦୟାଃ !
ସାଦରଂ ବନ୍ଦନାନି !
ଚଳିତଃ ମାଘମାସସ୍ୟ ଶୁକ୍ଳପଞ୍ଚମାଂ ତିର୍ଥେ ଚନ୍ଦ୍ରବାସରେ ଆଫ୍ଲୋ ୦୭-୦୨-୧୬ ମିତେ ଦିନାଙ୍କ ଅସ୍ଵାକଂ ମହାବିଦ୍ୟାଳୟେ ଶ୍ରୀ ଶ୍ରୀ ବାଗ୍‌ଦେବ୍ୟା ପୂଜନୋସତଃ ଯଥାବିଧ ସମ୍ପତ୍‌ସ୍ୟତେ । ଶୁଭାବସରେଽସ୍ମିନ୍ ଭବତାଂ ସଦୟୋପସ୍ଥିତିଃ ସାଗ୍ରହଂ ପ୍ରାର୍ଥୀତେ ।

ଇତି ଭବତାଂ ଦର୍ଶନାଭିଳାଷିଣ୍ୟଃ
ମହାବିଦ୍ୟାଳୟସ୍ୟ ଚ୍ଛାତ୍ରୀଚ୍ଛାତ୍ରାଶୃ
ଅସୁରେଶ୍ଵର ମହାବିଦ୍ୟାଳୟଃ
କଟକମ୍

କାର୍ଯ୍ୟସ୍ଵରୀ
ଦିବା ୯.୦୦ ବାଦନେ ପୂଜାରମ୍ଭ
ଦିବା ୧.୦୦ ବାଦନେ ପୁଷ୍ପାଞ୍ଜଳି
ଦିବା ୨.୦୦ ବାଦନେ ପ୍ରସାଦ ସେବନମ୍

अन्य संकीणर्णपत्राणि

4. वर्धापनपत्रम् ।
Answer:
ପ୍ରିୟମିତ୍ର,
+୨ କକ୍ଷାୟାଂ ଶୋଭନଙ୍କି ସମୁତ୍ତୀର୍ଣ୍ଣାୟ
ଭବେତେ ହାର୍ଦ୍ଦିକାନି ବର୍ଧାପନାନି ।
ଗିରାଂ ଦେବୀ ସମାରାଧ୍ୟ ତସ୍ୟାଃ ପ୍ରାପ୍ନାଦନୁଗ୍ରହମ୍
ଭବାପତ୍ୟ ସରସ୍ଵତ୍ୟାଃ ପୂଜିତା ବରଦା ହି ସା ||

ପ୍ରେଷକଃ
ଶ୍ରୀ ସୁଧାକର ପଣ୍ଡା

5. शुभाशेसा शुभकामना वा ।
Answer:
ଶରୀରମାଦ୍ୟଂ ଖଳୁ ଧର୍ମସାଧନମ୍
ଶ୍ରୁତଂ ମୟା, ରୁଗଣେ ଭବାନ୍ ।
ଭଗବତଃ ଅନୁଗ୍ରହେଣ ଶୀଘ୍ର ସ୍ବାସ୍ଥ୍ୟଲାଭାଂ କରୋତୁ ଭବାନ୍ ।

ଶୁଭୌଷୀ
ଶ୍ରୀକାନ୍ତ ସାହୁଃ

6. शुभाशेसा ।
Answer:
ଶୁଭାସ୍ତେ ସନ୍ତୁ ପନ୍ତାନଃ
ଭବତାଂ ବିଦେଶଗମନଂ ବିଜ୍ଞାୟ ପ୍ରସନ୍ନତା ଜାତା
ଯାତ୍ରେୟଂ ନିର୍ବିଘ୍ନ ସୁଖଦା ଶୁଭଦା ଚ ଭୂୟାତ୍ ॥

ଶୁଭୌଷୀ
ଅଦ୍ଵୈତ ବରାଳଃ

7. नववर्षाभिनन्दनपत्रम् ।
Answer:
ସେବାୟାମ୍/ପ୍ରତିଷ୍ଠାୟାମ୍
ସଚ୍ଚିଦାନନ୍ଦଃ
ନବବର୍ଷ ସୁଖଦଂ, ମଙ୍ଗଳମୟଂ, ସୌଭାଗ୍ୟକରଂ ଚ ଭୂୟାତ୍ ।
ନବବର୍ଷେଽୟଂ ଭବତୁ ଶୁଭପ୍ରଦଂ
ପରିପୂର୍ଣ୍ଣା ଭବତୁ ଭବତାମିଚ୍ଛା
ଅଗ୍ନିନ୍ନବସରେ ମମାପି ପ୍ରାର୍ଥନା
ପ୍ରଭୋ ! ପୂରୟତୁ ସର୍ବେକ୍ଷାଂ କାମନାମ୍ ॥

ପ୍ରେଷତଃ
ଶ୍ରୀମାନ୍‌ ଶୁଭଙ୍କର ପତିଃ ।