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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 BSE Odisha 10th Class Maths Solutions Geometry Chapter 1 ଜ୍ୟାମିତିରେ ସାଦୃଶ୍ୟ Ex 1(d) Textbook Exercise Questions and Answers.

BSE Odisha Class 10 Maths Solutions Geometry Chapter 1 ଜ୍ୟାମିତିରେ ସାଦୃଶ୍ୟ Ex 1(d)

Question 1.
ବନ୍ଧନୀ ମଧ୍ଯରୁ ଠିକ୍ ଉତ୍ତର ବାଛି ଶୂନ୍ୟସ୍ଥାନ ପୂରଣ କର ।

(i) ପାର୍ଶ୍ୱସ୍ଥ ଚିତ୍ରରେ ଥିବା △ABC ରେ ∠ABC = 90°
ଏବଂ \(\overline{\mathrm{BD}}\) ⊥ \(\overline{\mathrm{AC}}\), m∠ABD = ………
[m∠BAD, m∠DBC, m∠DCB, 2m∠BAD]

Solution:
m∠DCB

(ii) ଉପ6ର।କ୍ତ ରିତ୍ର6ର ଥିବା △ABC ରେ ∠ABC ସମ6କାଶ ଏବଂ BD AC ହେଲେ,
(a) AB2 = AD × ……….. [BC, CD, AC, BD]
(b) BC2 = AC × ……….. [DC, AD, BD, AB]
(c) BD2 = DC × ………. [AC, BC, AB, AD]
Solution:
(a) AC
(b) DC
(c) AD

Question 2.
ବମ୍ନ ଟିତ୍ର6ର ଥିର୍। △PQR ର m∠PQR = 90° ଏର୍ବ \(\overline{\mathbf{QM}}\) ⊥ \(\overline{\mathbf{PR}}\)
(i) QM = 12 6ସ.ମି. ଏବଂ PM = 6 6ସ.ମି. 6ଦ୍ର6କ , PR ନିଣ୍ଟଯ କର |
(ii) PQ = 6 6ସ.ମି. ଏବଂ PM = 3 6ସ.ମି. 6ଦ୍ର6କ , PR ନିଣ୍ଟଯ କର |
(iii) QR = 12 6ସ.ମି. ଏବଂ MR = 3 6ସ.ମି. 6ଦ୍ର6କ , PM ନିଣ୍ଟଯ କର |
(iv) PQ = 12 6ସ.ମି. ଓ RM = 3 6ସ.ମି. 6ଦ୍ର6କ , PM ନିଣ୍ଟଯ କର |
(v) PQ = 8 6ସ.ମି. ଓ QR = 15 6ସ.ମି. 6ଦ୍ର6କ , QM ଓ MR ନିଣ୍ଟଯ କର |

Solution:
(i) QM² = PM × RM ⇒ 12² = 6 × RM
⇒ RM = \(\frac{12^2}{6}\) 6ସ.ମି. = 24 6ସ.ମି.
PR = PM + RM = 6 6ସ.ମି. + 24 6ସ.ମି. = 30 6ସ.ମି.

(ii) PQ² = PM × PR
⇒ PR = \(\frac{\mathrm{PQ}^2}{\mathrm{PM}}\) = \(\frac{6^2}{3}\) 6ସ.ମି. = 12 6ସ.ମି.

(iii) QR² = MR × PR
⇒ PR = \(\frac{\mathrm{QR}^2}{\mathrm{MR}}\) = \(\frac{12^2}{9}\) 6ସ.ମି. = 16 6ସ.ମି.
PM = PR – MR = 16 6ସ.ମି. – 9 6ସ.ମି. = 7 6ସ.ମି. |

(iv) ମ6ନଜର PM = x 6ସ.ମି.
∴ PR = PM + MR = (x + 7) 6ସ.ମି. |
PQ² = PM . PR ⇒ 12² = x(x + 7)
⇒ x² + 7x – 144 = 0 ⇒ x² + 16x – 9x – 144 = 0
⇒ x(x + 16) -9(x + 16) = 0 ⇒ (x – 9) (x + 16) = 0
⇒ x = 9 ବ।, x = -16 (ଅସମ୍ମତ)
∴ PM = x 6ସ.ମି. = 9 6ସ.ମି.

(v) PR = \(\sqrt{\mathrm{PQ}^2+\mathrm{QR}^2}\) = \(\sqrt{8^2+15^2}\) = \(\sqrt{64+225}\) = \(\sqrt{289}\) = 17 6ସ.ମି.
QR² = MR × PR
⇒ MR = \(\frac{\mathrm{QR}^2}{\mathrm{PR}}\) = \(\frac{15 \times 15}{17}\) = \(\frac { 225 }{ 17 }\) = 13\(\frac { 4 }{ 17 }\) 6ସ.ମି.
QM² = PM × MR = (PR – MR) × MR
= (17 – \(\frac { 225 }{ 17 }\)) × \(\frac { 225 }{ 17 }\) = \(\frac { 64 }{ 17 }\) × \(\frac { 225 }{ 17 }\) ଦଙ୍ଗ6ସ.ମି.
∴ QM = \(\sqrt{\frac{64 \times 225}{17 \times 17}}\) = \(\frac{8 \times 15}{17}\) = \(\frac { 120 }{ 17 }\) = 7\(\frac { 1 }{ 17 }\) 6ସ.ମି. |

Question 3.
ବମ୍ନ ଟିତ୍ର6ର m∠ABC = m∠DCB = 90°, \(\overline{\mathbf{AC}}\) ଓ \(\overline{\mathbf{BD}}\) ର ଛେଦଦିନ୍ଦୁ O ଏବଂ \(\overline{\mathbf{AC}}\) ⊥ \(\overline{\mathbf{BD}}\) | OC = 6 6ସ.ମି. ଦଙ୍ଗ6ସ.ମି.
(i) BO ନିଶ୍ଚୟ କର;
(ii) OA ନିର୍ଣ୍ଣୟ କର;
(iii) BC ନିର୍ଣ୍ଣୟ କର;
(iv) AB ନିର୍ଣ୍ଣୟ କର; ଏବଂ
(v) CD ନିର୍ଣ୍ଣୟ କର ।

Solution:

Question 4.
△ABC ଭେ ∠ABC ସପ6କାଣ ଏବଂ \(\overline{\mathbf{B D}}\) ⊥ \(\overline{\mathbf{AC}}\) | AD = p ଏକକ ପ୍ରତେକ, ପ୍ତମାଣ କର : BC = \(\frac{q \sqrt{\mathbf{p}^2+q^2}}{p}\)
Solution:

Question 5.
△ABC ଭେ, m∠ABC = 90° ଏବଂ BD ⊥ AC ଦ୍ରେଲେ , ପ୍ରତମାଣ କର ଯେ , AB² : BC² = AD : DC |
Solution:

Question 6.
△ABC ଭେ, ∠ABC ସପ6କାଶ , \(\overline{\mathrm{BD}}\) ⊥ \(\overline{\mathrm{AC}}\) ଏବଂ BC² = AC.BD କ୍ତେଲେ , ପ୍ରମାଣ ଦର ଯେ \(\overline{\mathrm{BD}}\) 6ହର୍ଛି ∠ABC ତ ସପଦ୍ଦିଖଣ୍ଡକ |
Solution:

ଦର : △ABC ରେ , ∠B = 90° | \(\overline{\mathrm{BD}}\) ⊥ \(\overline{\mathrm{AC}}\) ଏବଂ BC² = AC.BD
ପ୍ରମ।ଣu : \(\overline{\mathrm{BD}}\) , ∠ABC ର ସମଦ୍ୱିଖଣ୍ଡକ |
ପ୍ରମ।ଣ : △BDC ~ △ABC ⇒ BC² = AC . AD ….(i)
କକୁ ଦଣ BC² = AC . AD ….(ii)
(i) ଓ (ii) ରୁ AC . AD = AC . BD ⇒ CD = BD ⇒ m∠DBC = m∠DCB … (iii)
କନ୍ତି △ABD ~ △BCD ⇒ m∠ABD = m∠DCB …(iv)
(iii) ଓ (iv) ର m∠DBC = m∠ABD ⇒ \(\overline{\mathrm{BD}}\) , ∠ABC ର ସମଦ୍ୱିଖଣ୍ଡ କ | (ପ୍ରମାଣିତ)

Question 7.
ପାଣ୍ଡଣ୍ ଚିତ୍ରରେ ଥ୍ରନା ରହୁକ୍ତିଙ ABCD ରେ m∠ABC = m∠ADC = 90° ଏବଂ AB = AD | କଣ୍ତଦ୍ରଯତ 6ଛଦଦିନ୍ଦୁ M ଦ୍ତେକେ , ପ୍ରମାଣ କର ଯେ AM × MC = DM | (ପ୍ରମେଯ 1.4 ର ସ୍ତ ଯୋଗ କରି ପ୍ରମାଣ କମା | )
Solution:

Question 8.
△ABC ରେ m∠ABC = 90°, \(\overline{\mathrm{BD}}\) ⊥ \(\overline{\mathrm{AC}}\) ଏବଂ ∠ABC ର ସମଦ୍ୱିଖଣ୍ଡ \(\overline{\mathrm{AC}}\) କ୍ E ଦିନ୍ଦୁରେ 6ଛଦ କଲୋ ପ୍ରମାଣ କର ଯେ AE² : EC² = AD : DC |
Solution:
ଦର : △ABC ରେ m∠ABC = 90° , \(\overline{\mathrm{BD}}\) ⊥ \(\overline{\mathrm{AC}}\) ଏବଂ ∠ABC ର ସମଦ୍ୱିଖଣ୍ଡ \(\overline{\mathrm{AC}}\) କ୍ E ଦିନ୍ଦୁରେ 6ଛଦ କଲୋ

ପ୍ତାମାଣଧ୍ୟ : \(\frac{\mathrm{AE}^2}{\mathrm{EC}^2}\) = \(\frac { AD }{ DC }\)
ପ୍ତମାଣ : △ABC ରେ ∠ABC ର ସମଦ୍ୱିଖଣ୍ଡ \(\overline{\mathrm{AC}}\) କ୍ E ଦିନ୍ଦୁରେ 6ଛଦ କଲୋ
⇒ \(\frac { AB }{ BC }\) = \(\frac { AE }{ EC }\) ⇒ \(\frac{\mathrm{AE}^2}{\mathrm{EC}^2}\) = \(\frac{\mathrm{AB}^2}{\mathrm{BC}^2}\)
△ABC ରେ m∠ABC = 90° ଓ \(\overline{\mathrm{BD}}\) ⊥ \(\overline{\mathrm{AC}}\)
⇒ AB² = AD . AC ଓ BC² = CD . AC
∴ \(\frac{\mathrm{AE}^2}{\mathrm{EC}^2}\) = \(\frac{\mathrm{AB}^2}{\mathrm{BC}^2}\) = \(\frac{\mathrm{AD} \times \mathrm{AC}}{\mathrm{CD} \times \mathrm{AC}}\) = \(\frac { AD }{ CD }\) (ପ୍ରମାଣିତ)

Question 9.
△ABC ରେ , m∠BAC = 90° ଏବଂ \(\overline{\mathrm{AD}}\) ⊥ \(\overline{\mathrm{BC}}\) | ପ୍ରମାଣ କର ଯେ △ADC ର ଷ୍ଟ୍ରେତୃଫକ = \(\frac{\mathbf{A B} \times \mathbf{A C} \mathbf{C}^3}{2 \mathbf{B C} C^2}\) |

Solution:
ଦଉ : △ABC 6ର m∠BCA = 90° ଏବଂ \(\overline{\mathrm{AD}}\) | \(\overline{\mathrm{BC}}\) |
ପ୍ର।ମାଣ୍ୟ : △ADC ର ଷ୍ଟେର୍ଫଳ = \(\frac{\mathrm{AB} \times \mathrm{AC}^3}{2 \mathrm{BC}^2}\) |
ପ୍ରମାଣ : ABC ରେ m∠BAC = 90° ଓ \(\overline{\mathrm{AD}}\) ⊥ \(\overline{\mathrm{BC}}\)
⇒ AC² = CD . BC
△ABC ର ଷ୍ଟେର୍ଫଳ = \(\frac { 1 }{ 2 }\) AC × AB = \(\frac { 1 }{ 2 }\) BC × AD 

Question 10.
△ABC ର m∠ABC ସମକୋଣ,, \(\overline{\mathrm{BD}}\) ⊥ \(\overline{\mathrm{AC}}\) ଏବଂ ∠BAC ର ଛେଦକରେ \(\overline{\mathrm{BD}}\) କୁ E ଦିନ୍ଦୁରେ ଛେଦକରେ | ପ୍ତପାଣ କର ଯେ BE² : DE² = AC : AD |
Solution:
ଦତ୍ତ : △ABC ରେ ∠ABC ସମକୋଣ, \(\overline{\mathrm{BD}}\) ⊥ \(\overline{\mathrm{AC}}\) ଏବଂ ∠BAC ର ସମଦ୍ବିଖଣ୍ଡକ \(\overline{\mathrm{BD}}\) କୁ E ବିନ୍ଦୁରେ ଛେଦକରେ ।
ପ୍ରାମାଣ୍ୟ : \(\frac{\mathrm{BE}^2}{\mathrm{DE}^2}\) : \(\frac{\mathrm{AC}}{\mathrm{AD}}\)
ପ୍ରମାଣ : △ABC ରେ ∠B ସମକୋଣ ଓ \(\overline{\mathrm{BD}}\) ⊥ \(\overline{\mathrm{AC}}\) |
⇒ AB² = AD × AC
△ABD ରେ ∠BAD ର ସମଦ୍ଦିଖଣ୍ଡକ ରଶ୍ମି \(\overline{\mathrm{BD}}\) କୁ E ରେ ଛେଦକରେ ।
⇒ \(\frac { AB }{ AD }\) = \(\frac { BE }{ DE }\)
∴ \(\frac{\mathrm{BE}^2}{\mathrm{DE}^2}\) = \(\frac{\mathrm{AB}^2}{\mathrm{AD}^2}\) = \(\frac { AD × AC }{ AD × AD }\) = \(\frac { AC }{ AD }\) (∵ AB² = AD × AC) (ପ୍ରମାଣିତ)

Odisha State Board BSE Odisha 10th Class Maths Solutions Algebra Chapter 2 ଦ୍ବିଘାତ ସମୀକରଣ Ex 2(b) Textbook Exercise Questions and Answers.

BSE Odisha Class 10 Maths Solutions Algebra Chapter 2 ଦ୍ବିଘାତ ସମୀକରଣ Ex 2(b)

Question 1.
ନିମ୍ନଲିଖତ ପ୍ରଶ୍ନମାନଙ୍କର ଉତ୍ତର ଦିଅ ।
(i) ଗୋଟିଏ ସଂଖ୍ୟା ଓ ଏହାର ବ୍ୟତ୍‌କ୍ରମର ସମଷ୍ଟି 2 । ସଂଖ୍ୟାଟିକୁ x ନେଇ ଏକ ଦ୍ଵିଘାତ ସମୀକରଣ ଗଠନ କର ।
(ii) ଦୁଇଗୋଟି କ୍ରମିକ ପୂର୍ଣ୍ଣ ସଂଖ୍ୟାର ଗୁଣଫଳ 20 । ସଂଖ୍ୟାଦ୍ଵୟ ମଧ୍ୟରୁ ଗୋଟିକୁ y ନେଇ ଆବଶ୍ୟକୀୟ ଦ୍ଵିଘାତ ସମୀକରଣ ଗଠନ କର ।
(iii) ଦୁଇଟି ସଂଖ୍ୟାର ସମଷ୍ଟି 18 ଏବଂ ସେମାନଙ୍କର ଗୁଣଫଳ 72 । ଗୋଟିଏ ସଂଖ୍ୟାକୁ x ନେଇ ଏକ ଦ୍ଵିଘାତ ସମୀକରଣ ଗଠନ କର ।
(iv) କୌଣସି ସଂଖ୍ୟା, ତାହାର ବର୍ଗ ସମାନ ହେଲେ ସଂଖ୍ୟାଟି ନିର୍ଣ୍ଣୟ କର ।
(v) ପ୍ରଥମ n ସଂଖ୍ୟକ ଗଣନ ସଂଖ୍ୟାର ସମଷ୍ଟି S = \(\frac{n(n+1) }{2}\)। ଯଦି S = 120 ହୁଏ, ତେବେ nର ମୂଲ୍ୟ ନିରୂପଣ କରିବା ପାଇଁ ଏକ ଦ୍ଵିଘାତ ସମୀକରଣ ଗଠନ କର ।
(vi) \(\sqrt{x}\) + x = 6 କୁ ଏକ ଦ୍ବିଘାତ ସମୀକରଣ ରୂପେ ପ୍ରକାଶ କର ।
(vii) \(\sqrt{x+9}\) + 3 = x କୁ ଏକ ସ୍ଵିଘାତ ସମୀକରଣ ରୂପେ ପ୍ରକାଶ କର ।
(viii) x – 2\(\sqrt{2}\) – 6 = 0 ସମୀକରଣକୁ ଏକ ଦ୍ଵିଘାତ ସମୀକରଣ ରୂପେ ପ୍ରକାଶ କର ।
ସମାଧାନ :
(i) ମନେକର ସଂଖ୍ୟାଟି x । ସଂଖ୍ୟାଟିର ବ୍ୟକ୍ରମ \(\frac{1}{x}\)
ପ୍ରଶ୍ନନୁସାରେ, x + \(\frac{1}{x}\) = 2 ⇒ x² + 1 = 2x ⇒ x² – 2x + 1 = 0
∴ ନିଶ୍ଚେୟ ଦ୍ବିଘାତ ସମୀକରଣ x² – 2x + 1 = 0

(ii) ମନେକର ଗୋଟିଏ ସଂଖ୍ୟା y । ଅନ୍ୟଟି = (y + 1)
ପ୍ରଶ୍ନନୁସାରେ, y (y + 1) = 20 ⇒ y² + y – 20 = 0
∴ ନିଶ୍ଚେୟ ଦ୍ବିଘାତ ସମୀକରଣ y + y – 20 = 0

(iii) ମନେକର ପ୍ରଥମ ସଂଖ୍ୟାଟି x ଓ ଦ୍ବିତୀୟ ସଂଖ୍ୟାଟି 18 – x ।
ପ୍ରଶ୍ନନୁସାରେ, x (18 – x) = 72 ⇒ 18x – x² = 72 = x² – 18x + 72 = 0
∴ ନିଶ୍ଚେୟ ଦ୍ବିଘାତ ସମୀକରଣ, x² – 18x + 72 = 0

(iv) ମନେକର ସଂଖ୍ୟାଟି x।
ପ୍ରଶ୍ନନୁସାରେ ସଂଖ୍ୟାଟି ତା’ର ବର୍ଗ ସହ ସମାନ ହେବ । ଅର୍ଥାତ୍ x = x²
⇒ x² – x = 0 ⇒ x ( x – 1 ) = 0
⇒ x = 0 ବା x – 1 = 0 ⇒ x= 0 ବା x = 1
∴ ସଂଖ୍ୟାଟି 0 କିମ୍ବା 1 ।

(v) ପ୍ରଶ୍ନନୁସାରେ \(\frac{n(n+1) }{2}\) = 120 ⇒ n² + n = 240 ⇒ n² + n – 240 = 0

(vi) √x + x = 6 ⇒ √x = 6 – x
⇒ (√x)² =(6 – x)² ⇒ x = 36 + x² – 12x
⇒ x² – 12x – x + 36 = 0 ⇒ x² – 13x + 36 = 0

(vii) √(x + 9) + 3 = x ⇒ √(x + 9) = x – 3
⇒ (\(\sqrt{x+9}\))² = (x – 3)² ⇒ x² – 6x + 9 = x + 9
⇒ x² – 6x – x + 9 – 9 = 0 = x² – 7x = 0
ବିକଳ୍ପ ସମାଧାନ :
ମନେକର \(\sqrt{x+9}\) = t
⇒ x + 9 = t² ⇒ x = t² – 9
xର ମାନ ସମୀକରଣରେ ବସାଇଲେ k + 3 = t² – 9
⇒ t – t – 9 – 3 = 0 ⇒ t – t – 12 = 0

(viii) x – 2√2 – 6 = 0 ⇒ x – 6 = 2√2
⇒ (x – 6)² = (2√2)² ⇒ x² – 12x + 36 = 8
⇒ x² – 12x + 36 – 8 = 0 ⇒ x² – 12x + 28 = 0

Question 2.
ନିମ୍ନଲିଖ ପ୍ରଶ୍ନମାନଙ୍କର ଉତ୍ତର ଦିଅ ।
(i) ଗୋଟିଏ ଧନାତ୍ମକ ସଂଖ୍ୟା ତାହାର ବର୍ଗମୂଳ ଅପେକ୍ଷା 12 ଅଧିକ ହେଲେ, ସଂଖ୍ୟାଟି ନିରୂପଣ କର ।
(ii) ଗୋଟିଏ ସଂଖ୍ୟା ଏବଂ ତାହାର ବ୍ୟତ୍‌କ୍ରମର ସମଷ୍ଟି \(\frac{41}{20}\) ହେଲେ ସଂଖ୍ୟାଟି ସ୍ଥିର କର ।
(iii) ଦୁଇଟି କ୍ରମିକ ପୂର୍ଣ ସଂଖ୍ୟାର ବ୍ୟୁତ୍‌କ୍ରମ ଭଗ୍ନସଂଖ୍ୟା ଦ୍ବୟର ଯୋଗଫଳ \(\frac{11}{30}\) ଦ୍ଵୟକୁ ନିରୂପଣ କରିବା ପାଇଁ ଆବଶ୍ୟକ ଦ୍ଵିଘାତ ସମୀକରଣଟି ଗଠନ କରି ସଂଖ୍ୟାଦ୍ଵୟ ନିରୂପଣ କର ।
(iv) ଦୁଇଟି କ୍ରମିକ ପୂର୍ଣ୍ଣ ସଂଖ୍ୟାର ବ୍ୟୁତ୍‌କ୍ରମର ସମଷ୍ଟି \(\frac{25}{123}\) ହେଲେ ସଂଖ୍ୟାଦ୍ଵୟ ନିର୍ଣ୍ଣୟ କର ।
(v) ଯଦି 51 କୁ ଦୁଇଭାଗ କଲେ ସେମାନଙ୍କର ଗୁଣଫଳ 378 ହୁଏ, ତେବେ ସଂଖ୍ୟା ଦ୍ଵୟ ନିର୍ଣ୍ଣୟ କର ।
ସମାଧାନ :
(i) ମନେକର ଧନାତ୍ମକ ସଂଖ୍ୟାଟି x² । ଏହାର ବର୍ଗମୂଳ = √x² = x
ପ୍ରଶ୍ନନୁସାରେ, x² = x + 12 = x² -x – 12 = 0
⇒ x² – 4x + 3x – 12 = 0 ⇒ x (x – 4) + 3 (x – 4) = 0
⇒ (x – 4) (x + 3) = 0 ⇒ x – 4 = 0 ବା x + 3 = 0
⇒ x = 4 ବା x = -3 (ଅସମ୍ଭବ)
∴ ଧନାତ୍ମକ ସଂଖ୍ୟାଟି x² = 4² = 16

(ii) ମନେକର ସଂଖ୍ୟାଟି x । ତାହାର ବ୍ୟତ୍‌କ୍ରମ \(\frac{1}{x}\)
ପ୍ରଶ୍ନନୁସାରେ, x + \(\frac{1}{x}=\frac{41}{20}\) = \(\frac{x²+1}{x}=\frac{41}{20}\)
⇒ 20x² + 20 = 41x ⇒ 20x² – 41x + 20 = 0
⇒ 20x² – 25x – 16x + 20 = 0 ⇒ 5x (4x – 5) – 4 (4x – 5) = 0
⇒ (4x – 5) (5x – 4)=0 = 4x – 5 = 0 ବା 5x – 4 = 0
⇒ x = \(\frac{5}{4}\)ବା x = \(\frac{4}{5}\)
∴ ସଂଖ୍ୟାଟି \(\frac{4}{5}\) ହେଲେ ଏହାର ବ୍ୟତ୍‌କ୍ରମ \(\frac{5}{4}\) ହେବ ।
ଅଥବା x = \(\frac{5}{4}\) ହେଲେ ଏହାର ବ୍ୟତ୍‌କ୍ରମ \(\frac{4}{5}\) ହେବ ।
∴ ସଂଖ୍ୟାଟି \(\frac{4}{5}\) ବା \(\frac{5}{4}\) ହେବ ।

(iii) ମନେକର କ୍ରମିକ ପୂର୍ଣ୍ଣ ସଂଖ୍ୟାଦ୍ୱୟ x ଓ x + 1
x ର ବ୍ୟୁତ୍‌କ୍ରମ \(\frac{1}{x}\) ଓ x + 1 ର ବ୍ୟକ୍ରମ \(\frac{1}{x+1}\)
ପ୍ରଶ୍ନନୁସାରେ, \(\frac{1}{x}+\frac{1}{x+1}=\frac{11}{30}\) ⇒ \(\frac{x+1+x}{x(x+1)}=\frac{11}{30}\)
⇒ \(\frac{2x+1}{x²+x)}=\frac{11}{30}\) ⇒ 11x² + 11x = 60x + 30
⇒ 11x² + 11x – 60x – 30 = 0 ⇒ 11x² – 49x – 30 = 0
⇒ 11x² – 55x + 6x – 30 = 0 ⇒ 11x (x – 5) + 6 (x – 5) = 0 ⇒ (x – 5) (11x + 6) = 0 ⇒ x – 5 = 0 ବା 11x + 6 = 0
⇒ x = 5 ବା x = \(\frac{-6}{11}\) (ଅସମ୍ଭବ)
∴ x = 5 ହେଲେ ଏହାର ପରବର୍ତ୍ତୀ
∴ ନିଶ୍ଚେୟ କ୍ରମିକ ପୂର୍ଣ୍ଣ ସଂଖ୍ୟାଦ୍ଵୟ 5 ଓ 6 । କ୍ରମିକ ସଂଖ୍ୟା x + 1 = 6

(iv) ମନେକର କ୍ରମିକ ପୂର୍ବସଂଖ୍ୟାଦ୍ୱୟ x ଓ x + 1 ।
ପ୍ରଶାନୁସାରେ, \(\frac{1}{x}+\frac{1}{x+1}=\frac{23}{132}\) ⇒ \(\frac{x+1+x}{x(x+1)}=\frac{23}{132}\)
⇒ \(\frac{2x+1}{x²+x)}=\frac{23}{132}\) ⇒ 23x² + 23x = 264x + 132
⇒ 23x² + 23x – 264x – 132 = 0 ⇒ 23x² – 241x – 132 = 0
⇒23x² – 253x + 12x – 132 = 0 ⇒ 23x (x – 11) + 12 (x – 11) = 0
⇒ (x – 11) (23x + 12) = 0 ⇒ x – 11 = 0 ବା 23x + 12 = 0
⇒ x = 11 ବା x = \(– \frac{12}{23}\) (ଅସମ୍ଭବ) ⇒ x + 1 = 11 + 1 = 12
∴ ପୂର୍ଣ୍ଣସଂଖ୍ୟାଦ୍ଵୟ 11 ଓ 12 ।

(v) ମନେକର ଗୋଟିଏ ସଂଖ୍ୟା x ଓ ଅନ୍ୟଟି 51 – x ।
ପ୍ରଶ୍ନନୁସାରେ, x (51 – x) = 378 ⇒ 51x – x² = 378
⇒ x – 51x + 378 = 0 ⇒ x² – 42x – 9x + 378 = 0
⇒x (x-42) – 9 (x – 42) = 0 ⇒ (x – 42) (x – 9) = 0
⇒ x – 42 = 0 ବା x – 9 = 0 ⇒ x = 42 ବା x = 9
∴ ଯଦି x = 9 ହୁଏ ତେବେ ଅନ୍ୟ ଭାଗଟି 42 ହେବ ।
∴ ସଂଖ୍ୟାଦ୍ୱୟ 9 ଓ 42 ସେହିପରି x = 42 ହୁଏ ତେବେ ଅନ୍ୟ ଭାଗଟି 9 ହେବ ।

Question 3.
ଏକ ଦୁଇଅଙ୍କ ବିଶିଷ୍ଟ ସଂଖ୍ୟା, ତାହାର ଅଙ୍କ ଦ୍ଵୟର ଗୁଣଫଳର 3 ଗୁଣ । ଏକକ ସ୍ଥାନରେ ଥ‌ିବା ଅଙ୍କଟି ଦଶକ ସ୍ଥାନରେ ଥ‌ିବା ଅଙ୍କଠାରୁ 2 ବୃହତ୍ତର । ସଂଖ୍ୟାଟି ନିରୂପଣ କର ।
ସମାଧାନ :
ମନେକର ଦୁଇଅଙ୍କ ବିଶିଷ୍ଟ ସଂଖ୍ୟାର ଏକକ ସ୍ଥାନୀୟ ଅଙ୍କ x ଓ ଦଶକ ସ୍ଥାନୀୟ ଅଙ୍କ y ।
∴ ସଂଖ୍ୟାଟି = 10y + x
ପ୍ରଶ୍ନନୁସାରେ 10y + x = 3xy …… (i) ଏବଂ x = y + 2 ……. (ii)
ସମୀକରଣ (ii)ରୁ x ର ମାନ ସମୀକରଣ (i) ରେ ବସାଇଲେ, 10y + y + 2 = 3y (y + 2)
⇒ 11y + 2 = 3y² + 6y ⇒ 3y² + 6y – 11y – 2 = 0
⇒ 3y² – 5y – 2 = 0 ⇒ 3y² – 6y+ y – 2 = 0
⇒ 3y (y – 2) + 1 (y – 2) = 0 ⇒ (y – 2) (3y + 1) = 0
⇒ y – 2 = 0 ବା 3y + 1 = 0 = y = 2 ବା y = (ଅସମ୍ଭବ)
∴ x = y + 2 = 2 + 2 = 4
ସଂଖ୍ୟାଟି = 10y + x = 10 × 2 + 4 = 24

Question 4.
ଗୋଟିଏ ପରିବାରରେ ଆଲ୍‌ଫାର ବୟସ, ବିଟା ଓ ଗାମାର ବୟସ୍‌ର ଗୁଣଫଳ ସହ ସମାନ । ଯଦି ବିଟା, ଗାମାଠାରୁ 1 ବର୍ଷ ବଡ଼ ହୁଏ ଏବଂ ଆଲ୍‌ଫାର ବୟସ 42 ହୁଏ, ତେବେ 5 ବର୍ଷ ପରେ ବିଟାର ବୟସ କେତେ ହେବ ?
ସମାଧାନ :
ମନେକର ପରିବାରରେ ଗାମାର ବୟସ = x ବର୍ଷ
∴ ବିଟାର ବୟସ = (x + 1) ବର୍ଷ । ଆଲ୍‌ଫାର ବୟସ = 42।
ପ୍ରଶ୍ନନୁସାରେ, x(x + 1) = 42 ⇒ x² + x = 42
⇒ x² + x – 42 = 0 ⇒ x² + 7x – 6x – 42 = 0
⇒ x(x + 7) – 6(x + 7) = 0 ⇒ (x + 7) (x – 6) = 0
⇒ x + 7 = 0 ବା x – 6 = 0 = x = -7 ବା x = 6
x = – 7 ବୟସ ଋଣାତ୍ମକ ହେବନାହିଁ ତେଣୁ ଏହା ଅସମ୍ଭବ। ତେବେ ଏଠାରେ x = 6 ଗ୍ରହଣୀୟ ।
∴ ଗାମାର ବୟସ = x ବର୍ଷ = 6 ବର୍ଷ ଏବଂ ବିଟାର ବୟସ = 6 + 1 = 7 ବର୍ଷ ।
∴ 5 ବର୍ଷ ପରେ ବିଟାର ବୟସ ହେବ = 7 + 5 = 12 ।

Question 5.
କୌଣସି ଏକ ଅରଣ୍ୟରେ ବାସ କରୁଥିବା ମର୍କଟମାନଙ୍କ ମଧ୍ୟରୁ ସେମାନଙ୍କ ସଂଖ୍ୟାର ଏକ ଅଷ୍ଟମାଂଶର ବର୍ଗ କ୍ରୀଡ଼ାରତ ଏବଂ ଅବଶିଷ୍ଟ ବାରଟି ମର୍କଟ ଏକ ଶୃଙ୍ଗ ଉପରେ ବସିଥିଲେ । ଅରଣ୍ୟରେ ସମ୍ଭବତଃ କେତେ ମର୍କଟ ଥିଲେ ?
ସମାଧାନ :
ମନେକର ଅରଣ୍ୟରେ ସମ୍ଭବତଃ xଟି ମର୍କଟ ଥିଲେ ।
ସେମାନଙ୍କ ଏକ- ଅଷ୍ଟମାଂଶର ବର୍ଗ କ୍ରିଡ଼ାରତ ଥିଲେ । କ୍ରୀଡ଼ାରତ ଥିବା ମର୍କଟଙ୍କ ସଂଖ୍ୟା = (\(\frac{x}{8}\))² = \(\frac{x²}{64}\)
ଅବଶିଷ୍ଟ 12 ଟି ମର୍କଟ ଶୃଙ୍ଗ ଉପରେ ବସିଥିଲେ ।
ପ୍ରଶ୍ନନୁସାରେ, \(\frac{x²}{64}+12\) = x = \(\frac{x²+768}{64}=x\)
⇒ x² + 768x = 64 x ⇒ x² – 64x + 768 = 0
⇒ x² -48x – 16x + 768 = 0 ⇒ x (x – 48) – 16 (x – 48) = 0
⇒ (x – 48) (x – 16)=0 = x – 48 = 0 ବା x – 16 = 0
⇒ x = 48 ବା x = 16
∴ ଅରଣ୍ୟରେ ସମ୍ଭବତଃ 48 କିମ୍ବା 16ଟି ମର୍କଟ ଥିଲେ ।

Question 6.
ଗୋଟିଏ ତ୍ରିଭୁଜର କ୍ଷେତ୍ରଫଳ 30 ବ. ସେ.ମି. । ତ୍ରିଭୁଜର ଉଚ୍ଚତା, ଭୂମିର ଦୈର୍ଘ୍ୟଠାରୁ 7 ସେ.ମି. ଅଧ୍ଵ ହେଲେ, ଭୂମିର ଦୈର୍ଘ୍ୟ ସ୍ଥିର କର ।
ସମାଧାନ :
ଗୋଟିଏ ତ୍ରିଭୁଜର କ୍ଷେତ୍ରଫଳ = 30 ବର୍ଗ ସେ.ମି.
ତ୍ରିଭୁଜର ଉଚ୍ଚତା, ଭୂମିର ଦୈର୍ଘ୍ୟଠାରୁ 7 ସେ.ମି. ଅଧ୍ଵ ।
ମନେକର ଭୂମିର ଦୈର୍ଘ୍ୟ X ସେ.ମି. । ଉଚ୍ଚତା = x + 7 ସେ.ମି.
∴ ତ୍ରିଭୁଜର କ୍ଷେତ୍ରଫଳ = \(\frac{1}{2}\) ଭୂମି × ଉଚ୍ଚତା = \(\frac{1}{2}\)x(x+7)
⇒ 30 = \(\frac{1}{2}\)(x² + 7x) ⇒ x² + 7x = 60
⇒ x²+7x-60 = 0 ⇒ x² + 12x – 5x – 60 = 0
⇒ x (x + 12) – 5 (x + 12) = 0 ⇒ (x + 12) (x – 5)= 0
⇒ x + 12 = 0 ବା x – 5 = 0 = x=- 12 (ଅସମ୍ଭବ) ବା x = 5
∴ ତ୍ରିଭୁଜର ଭୂମିର ଦୈର୍ଘ୍ୟ 5 ସେ.ମି. ।

Question 7.
ଏକ ସମକୋଣୀ ତ୍ରିଭୁଜର ସମକୋଣର ସଂଲଗ୍ନ ବାହୁଦ୍ୱୟର ଦୈର୍ଘ୍ୟ 5x ସେ.ମି. ଓ (3x – 1) ସେ.ମି. ଓ କ୍ଷେତ୍ରଫଳ 60 ବର୍ଗ ସେ.ମି. । ତେବେ ବାହୁଦ୍ୱୟର ଦୈର୍ଘ୍ୟ ନିର୍ଣ୍ଣୟ କର ।
ସମାଧାନ :
ସମକୋଣ ସଂଲଗ୍ନ ବାହୁଦ୍ୱୟର ଦୈର୍ଘ୍ୟ 5x ସେ.ମି. ଏବଂ (3x – 1) ସେ.ମି. ।
ସମକୋଣୀ ତ୍ରିଭୁଜର କ୍ଷେତ୍ରଫଳ = 60 ବର୍ଗ ସେ.ମି. (ଦତ୍ତ)
⇒ \(\frac{1}{2}\) × ସମକୋଣ ସଂଲଗ୍ନ ଦୈର୍ଘ୍ୟର ବାହୁଦ୍ୱୟର ଗଣଫଳ = 60
⇒ \(\frac{1}{2}\) . 5x . (3x – 1) = 60 ⇒ x(3x – 1) = 60 × \(\frac{2}{5}\)
⇒ 3x² – x = 24 ⇒ 3x² – x – 24 = 0
ଏଠାରେ a = 3, b = -1, c = – 24

∴ ସମକୋଣୀ ତ୍ରିଭୁଜର ବାହୁ ଦ୍ୱୟର ଦୈର୍ଘ୍ୟ 5x = 5 × 3 = 15 ସେ.ମି. ।
ଓ 3x – 1 = 3 × 3 – 1 = 8 ସେ.ମି. ।
ବିକଳ୍ପ ପ୍ରଣାଳୀ : 3x² – x – 24 = 0 ର ସମାଧାନ ଉତ୍ପାଦକୀକରଣ ଦ୍ଵାରା ସମ୍ଭବ ।

Question 8.
n ବାହୁ ବିଶିଷ୍ଟ ବହୁଭୁଜର କର୍ଡ ସଂଖ୍ୟା \(\frac{n}{2}\)n(n-3)। ଯଦି ବହୁଭୁଜର 54ଟି କର୍ଣ୍ଣ ରହିବ, ତେବେ ବହୁଭୁଜର ବାହୁର ସଂଖ୍ୟା କେତେ ?
ସମାଧାନ :
n ବାହୁ ବିଶିଷ୍ଟ ବହୁଭୁଜର କର୍ଣ୍ଣ ସଂଖ୍ୟା = \(\frac{n}{2}\)n(n-3)
ଏହାର କର୍ଣ୍ଣ ସଂଖ୍ୟା 54 1
ପ୍ରଶ୍ବାନୁସାରେ, \(\frac{n}{2}\)n(n-3) = 54 ⇒ n² – 3n = 108
→n² – 3n-108 = 0 ⇒ n² – 12n + 9n – 108 = 0
→ n (n – 12) + 9 (n – 12) = 0 → (n – 12) (n + 9) = 0
→ n – 12 = 0 ବା n + 9 = 0
⇒ n =12 ବା n = -9 (ଅସମ୍ଭବ)
∴ ବହୁଭୁଜର ବାହୁ ସଂଖ୍ୟା 12 ।

Question 9.
ଦୁଇଟି ବର୍ଗକ୍ଷେତ୍ରର କ୍ଷେତ୍ରଫଳର ସମଷ୍ଟି 468 ବ.ମି. ଏବଂ ପରିସୀମାଦ୍ବୟର ଅନ୍ତର 24 ମି. ହେଲେ ପ୍ରତ୍ୟେକର ବାହୁର ଦୈର୍ଘ୍ୟ ନିରୂପଣ କର ।
ସମାଧାନ :
ଦୁଇଟି ବର୍ଗକ୍ଷେତ୍ରର କ୍ଷେତ୍ରଫଳର ସମଷ୍ଟି 468 ବର୍ଗ ମି. । ପରିସୀମାଦ୍ବୟର ଅନ୍ତର = 24 ମିଟର
ମନେକର ଗୋଟିଏ ବର୍ଗକ୍ଷେତ୍ରର ବାହୁର ଦୈର୍ଘ୍ୟ a ମି. ଓ ଅନ୍ୟଟିର ଦୈର୍ଘ୍ୟ b ମି. ।
ପ୍ରଶ୍ବାନୁସାରେ, a² + b² = 468 ଏବଂ 4a – 4b = 24
⇒ 4(a – b) = 24 ⇒ a – b = \(\frac{24}{4}\) = 6 ………(i)
2ab = (a² + b²) – (a – b)² = 468 – 36 = 432
(a + b)² = a² + b² + 2ab = 468 + 432 = 900
⇒ a + b = 30 ……..(ii)
ସମୀକରଣ (i) ଓ (ii)କୁ ଯୋଗକଲେ a – b + a + b = 6 + 30 = 36
⇒ 2a = 36 ⇒ a = = 18 ମି., b = 30 a 30 18 = 12 ମି.
∴ ବହୁଭୁଜର ବାହୁ ଦୈର୍ଘ୍ୟ 18 ମି. ଓ 12 ମି. ।

ବିକଳ୍ପ ପ୍ରଣାଳୀ : a² + b² = 468 ଏବଂ a – b = 6
a – b = 6 = b = a – 6
∴ a² + (a – b)² = 468⇒ 2a² – 12a + 36 = 468
2a² – 12a – 432 = 0 = a² – 6a – 216 = 0
⇒ (a – 18) (a + 12) = 0 ⇒ a = 18 ବା -12
ଏଠାରେ a = 18 ଗ୍ରହଣୀୟ ଏବଂ a = -12 ଅଗ୍ରହଣୀୟ
ଯଦି a = 18 ହୁଏ, ଦେବେ b = 18 – 6 = 12
∴ ବର୍ଗକ୍ଷେତ୍ର ବାହୁର ଦୈର୍ଘ୍ୟ 18 ମି. ଓ ଅନ୍ୟଟିର ଦୈର୍ଘ୍ୟ 12 ମି. ।

Question 10.
ଜଣେ ବ୍ୟକ୍ତି ତାଙ୍କ ଚାଲିବାର ବେଗକୁ ଯଦି ଘଣ୍ଟା ପ୍ରତି 1 କି.ମି. ବୃଦ୍ଧି କରେ, ତେବେ 2 କି.ମି. ରାସ୍ତା ଅତିକ୍ରମ କରିବା ପାଇଁ 10 ମିନିଟ୍ କମ୍ ସମୟ ନେଇଥା’ନ୍ତା । ତେବେ ବ୍ୟକ୍ତିର ଚାଲିବାର ଘଣ୍ଟା ପ୍ରତି ବେଗ ସ୍ଥିର କର ।
ସମାଧାନ :
ମନେକର ବ୍ୟକ୍ତିଙ୍କର ଚାଲିବାର ଘଣ୍ଟାପ୍ରତି ବେଗ = x କି.ମି. ।
ଏହି ବେଗରେ 2 କି.ମି. ରାସ୍ତା ଅତିକ୍ରମ କରିବାପାଇଁ ସମୟ ଲାଗିଲା = \(\frac{2}{x}\) ଘଣ୍ଟା ।
ଯଦି ତାଙ୍କର ଚାଲିବାର ଘଣ୍ଟାପ୍ରତି ବେଗ 1 କି.ମି. ବୃଦ୍ଧି ହୁଏ, ତେବେ ତାଙ୍କର ବେଗ = (x+1) କି.ମି. ।
ବେଗ ବୃଦ୍ଧି ହେଲେ 2 କି.ମି. ରାସ୍ତା ଅତିକ୍ରମ କରିବାକୁ ସମୟ ଲାଗିବ = \(\frac{2}{x+1}\) ଘଣ୍ଟା ।
ପ୍ରଶ୍ବାନୁସାରେ, \(\frac{2}{x}-\frac{2}{x+1}=\frac{10}{60}\) (∵ 10 ମିନିଟ = \(\frac{10}{60}\) ଘଣ୍ଟା ।)

x = – 4 ଗ୍ରହଣୀୟ ନୁହେଁ (କାରଣ ବେଗ ଋଣାତ୍ମକ ହେବା ଅସମ୍ଭବ) ।
∴ ବ୍ୟକ୍ତିଙ୍କର ଚାଲିବାର ଘଣ୍ଟାପ୍ରତି ବେଗ 3 କି.ମି. ।

Question 11.
ଏକ ନୌକାର ବେଗ ସ୍ଥିର ଜଳରେ 15 କି.ମି. ପ୍ରତି ଘଣ୍ଟା । ଏହା ସ୍ରୋତର ପ୍ରତିକୂଳରେ 30 କି.ମି. ଅତିକ୍ରମ କରି ପୁନଶ୍ଚ (ଅନୁକୂଳରେ) ଫେରି ଆସିବାକୁ 4 ଘଣ୍ଟା 30 ମି. ସମୟ ନେଲା । ତେବେ ସ୍ରୋତର ଘଣ୍ଟା ପ୍ରତି ବେଗ ନିର୍ଣ୍ଣୟ କର ।
ସମାଧାନ :
ଗୋଟିଏ ନୌକାର ବେଗ ସ୍ଥିର ଜଳରେ 15 କି.ମି.|ଘଣ୍ଟା ।
ମନେକର ସ୍ରୋତର ବେଗ x କି.ମି./ଘଣ୍ଟା ।
ସ୍ରୋତର ଅନୁକୂଳରେ ନୌକାଟି 1 ଘଣ୍ଟାରେ ଯିବ (15 + x) କି.ମି. ।
ସ୍ରୋତର ପ୍ରତିକୂଳରେ ନୌକାଟି 1 ଘଣ୍ଟାରେ ଯିବ (15 – x) କି.ମି. ।
ସ୍ରୋତର ଅନୁକୂଳରେ 30 କି.ମି. ଯିବାକୁ ସମୟ ଲାଗିବ = \(\frac{30}{15+x}\) ଘଣ୍ଟା । ଏବଂ
ସ୍ରୋତର ପ୍ରତିକୂଳରେ 30 କି.ମି. ଯିବାକୁ ସମୟ ଲାଗିବ = \(\frac{30}{15-x}\) ଘଣ୍ଟା ।
ପ୍ରଶ୍ନ ନୁସାରେ, \(\frac{30}{15+x}+\frac{30}{15-x}=4\frac{1}{2}\)
⇒ 30(\(\frac{15-x+15+x}{(15+x)(15-x)}\)) = \(\frac{9}{2}\)
⇒ \(\frac{30×30}{225-x²}=\frac{9}{2}\) ⇒ 1800 = 9 (225 – x²)
⇒ 225 – x² = \(\frac{1800}{9}\) = 200 ⇒ 225 – 200 = x² ⇒ x² = 25 ⇒ x= √25 = 5
∴ ସ୍ରୋତର ବେଗ 5 କି.ମି./ ଘଣ୍ଟା ।

Question 12.
ଗୋଟିଏ ଶ୍ରେଣୀର ଏକ ନିର୍ଦ୍ଦିଷ୍ଟ ସଂଖ୍ୟକ ଛାତ୍ରଙ୍କ ମଧ୍ୟରେ 250 ଟଙ୍କାକୁ ସମାନ ଭାଗରେ ବଣ୍ଟାଗଲା । ଯଦି 25 ଜଣ ଛାତ୍ର ଅଧ‌ିକ ହୋଇଥା’ନ୍ତେ, ତେବେ ପ୍ରତ୍ୟେକ 0.50 ଟଙ୍କା ଲେଖାଏଁ କମ୍ ପାଇଥା’ନ୍ତେ । ଶ୍ରେଣୀର ଛାତ୍ର ସଂଖ୍ୟା ସ୍ଥିର କର ।
ମନେକର ଶ୍ରେଣୀର ଛାତ୍ର ସଂଖ୍ୟା x ଜଣ ।
x ଜଣ ଛାତ୍ରଙ୍କ ମଧ୍ୟରେ 250 ଟଙ୍କାକୁ ସମାନ ଭାଗରେ ବାଣ୍ଟିଲେ ଜଣକା ପାଇବେ = \(\frac{250}{x}\) ଟଙ୍କା
25 ଜଣ ଛାତ୍ର ଅଧ୍ଵ ହେଲେ ଛାତ୍ର ସଂଖ୍ୟା = (x + 25) ଜଣ ।
25 ଜଣ ଛାତ୍ର ଅଧ୍ୟା ହେଲେ ଜଣକା ପାଇବେ = (\(\frac{250}{x}\) – 0.50) ଟଙ୍କା
ପ୍ରଶ୍ନନୁସାରେ, (x + 25)(\(\frac{250}{x}\) – 0.50) = 250
⇒ (x + 25)(500 – x) = 500x ⇒ 500x – x² + 12500 – 25x = 500x
⇒ x² + 25x – 12500 = 0

x = – 125 ଗ୍ରହଣୀୟ ନୁହେଁ, କାରଣ ଛାତ୍ର ସଂଖ୍ୟା ଋଣାତ୍ମକ ହେବନାହିଁ ।
⇒ x = 100 ∴ ଶ୍ରେଣୀର ଛାତ୍ରସଂଖ୍ୟା 100 ।

Question 13.
ଗୋଟିଏ ଆୟତକ୍ଷେତ୍ରର ଦୈର୍ଘ୍ୟ, ପ୍ରସ୍ଥ ଅପେକ୍ଷା 8 ମିଟର ଅଧୀକ । କ୍ଷେତ୍ରର କ୍ଷେତ୍ରଫଳ 240 ବର୍ଗ ମିଟର ହେଲେ କ୍ଷେତ୍ରଟିର ପରିସୀମା କେତେ ?
ସମାଧାନ :
ଗୋଟିଏ ଆୟତକ୍ଷେତ୍ରର ଦୈର୍ଘ୍ୟ, ପ୍ରସ୍ଥ ଅପେକ୍ଷା 8 ମି. ଅଧ୍ବକ ।
ମନେକର ଆୟତକ୍ଷେତ୍ରର ପ୍ରସ୍ଥ = x ମି., ଦୈର୍ଘ୍ୟ = (x + 8) ମିଟର,
ଆୟତକ୍ଷେତ୍ରର କ୍ଷେତ୍ରଫଳ = (x + 8) x ବର୍ଗ ମିଟର ।
ପ୍ରଶ୍ନନୁସାରେ, (x + 8) x = 240
⇒x² + 8x = 240 ⇒ x² + 8x – 240 = 0
⇒ x² + 20x – 12x – 240 = 0
⇒ x(x + 20) – 12(x + 20) = 0
⇒ (x + 20) (x – 12) = 0
⇒ x + 20 = 0 ବା x – 12 = 0
⇒x = -20 (ଅସମ୍ଭବ) ବା x = 12
∴ ଆୟତକ୍ଷେତ୍ରର ପ୍ରସ୍ଥ = 12 ମି., ଦୈର୍ଘ୍ୟ = x + 8 = 12 + 8 = 20 ମି.
ଆୟତକ୍ଷେତ୍ରର ପରିସୀମା = 2 (ଦୈର୍ଘ୍ୟ + ପ୍ରସ୍ଥ) = 2 (20 + 12) = 2 × 32 = 64 ମିଟର ।

Question 14.
ଏକ ରେଳଗାଡ଼ି 300 କି.ମି. ଦୀର୍ଘ ଯାତ୍ରା ପଥରେ ସମାନ ବେଗରେ ଗତି କରୁଥିଲା । ଯଦି ଗାଡ଼ିର ବେଗ ଘଣ୍ଟାପ୍ରତି 5 କି.ମି. ଅଧ୍ଵ ହୋଇଥା’ନ୍ତା, ତେବେ ଗାଡ଼ିଟି ନିର୍ଦ୍ଦିଷ୍ଟ ସମୟର 2 ଘଣ୍ଟା ପୂର୍ବରୁ ଯଥା ସ୍ଥାନରେ ପହଞ୍ଚିନ୍ତା । ତେବେ ଗାଡ଼ିର ଘଣ୍ଟାପ୍ରତି ବେଗ ନିରୂପଣ କର ।
ସମାଧାନ :
ମନେକର ଗାଡ଼ିର ଘଣ୍ଟାପ୍ରତି ବେଗ x କି.ମି. ।
ଏହି ବେଗରେ ଗାଡ଼ିକୁ 300 କି.ମି. ଯିବାକୁ ସମୟ ଲାଗିବ \(\frac{300}{x}\) ଘଣ୍ଟା ।
ମାତ୍ର ଘଣ୍ଟାକୁ x + 5 କି.ମି. ବେଗରେ 300 କି.ମି. ଯିବାକୁ ସମୟ ଲାଗିବ \(\frac{300}{x+5}\) ଘଣ୍ଟା ।
ପ୍ରଶ୍ନନୁସାରେ, \(\frac{300}{x}-\frac{300}{x+5}=2\) ⇒ 300(\(\frac{1}{x}-\frac{1}{x+5}\) = 2

Question 15.
ଏକ ଆୟତାକାର ପଡ଼ିଆର ଦୈର୍ଘ୍ୟ 25 ମିଟର, ପ୍ରସ୍ଥ 16 ମିଟର ଓ ପଡ଼ିଆର ଚତୁଃପାର୍ଶ୍ବରେ ସମାନ ଚଉଡ଼ାର ଏକ ରାସ୍ତା ଅଛି । ଯଦି ଚତୁଃପାର୍ଶ୍ଵରେ ଥ‌ିବା ରାସ୍ତାର କ୍ଷେତ୍ରଫଳ 230 ବର୍ଗମିଟର ହୁଏ, ତେବେ ରାସ୍ତାର ଚଉଡ଼ା ନିର୍ଣ୍ଣୟ କର ।
ସମାଧାନ :
ମନେକର ରାସ୍ତାର ଚଉଡ଼ା = x ମିଟର
ଆୟତାକାର ପଡ଼ିଆର ଦୈର୍ଘ୍ୟ = 25 ମିଟର, ପ୍ରସ୍ଥ = 16 ମିଟର ।
.. ବାହାର ଆୟତାକାର ପଡ଼ିଆର ଦୈର୍ଘ୍ୟ = (25 + 2x) ମି. ଓ ପ୍ରସ୍ଥ = (16 + 2x) ମି.
ପ୍ରଶ୍ନନୁସାରେ, (25 + 2x) (16 + 2x) – 25 × 16 = 230
⇒ 400 + 50x + 32x + 4x² – 400 = 230
⇒ 4x² + 82x – 230 = 0 ⇒ 2x² + 41x – 115=0
ଏଠାରେ a = 2, b = 41, c = -115

∴ ରାସ୍ତାର ଚଉଡ଼ା 2.5 ମିଟର ।
ବି.ଦ୍ର. : ଉତ୍ପାଦକୀକରଣଦ୍ୱାରା 2x² + 41x – 115 = 0 ର ସମାଧାନ ମଧ୍ୟ ସମ୍ଭବ ।

Question 16.
କେତେକ ଛାତ୍ରଛାତ୍ରୀ ଏକ ବଣଭୋଜିର ଆୟୋଜନ କଲେ । ଖାଦ୍ୟ ଅଟକଳ (Budget) 480 ଟଙ୍କା ଥିଲା । ସେମାନଙ୍କ ମଧ୍ୟରୁ 8 ଜଣ ବଣଭୋଜିକୁ ଗଲେ ନାହିଁ; ଯାହା ଫଳରେ ଖାଦ୍ୟ ବାବଦ ଖର୍ଚ୍ଚ ଜଣପିଛା 10 ଟଙ୍କା ବଢ଼ିଗଲା । ତେବେ କେତେଜଣ ଛାତ୍ରଛାତ୍ରୀ ବଣଭୋଜିକୁ ଯାଇଥିଲେ ?
ସମାଧାନ :
ମନେକର x ଜଣ ଛାତ୍ରଛାତ୍ରୀ ବଣଭୋଜିର ଆୟୋଜନ କରିଥିଲେ ।
ଖାଦ୍ୟ ଅଟକଳ = 480 ଟଙ୍କା ଥିଲା ।
ଜଣେ ପିଲା ପିଛା ଖର୍ଚ୍ଚ ପଡ଼ିଥାନ୍ତା = \(\frac{480}{x}\)
8 ଜଣ ବଣଭୋଜିକୁ ନଯିବାରୁ ପିଲା ସଂଖ୍ୟା = x – 8
∴ ବଣଭୋଜିକୁ ଯାଇଥବା ଛାତ୍ରଛାତ୍ରୀ ସଂଖ୍ୟା = x – 8
x – 8 ଜଣ ପିଲାଙ୍କ ପାଇଁ ଖର୍ଚ୍ଚ 480 ଟଙ୍କା ହେଲେ, ଜଣେ ପିଲା ପିଛା ଖର୍ଚ୍ଚ ପଡ଼ିଲା = \(\frac{480}{x-8}\) ଟଙ୍କା ।
ପ୍ରଶ୍ନନୁସାରେ, \(\frac{480}{x-8}-\frac{480}{x}=10\) ⇒ 480(\(\frac{x-x+8}{(x-8)x}\)) = 10
⇒ \(\frac{8}{x²-8x}=\frac{1}{48}\) ⇒ x² – 8x = 384
⇒ x² -8x – 384 = 0 ⇒ x² – 24x + 16x – 384 = 0
⇒ x (x – 24) + 16 (x – 24) = 0 ⇒ (x – 24) (x + 16) = 0 ⇒x-24=0 Ql x + 16 = 0
⇒x= 24 Q1 x = – 16
∴ ବଣଭୋଜିକୁ ଯାଇଥବା ଛାତ୍ରଛାତ୍ରୀ ସଂଖ୍ୟା = x – 8 = 24 – 8 = 16 ଜଣ।

Question 17.
ସମାଧାନ ଜର :
(i) (x + 1) (x + 2) (x + 3) (x + 4) = 120
(ii) \(5 \sqrt{\frac{3}{x}}+7 \sqrt{\frac{x}{3}}=22 \frac{2}{3}\)
(iii) 3x + \(\frac{5}{16x}\) – 2 = 0
(iv) \(\frac{2x+1}{x+1}^4-6\frac{2x+1}{x+1}^2+8=0\)
(v) (3x² – 8)² – 23 (3x² – 8) + 76 = 0
(vi) 5 (5^x + 5^-x) = 26
(vii) (x² – 2x)² – 4 (x² – 2x) + 3 = 0
(viii) x^-4 – 5x^-2 + 4 = 0
(ix) \(2\left(x^2+\frac{1}{x^2}\right)-3\left(x+\frac{1}{x}\right)-1=0\)
(x) \(\frac{3}{√2x}-\frac{√2x}{5}=5 \frac{9}{10}\)
(xi) \(\frac{x}{x+1}+\frac{x+1}{x}=\frac{34}{15}\) (x ≠ 0, x ≠ – 1)
(xii) x (2x + 1) (x – 2) (2x – 3) = 63
(xiii) \(\frac{x-3}{x+3}+\frac{x+3}{x-3}=6 \frac{6}{7}\) (x ≠ – 3, 3)
(xiv) \(3\left(x^2+\frac{1}{x^2}\right)+4\left(x-\frac{1}{x}\right)-6=0\)
(xv) (\(\frac{x+1}{x-1}\))² – (\(\frac{x+1}{x-1}\)) – 2 = 0
(xvi) \(\sqrt{2x+9}\) + x = 13
(xvii) \(\sqrt{2 x+\sqrt{2 x+4}}\) = 4
ସମାଧାନ :
(i) (x + 1) (x + 2) (x + 3) (x + 4) = 120
= (x+1) (x+4) (x + 2) (x + 3) = 120 ⇒ (x² + 5x + 4) (x² + 5x + 6) = 120
ମନେକର x² + 5x = p
ପ୍ରଭେ ସମୀକରଣରଟି (p + 4) (p + 6) = 120
⇒ p² + 10p+ 24 = 120 ⇒ p² + 10p + 24 – 120 = 0
⇒p² + 10p – 96 = 0 ⇒ p² + 16p – 6p – 96 = 0
⇒p(p + 16) – 6(p + 16) = 0 ⇒ (p + 16) (p – 6) = 0
⇒p + 16 = 0 କିମ୍ବା p – 6 = 0 ⇒ p = -16 କିମ୍ବା p=6
p ପରିବର୍ତ୍ତେ x + 5x ନେଲେ x² + 5x = – 16
⇒ x² + 5x + 16 = 0 ଏଠାରେ D = b² – 4ac = 25 – 64 = -39
⇒ D < 0 ବୀଜଦ୍ଵୟ ଅବାସ୍ତବ, ଗ୍ରହଣୀୟ ନୁହେଁ ।
ପୁନଶ୍ଚ x² + 5x = 6
⇒ x²+5x – 6 = 0, ଏଠାରେ a = 1, b = 5, c = – 6

∴ ନିର୍ଦେୟ ସମାଧାନ – 6 ଓ 1 ।

(ii) \(5 \sqrt{\frac{3}{x}}+7 \sqrt{\frac{x}{3}}=22 \frac{2}{3}\)
ମନେକର \(\sqrt{\frac{3}{x}}\) = p = \(\sqrt{\frac{x}{3}}=\frac{1}{p}\)
ପ୍ରଭେ ସମୀକରଣରଟି 5p + \(\frac{7}{p}\) = \(\frac{68}{3}\) ⇒ \(\frac{5p²+7}{p}=\frac{68}{3}\)
⇒ 15p² + 21 = 68p ⇒ 15p² – 68p + 21 = 0, ଏଠାରେ a = 15, b = -68, c = 21

ବି.ଦ୍ର. : x ର ମାନ \(\frac{25}{147}\) ଦେଇ ଦତ୍ତ ସମୀକରଣଟି ସିଦ୍ଧ ହେଉଛି କି ନାହିଁ ପରୀକ୍ଷା କରି ଦେଖ । ଯଦି ନ ହେଉଥାଏ ତେବେ ସମାଧାନ କେବଳ 27 ହେବ । ଏଠାରେ \(\frac{25}{147}\) ଅଗ୍ରହଣୀୟ ।

(iii) 3x + \(\frac{5}{16x}\) – 2 = 0
⇒ \(\frac{48x²+5-32x}{16x}=0\) ⇒ 48x² + 5 – 32x = 0

(iv) \(\frac{2x+1}{x+1}^4-6\frac{2x+1}{x+1}^2+8=0\)
ମନେକର (2x+1)= p ଦଉ ସମୀକରଣ p⁴ – 6p² + 8 = 0 => p⁴ – 4p² – 2p² + 8 = 0
⇒ p²(p² – 4) – 2(p² – 4) = 0 ⇒ (p² – 4)(p² – 2) = 0
⇒ p² – 4 = 0 ବା p² – 2 = 0 ⇒ p² = 4 ବା p² = 2 ⇒ p = ±2 ବା p = ±√2
\(\frac{2x+1}{x+1}=2\), x ର ମାନ ନିର୍ଦୟ ଅସମ୍ଭବ |

ବି.ଦ୍ର. :
(\(\frac{2x+1}{x+1}\))² = p ମନେକର
ଦଉ ସମୀକରଣ p² – 6p + 8 = 0 ⇒ p = 4 ବା p = 2
\(\frac{2x+1}{x+1}\) = 4 ⇒ \(\frac{2x+1}{x+1}\) = ±2 ନେଇ x ର ମାନ ସ୍ଥିର କରାଯାଇପାରେ ।
ପୁନଶ୍ଚ (\(\frac{2x+1}{x+1}\))² = 2 = \(\frac{2x+1}{x+1}\) = ±√2 ନେଇ
x ର ମାନ ସ୍ଥିର କରାଯାଇପାରେ ।

(v) ମନେକର 3x² – 8 = p
ପ୍ରଭେ ସମୀକରଣ p² – 23p + 76 = 0 । ଏଠାରେ a = 1, b = 23, c = 76

(vi) ମନେକର 5^x = p
ପ୍ରଭେ ସମୀକରଣ 5(p + \(\frac{1}{p}\)) = 26 ⇒ \(\frac{5(p²+1)}{p}\) = 26
⇒ 5p² + 5 = 26p ⇒ 5p² – 26p + 5 = 0
⇒ 5p² – 25p – p + 5 = 0 ⇒ 5p (p -5) – 1(p – 5) = 0
⇒ (p – 5) (5p – 1) = 0 ⇒ p – 5 = 0 କିମ୍ବା 5p – 1 = 0
⇒p = 5 କିମ୍ବା p = \(\frac{1}{5}\) ⇒5^x = 5¹ କିମ୍ବା 5^x = 5^-1
⇒ x = 1 କିମ୍ବା x = -1
∴ ନିର୍ଦେୟ ସମାଧାନ 1 ଓ -1 ।

(vii) ମନେକର x² – 2x = p
ପ୍ରଭେ ସମୀକରଣ, p² – 4p + 3 = 0
p² – 3p – p + 3 = 0 p(p – 3) – 1(p – 3) = 0
⇒(p – 3) (p – 1) = 0 ⇒ p – 3 = 0 ବା p – 1 = 0
⇒ p = 3 ବା p = 1
x² – 2x = 3 ⇒ x² – 2x – 3 = 0
⇒ x – 3x + x – 3 = 0 ⇒ x (x – 3) + 1 (x – 3) = 0
⇒ (x – 3) (x + 1) = 0 ⇒ x – 3 = 0 ବା x + 1 = 0
⇒ x = 3 ବା x = – 1
ପୁନଶ୍ଚ x² – 2x = 1 ⇒ x² – 2x – 1 = 0 । ଏଠାରେ a = 1, b = – 2, c = – 1 ବା

(viii) ମନେକର x^-2 = p। ଦଉ ସମୀକରଣ, p² – 5p + 4 = 0

(ix) \(2\left(x^2+\frac{1}{x^2}\right)-3\left(x+\frac{1}{x}\right)-1=0\)

ପୁନଶ୍ଚ x + \(\frac{1}{x}\) = -1 ⇒ \(\frac{x²+1}{x}\) = -1 ⇒ x² + 1 = -x ⇒ x² + x + 1 = 0
ସମୀକରଣ x² + x + 1 = 0 ରେ D < 0 ହେତୁ ବାସ୍ତବ ବୀଜ ସମ୍ଭବ ନୁହେଁ ।
∴ ନିର୍ଦେୟ ସମାଧାନ 2 ଓ \(\frac{1}{2}\) ।

(x) ମନେକର \(\frac{1}{√2x}\) = p; √2x = \(\frac{1}{p}\)
ପ୍ରଭେ ସମୀକରଣ, 3p – \(\frac{1}{5p}=\frac{59}{10}\)
⇒ \(\frac{15p²-1)}{5p}=\frac{59}{10}\) ⇒ \(\frac{15p²-1)}{p}=\frac{59}{2}\)
⇒ 30p² – 2 = 59p ⇒ 30p² – 59p – 2 = 0
⇒30p² – 60p + p – 2 = 0 ⇒ 30p (p – 2) + 1 (p – 2) = 0
⇒ (p – 2) (30p+1) = 0 ⇒ p – 2 = 0 ବା 30p + 1 = 0
⇒p = 2 ବା p = \(– \frac{1}{30}\)
p = 2 ⇒ \(\frac{1}{√2x}\) = 2 ⇒ \(\frac{1}{2x}\) = 4 ⇒ x = \(\frac{1}{8}\)
p = \(– \frac{1}{30}\) ⇒ \(\frac{1}{√2x}=- \frac{1}{30}\) = \(\frac{1}{2x}=\frac{1}{900}\)
⇒ 2x = 900 ⇒ x = 450
x = 450 ପୁନଶ୍ଚ ସମୀକରଣରେ ବସାଇଲେ ଏହା ସମୀକରଣକୁ ସିଦ୍ଧ କରିବ ନାହିଁ ।
ତେଣୁ ଏହା ଗ୍ରହଣୀୟ ନୁହେଁ ।
∴ xର ଏକମାତ୍ର ମୂଲ୍ୟ \(\frac{1}{8}\) ଅଟେ ।

(xi) \(\frac{x}{x+1}+\frac{x+1}{x}=\frac{34}{15}\)
⇒ \(\frac{x²+(x+1)²}{x(x+1)}=\frac{34}{15}\) ⇒ \(\frac{x²+x²+2x+1}{x²+x)}=\frac{34}{15}\)
⇒ \(\frac{2x²+2x+1}{x²+x}=\frac{34}{15}\) ⇒ 34 (x² + x) = 15 (2x² + 2x + 1)
⇒ 34x² + 34x = 30x² + 30x + 15
⇒ 34x² – 30×2 + 34x – 30x – 15 = 0 ⇒ 4x² + 4x – 15 = 0
⇒ 4x² + 10x – 6x – 15=0 ⇒ 2x (2x + 5) – 3 (2x + 5) = 0
⇒ (2x + 5) (2x-3)=0 ⇒ x = \(\frac{-5}{2}\) ବା x = \(\frac{3}{2}\)
∴ ନିର୍ଦେୟ ସମାଧାନ \(\frac{3}{2}\) ଓ \(\frac{-5}{2}\) ।

(xii) x (2x + 1) (x – 2) (2x – 3) = 63 ⇒ {x (2x – 3)} {(2x + 1) (x – 2)} = 63
⇒ (2x² – 3x) (2x² -3x – 2) = 63
ମନେକର 2x² – 3x = p
∴ p (p – 2) = 63 ⇒ p² – 2p – 63 = 0
⇒p² – 9p + 7p – 63 = 0 ⇒ p (p – 9) + 7 (p – 9) = 0
(p – 9) (p + 7) =0 p – 9 = 0 p + 7 = 0
⇒ p = 9 ବା p = -7
∴ p = 9 ⇒ 2x² – 3x = 9 ⇒ 2x² – 3x – 9 = 0
⇒ 2x² – 6x + 3x – 9 = 0 ⇒ 2x (x – 3) + 3 (x – 3) = 0
⇒ (x – 3) (2x + 3) = 0 ⇒ x = 3 ବା x = \(\frac{-3}{2}\)
p = -7 ⇒ 2x² – 3x + 7 = 0
ଏଠାରେ b² – 4ac = 9 – 56 = – 47
D < 0 xର ଅବାବ ମାନ ଗ୍ରହଣୀୟ ନୁହେଁ ।
∴ ନିର୍ଦେୟ ସମାଧାନ 3 ଓ \(\frac{3}{2}\) ।

(xiii) ମନେକର \(\frac{x-3}{x+3}=p\), ପ୍ରଭଦ ସମୀକରଣଟି p – \(\frac{1}{p}=\frac{48}{7}\)
⇒ \(\frac{p²-1}{p}=\frac{48}{7}\) ⇒ 7(p²-1) = 48p ⇒ 7p² – 48p – 7 = 0
ଏଠାରେ a = 7, b = -48, c = -7

∴ ନିର୍ଦେୟ ସମାଧାନ -4 ଓ \(\frac{9}{4}\) ।

(xiv) \(3\left(x^2+\frac{1}{x^2}\right)+4\left(x-\frac{1}{x}\right)-6=0\)

(xv) (\(\frac{x+1}{x-1}\))² – (\(\frac{x+1}{x-1}\)) – 2 = 0
ମନେକର \(\frac{x+1}{x-1}=p\)
ଦଉ ସମାଜରଣ p² – p – 2 = 0
ଏଠାରେ a = 1, b = -1, c = -2

ଯଦି p = 2 ⇒ \(\frac{x+1}{x-1}\) = 2 ⇒ x + 1 = 2(x – 1) ⇒ x + 1= 2x – 2
⇒ x – 2x = -2 -1 ⇒ -x = -3 ⇒ x = 3
ଯଦି p = 1 ⇒ \(\frac{x+1}{x-1}\) = -1 ⇒ x + 1 = -1(x – 1) ⇒ x+ 1 = -x + 1
⇒ x + x = 1 – 1 ⇒ 2x = 0 ⇒ x = 0

(xvi) \(\sqrt{2x+9}\) + x = 13

x = 8 ଦ୍ବାରା ଦତ୍ତ ସମୀକରଣଟି ସିଦ୍ଧ ହେଉଛି; ମାତ୍ର x = 20 ହେଲେ, \(\sqrt{2x+9}\) + x ≠ 13
∴ ଦତ୍ତ ସମୀକରଣର ସମାଧାନ x = 8 ।

(xvii) \(\sqrt{2 x+\sqrt{2 x+4}}\) = 4

x = 6 ଦ୍ଵାରା ଦତ୍ତ ସମୀକରଣଟି ସିଦ୍ଧ ହେଉଛି; ମାତ୍ର x = \(\frac{21}{2}\) ହେଲେ,
\(\sqrt{2 x+\sqrt{2 x+4}}\) ≠ 4 (ଦତ୍ତ ସମୀକରଣଟି ସିଦ୍ଧ ହେଉନାହିଁ)
∴ ନିଶ୍ଚେ ସମୀକରଣର ସମାଧାନ 6।

Odisha State Board CHSE Odisha Class 12 Invitation to English 4 Solutions Grammar Tense Patterns Unit 6 The Past Simple and the Past Perfect Textbook Activity Questions and Answers.

CHSE Odisha 12th Class English Grammar Tense Patterns Unit 6 The Past Simple and the Past Perfect

SECTION – 1

Look at the sentences below.
(a) I reached the hostel in the morning and found that somebody had broken into my room during the night.
(b) She said that her friend had published a book.
(c) He had lived in this town for ten years; then he migrated to Japan.
Can you find the Past Perfect Tense in each sentence? Note that the sentence in which it occurs refers to two actions — the action expressed by the Past Perfect and another action expressed by the Past Simple. Of the two actions, which takes place earlier and which takes place later? List them below.

(a) (1) I reached the hotel in the morning and found. (later).
(2) that somebody had broken (earlier action) into my room during the night.
(b) (1) She said _________(later action).
(2) that her friend had published a book _________(earlier action).
(c) (1) _________then he migrated to Japan _________(later action).
(2) He had lived in this town for ten years, (earlier action).
Can you answer now? Which action does the Past Perfect refer to — the earlier one or the later one? Which action does the Past Simple refer to?
Answer:
The earlier action refers to Past Perfect and the later one refers to Past Simple Tense.

Activity – 23
Combine each pair of sentences below into a single sentence, using the Past Perfect to show which action took place earlier. (You may have to use words like after; when etc.
(a) (i) I finished my homework.
(ii) Then I went to buy a pen.
_____________________________________.
(b) (i) Then the doctor gave some medicine to the patient,
(ii) Then the patient regained his senses.
_____________________________________.
(c) (i) I read a few pages from the book.
(ii) After that I returned it to the librarian.
_____________________________________.
(d) (i) I worked in the garden for some time.
(ii) After that I had my breakfast.
____________________________
(e) (i) He left the place in a hurry.
(ii) After that his friend arrived.
____________________________
(f) (i) The young girl finished shopping.
(ii) Then she met with an accident.
____________________________
(g) (i) The thief ran away with the gold.
(ii) After that the police arrived.
____________________________
Answer:
(a) After I had finished my homework, I went to buy a pen.
(b) After the doctor had given some medicine to the patient, the patient regained his senses.
(c) After I had read a few pages from the book, I returned to the librarian.
(d) I had worked in the garden for some time before I had my breakfast.
(e) When he had left the place in a hurry. his friend arrived.
(f) After the young girl had finished shopping. she met with an accident.
(g) The thief had run away with the gold before the police arrived.

Activity — 24
Jatin arrived late at different places yesterday. What did he find when he arrived
at each place?
Example — When he arrived at the cricket stadium, the game had ended.
(a) the hank it / already I close.
____________________________
(b) his uncle’s house his uncle I go the sleep.
____________________________
(c) the bus stops the bus I already / leave.
____________________________
(d) book shop the book he wanted/sold out already.
_________________________________
(e) the club his friends/leave.
_________________________________
(f) the hostel everyone / go to bed.
_________________________________
Answer:
(a) When he arrived at the bank, it had already closed.
(b) When he got to his uncle’s house, his uncle had gone to sleep.
(c) When he reached the bus stop, the bus had already left.
(d) When he came to the bookshop, the book he wanted had been sold out already.
(e) When he arrived at the club, his friends had left.
(f) When he came to the hostel, everyone had gone to bed.

Activity – 25
Use the verb supplied in brackets in the appropriate form.
(a) We went to Anil’s house and _____________(knock) on the door but there _____________ (be) no answer. Either he _____________ (go) out or he _____________ (not want) to see anyone.
(b) Sadhan _____________(go) for a walk yesterday because the doctor _____________(tell) him last week that he _____________(need) exercise.
(c) A : _____________(Seema / arrive) at the party in time last night ?
B: No, she was late. By the time, we got there, everyone _____________(leave).
Answer:
(a) We went to Anil’s house and knocked on the door but there was no answer. Either he had gone out or he did not want to see anyone.
(b) Sadhan went for a walk yesterday because the doctor told him last week that he needed exercise.
(c) A: Did Seema arrive at the party in time last night?
B: No, she was late. By the time, we got there, everyone had left.

Odisha State Board CHSE Odisha Class 12 Invitation to English 4 Solutions Grammar Tense Patterns Unit 5 Past Simple and Past Progressive Textbook Activity Questions and Answers.

CHSE Odisha 12th Class English Grammar Tense Patterns Unit 5 Past Simple and Past Progressive

SECTION – 1

Study the sentences below :
(a) It started to rain while we were walking home.
(b) My sister was tidying my room when I saw your letter.
(c) Anita was walking along the road when suddenly she heard footsteps behind her. Someone was following her. She was frightened and started to run.
What do you think the use of the Past Simple and the Past Progressive indicates in these sentences?
(Hint: Think of a point of time and a duration of time in the past and relate them to the action.)

Note:
The Past Simple tells us that the work/action started and finished in the past. The speaker has a definite time in mind. But in Past Progressive the time of beginning or completion of the activity is not mentioned. The activity was in progress for that hour.

Activity – 21
Put the verbs into the correct form, Past Progressive or Past Simple.
(a) My friend ______________(meet) Anima and Amiya at the bus stop four days ago. They ______________(go) to Paradeep and my friend ______________ (go) to Bolangir. They ______________(have) a chat while they ______________(wait) for their buses.
(b) My brother ______________ (cycle) to school last Monday when suddenly an old woman ______________(step) out into the road in front of him. He ______________(go) quite fast but luckily he ______________ (manage) to stop in time and ______________(not / hit) her.
Answer:
(a) My friend met Anima and Amiya at the bus stop four days ago. They were going to Paradeep and my friend was going to Bolangir. They had a chat while they were waiting for their buses.
(b) My brother was cycling to school last Monday when suddenly an old woman stepped out into the road in front of him. He was going quite fast but luckily he managed to stop in time and did not hit her.

Activity – 22
Here is a true story.
An old couple …………………………living in a flat in Bhubaneswar. ………………………… locked up in one room ………………………….. Some unknown people took away everything ………………………… police arrived ………………………… climbed ………………………… rescued ………………………… broke open a door ………………………… one dacoit was killed ………………………… detective was called …………………………interviewed a witness.
Imagine that you are being questioned by the police as if you were a witness to the crime. A police Inspector is recording your statements in a notebook. Think about the situation and write the appropriate answers.
Inspector: Where were you standing at that time?
Answer: _____________________________.
Inspector: Why did you come here?
Answer: _____________________________.
Inspector: What was the old man doing at the time?
Answer: _____________________________.
Inspector: How did you see that?
Answer: _____________________________.
Inspector: How long were you standing there?
Answer: _____________________________.

Answer:
An old couple _____ living in a flat in Bhubaneswar, _____ Orissa locked up in one room _____. Some unknown people took away everything. _____ police arrived _____, limbed _____ rescued _____ , broke open a door _____one dacoit killed, _____detective was called _____ interviewed a witness.
Inspector: Where were you standing at that time?
Answer: I was standing near the flat.
Inspector: Why did you come here?
Answer: I came here to play.
Inspector: What was the old man doing at that time?
Answer: The old man was shouting and trembling out of fear.
Inspector: How did you see that?
Answer: I heard him shouting and saw him trembling.
Inspector: How long were you standing there?
Answer: I stood there till your arrival.

Odisha State Board CHSE Odisha Class 12 Invitation to English 4 Solutions Grammar Tense Patterns Unit 4 Present Perfect and Past Simple Textbook Activity Questions and Answers.

CHSE Odisha 12th Class English Grammar Tense Patterns Unit 4 Present Perfect and Past Simple

Study the dialogue given below.
Susant: Have you ever ridden a horse?
Subir: Yes, I have.
Susant: When was that?
Subir: I rode one last summer.
Susant : What was it like?
Subir: Oh, it was awful.
Susant : Why? What happened?
Subir: I fell off and hurt my back.
Identify the Present Perfect and Past Simple sentences and examine their use carefully. How are they different in meaning?
{Hint: One of them answers the question ‘When’? and the other does not)
Except for the first two i.e. “Have you ever ridden a horse ?” and “Yes, I have”, the rest of the sentences in the above dialogue belong to Past Simple constructions.
When an action/event took place in the past but its result is still operative at the present moment of speaking/time, we generally use a Present Perfect tense and the Past Simple means that the action/happening occurred before the present moment.

Activity – 18
1. Complete the dialogue using the hints given.
(i) A: ever / see /a lion _______________?
B: Yes, _______________.
A: Where _______________?
B: In the zoo _______________.
A: What/look _______________?
B: terrible _______________.
A: You / afraid _______________?
B: No, _______________?

(ii) A : ever / be to / Dhauligiri ………………………?
B: Yes, _______________.
A: What! see /there _______________?
B: A temple I top/hill _______________.
A See / the inscriptions _______________?
B: Yes, _______________?
A: Able to read the inscriptions _______________?
B: No, _______________.

Answer:
(i) A: Have you ever seen a lion?
B: Yes, I have.
A: Where did you see it?
B: I saw it in the zoo.
A: What did it look like?
B: Yes, it was very terrible to look at.
A: Were you afraid?
B: No, I wasn’t.

(ii) A: Have you ever been to Dhauligiri?
B: Yes, I have been two times.
A: What did you see there?
B: I saw a temple at the top of the hill.
A: Did you see the inscriptions there?
B: Yes, I saw the inscriptions there.
A: Were you able to read the inscriptions there?
B: No, I wasn’t.

Activity – 19
Choose the right verb for each blank space and put it into the correct tense.
(do, wear, carry, ask, say, think)
A : _______________your grandfather _______________ something really crazy ?
B: He _______________ something really silly last summer. One one of the hottest days he _______________ a raincoat and _______________ an umbrella. Everyone _______________ him why. He _______________he _______________ it was going to rain.
Answer:
A: Did your grandfather wear something really crazy?
B: He wore something really silly last summer. On one of the hottest days, he wore a raincoat and carried an umbrella. Everyone asked him why. He said he thought it was going to rain.

Activity – 20
Complete the sentences, using the verbs in brackets either in Past Simple or Present Perfect form.
(a) She _______________ up her mind (made). She’s going to look for another college.
(b) Amulya : _______________me his pen but I’m afraid I _______________ it. (give, lose)
(c) A: It’s a little bit noisy in here, isn’t it?
B: Pardon? I can’t hear. What _______________ you _______________? (say)
(d) Where is my bike? It _______________ outside the classroom. It _______________! (be, disappear)
(e) Did you know that Umesh _______________ a new scooter? (buy)
(f) I did Sanskrit at school but I _______________ most of it. (forget)
(g) A : Sima, this is Rajesh.
B: Hello, Rajesh. Actually, we know each other. We _______________ already ___________ (meet).

Answer:
(a) She has made up her mind. She’s going to look for another college.
(b) Amulya gave me his pen but I’m afraid I have lost it.
(c) A: It’s a little bit noisy in here, isn’t it?
B: Pardon? I can’t hear. What did you say?
(d) Where is my bike? It was outside the classroom. It has disappeared!
(e) Did you know that Umesh has bought a new scooter?
(f) I did Sanskrit at school but I have forgotten most of it.
(g) A : Sima, this is Rajesh.
B: Hello, Rajesh. Actually, we know each other. We have already met.

Odisha State Board CHSE Odisha Class 12 Invitation to English 4 Solutions Grammar Tense Patterns Unit 3 Past Simple Textbook Activity Questions and Answers.

CHSE Odisha 12th Class English Grammar Tense Patterns Unit 3 Past Simple

SECTION – 1
Read the passage below carefully.
Ramesh was bon in Baleswar in 1960. He was brought up in Cuttack by his uncle because of his parents in Bangalore. Then he went to Hyderabad to complete his studies. He got a first-class M.Sc. degree in Physics and became a lecturer at a college in Orissa. In 1985 he joined a university but soon went to the U.S.A. on a scholarship. He came back in 1990 and got married. He bought a house in Bhubaneswar in 1993. The happenings above took place in the past time. The basic element of meaning is: “the happening takes place before the present moment.” This means that the present moment is excluded.

Activity – 16
Answer the following questions which relate to the things you do every day. Answer in full sentences.
1. When do you wake up?
2. What do you eat before you go to college?
3. When do you leave home?
4. How do you get to college?
5. What do you pass on the way?
6. How long does it take you?
7. When do your classes start?
Answer:
1. I wake up early at 6 a.m. in the morning.
2. I usually eat rice, dal, and curry before I go to college.
3. I leave home at half past 9 (o’clock).
4. I get to college by bicycle.
5. I pass people, motor cars, etc. on the way to college.
6. It takes me twenty minutes time.
7. Our classes start at 10 o’clock.

Imagine that a friend of yours wants to know from you what you did last Wednesday, which was a very typical day in your life. What kind of questions would he ask you and what answers would you give him? Here are a few questions and answers for you to write.

Question 1.
When did you wake up last Wednesday?
Answer:
I woke up at 6. a.m. last Wednesday.

Question 2.
How did you enjoy the day?
Answer:
I went on merry-making with my friends as it was my birthday.

Question 3.
What did you do in the morning that day?
Answer:
I took a clean bath first and put on a new pair of clothes.

Question 4.
What did you do after wearing the pair of new clothes?
Answer:
I went to the temple to worship God.

Question 5.
What did you give your friends to eat?
Answer:
I gave them some cake, sweets, and other sumptuous food to eat.

Question 6.
What did they give you on the day?
Answer:
They presented me with fabulous gifts.

Question 7.
Were you really happy on that day?
Answer:
Yes. I was very happy on that day.

Activity – 17
The following years were related to important events in Gandhiji’s life. Can you write a sentence on each of these years? One has been done for you.
1. (1869) Gandhiji was born.
2. (1888) ____________________________________________.
3. (1891) ____________________________________________.
4. (1893) ____________________________________________.
5. (1906) ____________________________________________.
6. (1915) ____________________________________________.
7. (1917) ____________________________________________.
8. (1931) ____________________________________________.
9. (1942) ____________________________________________.
10. (1948) ____________________________________________.
Answer:
1. (1869) Gandhiji was bom.
2. (1888) He went to London to study law.
3. (1891) He passed the law examination and was admitted to the bar.
4. (1893) He went to South Africa.
5. (1906) He wrote a significant letter to his brother Laxmidas Gandhi.
6. (1915) He left for Rajkot and Porbandar to meet his relations.
7. (1917) He received a summons to appear before the sub-divisional officer on April 18.
8. (1931) Gandhi-Irwin pact was signed in Delhi.
9. (1942) He attended the All-India Congress Committee meeting at Wardha.
10. (1948) He was shot dead by Nathuram Binayak Godse an R.S.S. member.

Activity – 18
Imagine that you went for a picnic last Sunday with some friends. Write a letter to your friend telling him/her about the picnic. You may follow the hints given below.
1. when and how it was planned
2. the place selected
3. how you went there
4. what you saw on the way
5. what you did there
6. what you enjoyed most
7. when you returned

Dear ___________,
It was really nice to hear from you again. Thanks for telling me about your plans for an excursion. I’m afraid I won’t be able to make it. I have exams to sit for. But I have already made up for the loss.
Yesterday we went on a picnic, and I feel I must share the excitement of it all with you.
______________________________________________________
______________________________________________________
______________________________________________________
______________________________________________________

Yours truly,
___________.

Answer:
Dear Suresh,
It was really nice to hear from you again. Thanks for telling me about your plans for an excursion. I’m afraid I won’t be able to make it. I have exams to sit for. But I have already made up for the loss.
Yesterday we went on a picnic, and I feel I must share the excitement of it all with you.
Our classmates sat together a week ahead and decided to go on a picnic to Nandankanan. We got the permission of the Principal and gave each one hundred rupees for the purpose. We were 50 students and three lecturers too consented to go with us. All of us were of the opinion to see Nandankanan, the zoo. We hired a bus and it was about two hours journey from our college. We saw a lot of buses, cars, and trees full of flowers on our way to Nandankanan. We arrived at Nandankanan by 9 o’clock in the morning. On getting there, we first had our breakfast. Mohan, Sanjay, and Pravakar voluntarily agreed to take charge of cooking. They discharged their duty pretty well. Others went to see the beautiful march of colorful sights and sounds. Our menu was very simple. We had rice, mutton curry, and fruit salad. The preparation of the food was superb. We all enjoyed the food to our heart’s content. After finishing our lunch, we took a little rest and then went round the zoo again. We first returned to our college at 6 p.m. Then the day scholars went to their houses and we went to our hostel. It was really a pleasant outing. I felt your absence there.

With love
Yours truly
Sandeep.

Note:
We can use sentences in the past simply when ‘the speaker/writer has a definite time in mind’.
Examples :
Once this town was a beauty spot.
Mahatma was bom in 1869.
“Past simple” can also be used for “habit in the past”.

Example :
He always carried an umbrella.
They never drank tea.
Note that in most cases we use used to if we wish to emphasize that the habit has been discontinued.

Example :
He used to smoke, (which means that he smoked at one time but he doesn’t smoke now.)
Past Simple can also be used for hypothetical meaning.

Example :
It’s time we had a holiday.
If you caught the 9 o’clock train, you would get there by lunchtime, etc.

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 BSE Odisha 10th Class Maths Solutions Chapter 3 ସମାନ୍ତର ପ୍ରଗତି Ex 3(a) Textbook Exercise Questions and Answers.

BSE Odisha Class 10 Maths Solutions Algebra Chapter 3 ସମାନ୍ତର ପ୍ରଗତି Ex 3(a)

Question 1.
ନିମ୍ନଲିଖତ ପ୍ରତ୍ୟେକ କ୍ଷେତ୍ରରେ ସମ୍ଭାବ୍ୟ ଉତ୍ତର ମଧ୍ୟରୁ ସଠିକ୍ ଉତ୍ତରଟି ବାଛ ।
(i) 1, 2, 3, 4, …….. ଅନୁକ୍ରମରେ t₈ = …………..
(a) 6
(b) 7
(c) 8
(d) 9
ଉ-
(c) 8

(ii) 2, 4, 6, 8, …….. ଅନୁକ୍ରମରେ t₇ = …………..
(a) 12
(b) 14
(c) 16
(d) 18
ଉ-
(b) 14

(iii) – 5, – 3, – 1, 1, ……. ଅନୁକ୍ରମରେ t₁₁ = …………..
(a) 13
(b) 15
(c) 17
(d) 19
ଉ-
(b) 15

(iv) 3, 6, 9, ……… ରେ ସାଧାରଣ ଅନ୍ତର d = …………..
(a) 3
(b) 4
(c) 5
(d) 6
ଉ-
(a) 3

(v) -4, -2, 0, 2, A.P. ରେ ସାଧାରଣ ଅନ୍ତର d = …………..
(a) – 2
(b) -3
(c) 2
(d) 3
ଉ-
(c) 2

(vi) 10.2, 10.4, 10.6, 10.8, …. ରେ ସାଧାରଣ ଅନ୍ତର t₅ = …………..
(a) 11.0
(b) 11.2
(c) 11.4
(d) 11.6
ଉ-
(a) 11.0

(vii) 2.5, 2.9, 3.3, 3.7, ….. A.P. ରେ ସାଧାରଣ ଅନ୍ତର d = …………..
(a) 1.5
(b) 1.4
(c) 0.5
(d) 0.4
ଉ-
(d) 0.4

(viii) 3, x, 9, ଏକ A.P. ହେଲେ x = …………
(a) 4
(b) 5
(c) 6
(d) 7
ଉ-
(c) 6

(ix) 1.01, 1.51, 2.01, 2.51, ……. A.P. ରେ ସାଧାରଣ ଅନ୍ତର d = …………..
(a) 1
(b) 0.5
(c) 1.5
(d) 1.05
ଉ-
(b) 0.5

(x) 5, 0, -5, 10, ………. A.P. ରେ ସାଧାରଣ ଅନ୍ତର d = …………..
(a) – 5
(b) 5
(c) – 10
(d) 10
ଉ-
(a) – 5

ବ୍ୟାଖ୍ୟା ସହ ଉତ୍ତର:
(i) 1, 2, 3, 4, …….. ଅନୁକ୍ରମର t₈ = 8
(କାରଣ ଏଠାରେ t₁ = 1, t₂ = 2, t₃ = 3, t₈ = 8)

(ii) 2, 4, 6, 8 ……. ଅନୁକ୍ରମର t = 14
(କାରଣ t₁ = 2 × 1, t₂ = 2 × 2, t₃ = 2 × 3, t₇ = 2 × 7 = 14)

(iii) -5, -3, – 1, 1 ଅନୁକ୍ରମର t₁₁ = 15
(t₁₁ = 5 + (11 – 1) 2 = -5 + 20= 15).

(iv) 3,6,9, 60 ରେ ସାଧାରଣ ଅନ୍ତର d = 3
(କାରଣ 6 – 3 = 9 – 6 = 3)

(v) -4, -2, 0, 2 ……. A.P. ରେ ସାଧାରଣ ଅନ୍ତର d = 2
(କାରଣ -2 – (-4) = 0 -(-2) = 2)

(vi) 10.2, 10.4, 10.6, 10.8…… ରେ t₂ = 11
(କାରଣ 10.2 + (5 – 1) × 0.2 = 10.2 + 0.8 = 11)

(vii) 2.5, 2.9, 3.3, 3.7 …….. A.P. ରେ ସାଧାରଣ ଅନ୍ତର d = 0.4
(କାରଣ 2.9 – 2.5 = 3.3 – 2.9 = 0.4)

(viii) 3, x, 9 ….. A.P. ହେଲେ x = 6
(କାରଣ x = \(\frac{3+9}{2}\) = \(\frac{12}{2}\) =6)

(ix) 1.01, 1.51, 2.01, 2.51, ……. A.P. ରେ ସାଧାରଣ ଅନ୍ତର d = …………..
(କାରଣ 1.51 – 1.01 = 2.01 – 1.51 = 0.50)

(x) 5, 0, -5, 10, ………. A.P. ରେ ସାଧାରଣ ଅନ୍ତର d = …………..
(କାରଣ 0 – 5 = -5 – 0 = -5)

Question 2.
ନିମ୍ନଲିଖୂତ ଅନୁକ୍ରମ ମଧ୍ୟରୁ କେଉଁଗୁଡ଼ିକ A.P, ସେଗୁଡ଼ିକୁ ଚିହ୍ନଟ କର :
(i) 1, 4, 7, 10, 15, 16, 19, 22
(ii) 1, 8, 15, 22, 29, 36, 43, 50
(iii) 1, 6, 11, 15, 22, 28, 34, 40
(iv) 1, 4, 7, 9, 11, 14, 17, 20
(v) – 5, -3, -1, 0, 2, 4, 6, 8
(vi) a, a + d, a + 2d, a + 3d, a + 4d, a + 5d, a + 6d, a + 7d
(vii) 0.6, 0.8, 1.0, 1.5, 1.7, 1.8, 1.9, 2.0
(viii) -7,-4,- 1, 2, 5, 8, 11, 14
ଉ –
ନିମ୍ନଲିଖୂତ ଅନୁକ୍ରମ ମଧ୍ୟରୁ (ii), (vi) ଏବଂ (viii) ଗୁଡ଼ିକ A.P. ଅଟନ୍ତି ।

ବ୍ୟାଖ୍ୟା ସହ ଉତ୍ତର:
(i) 1, 4, 7, 10, 15, 16, 19, 22 ଅନୁକ୍ରମଟି A.P. ନୁହେଁ ।
(କାରଣ 4 – 1 ≠ 16 – 15, ଏଠାରେ ଯେକୌଣସି ପଦରୁ ତା’ର ପୂର୍ବ ପଦକୁ ବିୟୋଗ କଲେ ବିୟୋଗଫଳ ସମାନ ରହୁନାହିଁ ।)

(ii) 1, 8, 15, 22, 29, 36, 43, 50 ଅନୁକ୍ରମଟି A.P. ଅଟେ ।
(କାରଣ ଏଠାରେ ସାଧାରଣ ଅନ୍ତର (d) = 7 ଅଟେ ।)

(iii) 1, 6, 11, 15, 22, 28, 34, 40 ଅନୁକ୍ରମଟି A.P. ନୁହେଁ ।
(କାରଣ 6 – 1 ≠ 15 – 11, ଏଠାରେ d ସମାନ ନୁହେଁ ।)

(iv) 1, 4, 7, 9, 11, 14, 17, 20 ଅନୁକ୍ରମଟି A.P. ନୁହେଁ ।
(କାରଣ 4 – 1 ≠ 11 – ୨ ଏଠାରେ d ସମାନ ନୁହେଁ ।)

(v) -5, -3, -1, 0, 2, 4, 6, 8 ଅନୁକ୍ରମଟି A.P. ନୁହେଁ ।
(କାରଣ – 3 – (- 5) = 2 କିନ୍ତୁ 0 – (- 1) = 1, ଏଠାରେ d ସମାନ ନୁହେଁ ।)

(vi) a, a + d, a + 2d, a + 3d, a + 4d, a + 5d, a + 6d, a + 7d ଅନୁକ୍ରମଟି A.P. ଅଟେ ।
ଅନୁକ୍ରମର t₁ = a ଏବଂ ସାଧାରଣ ଅନ୍ତର = d)

(vii) 0.6, 0.8, 1.0, 1.5, 1.7, 1.8, 1.9, 2.0 ଅନୁକ୍ରମଟି A.P. ନୁହେଁ ।
(କାରଣ d ସବୁ କ୍ଷେତ୍ରରେ ସମାନ ନୁହେଁ ।)

(viii) -7,-4,-1, 2, 5, 8, 11, 14 ଅନୁକ୍ରମଟି A.P. ନୁହେଁ ।
(କାରଣ d = -4 – (-7) = -1 – (-4) = 2 – (-1) = 8 – 5 = 11 – 8 = 14 – 11 = 3)

Question 3.
ପ୍ରଶ୍ନ 2ରେ ଯେଉଁଗୁଡ଼ିକ A.P. ସେମାନଙ୍କ କ୍ଷେତ୍ରରେ ସାଧାରଣ ଅନ୍ତର ନିରୂପଣ କର ।
ସମାଧାନ :
(ii) 1, 8, 15, 22, 29, 36, 43, 50 ଅନୁକ୍ରମଟି A.P. ଅଟେ ।
ଏହି ଅନୁକ୍ରମର ସାଧାରଣ ଅନ୍ତର (d) = 8 – 1 = 15 – 8 = 7

(vi) , a + d, a + 2d, a + 3d, a + 4d, a + 5d, a + 6d, a + 7d ଅନୁକ୍ରମଟି A.P. ଅଟେ ।
ଏଠାରେ ସାଧାରଣ ଅନ୍ତର (d) = a + 4d – a – 3d = d

(viii) -7,-4,-1, 2, 5, 8, 11, 14 ଅନୁକ୍ରମଟି A.P. ଅଟେ ।
ଅନୁକ୍ରମଟିର ସାଧାରଣ ଅନ୍ତର (d) = – 4 – (- 7) = 3

Question 4.
ପ୍ରଥମ ପଦ a = 5 ନେଇ A.P.ର ପ୍ରଥମ ଚାରିଗୋଟି ପଦ ଲେଖ ଯେପରିକି ସାଧାରଣ ଅନ୍ତର
(i) d = 5
(ii) d = 4
(iii) d = 2
(iv) d =-2
(v) d=-3 ହେବ |
ସମାଧାନ :
(i) ପ୍ରଥମ ପଦ a = 5, d = 5 ହେଲେ,
t₁ = a = 5, t₂ = a + d = 5 + 5 = 10, t₃ = a + 2d = 5 + 2 × 5 = 15
t₄ = a + 3d = 5 + 3 × 5 = 20
∴ A.P.ର ପ୍ରଥମ ଚାରିଗୋଟି ପଦ 5, 10, 15 ଓ 20 1

(ii) a = 5, d = 4 ହେଲେ,
t₁ = a = 5, t₂ = a + d = 5 + 4 = 9, t₃ = a + 2d = 5 + 2 × 4 = 13
t₄ = a + 3d = 5 + 3×4 = 17
∴ A.P.ର ପ୍ରଥମ ଚାରିଗୋଟି ପଦ 5, 9, 13 ଓ 17 !

(iii) a = 5, d = 2 ହେଲେ,
t₁ = a = 5, t₂ = a + d = 5 + 2 = 7, t₃ = a + 2d = 5 + 2 × 2=9
t₄ = a + 3d = 5 + 3 × 2 = 11
∴ A.P.ର ପ୍ରଥମ ଚାରିଗୋଟି ପଦ 5, 7, 9 ଓ 11 1

(iv) a = 5, d= – 2 ହେଲେ,
t₁ = a = 5, t₂ = a + d = 5 + (- 2) = 3, t₃ = a + 2d = 5 + 2 (- 2) = 1
t₄ = a + 3d = 5 + 3 (- 2) = – 1
∴ A.P.ର ପ୍ରଥମ ଚାରିଗୋଟି ପଦ 5, 2, – 1 ଓ – 4 ।

(v) a = 5, d = -3 ହେଲେ,
t₁ = a = 5, t₂ = a + d = 5 + (- 3) = 2, t₃ = a + 2d = 5 + 2 (- 3) = – 1
t₄ = a + 3d = 5 + 3 (- 3) = – 4
∴ A.P.ର ପ୍ରଥମ ଚାରିଗୋଟି ପଦ 5, 2, – 1 ଓ – 4 ।

Question 5.
ଏକ A.P.ର n ତମ ପଦ । ନିମ୍ନରେ ପ୍ରଦତ୍ତ ହୋଇଛି । ପ୍ରତ୍ୟେକ କ୍ଷେତ୍ରରେ t₅, t₈ ଓ t₁₀ କେତେ ନିରୂପଣ କର।
(i) t_n = \(\frac{n+1}{2}\)
(ii) t_n = -10 + 2n
(iii) t_n = 10n + 5
(iv) t_n = 4n – 6
ସମାଧାନ :
(i) ଏକ A.P. ର t_n = \(\frac{n+1}{2}\)
t₅ = \(\frac{6+1}{2}=\frac{6}{2}=3\)

t₈ = \(\frac{8+1}{2}=\frac{9}{2}=4.5\)

t₁₀ = \(\frac{10+1}{2}=\frac{11}{2}=5.5\)

(ii) t_n = -10 + 2n
t₅ = -10 +2 × 5
= – 10 + 10 = 0,

t₈ = -10 + 2 × 8
= -10 + 16 = 6

t₁₀ = -10 + 2 × 10
= 10 + 20= 10

(iii) t_n = 10n + 5
t₅ = 10 × 5 + 5 = 55,
t₈ = 10 × 8+ 5 = 85,
t₁₀ = 10 x 10 + 5 = 105

(iv) t_n = 4n – 6,
t₅ = 4 × 5 – 6
= 20 – 6 = 14

t₈ = 4 × 8 – 6
= 32 – 6 = 26

t₁₀ = 4 × 10 – 6
= 40 – 6 = 34

Question 6.
ନିମ୍ନଲିଖୂ A.P. ଗଠନ କର (କେବଳ ଦ୍ବିତୀୟ, ତୃତୀୟ ଓ ଚତୁର୍ଥ ପଦ ତ୍ରୟ ଆବଶ୍ୟକ) ଯେଉଁଠାରେ :
(i) ପ୍ରଥମ ପଦ a = 4, ସାଧାରଣ ଅନ୍ତର d =3
(ii) ପ୍ରଥମ ପଦ a = -8, ସାଧାରଣ ଅନ୍ତର d = – 2
(iii) ପ୍ରଥମ ପଦ a = 7, ସାଧାରଣ ଅନ୍ତର d = – 4
(iv) ପ୍ରଥମ ପଦ a = 10, ସାଧାରଣ ଅନ୍ତର d = 5
(v) ପ୍ରଥମ ପଦ a = \(\frac{1}{2}\), ସାଧାରଣ ଅନ୍ତର d = \(\frac{3}{2}\)
(vi) ପ୍ରଥମ ପଦ a = \(\frac{1}{2}\), ସାଧାରଣ ଅନ୍ତର d = -1
ସମାଧାନ :
(i) ଏକ A.P.ର ପ୍ରଥମ ପଦ a = 4, ସାଧାରଣ ଅନ୍ତର d = 3
t₂ = a + d = 4 + 3 = 7,
t₃ = a + 2d = 4 + 2 × 3 = 10
t₄= a + 3d = 4 + 3 × 3 = 13

(ii) a = -8, d = -2
t₂ = a + d = -8 + (- 2) = – 10
t₃ = a + 2d = – 8 + 2 × (- 2) = – 8 – 4 = – 12
t₄ = a + 3d = -8 + 3 (- 2) = – 8 – 6=-14

(iii) a = 7, d=-4
t₂ = a + d = 7+ (- 4) = 3,
t₃ = a + 2d = 7 + 2 (- 4) = 7 – 8 = -1
t₄ = a + 3d = 7+ 3 (- 4) = 7 – 12 = – 5

(iv) a = 10, d = 5
t₂ = a + d = 10 + 5 = 15,
t₃ = a + 2d = 10 + 2 × 5 = 20
t₄ = a + 3d = 10 + 3 × 5 = 10 + 15 = 25

(v) a = \(\frac{1}{2}\), d = \(\frac{3}{2}\)
t₂ = a + d = \(\frac{1}{2}+\frac{3}{2}=\frac{4}{2}=2\)
t₃ = a + 2d = \(\frac{1}{2}+2 × \frac{3}{2}=\frac{1}{2}+3=\frac{7}{2}\)
t₄ = a + 3d = \(\frac{1}{2}+3 × \frac{3}{2}=\frac{1}{2}+\frac{9}{2}=\frac{10}{2}=5\)

(vi) a = \(\frac{1}{2}\), d = -1
t₂ = a + d = \(\frac{1}{2}-1=-\frac{1}{2}\)
t₃ = a + 2d = \(\frac{1}{2}+2(-1)=\frac{1}{2}-2=\frac{1-4}{2}=\frac{-3}{2}\)
t₄ = a + 3d = \(\frac{1}{2}+3(-1)=\frac{1}{2}-3=\frac{1-6}{2}=\frac{-5}{2}\)

Question 7.
ନିମ୍ନରେ ପ୍ରଦତ୍ତ ଉକ୍ତିଗୁଡ଼ିକ ଭୁଲ୍ ବା ଠିକ୍ ଲେଖ ।
(a) 1, 2, 3, 4 ……. ସମାନ୍ତର ପ୍ରଗତି ସୃଷ୍ଟି କରନ୍ତି ।
(b) 1, – 1, 1, -1, -1, ……. ଅନୁକ୍ରମଟି ସମାନ୍ତର ପ୍ରଗତି ଅଟେ ।
(c) 2, 1, – 1, – 2 ସଂଖ୍ୟା ଚାରିଗୋଟି ସମାନ୍ତର ପ୍ରଗତିରେ ବିଦ୍ୟମାନ ।
(d) ଯେଉଁ ଅନୁକ୍ରମର t = n – 1, ତାହା ଏକ A.P. ଅଟେ ।
(e) ଯେଉଁ ଅନୁକ୍ରମର S_n = \(\frac{n(n-1)}{2}\) ତାହା A.P. ଅଟେ ।
(f) ଯଦି କୌଣସି ତ୍ରିଭୁଜର କୋଣ ତ୍ରୟର ପରିମାଣର ଅନୁପାତ 2 : 3 : 4 ହୁଏ, ତେବେ କୋଣତ୍ରୟର ପରିମାଣ ଗୋଟିଏ A.P. ଗଠନ କରିବେ ।
(g) ଗୋଟିଏ ସମକୋଣୀ ତ୍ରିଭୁଜର ବାହୁତ୍ରୟର ଦୈର୍ଘ୍ୟ ଗୋଟିଏ A.P.ରେ ରହିପାରିବେ ।
(h) ଅଯୁଗ୍ମ ସଂଖ୍ୟାମାନେ A.P. ଗଠନ କରନ୍ତି ନାହିଁ ।
(i) 5 ଦ୍ବାରା ବିଭାଜ୍ୟ ସମସ୍ତ ଗଣନ ସଂଖ୍ୟା ଏକ A.P. ଅଟନ୍ତି ।
(j)5, x, ୨ ସଂଖ୍ୟାତ୍ରୟ ସମାନ୍ତର ପ୍ରଗତିରେ ରହିଲେ x = 6 I
ଉ –
(a) ଠିକ୍, (b) ଭୁଲ୍, (c) ଭୁଲ୍, (d) ଠିକ୍, (e) ଠିକ୍, (f) ଭୁଲ୍, (g) ଭୁଲ୍, (h) ଭୁଲ୍, (i) ଠିକ୍, (j) ଭୁଲ

ବ୍ୟାଖ୍ୟା ସହ ଉତ୍ତର :
(a) 1, 2, 3, 4 ………. ସମାନ୍ତର ପ୍ରଗତି ସୃଷ୍ଟି କରନ୍ତି । (ଠିକ୍)
(କାରଣ ଏହି ଅନୁକ୍ରମର ସାଧାରଣ ଅନ୍ତର (d) = 1)
(b) 1, – 1, 1, – 1, ……. ଅନୁକ୍ରମଟି ସମାନ୍ତର ପ୍ରଗତି ଅଟେ । (ଭୁଲ୍)
(କାରଣ ଏହି ଅନୁକ୍ରମର ଓଁ ସମାନ ନୁହେଁ ।)
(c) 2, 1, – 1, – 2 ସଂଖ୍ୟାଟି ଚାରିଗୋଟି ସମାନ୍ତର ପ୍ରଗତିରେ ବିଦ୍ୟମାନ । (ଭୁଲ୍)
(କାରଣ 1 – 2 = -1, -1 – 1 = -2 ଏଠାରେ d ଅସମୀନ)
(d) t = n – 1 ଏକ A.P. ଅଟେ । (ଠିକ୍)
(କାରଣ ଅନୁକ୍ରମର ସାଧାରଣ ପଦ t_n = n – 1)
(e) ଯେଉଁ ଅନୁକ୍ରମର S_n = \(\frac{n(n-1)}{2}\) ତାହା A.P. ଅଟେ । (ଠିକ୍)
(f) ଯଦି କୌଣସି ତ୍ରିଭୁଜର କୋଣ ତ୍ରୟର ପରିମାଣର ଅନୁପାତ 2 : 3 : 4 ହୁଏ, ତେବେ କୋଣତ୍ରୟର ପରିମାଣ ଗୋଟିଏ A.P. ଗଠନ କରିବେ । (ଠିକ୍)
(କାରଣ କୋଣତ୍ରୟ 40°, 60°, 80° ଏହା A.P. ଅଟେ ।)
(g) ଗୋଟିଏ ସମକୋଣୀ ତ୍ରିଭୁଜର ବାହୁତ୍ରୟର ଦୈର୍ଘ୍ୟ ଗୋଟିଏ A.P.ରେ ରହିପାରିବେ । (ଭୁଲ୍)
(କାରଣ 3, 4, 5; 6, 8, 10 ଇତ୍ୟାଦି ସମକୋଣୀ ତ୍ରିଭୁଜର ବାହୁ; କିନ୍ତୁ ଯେକୌଣସି ସମକୋଣୀ ତ୍ରିଭୁଜ ପାଇଁ ଠିକ୍ ନୁହେଁ ।)
(h) ଅଯୁଗ୍ମ ସଂଖ୍ୟାମାନେ A.P. ଗଠନ କରନ୍ତି ନାହିଁ । (ଭୁଲ୍)
(3, 5, 7, 9 A.P. ଗଠନ କରନ୍ତି ।)
(i) 5 ଦ୍ବାରା ବିଭାଜ୍ୟ ସମସ୍ତ ଗଣନ ସଂଖ୍ୟା ଏକ A.P. ଅଟନ୍ତି । (ଠିକ୍)
(କାରଣ 5, 10, 15, 20 A.P. ଅଟନ୍ତି ।)
(j) 5, x, ୨ A.P. ରେ ରହିଲେ x = 6 । (ଭୁଲ୍) ( କାରଣ x = \(\frac{9+5}{2}=\frac{14}{2}=7\)

‘ଖ’ ବିଭାଗ

Question 8.
(a) 1 + 2 + 3 + ….. ରେ S₃₀ କେତେ ?
(b) 1 + 3 + 5 + …….. ରେ S₁₀ କେତେ ?
(c) 2 + 4 + 6 + …….ରେ S₁₅ କେତେ ?
(d) 1 – 2 + 3 – 4 + ….. ରେ S₃₀ କେତେ ?
(e) 1 – 2 + 3 – 4 + …….. ରେ S₄₁ କେତେ ?
(f) 1 + 1 + 2 + 2 + 3 + 3 ……… ରେ S₁₇ କେତେ ?
(g) 1 + 2 + 3 + 2 + 3 + 4 + 3 + 4 + 5 ….. ରେ S₃₉ କେତେ ?
(h) -7 – 10 – 13 – ……… ରେ S₂₁ କେତେ ?
(i) 10 + 6 + 2 + ……. ରେ S₁₅ କେତେ ?
(j) 20 + 9 – 2 + …….. ରେ S₂₅ କେତେ ?
(k) n + (n – 1) + (n – 2) + ……… ରେ S_n କେତେ ?
(l) \(5+4 \frac{1}{3}+3 \frac{2}{3}\) + ……… ରେ S₂₀ କେତେ ?
ସମାଧାନ :
(a) S_n = \(\frac{n}{2}\) {2a + (n – 1) d}
ଏଠାରେ a = 1, d= 3 – 2 = 2 – 1 = 18 ଓ n = 30
S_n = \(\frac{n}{2}\) {2a + (n – 1) d}
S₃₀ = \(\frac{30}{2}\) {2.1 + (30 – 1) .1} = 15(2 + 29) = 15 × 31 = 465

(b) ଏଠାରେ a = 1, d= 3 – 1 = 5 – 3 = 2, n = 10
S_n = \(\frac{n}{2}\) {2a + (n – 1) d}
⇒ S₁₀ = \(\frac{10}{2}\) {2.1 + (10 – 1) 2} = 15(2 + 29) = 5 × 20 = 100

(c) ଏଠାରେ a = 2, d= 4 – 2 = 6 – 4 = 2, n = 15
S₁₅ = \(\frac{15}{2}\) {2 × 2 + (15 – 1) 2} = \(\frac{15}{2}\)(4 + 28) = 15 × 16 = 240

(d) 1 – 2 + 3 – 4 + ……… S₃₀
ଦତ୍ତ ଅନୁକ୍ରମଟି A.P.ରେ ନାହିଁ ।
ଦୁଇଟି ପଦକୁ ଗୋଟିଏ ପଦରେ ପରିଣତ କଲେ ଏହା ଅନୁକ୍ରମ ହେବ ଓଏହା S₁₅ ହେବ ।
= (1 – 2) + (3 – 4) + (5 – 6) + ……… (29 – 30)
= (- 1) + (- 1) + (- 1) + ………. = -1 × 15 = -15

(e) 1 – 2 + 3 – 4 + ……… ରେ S₄₁
ପଦସଂଖା 41 ହେତୁ ଅନୁକ୍ରମଟି
(1 – 2) + (3 – 4) + ……. (39 – 40) + 41
= ( – 1 ) + ( – 1) + (- 1) + ……. S₂₀ + 41 = -20 + 41 = 21

(f) 1 + 1 + 2 + 2 + 3 + 3 ……… ରେ S₁₇ ନିର୍ଣ୍ଣୟ କରାଯିବ ।
= (1 + 2 + 3 + ……. + 9) + (1 + 2 + 3 + ……. + 8)
= \(\frac{9×10}{2}+\frac{8×9}{2}\) (∵ Sn = \(\frac{n(n-1)}{2}\))
= 9 × 5 + 4 × 9 = 45 + 36 = 81

(g) 1 + 2 + 3 + 2 + 3 + 4 + 3 + 4 + 5 ….. ରେ S₃₉ କେତେ ?
ତିନୋଟି ପଦକୁ ଗୋଟିଏ ପଦରେ ପରିଣତ କଲେ, 39ଟି ପଦ \(\frac{39}{3}\) = 13ଟି ପଦ ହେବ ।
(1 + 2 + 3) + (2 + 3 + 4) + (3 + 4 +5) ………. ରେ S₁₃
6 + 9 + 12 + ……. ରେ S₁₃
= \(\frac{13}{2}\) {2 × 6 + (13 – 1) 3} = \(\frac{13}{2}\) {12 + 36} = \(\frac{13}{2}\) × 48 = 13 × 24 = 312

(h) S_n = \(\frac{n}{2}\) {2a + (n – 1) d}
ଏଠାରେ a = – 7, d = -10 – (-7) = – 10 + 7 = – 3, n = 21
S₂₁ = \(\frac{21}{2}\) {2 · (- 7) + (21 – 1) (- 3)}
= \(\frac{21}{2}\) { – 14 – 60} = = \(\frac{21}{2}\) { -74) = 21 × (-37) =-777

(i) ଏଠାରେ a = 10, d = 6 – 10 = 2 – 6 = – 4, n = 15
S₁₅ = \(\frac{15}{2}\) {2 × 10 + (15 – 1) (- 4)}
= \(\frac{15}{2}\) {20+ (-56)} = \(\frac{15}{2}\) × -36 = 15 × (-18) = -270

(j) S_n = \(\frac{n}{2}\) {2a + (n – 1) d}
ଏଠାରେ a = 20, d = 9 – 20 = -2 – 9 = – 11, n = 25
S₂₅ = \(\frac{25}{2}\) {2 × 20+ (25 – 1) (- 11)}
= \(\frac{25}{2}\) {40 + 24 × (-11)} = \(\frac{25}{2}\) (40 – 264)
= \(\frac{25}{2}\) × – 224 = 25 × (- 112) = – 2800

(k) ଏଠାରେ a = n, d = (n – 1) – n = -1, n = n
S_n = \(\frac{n}{2}\) {2a + (n – 1) d} = \(\frac{n}{2}\) {2n + (n – 1)(-1)}
=\(\frac{n}{2}\) (2n – n + 1) = \(\frac{n}{2}\) (n + 1)

(l) ଏଠାରେ a = 5, d = \(4 \frac{1}{3}-5=\frac{13}{3}-5=\frac{13-15}{3}=-\frac{2}{3}\), n = 20
S₂₀ = \(\frac{20}{2}\) {2 × 5 + (20 – 1) (-3)} [∵ S_n = \(\frac{n}{2}\) {2a + (n – 1) d}]
= 10 {10 – \(\frac{38}{3}\)} = 10 (\(\frac{30-38}{3}\)) = 10 × \(\frac{-8}{3}\) = \(\frac{-80}{3}\) = \(-26 \frac{2}{3}\)

Question 9.
(a) ଯଦି a = 3, d = 4, n = 10, ତେବେ S_n କେତେ ?
(b) ଯଦି a = – 5, d = – 3, ତେବେ S₁₇ କେତେ ?
(c) ଯଦି t_n = 2n – 1, ତେବେ ପ୍ରଥମ 5ଟି ପଦ ଲେଖ ।
(d) ଯଦି t_n = 3n + 2, S₆₁ ନିର୍ଣ୍ଣୟ କର ।
(e) ଯଦି t_n = 3n – 5, ତେବେ S₅₀ ନିର୍ଣ୍ଣୟ କର ।
(f) ଯଦି t_n = 2 – 3n, ତେବେ S_n ନିର୍ଣ୍ଣୟ କର ।
(g) ଯଦି S_n = n², ତେବେ t₁₅ କେତେ ?
(h) ଏକ A.P.ର a = 3, d = 4, S_n = 903, ତେବେ n କେତେ?
(i) ଏକ A.P. ର d = 2, S₁₅ = 285, ତେବେ a କେତେ?
(j) ଏକ A.P. ର t₁₅ = 30, t₂₀ = 50, ତେବେ S₁₇ କେତେ?
ସମାଧାନ :
MBD
(a) S_n = \(\frac{n}{2}\) {2a + (n – 1) d}
ଏଠାରେ a = 5, d = 4, n = 10
S₁₀ = \(\frac{10}{2}\) {2 × 3 + (10 – 1) 4} = 5 {6 + 36} = 5 × 42 = 210

(b) S_n = \(\frac{n}{2}\) {2a + (n – 1) d}
ଏଠାରେ a = -5, d = -3, n = 17
S₁₇ = \(\frac{17}{2}\) {2 × (-5) + (17 – 1) (-3)} = \(\frac{17}{2}\) {-10 – 48} = \(\frac{17}{2}\) × (-58) = 17 × (-29) = – 493

(c) t_n = 2n – 1,
t₁ = a = 2 × 1 – 1 = 2 – 1 = 1
t₂ = a = 2 × 2 – 1 = 4 – 1 = 3,
t₃ = a = 2 × 3 – 1 = 6 – 1 = 5,
t₄ = a = 2 × 4 – 1 = 8 – 1 = 7,
t₅ = a = 2 × 5 – 1 = 10 – 1 = 9,

(d) t_n = 3n + 2
t₁ = a = 3 × 1 + 2 = 5
t₂ = a = 3 × 2 + 2 = 8
t₃ = a = 3 × 3 + 2 = 11
ଏଠାରେ a = 5, d = 8 – 5 = 11 – 8 = 3, n = 61
S_n = \(\frac{n}{2}\) {2a + (n – 1) d}
S₆₁ = \(\frac{61}{2}\) {2 × 5+ (61 – 1) 3} = \(\frac{61}{2}\) {10 + 180} = 61 × 95 = 5795

(e) t_n = 3n – 5
t₁ = 3 × 1 – 5 = -2
t₂ = 3 × 2 – 5 = 1
t₂ = 3 × 3 – 5 = 4
a = -2, d = 1 – (- 2) = 4 – 1 = 3, n = 50
S₅₀ = \(\frac{50}{2}\) {2 × (-2) + (50 – 1) 3} = 25 × (- 4 + 147) = 25 × 143 = 3575

(f) t_n = 2 – 3n
t₁ = 2 – 3 × 1 = – 1
t₂ = 2 – 3 × 2 = -4
t₃ = 2 – 3 × 3 = -7 ଇତ୍ୟାଦି
ଏଠାରେ a = -1, d= – 4 – (- 1) = -7 – (- 4) = -3
S_n = \(\frac{n}{2}\) {2a + (n – 1) d} = \(\frac{n}{2}\) (2 × (- 1) + (n – 1) (- 3)}
= \(\frac{n}{2}\) (-2 – 3n + 3} = \(\frac{n}{2}\) (1 – 3n)

(g) t_n = S_n – S_n-1
S_n = n², S₁₅ = 15², S₁₄ = 14²
t₁₅ = S₁₅ – S₁₄ = 15² – 14² = 225 – 196 = 29

(h) ଏକ A.P.ର a = 3, d = 4, S_n = 903

∴ n = 2

(i) A.P.ର d = 2, S₁₅ = 285, n = 15

(j) ମନେକର A.P.ର ପ୍ରଥମ ପଦ = a ଓ ସାଧାରଣ ଅନ୍ତର = d
t₁₅ = 30 ⇒ a + (15 – 1) d = 30 ⇒ a + 14d = 30 …(i)
t₂₀ = 50 ⇒ a + (20 – 1) d = 50 ⇒ a + 19d = 50 …(ii)
ସମୀକରଣ (i)ରୁ (ii)କୁ ବିୟୋଗ କଲେ

d ର ମାନ ସମୀକରଣ (i)ରେ ପ୍ରୟୋଗ କଲେ
a + 14d = 30 = a + 14 × 4 = 30
= a + 56 = 30 = a = 30 – 56 = -26
S₁₇ = \(\frac{17}{2}\) {-52 + (17 – 1)4} = \(\frac{17}{2}\) {-52 + 64}= \(\frac{17}{2}\) × 12 = 17 × 6 = 102

Question 10.
(i) ‘ଓଲଟାଇ ମିଶାଇବା କୌଶଳରେ’ ଯୋଗଫଳ ନିର୍ଣ୍ଣୟ କର ।
(a) 1 ଠାରୁ 105 ପର୍ଯ୍ୟନ୍ତ ସମସ୍ତ ଗଣନ ସଂଖ୍ୟା ।
(b) 25 ଠାରୁ 93 ପର୍ଯ୍ୟନ୍ତ ସମସ୍ତ ଗଣନ ସଂଖ୍ୟା ।
(c) 111 ଠାରୁ 222 ପର୍ଯ୍ୟନ୍ତ ସମସ୍ତ ଗଣନ ସଂଖ୍ୟା ।

(ii) 1, 2, 3, …….. ଅନୁକ୍ରମର।
(a) S₂₀ ନିଶ୍ଚୟ କର (b) S₅₀ ନିର୍ଣ୍ଣୟ କର ।
(iii) 32 ଠାରୁ 85 ପର୍ଯ୍ୟନ୍ତ ସମସ୍ତ ଗଣନ ସଂଖ୍ୟାର ସମଷ୍ଟି ନିର୍ଣ୍ଣୟ କର ।
(iv) 100 ଠାରୁ କ୍ଷୁଦ୍ରତର ସମସ୍ତ ଧନାତ୍ମକ ଯୁଗ୍ମ ସଂଖ୍ୟାର ସମଷ୍ଟି ନିର୍ଣ୍ଣୟ କର ।
(v) 150 ଠାରୁ କ୍ଷୁଦ୍ରତର ସମସ୍ତ ଧନାତ୍ମକ ଅଯୁଗ୍ମ ସଂଖ୍ୟାର ସମଷ୍ଟି ନିର୍ଣ୍ଣୟ କର ।
ସମାଧାନ :
(i) (a) 1 ଠାରୁ 105 ପର୍ଯ୍ୟନ୍ତ ଗଣନ ସଂଖ୍ୟାମାନଙ୍କର ମୋଟ ସଂଖ୍ୟା = 105
ମନେକର 1 ଠାରୁ 105 ପର୍ଯ୍ୟନ୍ତ ଗଣନ ସଂଖ୍ୟାମାନଙ୍କର ଯୋଗଫଳ = S

∴ 1 ଠାରୁ 105 ପର୍ଯ୍ୟନ୍ତ ଗଣନ ସଂଖ୍ୟାର ଯୋଗଫଳ = 5565

(b) 25 ଠାରୁ 93 ପର୍ଯ୍ୟନ୍ତ ଗଣନ ସଂଖ୍ୟାମାନଙ୍କର ମୋଟ ସଂଖ୍ୟା = 93 – 24 = 69
ମନେକର 25 ଠାରୁ 93 ପର୍ଯ୍ୟନ୍ତ ଗଣନ ସଂଖ୍ୟାମାନଙ୍କର ଯୋଗଫଳ = S

∴ 25 ଠାରୁ 93 ପର୍ଯ୍ୟନ୍ତ ଗଣନ ସଂଖ୍ୟାର ସମଷ୍ଟି = 4071

(c) 111 ଠାରୁ 222 ପର୍ଯ୍ୟନ୍ତ ଗଣନ ସଂଖ୍ୟାମାନଙ୍କର ସଂଖ୍ୟା = 222 – 110 = 112
ମନେକର 111 ଠାରୁ 222 ପର୍ଯ୍ୟନ୍ତ ଗଣନ ସଂଖ୍ୟାମାନଙ୍କର ଯୋଗଫଳ = S

⇒ 2S = 333 × 112 ⇒ S = \(\frac{333×112}{2}\) = 333 × 56 = 18648
∴ ମନେକର 111 ଠାରୁ 222 ପର୍ଯ୍ୟନ୍ତ ଗଣନ ସଂଖ୍ୟାର ସମଷ୍ଟି = 18648

(ii) S_n = \(\frac{n(n-1)}{2}\)

(a) 1, 2, 3 ଅନୁକ୍ରମର S₂₀ = \(\frac{20×21}{2}\) = 10 × 21 = 210

(b) S₅₀ = \(\frac{50×51}{2}\) = 25 × 51 = 1275

(iii) S = 32 +33 +34 …….. + 85
ଏଠାରେ a = 32, d = 33 – 32 = 1, n = 85 – 31 = 54
S_n = \(\frac{n}{2}\) {2a + (n – 1) d} = \(\frac{54}{2}\) {2 × 32 + (54 – 1) × 1}
= 27 × {64 +53} = 27 × 117 = 3159
∴ ମନେକର 111 ଠାରୁ 222 ପର୍ଯ୍ୟନ୍ତ ଗଣନ ସଂଖ୍ୟାର ସମଷ୍ଟି 3159।

(iv) S_n = \(\frac{n(n-1)}{2}\)
100 ଠାରୁ କ୍ଷୁଦ୍ରତର ସମସ୍ତ ଧନାତ୍ମକ ଯୁଗ୍ମ ସଂଖ୍ୟାମାନଙ୍କର ସମଷ୍ଟି
S = 2 + 4 + 6 + 8 …… + 98
= 2 (1 + 2 + 3 + 4 ……. + 49)
= 2 × \(\frac{49 (49 + 1)}{2}\) = \(\frac{49×50}{2}\) = 49 × 50 = 2450
∴ 100 ଠାରୁ କ୍ଷୁଦ୍ରତର ସମସ୍ତ ଧନାତ୍ମକ ସଂଖ୍ୟାମାନଙ୍କର ସମଷ୍ଟି = 2450

(v) 150 ଠାରୁ କ୍ଷୁଦ୍ରତର ସମସ୍ତ ଅଯୁଗ୍ମ ସଂଖ୍ୟାମାନଙ୍କର ସମଷ୍ଟି
S = 1 + 3 + 5 + 7 ……. + 149
ଏଠାରେ a = 1, d = 3 – 1 = 5 – 3 = 2
t_n = 149
⇒ a + (n – 1) d = 149 ⇒ 1 + (n – 1) 2 = 149
⇒ n – 1 = \(\frac{149 – 1}{2}=\frac{148}{2}=74\) ⇒ n = 74 + 1 = 75
S_n = \(\frac{n}{2}\) {2a + (n – 1) d} = \(\frac{75}{2}\) {2 × 1 + (75 – 1) 2}
= \(\frac{75}{2}\) {2 + 148} = \(\frac{75×150}{2}\)
= 75 × 75 = 5625
∴ 150 ଠାରୁ କ୍ଷୁଦ୍ରତର ସମସ୍ତ ଧନାତ୍ମକ ଅଯୁଗ୍ମ ସଂଖ୍ୟାମାନଙ୍କର ସମଷ୍ଟି 5625 ।

Question 11.
ଯେଉଁ ସମାନ୍ତର ଅନୁକ୍ରମର ପ୍ରଥମ ପଦ 17 ଓ ସାଧାରଣ ଅନ୍ତର – 2 ତାହାର କେତୋଟି ପଦର ସମଷ୍ଟି 72 ହେବ ? ଏହାର ଦୁଇଟି ଉତ୍ତର ମିଳିବାର କାରଣ ଲେଖ ।
ସମାଧାନ :
ଏକ ଅନୁକ୍ରମର ପ୍ରଥମ ପଦ (a) = 17, ସାଧାରଣ ଅନ୍ତର (d) = – 2
ମନେକର nଟି ପଦର ସମଷ୍ଟି 72 ।
⇒ S_n = 72 ⇒ \(\frac{n}{2}\) {2a + (n – 1) d} = 72
⇒ \(\frac{n}{2}\) {2 × 17 + (n – 1) × (- 2)} = 72 ⇒ \(\frac{n}{2}\) {34 – 2n + 2} = 72
⇒ \(\frac{n}{2}\) {36 – 2n } = 72 ⇒ \(\frac{n}{2}\) × 2 (18 – 2n) = 72
⇒ 18n – 2n² – 72 = 0 ⇒ -2 (n² – 18n + 72) = 0
⇒ n² – 18n + 72 = 0 ⇒ n² – 12n – 6n + 72 = 0
⇒ n (n – 12) – 6 (n – 12) = 0 ⇒ (n – 6) (n – 12) = 0
n = 6 ବା n = 12
t₇ = a + (7 – 1) d = 17 + 6 (- 2) = 17 – 12 = 5
t₈ = a + (8 – 1) d = 17 + 7 (- 2) = 17 – 14 = 3
t₉ = a + (9 – 1) d = 17 + 8 (- 2) = 17 – 16 = 1
t₁₀ = a + (10 – 1) d = 17 + 9 (- 2) = 17 – 18 = – 1
t₁₁ = a + (11 – 1) d = 17 + 10 (- 2) = 17 – 20 = -3
t₁₂ = a + (12 – 1) d = 17 + 11 (- 2) = 17 – 22 = -5
∴ t₇ + t₁₁ ………. + t₁₂ = (5 + 3 + 1) – (1 + 3 + 5) = 0
∴ ସପ୍ତମ ପଦରୁ ଦ୍ଵାଦଶ ତମ ପଦ ପର୍ଯ୍ୟନ୍ତ ସଂଖ୍ୟାଗୁଡ଼ିକର ଯୋଗଫଳ 0 ହୋଇଥିବାରୁ
ପ୍ରଥମ 6ଟି ପଦର ସମଷ୍ଟି = ପ୍ରଥମ 12 ଟି ପଦର ସମଷ୍ଟି
ତେଣୁ ଆମେ ଦୁଇଟି ଉତ୍ତର ପାଇଲୁ ।

Question 12.
(i) ଏକ ସମାନ୍ତର ଅନୁକ୍ରମରେ ଅବସ୍ଥିତ ତିନୋଟି ରାଶିର ଯୋଗଫଳ 18 ଏବଂ ଗୁଣଫଳ 192 ହେଲେ, ସଂଖ୍ୟାଗୁଡ଼ିକ ସ୍ଥିର କର ।
(ii) ଏକ ସମାନ୍ତର ଅନୁକ୍ରମରେ ଅବସ୍ଥିତ ଛଅଟି ପଦ ମଧ୍ୟରୁ ପ୍ରାନ୍ତ ପଦଦ୍ୱୟର ଯୋଗଫଳ 16 ଏବଂ ମଧ୍ୟ ପଦଦ୍ୱୟର ଗୁଣଫଳ 63 ହେଲେ, ପଦଗୁଡ଼ିକ ସ୍ଥିର କର ।
ସମାଧାନ :
(i) ମନେକର ସମାନ୍ତର ଅନୁକ୍ରମର ଥ‌ିବା ପଦତ୍ରୟ a – d, a, a + d ।
ପ୍ରଶ୍ନନୁସାରେ ପଦତ୍ରୟର ଯୋଗଫଳ = 18
⇒ a – d + a + a + d = 18 ⇒ 3a = 18 ⇒ a = \(\frac{18}{3}\) ⇒ a = 6.
ପୁନଶ୍ଚ (a – d) × a (a + d) = 192 ⇒ a (a² – d²) = 192
⇒ 6 {(6)² – d²} = 192 ⇒ 36 – d² = \(\frac{192}{6}\) = 32
⇒ d² = 32 – 36 = -4 ⇒ d² = 4 ⇒ d= ±√4⇒ d = ±2
a = 6 ଓ d = 2 ହେଲେ A.P. ର ପଦତ୍ରୟ a – d = 6 – 2 = 4
a = 6 ଏବଂ a + d = 6 + 2 = 8
a = 6 ଓ d = – 2 ହେଲେ A.P.ର ପଦତ୍ରୟ a – d = 6 – (-2) = 6 + 2 = 8
a = 6, a – d = 6 + (- 2) = 6 – 2 = 4
∴ A.P.ର ପଦତ୍ରୟ 4, 6, 8 ବା 8, 6, 4 ।

(ii) ମନେକର ସମାନ୍ତର ଅନୁକ୍ରମର ଥ‌ିବା ଛଅଟି ପଦ a – 5d, a – 3d, a – d, a + d, a + 3d, a + 5d ।
ପ୍ରଶ୍ନନୁସାରେ, a – 5d + a + 5d = 16 ⇒ a = \(\frac{16}{2}\) = 8
ପୁନଶ୍ଚ (a – d) × (a + d) = 63
⇒ (a² – d²) = 63 ⇒ 64 – d² = 63
⇒ d² = 1 ⇒ d = ± √1 = ±1
a = 8 ଓ d = 1 ହେଲେ ପଦଗୁଡ଼ିକ a – 5d = 8 – 5 = 3, a – 3d = 8 – 3 = 5,
a – d = 8 – 1 = 7, a + d = 8 + 1 = 9, a + 3d = 8 + 3 = 11, a + 5d = 8+ 5 = 13,
a = 8 ଓ d = -1 ହେଲେ ପଦଗୁଡ଼ିକ a – 5d = 8 – 5 (-1) = 8 + 5 = 13, a – 3d = 8 – 3 (- 1)
= 8 + 3 = 11, a – d = 8 – (- 1) = 9
a + d = 8+ (-1) = 7, a + 3d = 8 + 3 (- 1) = 8 – 3 = 5
a + 5d = 8 +5(-1) = 8 – 5 = 3
∴ A.P.ର ପଦତ୍ରୟ 3, 5, 7, 9, 11, 13 ବା 13, 11, 9, 7, 5, 3 ।

Question 13.
ଏକ ସମାନ୍ତର ଅନୁକ୍ରମରେ ଅବସ୍ଥିତ ତିନୋଟି ପଦର ଯୋଗଫଳ 21 ଏବଂ ସେମାନଙ୍କ ବର୍ଗର ଯୋଗଫଳ 155; ପଦଗୁଡ଼ିକ କେତେ ?
ସମାଧାନ :
ମନେକର ସମାନ୍ତର ଅନୁକ୍ରମର ଥ‌ିବା ତିନୋଟି ପଦ a + d, a, a + 3d ।
ପ୍ରଶ୍ନନୁସାରେ, a – d + a + a + d = 21 ⇒ 3a = 21 ⇒ a = \(\frac{21}{3}\) = 7
ପୁନଶ୍ଚ a (a – d)² + a² + (a + d)² = 155 ⇒ a² + (a – d)² + (a + d)² = 155
⇒ a² + 2 (a² + d²) = 155 ⇒ 7² + 2 (7² + d²) = 155
⇒ 2 (49 + d²) = 155 – 49 = 106 ⇒ 98 + 2d² = 106
⇒2d² = 106 – 98 ⇒ d² = \(\frac{8}{2}\) = 4 ⇒ d = = ±√4 = ±2
a = 7 ଓ d = 2 ହେଲେ ପଦଗୁଡ଼ିକ a – d = 7 – 2 = 5, a = 7, a + d = 7 + 2 = 9
a = 7 ଓ d = -2 ହେଲେ ପଦଗୁଡ଼ିକ a – d = 7 – (- 2) = 7 + 2 = 9,
a = 7, a + d = 7 + (- 2) = 7 – 2 = 5
∴ ସମାନ୍ତର ଅନୁକ୍ରମର ପଦତ୍ରୟ 5, 7, 9 ବା 9, 7, 5 ।

Question 14.
ଗୋଟିଏ ସମକୋଣୀ ତ୍ରିଭୁଜର ବାହୁଗୁଡ଼ିକର ଦୈର୍ଘ୍ୟ ଏକ ସମାନ୍ତର ଅନୁକ୍ରମରେ ଥିଲେ ପ୍ରମାଣ କର ଯେ, ସେମାନଙ୍କର ଅନୁପାତ 3 : 4 : 5 ହେବ ।
ସମାଧାନ :
ଏକ ସମକୋଣୀ ତ୍ରିଭୁଜର ବାହୁତ୍ରୟ ସମାନ୍ତର ଅନୁକ୍ରମରେ ଅବସ୍ଥିତ ।
ମନେକର ବାହୁତ୍ରୟର ଦୈର୍ଘ୍ୟ a – d, a, a + d ।
ପ୍ରଶ୍ନନୁସାରେ, (a – d)² + a² = (a + d)² ⇒ (a + d)² – (a – d)² = a²
⇒ (a + d + a + d) (a + d – a + d) = a² ⇒ 2 × 2d = a² ⇒ a = 4d
⇒ ସମକୋଣୀ ତ୍ରିଭୁଜର ବାହୁତ୍ରୟର ଦୈର୍ଘ୍ୟ, a – d = 4d – d = 3d, a = 4d,
a + d = 4d + d = 5d
ବାହୁଗୁଡ଼ିକର ଦୈର୍ଘ୍ୟ ଅନୁପାତ = 3d : 4d : 5d = 3 : 4 : 5 (ପ୍ରମାଣିତ)

Question 15.
100 ରୁ କ୍ଷୁଦ୍ରତର ଏବଂ 5 ଦ୍ୱାରା ବିଭାଜ୍ୟ ସମସ୍ତ ଧନାତ୍ମକ ପୂର୍ଣ୍ଣସଂଖ୍ୟାମାନଙ୍କର ଯୋଗଫଳ ନିର୍ଣ୍ଣୟ କର ।
ସମାଧାନ :
ମନେକର 100 ଠାରୁ କ୍ଷୁଦ୍ରତର ଏବଂ 5 ଦ୍ବାରା ବିଭାଜ୍ୟ ସମସ୍ତ ଧନାତ୍ମକ ପୂର୍ଣ୍ଣ ସଂଖ୍ୟାର ଯୋଗଫଳ = S
S = 5 + 10 + 15 ……… + 95
ଏଠାରେ a = 5, d = 5, n = \(\frac{95}{5}\) = 19
S_n = \(\frac{n}{2}\) {2a + (n – 1) d}
= \(\frac{19}{2}\) {2 × 5+ (19 – 1) 5} = \(\frac{19}{2}\) (10 + 90) = \(\frac{19}{2}\) × 100 = 19 × 50 = 950
∴ ନିର୍ଦେୟ ଯୋଗଫଳ = 950

Question 16.
200 ରୁ କ୍ଷୁଦ୍ରତର ଓ 3 ଦ୍ଵାରା ଅବିଭାଜ୍ୟ ସମସ୍ତ ଧନାତ୍ମକ ପୂର୍ବସଂଖ୍ୟାମାନଙ୍କର ଯୋଗଫଳ ନିର୍ଣ୍ଣୟ କର ।
ସମାଧାନ :
ମନେକର 1 ଠାରୁ ଆରମ୍ଭ କରି 200ରୁ କ୍ଷୁଦ୍ରତର ସମସ୍ତ ଧନାତ୍ମକ ପୂର୍ଣ୍ଣ ସଂଖ୍ୟାର ଯୋଗଫଳ = S₁
S₁ = 1 + 2 + 3 + 4 ……. + 199 = \(\frac{199×200}{2}\) = 199 × 100 = 19900
ପୁନଶ୍ଚ 200 ଠାରୁ କ୍ଷୁଦ୍ରତର 3 ଦ୍ଵାରା ବିଭାଜ୍ୟ ସମସ୍ତ ଧନାତ୍ମକ ପୂର୍ଣ୍ଣ ସଂଖ୍ୟାର ଯୋଗଫଳ
S₂ = 3 + 6 + 9 ………. + 198
ଏଠାରେ a = 3, d = 6-3 = 3, n = \(\frac{198}{3}\) = 66
S₂ = \(\frac{n}{2}\) {2a + (n – 1) d} = \(\frac{66}{2}\) {2 × 3 + (66 – 1) × 3}
= 33 {6 + 65 × 3) = 33 × (6 + 195) = 33 × 201 = 6633
200 ଠାରୁ କ୍ଷୁଦ୍ରତର 3 ଦ୍ଵାରା ଅବିଭାଜ୍ୟ ସମସ୍ତ ଧନାତ୍ମକ ପୂର୍ଣ୍ଣ ସଂଖ୍ୟାର ଯୋଗଫଳ
= S = S₁ – S₂ = 19900 – 6633 = 13267

Question 17.
15 କୁ ଏପରି 3 ଭାଗରେ ବିଭକ୍ତ କର ଯେପରିକି ସେମାନେ ଏକ ସମାନ୍ତର ଅନୁକ୍ରମରେ ରହିବେ ଓ ସେମାନଙ୍କର ଗୁଣଫଳ 120 ହେବ ।
ସମାଧାନ :
15କୁ ଏପରି ତିନି ଭାଗରେ ବିଭକ୍ତ କରାଯିବ ଯେପରି ସେମାନେ A.P.ରେ ରହିବେ ।
ମନେକର ସଂଖ୍ୟା ତ୍ରୟ a – d, a, a + d ।
ପ୍ରଶାନୁସାରେ, a – d + a + a + d = 15 ⇒ 3a = 15 ⇒ a = 5
ପୁନଶ୍ଚ (a – d) a (a + d) = 120
⇒ 5 (5² – d²) = 120 ⇒ 25 – d² = \(\frac{120}{5}\) ⇒ d² = 25 – 24
⇒ d² = 1 = d = +1
a=5 ଓ d = 1 ହେଲେ a – d = 5 – 1 = 4, a = 5, a + d = 5 + 1 = 6
a = 5 ଓ d = -1 ହେଲେ a – d = 5 – (- 1) = 5 + 1 = 6, a= 5,
a + d = 5 + (- 1) = 5 – 1= 4
∴ ସଂଖ୍ୟାତ୍ରୟ 4, 5, 6 ବା 6, 5,4 ।

Question 18.
A.P. ରେ ଥ‌ିବା ତିନୋଟି ପଦର ଯୋଗଫଳ 15 ଏବଂ ପ୍ରାନ୍ତ ପଦଦ୍ୱୟର ବର୍ଗର ଯୋଗଫଳ 58 ହେଲେ ପଦତ୍ରୟ ନିର୍ଣ୍ଣୟ କର ।
ସମାଧାନ :
ମନେକର A.P. ରେ ଥ‌ିବା ପଦ ତିନୋଟି a – d, a, a + d ।
ପ୍ରଶ୍ନନୁସାରେ, a – d + a + a + d = 15 ⇒ 3a = 15 ⇒ a = \(\frac{15}{3}\) = 5
ପୁନଶ୍ଚ (a – d)² + (a + d)² = 58 ⇒ 2 (a² + d²) = 58
⇒ 5²+ d² = \(\frac{58}{2}\) ⇒ d² = 29 – 25 = 4 ⇒ d = ± √4 = ±2
a = 5 ଓ d = 2 ହେଲେ ପଦତ୍ରିୟ a – d = 5 – 2 = 3, a = 5, a + d = 5 + 2 = 7 ̧
a = 5 ଓ d = -2 ହେଲେ ପଦତ୍ରିୟ a – d = 5 – (- 2) = 5 + 2 = 7, a = 5,
a + d = 5 + (-2) = 3 ।
∴ A.P.ର ପଦତ୍ରୟ 3, 5, 7 ବା 7, 5, 3 ।

Question 19.
A.P. ରେ ଥିବା ଚାରୋଟି ପଦ ମଧ୍ୟରୁ ପ୍ରାନ୍ତ ପଦ ଦ୍ଵୟର ଯୋଗଫଳ 8 ଏବଂ ମଧ୍ୟ ପଦ ଦ୍ଵୟର ଗୁଣଫଳ 15 ହେଲେ ପଦଗୁଡ଼ିକ ସ୍ଥିର କର ।
ସମାଧାନ :
ମନେକର A.P.ରେ ଥ‌ିବା ଚାରୋଟି ପଦ a – 3d, a – d, a + d ଓ a + 3d ।
ପ୍ରଶ୍ନନୁସାରେ, a – 3d + a + 3d = 8⇒ 2a = 8 ⇒ a = \(\frac{8}{2}\) = 4
ପୁନଶ୍ଚ (a – d) (a + d) = 15 ⇒ a² – d² = 15 ⇒ 4² – 15 = d²
⇒ d² = 16 – 15 = 1 ⇒ d = ±√1 = ±1
a = 4 ଓ d = 1 ହେଲେ ପଦତ୍ରିୟ a – 3d = 4 – 3 (1) = 4 – 3 = 1
a – d = 4 – 1 = 3, a + d = 4 + 1 = 5
a + 3d = 4 + 3 · 1 = 4 + 3 = 7
a =4 ଓ d = 1 ହେଲେ ପଦତ୍ରିୟ a – 3d = 4 – 3 (-1) = 4 + 3 = 7,
a – d = 4 – (-1) = 4 + 1 = 5,
a + d = 4 + (-1) = 3 ଏବଂ a + 3d = 4 + 3 (- 1) = 4 – 3 = 1
∴ A.P.ର ଥ‌ିବା ପଦତ୍ରୟ 1, 3, 5, 7 ବା 7, 5, 3, 1 ।

Question 20.
A.P. ରେ ଥ‌ିବା ତିନୋଟି ରାଶିମାଳାର n ସଂଖ୍ୟକ ପଦମାନଙ୍କର ସମଷ୍ଟି S₁ S₂ ଏବଂ S₃ । ପ୍ରତ୍ୟେକ ରାଶିମାଳାର ପ୍ରଥମ ପଦ 1 ଏବଂ ସାଧାରଣ ଅନ୍ତର ଯଥାକ୍ରମେ 1, 2, 3 ହେଲେ ପ୍ରମାଣ କର ଯେ, S₁ + S₃ = 2S₂ ।
ସମାଧାନ :
A.P. ରେ ଥିବା ତିନୋଟି ରାଶିମାଳାର n ସଂଖ୍ୟକ ପଦମାନଙ୍କର ସମଷ୍ଟି S₁ S₂ ଏବଂ S₃ ।
ପ୍ରତ୍ୟେକ ରାଶିମାଳାର ପ୍ରଥମ ପଦ = a
ପ୍ରଥମ ରାଶିମାଳାର d = 1, ଦ୍ୱିତୀୟ ରାଶିମାଳାର d = 2, ତୃତୀୟ ରାଶିମାଳାର d = 3
S₁ = \(\frac{n}{2}\) {2 ×1 + (n – 1) 1} = \(\frac{n}{2}\) (n + 1)
S₂ = \(\frac{n}{2}\) {2 ×1 + (n – 1) 2} = n²
ଏବଂ S₃ = \(\frac{n}{2}\) {2 ×1 + (n – 1) 3} = \(\frac{n}{2}\) (3n – 1)
L.H.S. = S₁ + S₃ = \(\frac{n}{2}\) (n + 1) + \(\frac{n}{2}\) (3n – 1)
= \(\frac{n}{2}\) {n + 1 + 3n – 1}
R.H.S. = 2S₂ = 2 × n² = 2n²
∴ L.H.S = R.H.S. (ପ୍ରମାଣିତ)

Question 21.
ଏକ A.P. ର p-ତମ, ୟୁ-ତମ ଏବଂ r-ତମ ପଦଗୁଡ଼ିକର ମାନ ଯଥାକ୍ରମେ a, b ଏବଂ c ହେଲେ ପ୍ରମାଣ 6, a (q – r) + b (r -p) + c (p – q) = 0 |
ସମାଧାନ :
ମନେକର A.P. ର ପ୍ରଥମ ପଦ = x, ସାଧାରଣ ଅନ୍ତର = d
t = x + (p – 1) d = a ……… (i)
t = x + (q – 1) d = b ……… (ii)
t = x + (r – 1) d = c …….(iii)
ସମୀକରଣ (i)ରୁ (ii)କୁ ବିୟୋଗ କଲେ,

⇒ d (p – q) = a – b ⇒ d = \(\frac{a – b}{p – q}\) …… (iv)
ସେହିପରି (ii)ରୁ (iii)କୁ ବିୟୋଗ କଲେ ଆମେ ପାଇବା d = \(\frac{b – c}{q – r}\) …… (v)
ସମୀକରଣ (iv) ଓ (v) ରୁ ⇒ \(\frac{a – b}{p – q}=\frac{b – c}{q – r}\) ⇒ (a – b) (q- r) = (b -c) (p – q)
⇒ a (q – r) – b(q – r) = b(p – q) – c (p – q)
⇒ a (q – r) – b(q – r) = b (p – q) + c (p – q) = 0
⇒ a (q – r) – b(q – r + p – q) + c (p – q) = 0
⇒ a (q – r) – b(p – r) + c (p – q) = 0
⇒ a (q – r) – b(r – p) + c (p – q) = 0 (ପ୍ରମାଣିତ)
ବି.ଦ୍ର. : (i) ରୁ a = x + (p – 1)d
∴ a(q – r) = x (q – r) + (p – 1) (q – r)d
ସେହିପରି b(r- p) = x (r – p) + (q – 1) (r – p)d ଏବଂ
c (p – q) = x (p – q) + (r – 1) (p – q) d
ଯୋଗକଲେ a (q – r) + b (r- p) + c (p -q) = 0 ପାଇବା

Question 22.
ତିନୋଟି ସଂଖ୍ୟା a, b, ୯ ସମାନ୍ତର ପ୍ରଗତିରେ ରହିଲେ ପ୍ରମାଣ କର ଯେ ନିମ୍ନରେ ପ୍ରଦତ୍ତ ସଂଖ୍ୟା ତ୍ରୟ ମଧ୍ୟ ସମାନ୍ତର ପ୍ରଗତିରେ ରହିବେ ।
(i) \(\frac{1}{bc},\frac{1}{ca},\frac{1}{ab}\)
(ii) b+c, c + a, a + b
(iii) b+c-a, c + a-b, a + b-c
(iv) \(\frac{1}{a}(\frac{1}{b}+\frac{1}{c}),\frac{1}{b}(\frac{1}{c}+\frac{1}{a}),\frac{1}{c}(\frac{1}{a}+\frac{1}{b})\)
(v) a² (b+c), b² (c + a), c²(a + b)
ସମାଧାନ :
(i) a, b, c ସମାନ୍ତର ପ୍ରଗତିରେ ଅବସ୍ଥିତ ।
ପ୍ରତ୍ୟେକ ପଦକୁ abc ଦ୍ବାରା ଭାଗକଲେ \(\frac{1}{abc},\frac{1}{abc},\frac{1}{abc}\) ସମାନ୍ତର ପ୍ରଗତିରେ ରହିବେ ।
⇒ \(\frac{1}{bc},\frac{1}{ca},\frac{1}{ab}\) (ପ୍ରମାଣିତ)

(ii) a, b, c ସମାନ୍ତର ପ୍ରଗତିରେ ଅବସ୍ଥିତ ।
(a + b + c) ବିୟୋଗ କଲେ ବିୟୋଗଫଳ A.P.ରେ ରହିବ ।
ଅର୍ଥାତ୍ (a+b+c), b – (a + b + c), c (a + b + c) A.P.ରେ ରହିବ ।
⇒ – (b + c), -(c + a), -(a + b) A.P.ରେ ରହିବ ।
⇒ b + c, c + a, a + b ସମାନ୍ତର ପ୍ରଗତିରେ ରହିବେ । (ପ୍ରତ୍ୟେକ ପଦକୁ –1 ଦ୍ଵାରା ଗୁଣିଲେ)

(iii) a, b, c A.P.ରେ ଅବସ୍ଥିତ ।
⇒ 2a, 2b, 2c ମଧ୍ଯ A.P. ରେ ରହିବେ (ପ୍ରତ୍ୟେକ ପଦରେ 2 ଗୁଣନ କଲେ ।)
ପ୍ରତ୍ୟେକରୁ (a + b + c) ପଦକୁ ବିୟୋଗ କଲେ ବିୟୋଗଫଳ A.P ରେ ରହିବ ।
⇒ 2a – (a+b+c), 2b – (a + b + c), 2c – (a + b + c) A.P.ରେ ରହିବ ।
⇒ (b + c – a), (c + a – b), (a + b – c) A.P.ରେ ରହିବ ।
⇒ b+c-a, c+a-b, a+b-c A.P.ରେ ରହିବ ।

(iv) a, b, c A.P.ରେ ଅବସ୍ଥିତ ।
A.P.ର ପ୍ରତ୍ୟେକ ପଦରେ (-1) ଗୁଣନ କଲେ, -a, -b, -c A.P.ରେ ରହିବ ।
A.P.ର ପ୍ରତ୍ୟେକ ପଦରେ a + b + c ଯୋଗକଲେ
a + b + c – a, a + b + c – b, a + b + c – c A.P.ରେ ରହିବ ।
∴ b + c, c + a, a + b A.P.ରେ ରହିବ ।
A.P.ର ପ୍ରତ୍ୟେକ ପଦକୁ abc ଦ୍ଵାରା ଭାଗକଲେ \(\frac{b + c}{abc},\frac{c + a}{abc},\frac{a + b}{abc}\) A.P.ରେ ରହିବ ।
⇒ \(\frac{1}{a}(\frac{b + c}{bc}),\frac{1}{b}(\frac{c + a}{ac}),\frac{1}{c}(\frac{a + b}{ab})\) A.P.ରେ ରହିବ ।
⇒ \(\frac{1}{a}(\frac{1}{b}+\frac{1}{c}),\frac{1}{b}(\frac{1}{c}+\frac{1}{a}),\frac{1}{c}(\frac{1}{a}+\frac{1}{b})\) A.P.ରେ ରହିବ ।

(v) a, b, c A.P.ରେ ଅବସ୍ଥିତ ।
A.P.ର ପ୍ରତ୍ୟେକ ପଦରେ ab + bc + ca ଗୁଣନ କଲେ ଗୁଣଫଳ A.P.ରେ ରହିବ ।
a (ab + bc + ca), b (ab + bc + ca), c (ab + bc + ca) A.P.ରେ ରହିବେ ।
a²b+ abc + ca², ab²+ b²c + abc, abc + bc² + c²a A.P.ରେ ରହିବ ।
= A.P.ର ପ୍ରତ୍ୟେକ ପଦରୁ abc ବିୟୋଗ କଲେ ଲବ୍‌ଧ ଅନୁକ୍ରମ A.P.ରେ ରହିବ ।
a²b + ca², ab² + b²c, bc² + c²a A.P.ରେ ରହିବ ।
a²(b+c), b²(c + a), c²(a + b), A.P.ରେ ରହିବ ।

Question 23.
(i) \(\frac{1}{a},\frac{1}{b},\frac{1}{c}\) ରେ ରହିଲେ ଏବଂ a + b + c ≠ 0 ହେଲେ, ପ୍ରମାଣ କର ଯେ \(\frac{b + c}{a},\frac{c + a}{b},\frac{a + b}{c}\) ମଧ୍ୟ A.P. ରେ ରହିବେ ।
(ii) \(\frac{a}{b + c},\frac{b}{c + a},\frac{c}{a + b}\) ଅନୁକ୍ରମ A.P.ରେ ରହିଲେ ଏବଂ a + b + c ≠ 0 ହେଲେ ପ୍ରମାଣ କର ଯେ, \(\frac{1}{b + c},\frac{1}{c + a},\frac{1}{a + b}\) A.P.ରେ ରହିବେ ।
ସମାଧାନ :
(i) \(\frac{1}{a},\frac{1}{b},\frac{1}{c}\) A.P.ରେ ଅବସ୍ଥିତ । (a + b + c ≠ 0)
⇒ A.P. ର ପ୍ରତ୍ୟେକ ପଦରେ a + b + c ଗୁଣନ କଲେ,
\(\frac{a+b+c}{a},\frac{a+b+c}{b},\frac{a+b+c}{c}\) A.P.ରେ ରହିବ ।
⇒ \(\frac{a}{a}+\frac{b+c}{a}, \frac{c+a}{b}+\frac{b}{b}, \frac{a+b}{c}+\frac{c}{c}\)
⇒ \(1+\frac{b+c}{a}, \frac{c+a}{b}+1, \frac{a+b}{c}+1\) A.P.ରେ ରହିବ ।
⇒ \(\frac{b+c}{a}, \frac{c+a}{b}, \frac{a+b}{c}\) A.P.ରେ ରହିବ । (ପ୍ରତ୍ୟେକ ପଦରୁ 1 ବିୟୋଗ କଲେ ।) (ପ୍ରମାଣିତ)

Question 24.
ଯଦି କୌଣସି A.P.ର ପ୍ରଥମ ପଦ a ଏବଂ ଶେଷ ପଦ l ହୁଏ ପ୍ରମାଣ କର ଯେ, ଅନୁକ୍ରମର ପ୍ରଥମରୁ r ତମ ପଦ ଏବଂ ଶେଷରୁ r ତମ ପଦର ସମଷ୍ଟି, ପ୍ରଥମ ଓ ଶେଷ ପଦର ସମଷ୍ଟି ସହିତ ସମାନ ।
ସମାଧାନ :
A.P.ର ପ୍ରଥମପଦ = a, ଶେଷ ପଦ = l
ମନେକର ସାଧାରଣ ଅନ୍ତର = d ।
∴ A.P. ଟି a, a + d, a + 2d, a + 3d, …….. l + 2d, l + d, l
ପ୍ରଥମରୁ rତମ ପଦ = a + (r – 1) d
ଶେଷ r ତମ ପଦର a = l, d = -d
t_r = l + (r – 1 ) (-d)
∴ ପ୍ରଥମରୁ r ତମ ପଦ + ଶେଷରୁ r ତମ ପଦ = [a + (r – 1) d] + [l + (r – 1) (- d)]
= a + (r – 1) d + l – (r – 1) d = a + l
∴ ପ୍ରଥମ ଓ ଶେଷ ପଦର ସମଷ୍ଟି = a + l
∴ ପ୍ରଥମରୁ r ତମ ପଦ ଓ ଶେଷରୁ 1 ତମ ପଦର ସମଷ୍ଟି = ପ୍ରଥମ ଓ ଶେଷ ପଦର ସମଷ୍ଟି ।
ବିକଳ୍ପ ପ୍ରଣାଳୀ : ପ୍ରଥମରୁ 1 ତମ ପଦ (t) = a + (r – 1)d
ଶେଷରୁ r ତମ ପଦ = ପ୍ରଥମରୁ (n – r + 1) ତମ ପଦ
= t_n-r+1 = a + {(n – r+1) – 1}d = a + (n – r)d
∴ t_r + _n-r + 1 = a = (r – 1 )d + a + (n – r) d = 2a + (n – 1)d = a + {a + (n – 1)d} = a + 1

Question 25.
ଗୋଟିଏ ସମାନ୍ତର ପ୍ରଗତିର ପ୍ରଥମ p ସଂଖ୍ୟକ ପଦର ସମଷ୍ଟି r, ପ୍ରଥମ q ସଂଖ୍ୟକ ପଦର ସମଷ୍ଟି s ଏବଂ ସାଧାରଣ ଅନ୍ତର । ହେଲେ ପ୍ରମାଣ କର ଯେ, \(\frac{r}{p}-\frac{s}{q}\) = (p – q) \(\frac{d}{2}\) ହେବ ।
ସମାଧାନ :
ଗୋଟିଏ ସମାନ୍ତର ପ୍ରଗତିର S_p = x ଏବଂ S_q = s
ସାଧାରଣ ଅନ୍ତର = d । ମନେକର ପ୍ରଥମ ପଦ = a
.. ପ୍ରଥମ p-ସଂଖ୍ୟକ ପଦର ସମଷ୍ଟି = S_p = \(\frac{p}{2}\) {2a + (p – 1) d}
ପ୍ରଶାନୁସାରେ, \(\frac{p}{2}\) {2a + (p – 1) d} = r ⇒ 2a + (p – 1) d = \(\frac{2r}{p}\) …….(i)
.. ପ୍ରଥମ q-ସଂଖ୍ୟକ ପଦର ସମଷ୍ଟି = S_q = \(\frac{q}{2}\) {2a + (q – 1) d}
ପୁନଶ୍ଚ, \(\frac{q}{2}\) {2a + (q – 1) d} = s ⇒ 2a + (q – 1) d = \(\frac{2s}{q}\) …….(ii)
ସମୀକରଣ (i)ରୁ (ii)କୁ ବିୟୋଗ କଲେ,

Question 26.
ଗୋଟିଏ ସମାନ୍ତର ଶ୍ରେଣୀର ପ୍ରଥମ p, q, r ସଂଖ୍ୟକ ପଦର ସମଷ୍ଟି a, b, c ହେଲେ ପ୍ରମାଣ କର ଯେ, \(\frac{a}{p}\) (q – r) + \(\frac{b}{q}\) (r – p) + \(\frac{c}{r}\) (p – q) = 0 ହେବ ।
ସମାଧାନ :
ଗୋଟିଏ ସମାନ୍ତର ଶ୍ରେଣୀର ପ୍ରଥମ p, q, r ସଂଖ୍ୟକ ପଦର ସମଷ୍ଟି a, b, c ।
ମନେକର A.P.ର ପ୍ରଥମ ପଦ = x ଓ ସାଧାରଣ ଅନ୍ତର = d
S_p = \(\frac{p}{2}\) {2a + (p – 1) d} = a ⇒ 2x + (p – 1) d = \(\frac{2a}{p}\) ……(i)
S_q = \(\frac{q}{2}\) {2a + (q – 1) d} = b ⇒ 2x + (q – 1) d = \(\frac{2b}{q}\) ……(ii)
S_r = \(\frac{r}{2}\) {2a + (r – 1) d} = c ⇒ 2x + (r – 1) d = \(\frac{2c}{r}\) ……(iii)
ସମୀକରଣ (i)ରୁ (ii)କୁ ବିୟୋଗ କଲେ,


ବି.ଦ୍ର. : (i) ରୁ 2 =x + \(\frac{1}{2}\)(p – 1)d ⇒ \(\frac{a}{p}\) (q – r) = x(q – r) + \(\frac{1}{2}\) (p – 1)(q – r)d
ସେହିପର \(\frac{b}{q}\)(r – p) = x(r – p) + \(\frac{1}{2}\) (q – 1)(r – p)d
c(p – q) = x(p – q) + 2(r – 1)(p – q)d
ଯୋଗକଲେ \(\frac{a}{p}\) (q – r) + \(\frac{b}{q}\) (r – p) + \(\frac{c}{r}\) (p – q) = 0 ହେବ ।

Question 27.
କୌଣସି A.P.ର t = q, t = p ହେଲେ ପ୍ରମାଣ କର ଯେ t = p + q – m ।
ସମାଧାନ :
ମନେକର A.P.ର ପ୍ରଥମ ପଦ = a ଓ ସାଧାରଣ ଅନ୍ତର = d
ଏକ A.P.ର t_p = q ⇒ a + (p – 1)d = q … (i)
⇒ t_q = p ⇒ a + (q – 1)d = p … (ii)
t_p – t_q = q – p ⇒ a + (p – 1)d – a + (q – 1)d = q – p
⇒ d(p – 1 – q + 1) = q – p ⇒ d(p – q) = q – p
⇒ d = \(\frac{q-p}{p-q}=\frac{-(p-q)}{p-q}=-1\)
‘d’ର ମାନ (i) ରେ ବସାଇଲେ a + (p – 1) d = q
⇒ a + (p – 1) (-1) = q ⇒ a – p + 1 = q
⇒ a = p + q – 1
:. t = a + (m – 1)d = p + q – 1 +(m – 1)(-1)
= p + q – 1 – m + 1 = p + q – m

Question 28.
କୌଣସି A.P.ର S_m = n, S_n = m ହେଲେ, ପ୍ରମାଣ କର ଯେ S_m+n = -(m + n) ହେବ ।
ସମାଧାନ :
ମନେକର A.P.ର ପ୍ରଥମ ପଦ = a, ସାଧାରଣ ଅନ୍ତର = d
S_m = n ⇒ \(\frac{m}{2}\) {2a + (m – 1) d} = n ⇒ am + (m² – m) \(\frac{d}{2}\) = n ……(i)
S_n = m ⇒ \(\frac{n}{2}\) {2a + (n – 1) d} = m ⇒ an + (n² – n) \(\frac{d}{2}\) = m ……(ii)
ସମୀକରଣ (i)ରୁ (ii)କୁ ବିୟୋଗ କଲେ

Odisha State Board CHSE Odisha Class 11 Invitation to English 1 Solutions Chapter 3 The Golden Touch Textbook Exercise Questions and Answers.

CHSE Odisha 11th Class English Solutions Chapter 3 The Golden Touch

CHSE Odisha Class 11 English The Golden Touch Text Book Questions and Answers

UNIT – I
Gist with Glossary

Gist:
The legendary king Midas’s madness for gold had no limit. To him, the most precious thing in the world was gold. The love that he had for his little daughter Marygold was no less. It was his earnest wish to see everything such as the golden light of the sun at the evening, a bunch of sweet yellow flowers, and the most beautiful roses in his garden turn into gold. Even the king’s intense love for music in his youth paled into insignificance before the sound of coins, one against another. He always dreamt about gold. He could not resist the temptation of counting his gold pieces.

One morning the sight of an unknown person caught his attention. His astonishment knew no bounds to see him in his treasure room which he himself had locked. He asked Midas, why he was not satisfied, despite being vastly wealthy. He gave vent to his wish that everything he touched would become gold. It would give him supreme happiness. The stranger fulfilled the king’s desire. The following day, at sunrise the latter’s dream came true. He had the Golden Touch at his disposal. Strangely enough, the stranger had vanished.

Glossary:
besides : in addition to (ଏତଦ୍‌ବ୍ୟତୀତ)
dearly : deeply (ଗଭୀର ଭାବରେ)
precious : valuable (ମୂଲ୍ୟବାନ)
treasure-room : here, the room where king Midas had kept bars of gold (ଗନ୍ତାଘର )
brightened : shone (ଉଜ୍ଜ୍ବଳି ଉଠିଲା)
vividly : clearly (ପ୍ରାଞ୍ଜଳ ଭାବରେ)

Think it:
Question 1.
What do you learn about King Midas from the first two paragraphs of the story?
Answer:
The first two paragraphs throw light on King Midas’s vast wealth and his obsession with gold.

Question 2.
What did he wish when saw the golden light of the evening sun?
Answer:
When he saw the golden light of the evening sun, he wished it could change everything into genuine gold.

Question 3.
Why didn’t he like the roses in his garden?
Answer:
He didn’t like the rose in his garden, because they were not made of gold.

Question 4.
How did he spend his time in his ‘treasure room’?
Answer:
He spent his time in his treasure room counting his gold pieces. Besides, he held the bars of gold and praised his gold cups and plates.

Question 5.
How did he come across the stranger?
Answer:
He came across the stranger when his treasure room was bathed in bright sunshine; he found the latter in his locked room.

Question 6.
How did King Midas answer the stranger’s question, “What else do you want ?”
Answer:
When the stranger asked him ‘What else do you want ?’, the king expressed his wish that everything he would touch could be changed into gold. Besides, he was sick of collecting his wealth at a very slow rate.

Question 7.
How did the stranger fulfill his wishes?
Answer:
The stranger fulfilled his ambition by telling him that the following day at sunrise, he would find the Golden Touch at his disposal.

UNIT – II
Gist with Glossary

Gist:
This unit begins with the king’s discovery that his desire for the golden touch had not been fulfilled. His bed had not changed into gold. His sadness did not last long. A sudden sight of the reflection of the golden light of the earliest sunrise on him worked wonders. The sheet on his bed turned into a cloth of gold. The first sunbeam had truly brought the Golden Touch. Driven by excitement, he touched everything – one of the legs of the bed, the curtain at the window, his dress, and spectacles. There was gold everywhere. The loss of spectacles didn’t matter to him much.

The gold was more precious than his pair of spectacles and his daughter Marygold would read to him. The magic touch turned everything into gold, beginning from the brass handle of the door, and the rose trees, the constant source of his pride and joy in the past. At the moment, he went into breakfast that morning, his little beautiful daughter Marygold came in crying bitterly. When he kissed her, she wore a terrible look, with her little face, beautiful hair, and her little body gone. She became a hard golden figure.

Glossary:
turned into gold: transformed into gold (ସୁନା ହୋଇଗଲା)
disappointed : sad (ଦୁଃଖ)
sunbeam : sunlight (ସୂର୍ଯ୍ୟକିରଣ )
put on : wore (ପିନ୍ଧିଲେ)
bitterly: in a painful or unhappy mood (ଦୁଃଖଦ ଅବସ୍ଥାରେ)
scent : fragrance (ସୁଗନ୍ଧ)
comfort : (here) console (ସାନା ଦେବା )
terrible : ferocious (ଭୟଙ୍କର )

Think it out:
Question 1.
When did the king discover that his desire for the golden touch had been fulfilled?
Answer:
The king discovered that his desire for the golden touch had been fulfilled at the sight of his bed sheet transformed into a cloth of gold.

Question 2.
Why was the king not sad when his spectacles turned into gold?
Answer:
The king was not sad when his spectacles turned into gold, because he felt that a pair of spectacles was not as precious as the Golden Touch. Besides, her daughter Marygold could read to him.

Question 3.
What was Marygold’s complaint about the golden rose?
Answer:
Marygold’s complaint about the golden rose was that it had become yellow and hard and lost its fragrance.

Question 4.
How did the king console his daughter?
Answer:
The king consoled his daughter stating that she should not think of it at all because these rose flowers that had turned into gold were invaluable. He lovingly said to her to sit and take her breakfast.

Question 5.
Why couldn’t the king enjoy his breakfast?
Answer:
The king couldn’t enjoy his breakfast which included eggs, fish, bread, butter, and a spoonful of coffee, because they all became gold.

Question 6.
What happened to Marygold when the king kissed her?
Answer:
When the king kissed her, the king saw before him a terrible figure instead of his little daughter. Her sweet face, her beautiful hair, and her little body had all gone. There stood a statue of solid gold.

UNIT-III
Gist with Glossary

Gist:
King Midas sank into despair at the loss of everything he loved including his dear little daughter Marygold. In the meantime, the stranger reappeared and enquired him about his Golden Touch. He narrated his tale of woe to the former. He now realized the futility of the Golden Touch. He was terribly thirsty and pined for a cup of cold water to taste. The stranger kept on asking him what he preferred – the Golden touch or a piece of bread and gold or his own little daughter. The grief-stricken father wanted his daughter to get back. His repentance made the stranger remark that he was wiser than before. The stranger knew human nature dies hard and the king was no different. Midas had now become a virulent hater of gold.

He trembled in fear at the sight of a fly perching on his nose and at once felt the ground turning into a small piece of gold. In order to get rid of the burden of the golden touch, the stranger advised the king to go to the end of his garden, and wash in the water of the river there. This was not all. He should bring some of the same water and sprinkle it over anything, he wished to change back again. The king rose to the occasion without wasting time. To his utter delight, he got his lost daughter back by sprinkling water on her. Marygold was kept in dark about this painful incident. The king started his life afresh in the midst of his loving daughter and his garden full of fragrant roses.

Glossary:
lingered (here) saw for some time (କିଛି ସମୟ ପାଇଁ ଦେଖୁବା)
despair : misery (ଦୁର୍ଦ୍ଦଶା)
passionately : the state of mind caused by passion (ଆଗ୍ର ହାନ୍ତି ତ ହୋଇ)
scrap: piece (ଖଣ୍ଡ)
weight : (here) the burden of greed for gold

Think it out:
Question 1.
How did the king realize that the golden touch was a useless gift for him?
Answer:
The king realized that the golden touch was a useless gift for him because it deprived him of everything that he loved, especially his sweet little daughter Marygold. The king had become a grief-stricken person.

Question 2.
‘You are wiser than you were’ – why did the stranger say so?
Answer:
The stranger said so because he marked that the king was seething with repentance for his obsession with gold. The king wished he had not given one hair of his daughter’s head in exchange for the power to transform the entire earth into gold.

Question 3.
What did the stranger advise the king to do to get rid of his golden touch?
Answer:
In order to get rid of his golden touch, the stranger advised the king to go to the end of his garden, and wash in the water of the river there. This was not all. He should bring some of the same water and sprinkle it over anything he wished to change back again.

Question 4.
How did the king get back his daughter?
Answer:
The king got his daughter back by going straight to the golden figure of his daughter and then sprinkling some water brought from the river on her.

Question 5.
Is the story a tragic or comic one? Give your reasons.
Answer:
The story, The Golden Touch is not a tragic one, because though the grief-stricken king could not look at Marygold, there is no death inevitable. Instead, the story is a comic one. King Midas, the protagonist of the story, in spite of his sadness, makes us laugh at his blind love of gold. How can a father make his loving little daughter a victim of his boundless greed? His reaction at the loss of everything he loved and his belated realization of his mistakes and above all, the stranger’s words contribute to the comic aspect of the story, although there are patches of pathos.

Post-Reading Activities:

A. Arrange the following sentences according to their logical order.
(a) Midas said, ‘I wish everything I touch could be turned into gold’.
(b) ‘The Golden Touch !’ exclaimed the stranger.
(c) Midas said, ‘It would give me perfect happiness’.
(d) The stranger said, ‘Tomorrow at sunrise you will find that you have the Golden Touch’.
(e) King Midas came across a stranger smiling at him.
(f) The stranger asked, ‘What do you want ?’
(g) He guessed that the stranger was no ordinary person.
Answer:
(e) King Midas came across a stranger smiling at him.
(g) He guessed that the stranger was no ordinary person.
(f) The stranger asked, ‘What do you want ?’
(a) Midas said, ‘I wish everything I touch could be turned into gold’.
(b) ‘The Golden Touch !’ exclaimed the stranger.
(c) Midas said, ‘It would give me perfect happiness’.
(d) The stranger said, ‘Tomorrow at sunrise you will find that you have the Golden Touch’.

B. Doing with words.
1. Write the antonyms of the following words :
love —
bright —
perfect —
wise —
please —
usual —
happiness —
common —
beautiful —
careful —
proud —
sincere —
Answer:
love — hate
bright —dull
perfect — imperfect
wise — fool
please — displease
usual — unusual
happiness — sadness
common — uncommon
beautiful — ugly
careful — careless
proud — humble
sincere — insincere

B. Match the expressions in column A with their one-word substitution in column B.

Answer:

3. Write the nouns derived from the following verbs :
collect        ______________
satisfy        ______________
exclaim      ______________
disappoint ______________
reflect        ______________
astonish    ______________
expect       ______________
Answer:
collect        — collection
satisfy        — satisfaction
exclaim      — exclamation
disappoint — disappointment
reflect        — reflection
astonish     — astonishment
expect       — expectation

4. Fill in the blanks with the adjectival forms of the following nouns:
gold        _____________
beauty    _____________
palace     _____________
magic      _____________
spectacle _____________
comfort   _____________
sorrow     _____________
passion    _____________
Answer:
gold        — golden
beauty    — beautiful
palace     — palatial
magic      — magical
spectacle — spectacular
comfort   — comfortable
Sorrow     — sorrowful
passion    — passionate

5. Fill in the blanks with the verbs from which the following nouns have been
speech        _______________
thought      _______________
excitement _______________
collection   ________________
service       _______________
Answer:
speech        — speak
thought      — think
excitement — excite
collection   — collect
service        — serve

CHSE Odisha Class 11 English The Golden Touch Fours Important Questions and Answers

I. Short Answer Type Questions with Answers

1. Read through the extract and answer the questions that follow.
Long ago, there lived a very rich man called Midas. Besides being rich, he was a king, and he had a little daughter called Marygold. King Midas loved gold more than anything else in the world. He liked being a king, chiefly because he loved his golden crown. He loved his daughter dearly too, and the more he loved her, the more gold he wanted for her sake. When King Midas saw the golden light of the sun in the evening, he wished it could turn everything into real gold. When Marygold came to him with a bunch of sweet yellow flowers, he would say, ‘If they were as golden as they look, they would be worth picking !’.

Even the roses in his garden did not please him anymore – the largest and sweetest and most beautiful roses ever seen – because they were not made of gold. And although the king was very fond of music in his youth, the only music he loved now was the sound of gold coins, one against another. At last, King Midas could not bear to touch anything that was not gold. He used to go down to a secret room under his palace where he kept his precious store. He would let himself in and count his gold pieces. He would hold the bars of gold and admire his gold cups and plates until he could hardly bear to leave them.

Now in those days, a great many wonderful things used to happen just as they do today. One morning King Midas was in his treasure room when he noticed that the sun was shining into the room more brightly than usual. Not only that, but a stranger stood there, smiling at him in the light of the sunbeam. King Midas knew that he had locked himself in as usual, and so he guessed that his visitor was no ordinary person. The stranger looked at the gold pieces that the king was counting. ‘You seem to be a very rich man’ he said. ‘But it has taken me a long time to collect this gold’, said King Midas. ‘If I could live a thousand years, I might have time to get richer.

Questions :
(i) Why did King Midas like being a king?
(ii) How did he respond, when his daughter came to him with a bunch of beautiful yellow flowers?
(iii) What did the king love deeply in his youth?
(iv) Why did he not want to touch anything at last?
(v) What led the king to guess that the stranger was not an ordinary person?

Answers :
(i) King Midas liked being a king, mainly because he was fond of his golden crown.
(ii) When his daughter came to him with a bunch of beautiful yellow flowers, he would pick them, if they were as golden as they looked.
(iii) The king loved music deeply in his youth.
(iv) His frenzied desire for gold refrained him from touching anything at last. In other words, he was not interested to touch anything that was not gold.
(v) In spite of his treasure room being locked by himself, to his astonishment the king found the stranger inside it. This led the king to guess that the stranger was not an ordinary person.

2. Read through the extract and answer the questions that follow.
The next morning, King Midas awoke before dawn. He looked eagerly to see if his bed had been turned into gold. But no; it was exactly as it had been before. He lay, very disappointed, looking around his room. Suddenly, the earliest sunbeam of the rising sun shone through the window and up to the ceiling above. It seemed to reflect its golden light toward him. Looking at the sheet on his bed, Midas was astonished to find that it had become cloth of gold. The Golden Touch had truly come to him, with the first sunbeam. King Midas got out of bed in excitement. He touched one of the legs of the bed as he did so – and it immediately became a golden pillar.

He pulled the curtain at the window, and at once it became golden, too. He put on his clothes and found himself dressed in golden cloth. He took up his spectacles and put them on – and he found he could see nothing at all. The glasses had turned into gold and he could not see through them. He took them off again. ‘Never mind’, he thought to himself. ‘The Golden Touch is worth more than a pair of spectacles, and Marygold will be able to read to me.’ King Midas went downstairs and into the garden. He noticed that even the brass handle of the door became gold as soon as he turned it. Then he went among the rose trees that had always been his pride and joy in the past.

When he went to breakfast that morning, he felt more hungry than usual. While he was waiting for his eggs to be ready, little Marygold came in crying bitterly. ‘Look, father !’ she cried, holding out a golden rose. ‘I went to pick you some roses and they are yellow and hard, and their sweet scent is gone.’ ‘Never mind, my dear’, said her father. ‘They are worth much more like that. Sit down and eat your breakfast.’ He poured himself a cup of coffee as he spoke. The coffee pot was a golden one when he put it back on the table. Then he tried a spoonful of coffee, to see if it was sweet enough. But it had become liquid gold.

Questions :
(i) When did King Midas get up the following morning?
(ii) How did he first make use of the Golden touch and what was the result?
(iii) What had always been his object of pride and happiness in the past?
(iv) ‘But it had become liquid gold.’ What does ‘it’ refer to?
(v) Suggest a suitable title for the extract.

Answers :
(i) The following morning, King Midas got up before the crack of dawn.
(ii) He made use of the Golden touch for the first time by touching one of the legs of the bed and at once it turned into a golden pillar.
(iii) The rose trees in his garden had always been the object of his pride and happiness in the past.
(iv) ‘It’ refers to a spoonful of coffee.
(v) King Midas and his irresistible temptation for gold.

3. Read through the extract and answer the questions that follow.
In despair, Midas looked about him. Suddenly he saw the stranger that had visited him the day before. ‘Well Midas’, said the stranger. ‘How do you like having the Golden touch ?’ ‘I have lost everything I really loved’, said King Midas. ‘I am full of sorrow and regret. Gold is of no use to me now.’ ‘So you have learnt something since yesterday ?’ asked the stranger. ‘Now which is worth more – the Golden Touch or a cup of cold water ?’ ‘Oh, blessed water !’ exclaimed Midas. ‘Will I ever taste it again?’ ‘The Golden Touch – or a piece of bread ?’ the stranger said. ‘A piece of bread’, answered Midas, ‘is worth all the gold on earth !’

‘Gold – or your own little daughter ?’ asked the stranger. ‘Oh – my child, my child !’ cried poor Midas. ‘I would not have given one hair of her head for the power to change the whole earth into gold !’ The stranger looked seriously at King Midas. ‘You are wiser than you were’, he said. ‘Your heart is still flesh and blood. You know truly that the common things of life, which are within everyone’s reach, are more valuable than riches. Tell me, do you want to keep the Golden Touch ?’ ‘No, it is hateful to me now’, said Midas, passionately. A fly settled on the king’s nose and immediately fell to the floor, a small scrap of gold.

Midas shuddered. ‘Then go down to the end of your garden’, said the stranger, ‘and wash in the water of the river there. Then bring some of the same water and sprinkle it over anything that you wish to change back again. If you do this, truly and sincerely, you can set right again the results of your greed of gold.’ King Midas bowed his head. When he looked up again, the stranger had vanished. The king ran at once to the river. Without waiting to take off his clothes, he dived in. In the coolness of the water, he felt at once that a weight had been lifted from his heart and body.

Questions :
(i) What was the king’s response to the stranger’s question concerning the possession of the Golden Touch?
(ii) “Well I ever taste it again ?” What does ‘it’ refer to?
(iii) Why did Midas tremble in fear?
(iv) What were the results of the king’s greed of gold?
(v) ‘The felt at once that a weight had been lifted from his heart and body.’ What does the ‘weight’ refer to?

Answers :
(i) The king’s response to the stranger’s question concerning the possession of the Golden Touch was only deep sorrow and regret. He was sad at the loss of everything he loved.
(ii) ‘It’ refers to a cup of cold water the king yearned for.
(iii) Midas trembled in fear at the sight of a fly that perched on his nose and at once fell to the floor, eventually turning into a small scrap of gold.
(iv) As a result, of his greed for gold, the king lost everything including precious water and his beloved little daughter Marygold.
(v) The ‘weight’ refers to Midas’s boundless greed for gold.

II. Multiple Choice Questions (MCQs) with Answers
Choose the correct option.

Unit – I
The text
Long ago ……… stranger had go.

Question 1.
Long long ago there lived a very rich man called :
(a) Midas
(b) Devdas
(c) Raidas
(d) Bidas
Answer:
(a) Midas

Question 2.
Besides being rich, Midas was a :
(a) merchant
(b) Minister
(c) Chief
(d) king
Answer:
(d) king

Question 3.
Midas had a little daughter called :
(a) Marygold
(b) Rose
(c) Sunshine
(d) Nainegold
Answer:
(a) Marygold

Question 4.
King Midas liked being a king, chiefly because he loved his :
(a) golden chair
(b) golden crown
(c) people
(d) name and fame
Answer:
(b) golden crown

Question 5.
King Midas was a lover of more than anything else in the world.
(a) Silver
(b) platinum
(c) gold
(d) Diamond
Answer:
(c) gold

Question 6.
King Midas wished everything to turn into :
(a) real platinum
(b) real diamond
(c) real silver
(d) real gold
Answer:
(d) real gold

Question 7.
The king was very fond of music in his youth, the only music he loved now was :
(a) the sound of birds
(b) the sound of gold coins
(c) the roar of wild animals
(d) the sound of drums
Answer:
(b) the sound of gold coins

Question 8.
King Midas could not bear to touch anything that was not:
(a) plastic
(b) Silver
(c) gold
(d) Diamond
Answer:
(c) gold

Question 9.
King used to go down to a secret room under his palace where he kept his :
(a) precious gold
(b) coal
(c) precious metals
(d) valuable books and maps
Answer:
(a) precious gold

Question 10.
One morning King Midas was in his treasure room and he noticed that:
(a) the sun was shining into the room more brightly than usual
(b) birds were singing a song beautifully
(c) the golds were being doubled magically
(d) the sun had not risen yet
Answer:
(a) the sun was shining into the room more brightly than usual

Question 11.
Who do you think, standing in front and smiling at the king in the light of a sunbeam
(a) king’s daughter Marygold
(b) the queen
(c) a stranger
(d) the minister
Answer:
(c) a stranger

Question 12.
King Midas knew that he had locked himself in as usual and so he guessed that his visitor was :
(a) no special person
(b) no ordinary person
(c) no poor person
(d) no rich person
Answer:
(b) no ordinary person

Question 13.
Midas thought carefully. This was a wonderful chance, and he felt that the stranger had :
(a) spiritual powers
(b) magical powers
(c) no power
(d) physical powers
Answer:
(b) magical powers

Question 14.
“I am tired of collecting my riches so slowly. I wish everything I touch could be turned into gold.” Who said this?
(a) daughter Marygold
(b) the minister
(c) the stranger
(d) the king
Answer:
(d) the king

Question 15.
The stranger granted his wish to be fulfilled, i.e.
(a) wish to have a son
(b) the Golden Touch
(c) three wishes
(d) none of the above
Answer:
(b) the Golden Touch

Unit – II
The text
The next morning ……..what had be done?

Question 16.
The next morning, King Midas awoke before dawn and looked eagerly to see :
(a) if his bed had been turned into gold
(b) if his bed had been turned into a bed of roses
(c) if his bed had been turned into a hanging swing
(d) none of these
Answer:
(a) if his bed had been turned into gold

Question 17.
The Golden Touch had truly come to the king :
(a) with his first touch
(b) with the first sunbeam
(c) with his first sight
(d) all of the above
Answer:
(b) with the first sunbeam

Question 18.
After a touch, the things turn into gold. And with this he found himself in :
(a) a sad mood
(b) an angry mood
(c) excitement
(d) a worrying situation
Answer:
(c) excitement

Question 19.
Marygold was holding out a golden rose.
(a) happy
(b) angry
(c) sad
(d) crying bitterly
Answer:
(d) crying bitterly

Question 20.
King Midas tried a spoonful of coffee, to see if it was sweet enough. But it had become :
(a) poison
(b) sour
(c) liquid gold
(d) bitter
Answer:
(c) liquid gold

Question 21.
But the eggs, the fish, the bread, the butter, and all the food was uneatable for the
(a) daughter Marygold
(b) king
(c) stranger
(d) Queen
Answer:
(b) king

Question 22.
King Midas turned annoyed, sad, and worried because :
(a) he was unable to eat anything because of his Golden Touch
(b) he was unable to rule over his state
(c) he was unable to see anything
(d) none of the above
Answer:
(a) he was unable to eat anything because of his Golden Touch

Question 23.
Midas bent down and kissed his :
(a) gold coins
(b) little daughter
(c) cups and plates
(d) all of the above
Answer:
(b) little daughter

Question 24.
What do you think, that might happen to Marygold after getting a kiss from his father?
(a) she became more affectionate toward her father
(b) she became happy
(c) she turns into an ugly girl
(d) she became a statue of gold
Answer:
(d) she became a statue of gold

Question 25.
What terrible change came over Marygold? Her sweet little face, lovely hair, and little body turned into.
(a) yellow gold, golden metal, and a figure of soid gold
(b) white diamond, shinning metal, and a hard substance
(c) shining platinum and precious metal
(d) none of the above
Answer:
(a) yellow gold, golden metal, and a figure of sold gold

Unit – III
The text
This story ………..roses.

Question 26.
King Midas felt so sad and sorrowful that he wished, he was the in all the world, if only his beloved daughter could be herself again.
(a) richest man
(b) happiest man
(c) poorest man
(d) honest man
Answer:
(c) poorest man

Question 27.
In despair, Midas looked about him and suddenly he saw that had visited him the day before.
(a) stranger
(b) Marygold
(c) known person
(d) none of the above
Answer:
(a) stranger

Question 28.
“I have lost everything I really loved; I am full of sorrow and regret. Gold is of no use to me now.” What does the expression show?
(a) the king is in excitement
(b) the king is sad
(c) the king is repenting for his deed
(d) the king is happy, what happened
Answer:
(c) the king is repenting for his deed

Question 29.
After having the joy of ‘the Golden Touch’, the king’s view changed :
(a) the Golden Touch is worthful than anything
(b) the Golden Touch is of no use if man’s need is not satisfied
(c) both (a) and (b)
(d) none of the above
Answer:
(b) the Golden Touch is of no use if man’s need is not satisfied

Question 30.
Midas wanted everything back to normal because :
(a) those were worthful
(b) those were his wants
(c) those were useless
(d) those were his needs
Answer:
(d) those were his needs

Question 31.
“I wouldn’t have given one hair of her head for the power to change the whole earth into gold!” This expression said by the king shows :
(a) his hate for his daughter
(b) duty towards his daughter
(c) love for his daughter
(d) all of the above
Answer:
(c) love for his daughter

Question 32.
“You are wiser than you were,” he said. “Your heart is still flesh and blood.” Here ‘you’ and ‘he’ stands for
(a) king and daughter
(b) king and stranger
(c) stranger and king
(d) daughter and stranger
Answer:
(b) king and stranger

Question 33.
Word ‘shuddered’ means
(a) tremble or shake violently
(b) rearrange
(c) avoid or reject
(d) past part
Answer:
(a) tremble or shake violently

Introducing the Author:
Nathaniel Hawthorne is an American novelist and short story writer. Much of Hawthorne’s writing centers on New England, with many works featuring moral allegories with a puritanical inspiration. His fiction works are considered part of the Romantic movement and more specifically, dark romanticism. His themes often center on the inherent evil and sin of humanity, and his works have deep psychological complexity.

About the Topic:
‘The Golden Touch, as the title implies, deals with King Midas’s boundless greed for gold. The inevitable happened. The legendary king sank into despair. At last his obsession with the yellow metal filled him with great repentance and changed his attitude.

Summary:
Hawthorne’s story, ‘The Golden Touch’, takes us back to a long past when there lived king Midas who was vastly wealthy. He was the father of a little daughter Marygold by name. His fascination in gold was more than anything else in the world. He also loved his daughter deeply. The spectacle of the golden light of the sun evoked his strong wish – everything could change into real gold. Even the largest and sweetest and most beautiful roses paled into significance before this precious yellow metal. In his youth, he loved music deeply, but now the sound of gold coins, one against another fascinated him most.

At last, kind Midas’s desire for gold became irresistible. He became a frequent visitor to a secret treasure room under his palace. He would allow himself in, count his gold pieces and hold the bars of gold. One morning, when the sun was shining brightly, he was in his treasure room which was locked inside. In the meantime, he noticed an unknown person, standing there, giving a smile at him in the sunlight. His amazement knew no bounds to see the stranger. The king did not make out how he came inside the locked room. As a result, he guessed the stranger was not an ordinary person.

The unknown person came to know of the king’s insatiable desire for gold; therefore, the former asked the latter what he wanted. The king expressed his wish that everything he touched could transform into gold. His wish to have the Golden Touch filled the stranger with surprise. He asked the king if he would not regret it. The king’s response was swift – it would give him perfect happiness. The stranger fulfilled his wish. He said to the king that the next day at sunrise, he would have his coveted Golden touch. The light of the sunbeam was too bright for Midas to see anything around him.

To his amazement, the stranger had vanished, when he opened his eyes. The following morning, the king discovered that his desire for the golden touch had not been fulfilled. His bed had not changed into gold. His sadness did not last long. A sudden sight of the reflection of the golden light of the earliest sunrise on him worked wonders. The sheet on his bed turned into a cloth of gold. The first sunbeam had truly brought the Golden Touch. Driven by excitement, he touched everything – one of the legs of the bed, the curtain at the window, his dress, and spectacles.

There was gold everywhere. The loss of spectacles didn’t matter to him much. The gold was more precious than his pair of spectacles and his daughter Marygold would read to him. The magic touch turned everything into gold beginning from the brass handle of the door, and the rose trees, the constant source of his pride and joy in the past. At the moment, he went into breakfast that morning, his little beautiful daughter Marygold came in crying bitterly. When he kissed her, she wore a terrible look, with her little face, beautiful hair, and her little body gone.

She became a hard golden figure. We find king Midas in a dejected mood. He plunged into deepening despair at the loss of everything he loved including his dear little daughter Marygold. In the meantime, the stranger reappeared and enquired him about his Golden Touch. He narrated his tale of woe to the former. He now realized the futility of the Golden Touch. He was terribly thirsty and pined for a cup of cold water to taste. The stranger kept on asking him what he preferred – the Golden touch or a piece of bread and gold or his own little daughter.

The grief-stricken father wanted his daughter to get back. His repentance made the stranger remark that he was wiser than before. The stranger knew human nature dies hard and the king was no different. Midas had now become a virulent hater of gold. He trembled in fear at the sight of a fly perching on his nose at once felt the ground turning into a piece of gold. In order to get rid of the burden of his Golden Touch, the stranger advised him to go down to the end of his garden, wash in the water of the river there, bring some of the same water and sprinkle it over anything he wished to change back again.

If the king does this sincerely and truly he can rectify his greed for gold. The king magnificently rose to the occasion. The story ends on a happy note. At first, the king sprinkled the water on the golden figure of his little daughter, Marygold. The inevitable happened. He got back his daughter again. Marygold was kept in dark about this unfortunate and painful incident. The king and his daughter lived happily.

ସାରାଂଶ:
ହଥର୍ୟଙ୍କ ଗଳ୍ପ ‘The golden Touch’ ଆମକୁ ଭସାଇଦିଏ ଏକ ସୁଦୂର ଅତୀତକୁ ଯେତେବେଳେ ଅହେତୁକଭାବେ ସୁନା ପ୍ରତି ଲୋଭ ଥିବା ଏକ ଶକ୍ତିଶାଳୀ ଓ ଧନୀ ରାଜା ବାସ କରୁଥିଲେ ଯାହାଙ୍କ ନାମ ଥିଲା ମିଦାସ୍ । Marygold ନାମକ ତାଙ୍କର ଗୋଟିଏ କୁନି ଝିଅ ଥିଲା। ପୃଥ‌ିବୀର ସବୁ ଜିନିଷଠାରୁ ସେ ସୁନାକୁ ବେଶି ଭଲ ପାଉଥିଲେ । ତା’ ସହିତ ସେ ତାଙ୍କର କନ୍ୟାକୁ ମଧ୍ୟ ଅତି ନିବିଡ଼ଭାବେ ଭଲ ପାଉଥିଲେ । ଅସ୍ଥାୟମାନ ସୂର୍ଯ୍ୟଙ୍କର ସୁନେଲି କିରଣକୁ ଦେଖ୍ ସେ ଭାବୁଥିଲେ ଏହା ସବୁ ଜିନିଷକୁ ସୁନାରେ ପରିଣତ କରିପାରନ୍ତା କି ? ତାଙ୍କ ଦୃଷ୍ଟିରେ ଏହି ମୂଲ୍ୟବାନ୍ ସୁନେଲୀ ଧାତୁର ମୂଲ୍ୟତୁଳନାରେ ସୁନ୍ଦର ଗୋଲାପଗୁଡ଼ିକର ମୂଲ୍ୟ କିଛି ନ ଥିଲା । ତାଙ୍କ ଝିଅ ଆଣିଥିବା ସୁନେଲି ରଙ୍ଗର ଫୁଲଟିକୁ ଦେଖି ରାଜା ଖୁସି ହୋଇ ନ ଥିଲେ କାରଣ ତାହା ସୁନାରେ ନିର୍ମିତ ନ ଥିଲା ।

ତାଙ୍କ ଯୁବାବସ୍ଥାରେ ସେ ସଙ୍ଗୀତକୁ ଗଭୀରଭାବେ ଭଲ ପାଉଥିଲେ ଏବଂ ଏବେ ଭଲ ପାଉଛନ୍ତି ସୁନାକୁ । ପରିଶେଷରେ ସୁନା ପ୍ରତି ତାଙ୍କର ଅହେତୁକ ଲୋଭ ବୃଦ୍ଧି ପାଇବାରେ ଲାଗିଲା । ତାଙ୍କ ପ୍ରାସାଦରେ ଥ‌ିବା ମୂଲ୍ୟବାନ ଜିନିଷର ଭଣ୍ଡାର ଏକ ଗୋପନ କୋଠରିକୁ ସେ ସୁନାର ମୁଦ୍ରା ଗଣିବାକୁ ବାରମ୍ବାର ପ୍ରବେଶ କରୁଥିଲେ । ତାଙ୍କ ପାଖରେ ସୁନାର ସ୍ତମ୍ଭ, ସୁନାର ପ୍ଲେଟ୍ ଓ ସୁନାର ପାଣିପାତ୍ର ସବୁ ରହିଥିଲା । ଦିନେ ସକାଳେ ଯେତେବେଳେ ସୂର୍ଯ୍ୟଙ୍କର ସୁନେଲି କିରଣ ବିଛେଇ ହୋଇ ପଡ଼ିଥିଲା, ଭିତର ପାଖରୁ ବନ୍ଦଥ‌ିବା ଗୁପ୍ତ କୋଠରି ଭିତରେ ଥାଇ ସେ ଦେଖିଲେ ଯେ ଜଣେ ଅପରିଚିତ ବ୍ୟକ୍ତ ସେଠାରେ ଠିଆ ହୋଇ ତାଙ୍କୁ ଚାହିଁ ସ୍ମିତହାସ୍ୟ କରୁଛନ୍ତି । ରାଜା ଅନୁମାନ କଲେ ଯେ ସେହି ଅପରିଚିତ ବ୍ୟକ୍ତିଜଣଙ୍କ ଜଣେ ସାଧାରଣ ମଣିଷ ହୋଇ ନପାରନ୍ତି । ଅପରିଚିତ ବ୍ୟକ୍ତିଜଣଙ୍କ ରାଜାଙ୍କର ସୁନା ପ୍ରତି ଥିବା ଅତୃପ୍ତ ଲୋଭ ବିଷୟରେ ଜାଣିପାରିଲେ ।

ରାଜା କ’ଣ ଚାହାନ୍ତି ବୋଲି ସେ ପ୍ରଶ୍ନ କଲେ । ରାଜା ଜାଣିଥିଲେ ସେ ଜଣେ ଅସାଧାରଣ ଅଲୌକିକ ବ୍ୟକ୍ତି ଥିଲେ । ତେଣୁ ରାଜା ବର ପ୍ରାର୍ଥନା କଲେ ଯେ ସେ ଯାହା ଛୁଇଁବେ ସେସବୁ ସୁନାରେ ପରିଣତ ହୋଇଯାଉ । ରାଜାଙ୍କର ଏହି ପ୍ରାର୍ଥନାରେ ଅପରିଚିତ ବ୍ୟକ୍ତିଜଣକ ଆଶ୍ଚର୍ଯ୍ୟ ହୋଇଗଲେ ଏବଂ ରାଜାଙ୍କୁ ପଚାରିଲେ ତାଙ୍କର ଏଥ‌ିରେ କୌଣସି ଅନୁଶୋଚନା ରହିବ ନାହିଁ ତ ! ରାଜା ଚଳଚଞ୍ଚଳ ମନରେ ଉତ୍ତର ଦେଲେ ଯେ ସେ ଏହି ବର ପାଇଲେ ଅତ୍ୟନ୍ତ ଖୁସି ହେବେ। ରାଜାଙ୍କର ଇଚ୍ଛା ପୂର୍ଣ ହେଉ ବୋଲି ସେ ବର ଦେଲେ । ସେ ରାଜାଙ୍କୁ କହିଲେ ଯେ ତା’ ପରଦିନ ସୂର୍ଯ୍ୟୋଦୟ ସମୟରେ ସେ ଏହି ସ୍ଵର୍ଣ୍ଣ ସ୍ପର୍ଶର ଫଳ ପ୍ରାପ୍ତ ହେବେ । ସୂର୍ଯ୍ୟଙ୍କ ରଶ୍ମି ଏତେ ଉଜ୍ଜ୍ଵଳ ଥିଲା ଯେ ରାଜା ତାଙ୍କ ଚତୁପାର୍ଶ୍ଵରେ ଥ‌ିବା କୌଣସି ଜିନିଷ ଦେଖି ପାରୁ ନ ଥିଲେ । ଯେତେବେଳେ ରାଜା ଆଖି ଖୋଲିଲେ, ସେ ସେହି ଅଦୃଶ୍ୟ ବ୍ୟକ୍ତିଜଣଙ୍କୁ ଦେଖିବାକୁ ପାଇଲେ ନାହିଁ ।

ସେ ଅଦୃଶ୍ୟ ହୋଇ ଯାଇଥିଲେ । ପରଦିନ ରାଜା ଶୀଘ୍ର ଶଯ୍ୟାତ୍ୟାଗ କରି ଦେଖିଲେ ଯେ ତାଙ୍କର ସ୍ଵର୍ଣ୍ଣ ସ୍ପର୍ଶର ବରଟି ପରିପୂର୍ଣ୍ଣ ହୋଇନାହିଁ । ତାଙ୍କର ଶଯ୍ୟା ସୁବର୍ଣ୍ଣରେ ପରିଣତ ହୋଇନାହିଁ । ସେ ଦୁଃଖରେ ଭାଙ୍ଗିପଡ଼ିଲେ । ତା’ପରେ ସୂର୍ଯ୍ୟଙ୍କର ସୁନେଲି କିରଣ ଝରକା ଦେଇ ତାଙ୍କ ଶଯ୍ୟାରେ ପଡ଼ିଲା । ତାଙ୍କ ବିଛଣା ଚାଦରଟି ସୁନାରେ ପରିଣତ ହୋଇଥିବା ଦେଖ୍ ସେ ଖୁସିରେ ଆତ୍ମହରା ହୋଇଗଲେ । ସେ ସବୁ ଜିନିଷକୁ ସ୍ପର୍ଶ କରିବାକୁ ଇଚ୍ଛାକଲେ । ତା’ପରେ ଖଟର ଗୋଟେ ଗୋଡ଼, ଝରକାର ପରଦା ଓ ନିଜ ପୋଷାକକୁ ଛୁଇଁଦେଲେ, ସବୁଯାକ ଜିନିଷ ସୁନା ପାଲଟିଗଲା । ତାଙ୍କର ସ୍ପର୍ଶରେ ତାଙ୍କର ଚଷମାଟି ମଧ୍ଯ ସୁନା ପାଲଟିଗଲା । ଏଥରେ ତାଙ୍କର ତିଳେମାତ୍ର ମନଦୁଃଖ ହେଲା ନାହିଁ । କାରଣ ଚଷମା ଅପେକ୍ଷା ସୁନା ଥିଲା ତାଙ୍କ ପାଇଁ ଅଧିକ ମୂଲ୍ୟବାନ୍ । ଏହି ଚମତ୍କାର ସ୍ପର୍ଶରେ ସବୁକିଛି ସୁନାରେ ପରିଣତ ହୋଇଗଲା ।

ତାଙ୍କ ଅତୀତର ଗର୍ବ ଓ ଖୁସିର ଉତ୍ସ ଗୋଲାପ ଗଛଗୁଡ଼ିକ ମଧ୍ୟ ସୁନାରେ ପରିଣତ ହୋଇଗଲା । ଏହି ସମୟରେ ତାଙ୍କ ପାଖକୁ ତାଙ୍କର ଝିଅ ଏକ ସୁନାର ଗୋଲାପ ଫୁଲ ଧରି କାନ୍ଦି କାନ୍ଦି ଆସିଲା ଏବଂ ଏହାର ବାସ୍ନା ଚାଲିଯାଇଥିବାରୁ ବ୍ୟସ୍ତ ହୋଇପଡ଼ିଲା । ରାଜା ନିଜେ କଫି ପିଇବାକୁ ଇଚ୍ଛା କରି, ସେ କଫି ଗ୍ଲାସ୍‌କୁ ସ୍ପର୍ଶ କରନ୍ତେ ତାହା ତରଳ ସୁନା ପାଲଟିଗଲା । ଭୋକିଲା ରାଜା ଜଳଖୁଆ ପାଇଁ ମାଛ, ଅଣ୍ଡା, ରୁଟି ଓ ଲହୁଣି ଖାଇବାକୁ ଚାହାନ୍ତେ, ସେସବୁ ତାଙ୍କ ସ୍ପର୍ଶ ପାଇବା ପରେ ସୁନାରେ ପରିଣତ ହୋଇଗଲା । ସେ ଖାଇ ନ ପାରିବାରୁ ମନ ଦୁଃଖ କଲେ । ଏହି ସମୟରେ ତାଙ୍କ ମନଦୁଃଖର କାରଣ ବିଷୟରେ ତାଙ୍କ କୁନି ଝିଅ ତାଙ୍କୁ ପଚାରିଲେ । ରାଜା ତାଙ୍କୁ ଚୁମ୍ବନ ଦେଲେ । ଫଳରେ ମେରିଗୋଲ୍ଡର ଶରୀରରେ ଏକ ଭୟଙ୍କର ପରିବର୍ତ୍ତନ ଘଟିଲା । ତାଙ୍କର କୁନି ସୁନ୍ଦର ମୁଖମଣ୍ଡଳ, ସୁନ୍ଦର କେଶରାଶି ଏବଂ କୁନି କୋମଳ ଶରୀରଟି କଠିନ ସୁନାରେ ପରିଣତ ହୋଇଯାଇଛି ।

ଯେତେବେଳେ Marygoldର ଶରୀରଟି ସୁନା ପାଲଟିଗଲା, ରାଜା ଆଶ୍ଚର୍ଯ୍ୟ ହୋଇଗଲେ ଏବଂ ଦୁଃଖ ଓ ଶୋକରେ ଭାଙ୍ଗିପଡ଼ିଲେ । ସବୁ ଜିନିଷ ସୁନାରେ ପରିଣତ ହୋଇଯିବା ହେତୁ ସେ ଗଭୀର ଦୁଃଖରେ ମର୍ମାହତ ହୋଇଗଲେ । ନିଜ ଝିଅର ପୂର୍ବ ଅବସ୍ଥା ଫେରି ପାଇବାପାଇଁ ସେ ବ୍ୟାକୁଳ ହୋଇ ଉଠିଲେ । ଏହି ସମୟରେ ସେହି ଅପରିଚିତ ବ୍ୟକ୍ତିଜଣକ ପୁନର୍ବାର ଆବିର୍ଭାବ ହେଲେ ଏବଂ ସେହି ବରର ପୂର୍ଣ୍ଣତା ବିଷୟରେ ଜାଣିବାକୁ ଚାହିଁଲେ । ରାଜା ତାଙ୍କ ଦୁଃଖର କାହାଣୀ ବର୍ଣ୍ଣନା କଲେ । ଏହି Golden Touchର ମୂଲ୍ୟହୀନତା ବିଷୟରେ ସେ ଅନୁଭବ କରିପାରିଲେ । ସେହି ଅପରିଚିତ ବ୍ୟକ୍ତିଜଣଙ୍କ ରାଜାଙ୍କୁ ପଚାରିଲେ ସୁନା କିମ୍ବା ଏକ ଗ୍ଲାସ୍ ପାଣି କେଉଁଟି ବିଶେଷ ଦରକାର ? ରାଜା ଉତ୍ତର ଦେଲେ ଜୀବନରେ ବଞ୍ଚିବାପାଇଁ ପାଣିର ଆବଶ୍ୟକତା ଅପରିହାର୍ଯ୍ୟ ।

ଅପରିଚିତ ବ୍ୟକ୍ତିଟି ପୁନର୍ବାର ରାଜାଙ୍କୁ ପଚାରିଲେ ସୁନା ଏବଂ ରୁଟି ମଧ୍ୟରୁ ତାଙ୍କ ପାଇଁ କେଉଁଟି ଅଧ୍ଵ ମୂଲ୍ୟବାନ୍ ? ତା’ପରେ ପଚାରିଲେ ସୁନା କିମ୍ବା ତାଙ୍କର କୁନି ଝିଅ ମଧ୍ୟରୁ ତାଙ୍କ ପାଇଁ କେଉଁଟି ମୂଲ୍ୟବାନ ? ଦୁଃଖରେ ଭାଙ୍ଗିପଡ଼ିଥିବା ରାଜା ତାଙ୍କ ଝିଅକୁ ପୁନର୍ବାର ଫେରି ପାଇବାପାଇଁ ବ୍ୟାକୁଳ ହୋଇଉଠିଲେ । ରାଜାଙ୍କର ଶୋଚନାରୁ ଜଣାଗଲା ଯେ ରାଜା ପୂର୍ବାପେକ୍ଷା ଅଧିକା ଜ୍ଞାନୀ ହୋଇ ପାରିଛନ୍ତି । ରାଜା ସୁନାକୁ ଘୃଣା କରୁଥିଲେ । ତାଙ୍କ ନାକରେ ବସିଥିବା ମାଛିଟି ତତ୍‌କ୍ଷଣାତ୍ କ୍ଷୁଦ୍ର ସୁନାଖଣ୍ଡଟିଏ ହୋଇ ମାଟିରେ ପଡ଼ିଲା । ଏହି ସ୍ପର୍ଷ ସ୍ପର୍ଶ ବରପ୍ରାପ୍ତିରୁ ମୁକ୍ତ ହେବା ପାଇଁ ରାଜା ବ୍ୟାକୁଳ ହୋଇପଡ଼ିଲେ । ଅପରିଚିତ ବ୍ୟକ୍ତିଜଣଙ୍କ ତାଙ୍କୁ ଉପଦେଶ ଦେଇ କହିଲେ ଯେ ବଗିଚାର ଶେଷରେ ଥିବା ନଦୀକୁ ଯାଇ ସ୍ନାନ କରିବେ ଏବଂ କିଛି ଜଳ ଆଣି ଯାହାକୁ ପୂର୍ବାବସ୍ଥାକୁ ଆଣିବାକୁ ଚାହୁଁଥ‌ିବେ ତାହାକୁ ତା’ ଉପରେ ସିଞ୍ଚନ କରିବେ ।

ତା’ପରେ ସେ ସେସବୁକୁ ତା’ର ପୂର୍ବ ଅବସ୍ଥାରେ ଫେରି ପାଇବେ । ସେ ତତ୍‌କ୍ଷଣାତ୍‌ ନଦୀରେ ସ୍ନାନ କଲେ ଓ ନଦୀରୁ ପାଣି ଆଣି ନିଜ ଝିଅ ଉପରେ ସିଞ୍ଚନ କଲେ ଓ ଅନ୍ୟାନ୍ୟ ଜିନିଷ ଉପରେ ମଧ୍ୟ ସିଞ୍ଚନ କଲେ । ଫଳରେ ସେଗୁଡ଼ିକ ପୂର୍ବ ଅବସ୍ଥାକୁ ଫେରିଆସିଲା। Marygoldଙ୍କୁ ଏହି ଦୁର୍ଭାଗ୍ୟ ଓ ଦୁଃଖଦ ଘଟଣା ବିଷୟରେ ଜଣାଇ ଦିଆଗଲା ନାହିଁ । ରାଜା ଅନୁଭବ କରିପାରିଲେ ଯେ ସୁନାର ସମୁଦ୍ର ଅପେକ୍ଷା ସେ ଆଣିଥିବା ପାଣି ମାଠିଆର ମୂଲ୍ୟ ବହୁତ ଅଧିକ । ଏହାପରେ ରାଜା ତାଙ୍କ ଝିଅକୁ ନେଇ ଖୁସିରେ ଜୀବନ ବିତାଇଲେ ।