Class 11

Odisha State Board CHSE Odisha Class 11 Math Solutions Chapter 3 Relations And Functions Ex 3(b) Textbook Exercise Questions and Answers.

CHSE Odisha Class 11 Math Solutions Chapter 3 Relations And Functions Exercise 3(b)

Question 1.
Let A = {a, b, c }, |B| = {1, 2}
(a) Determine all the relations from A to B and determine the domain, range, and inverse of each relation.
(b) Determine all the relations from B to A.
(c) Is there any relationship that is both a relation from A to B and B to A? How many?
(d) Of all the relations from A to B, identify which relations are many ones, one-many, and one-one and represent this diagrammatically.
Solution:
(a) A = {a, b, c}, B = {1, 2}
∴ A × B = {(a, 1), (a, 2), (b, 1), (b, 2), (c, 1), (c, 2)}
∴ |A × B| = 6
∴ |P(A × B)| = 2⁶ = 64
∴ There are 64 relations from A to B as any subset of A × B. The domain of these relations is any subset of A. The inverse of these relations is any subset of B × A.
(b) There are 64 relations from B to A as any sub-set of B x A is a relation from B to A.
(c) Φ is the only relation that is from A to B and from B to A.
(d) Some many-one relations are {(a, 1), (b, 1), (c, 1), (b, 2) (c, 2)}, {(a, 2), (b, 2), (c, 2)}.

Question 2.
Are the following sets related?
(i) Φ from A to B.
(ii) A × B from A to B.
(iii) A × Φ from A to Φ.
(iv) Φ × B from Φ to B.
(v) Φ × Φ from Φ to Φ.
(vi) Φ × C from A to B.
(vii) Φ × Φ from A to B.
Determine the domain range and inverse of each of the relations mentioned above
Solution :
(i) Φ from A to B is a relation.
(ii) A × B from A to B is a relation.
(iii) A × Φ from A to Φ is a relation.
(iv) Φ × B from Φ to B is a relation.
(v) Φ × Φ from Φ to Φ is a relation.
(vi) Φ × C from A to B is a relation.
(vii) Φ × Φ from A to B is a relation.
∴ Domain of Φ i.e. D_Φ = Φ
Range of Φ i.e., R_Φ = Φ
Similarly, D_{A × B} = A, R_{A × B} = β
D_{A × Φ} = Φ, R_{A × Φ} = Φ
D _{Φ × B} = Φ = Φ, R _{Φ × B} = Φ
D _{Φ × Φ} = Φ, R _{Φ × C} = Φ
D _{Φ × C} = Φ, R _{Φ × C} = Φ
D _{Φ × Φ} = Φ, R _{Φ × Φ} = Φ
The inverse of the above relations is Φ, B × A, Φ × A, B × Φ, Φ × Φ, C ×  Φ, and Φ × Φ respectively.

Question 3.
Express the following relations on A to B in each case in tabular form :
(i) A = {n ∈ N : n ≤ 10}, B = N
f = {(x, y) ∈ A × B : y = x²}
Solution:
A = {n ∈ N : n ≤ 10}
= {1, 2, 3,…..10}, B = N
∴ B = {1, 2, 3}
∴ f ={(x, y) ∈ A × B : y = x²}
= {(1, 1), (2, 4), (3, 9)…..(10, 100)}

(ii) A = B = R
∴ f = {(x, y) : x² + y² = 1 and |x – y| = 1}
Solution:
A = B = R
∴ f = {(x, y) : x² + y² = 1 and |x – y| = 1}
={(0, 1) (1, 0), (-1, 0), (0, -1)}

(iii) (1, 2, 3, 4), B = {1, 2, 3, 4, 5}
f = {x, y) : 2 divides 3x+y}
Solution:
A = {1, 2, 3, 4}, B = {1, 2, 3, 4, 5}
∴ f = {(x, y) : 2 divides 3x+y}
={(1, 1), (1, 3), (1, 5), (2, 2), (2, 4), (3, 1), (3, 3), (3,5), (4, 2), (4, 4)}

Question 4.
A and B are non-empty sets such that |A| = m, |B| = n. How many relations can be defined from A to B ? (Remember that the number of relations is the number of subsets of (A × B).
Solution:
|A| = m, |B| = n
⇒ |A × B| = mn
A relation is a subset of A to B
∴ Number of relations from A to B
= Number of subsets of A × B
= 2^mn (∴ |A × B| = mn)

Question 5.
Give an example of a relation f such that
(i) dom f – rng f (ii) dom f ⊂ rng f
(iii) dom f ⊃ rng f
(iv) f ∪ f^-1 = Φ
(v) f = f^-1
(vi) f ∪ f^-1 ≠ Φ
Solution:
Let A = { 1, 2, 3} = B
(i) Let f = {(x, y) ∈ A × B : x = y}
∴ Dom f = {1, 2, 3} = Range f

(ii) Let f = {(1, 1), (1, 2), (2, 3)}
on A = (1, 2, 3}
∴ Dom f = {1, 2} ⊂ { 1, 2, 3} = Range f

(iii) Do yourself

(iv) Let f = Φ
∴ f^-1 = Φ = f ∪ f^-1 = Φ

(v) Let f = {(x, y) ∈ A × B; x² + y² = 1}, where A = B = {1, – 1, 0}
= {(1, 0), (0, 1), (-1, 0), (0, -1)}
f^-1 = {(0, 1) (1, 0), (0, -1), (-1, 0)}
=f

(vi) Let f = {(1, 3), (3, 1)} on A = { 1, 2, 3}
∴ f^-1 = {3, 1), (1, 3)},
so that f ∩ f^-1 = Φ.

Question 6.
Let R = {(a, a³) I a is a prime number less than 10}
Fine (i) R, (ii) dom R, (iii) rng R (iv) R^-1 (v) dom R^-1 (vi) rng R^-1
Solution:
R = {(a, a³)} a is a prime number less than 10}
(i) R = {(2, 8), (3, 27), (5, 125), (7, 343)}
(ii) dom R = {2, 3, 5, 7}
(iii) rng R = {8, 27, 125, 343}
(iv) R^-1 = {(8, 2), (27, 3), (125, 5), (343, 7)}
(v) Dom R^-1 = {8, 27, 125, 343} = rng R
(vi) rng R^-1 = {2, 3, 5, 7} = dom R

Question 7.
Let A = {1, 2, 3, 4, 5, 6} and Let R be a relation on A defined by R = {(a, b)} a divides b
Find (i) R, (ii) dom R, (iii) rng R (iv) R^-1, (v) Dom R^-1 (vi) rng R^-1
Solution:
A = {1, 2, 3, 4, 6}
R on A is defined by
R = {(a, b) | a divides b}
(i) R = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 6) (2, 2), (2, 4), (2, 6), (3, 3), (3, 6), (4, 4), (6, 6)}
(ii) dom R = {1,2, 3, 4, 6} = A
(iii) rng R = {1, 2, 3, 4, 6} = A
(iv) R^-1 = {(1, 1), (2, 1), (3, 1), (4, 1), (6, 1), (2, 2) (4, 2), (6, 2), (3, 3), (6, 3), (4, 4), (6, 6)}
(v) dom R^-1 = {1, 2, 3, 4, 6} = A
(vi) rng R^-1 = {l, 2, 3, 4, 6} = A

Odisha State Board Elements of Mathematics Class 11 CHSE Odisha Solutions Chapter 4 Trigonometric Functions Ex 4(a) Textbook Exercise Questions and Answers.

CHSE Odisha Class 11 Math Solutions Chapter 4 Trigonometric Functions Exercise 4(a)

Question 1.
State which of the following is positive.
(i) cos 271°
Solution:
cos 271° is + ve as 271° lies in 4th quadrant.

(ii) sec 73°
Solution:
sec 73° is + ve as sec +ve in the 1st quadrant.

(iii) sin 302°
Solution:
sin 302° is- ve as sin is -ve in the 4th quadrant

(iv) cosec 159°
Solution:
cosec 159° is + ve as 159° lies in 2nd quadrant and cosec is +ve there.

(v) sec 199°
Solution:
sec 199° is – ve as 199° lies in the 3rd quadrant and sec is -ve there.

(vi) cosec 126°
Solution:
cosec 126° is + ve as cosec is +ve in 2nd quadrant.

(vii) cos 315°
Solution:
cos 315° is +ve as 315° lies in 4th quadrant and cos is +ve there.

(viii) cot 375°
Solution:
cot 375° is +ve as 375° lies in 1st quadrant.

Question 2.
Express the following as trigonometric ratios of some acute angles.
(i) sin 1185°
Solution:
sin 1185° = sin\(\left(13 \frac{\pi}{2}+15^{\circ}\right)\)
=(- 1) ^{\(\frac{13-1}{2}\)} cos 15° = cos 15°

(ii) tan 235°
Solution:
tan 235° = tan (180° + 45°) = tan 45°

(iii) sin (- 3333°)
Solution:
sin (-3333°) – -sin 3333°
= – sin\(\left(37 \frac{\pi}{2}+3^{\circ}\right)\)
= – (- 1) ^{\(\frac{27-1}{2}\)} cos 3° =- cos 3°

(iv) cot (- 3888°)
Solution:
cot (-3888°) = – cot 3888°
= – cot\(\left(43 \frac{\pi}{2}+18^{\circ}\right)\)
= – (- tan 18°) = tan 18°

(v) tan 458°
Solution:
tan 458° = tan\(\left(5 \frac{\pi}{2}+8^{\circ}\right)\) = – cot 8°

(vi) cosec (- 60°)
Solution:
cosec (- 60°) = – cosec 60°

(vii) cos 500°
Solution:
cos 500° = cos\(\left(5 \frac{\pi}{2}+50^{\circ}\right)\)
= – (-1) ^{\(\frac{5+1}{2}\)} sin 55° – sin 50°

(viii)sec 380°
Solution:
sec 380° = sec (360° + 20°)
= sec 20°

Question 3.
Find the domain of tangent and cotangent functions.
Solution:
Domain of tan x is R – \(\left\{\frac{(2 n+1) \pi}{2}, n \in Z\right\}\) as tangent is not defined for
x = \(\frac{(2 n+1) \pi}{2}\)
The domain of cot x is R – {nπ, n ∈ Z} as cotangent is not defined for x = nπ.

Question 4.
Determine the ranges of sine and cosine functions.
Solution:
The maximum and minimum values of sine and cosine are 1 and -1, respectively.
∴ Ranges of sine and cosine are [-1, 1].

Question 5.
Find a value of A when cos 2A = sin 3A
Solution:
cos 2A = sin 3A = cos (90° – 3A)
or, 2A = 90° – 3A
or, 5A = 90° or, A = 18°

Question 6.
Find the value of
cos 1°. cos 2° …..cos 100°
Solution:
cos 1° cos 2° …..cos 100°
= 0 as cos 90° is there which is zero.

Question 7.
Find the value of
cos 24° + cos 5° + cos 175° + cos 204° + cos 300°
Solution:
cos 24° + cos 5° + cos 175° + cos 204° + cos 300°
= cos 24° + cos 5° + cos (180° – 5°) + cos (180° + 24°) + cos (360°- 60°)
= cos 24° + cos 5° – cos 5° – cos 24° + cos 60° = cos 60° = 1/2

Question 8.
Evaluate
tan\(\frac{\pi}{20}\).tan\(\frac{3 \pi}{20}\).tan\(\frac{5 \pi}{20}\).tan\(\frac{7 \pi}{20}\).tan\(\frac{9 \pi}{20}\)
Solution:
tan\(\frac{\pi}{20}\).tan\(\frac{3 \pi}{20}\).tan\(\frac{5 \pi}{20}\).tan\(\frac{7 \pi}{20}\).tan\(\frac{9 \pi}{20}\)
= tan 9° tan 27° tan 45° tan 63° tan 81°
= tan 9°. tan 27°. 1 tan (90° – 27°). tan (90° – 9°)
= tan 9° tan 27° cot 27° cot 9°
= (tan 9°. cot 9°) x (tan 27°. cot 27°)
=1 × 1=1

Question 9.
Show that
\(\frac{\sin ^3\left(180^{\circ}+\mathbf{A}\right) \cdot \tan \left(360^{\circ}-\mathbf{A}\right) \sec ^2\left(180^{\circ}-\mathbf{A}\right)}{\cos ^2\left(90^{\circ}+\mathbf{A}\right){cosec}^2 A \cdot \sin \left(180^{\circ}-A\right)}\) = tan³ A
Solution:
L.H.S

= tan³A      (R.H.S)

Question 10.
If A = cos² θ + sin⁴ θ then prove that for all values of θ, 3/4 ≤ A ≤ 1.
Solution:
A = cos² θ + sin⁴ θ =1 – sin² θ sin⁴ θ
or, sin⁴ θ – sin² θ + (1 – A) = 0 …(1)
Eqn. (I) is quadratic in sin² θ.
∴ sin² θ = \(\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}\)
\(=\frac{1 \pm \sqrt{1-4(1-\mathrm{A})}}{2 \times 1}\)
Where a=1, b = – 1, c = 1 – A
∴ sin² θ = \(\frac{1 \pm \sqrt{4 A-3}}{2}\)
We know that sin20 is not negative and lies in [0, 1]
So, \(\sqrt{4 \mathrm{~A}-3}\) ≤ 1
⇒ 4A – 3 ≤ 1 ⇒ 4A ≤ 4 ⇒ A ≤ 1  …(2)
Again, since sin² θ is real,
b² – 4ac must be +ve
i.e., 4A – 3 ≥ 0 ⇒ A ≥ 3/4
∴ From (2) and (3),
We have 3/4 ≤ A ≤ 1          (Proved)

Odisha State Board CHSE Odisha Class 11 Math Solutions Chapter 6 Complex Numbers and Quadratic Equations Ex 6(a) Textbook Exercise Questions and Answers.

CHSE Odisha Class 11 Math Solutions Chapter 6 Complex Numbers and Quadratic Equations Exercise 6(a)

Question 1.
Multiply (2√-3 + 3√-2) by (4√-3 – 5√-2)
Solution:
(2√-3 + 3√-2) by (4√-3 – 5√-2)
= (2√3i + 3√2i) (4√3i – 5√2i)
= i² (2√3 + 3√2) (4√3 – 5√2)
= – 1(24 – 10√6 + 12√6 – 30)
= – 1(- 6 + 2√6) = 6 – 2√6

Question 2.
Multiply (3√-7 – 5√-2) (3√-2 + 5√-2)
Solution:
(3√-7 – 5√-2) (3√-2 + 5√-2)
= (3√7i – 5√-2i) (3√2i + 5√2i)
= i² (3√7- 5√2) (3√2 + 5√2)
= (- 1)8√2(3√7 – 5√2 )
= – 24√14 + 80

Question 3.
Multiply (√-1 +√-1) (a – b√-1)
Solution:
(√-1 +√-1) (a – b√-1)
= (i + i) (a – bi) = 2i(a – bi)
= 2ai – 2bi² = 2ai + 2b

Question 4.
Multiply (x – \(\frac{1+\sqrt{-3}}{2}\)) (x – \(\frac{1-\sqrt{3}}{2}\))
Solution:
(x – \(\frac{1+\sqrt{-3}}{2}\)) (x – \(\frac{1-\sqrt{3}}{2}\))
= (x + \(\frac{-1-i \sqrt{3}}{2}\)) (x + \(\frac{-1+i \sqrt{3}}{2}\))
= (x + ω) (x + ω²) = x² + ω²x + ωx + ω³
= x² + x (ω ² + ω) + 1 = x² –  x + 1

Question 5.
Express with rational denominator. \(\frac{1}{3-\sqrt{-2}}\)
Solution:
\(\begin{aligned}
& \frac{1}{3-\sqrt{-2}}=\frac{1}{3-\sqrt{2} \mathrm{i}}=\frac{3+\sqrt{2} \mathrm{i}}{(3-\sqrt{2} \mathrm{i})(3+\sqrt{2} \mathrm{i})} \\
& =\frac{3+\sqrt{2} \mathrm{i}}{9-2 \mathrm{i}^2}=\frac{3+\sqrt{2} \mathrm{i}}{9+2}=\frac{3+\mathrm{i} \sqrt{2}}{11}
\end{aligned}\)

Question 6.
\(\frac{3 \sqrt{-2}+2(-5)}{3 \sqrt{-2}-2 \sqrt{-2}}\)
Solution:

Question 7.
\(\frac{3+2 \sqrt{-1}}{2-5 \sqrt{-1}}+\frac{3-2 \sqrt{-1}}{2+5 \sqrt{-1}}\)
Solution:

Question 8.
\(\frac{a+x \sqrt{-1}}{a-x \sqrt{-1}}-\frac{a-x \sqrt{-1}}{a+x \sqrt{-1}}\)
Solution:

Question 9.
\(\frac{(x+\sqrt{-1})^2}{x-\sqrt{-1}}-\frac{\left(x-\sqrt{-1^2}\right)}{x+\sqrt{-1}}\)
Solution:

Question 10.
\(\frac{(a+\sqrt{-1})^3-(a-\sqrt{-1})^3}{(a+\sqrt{-1})^2-(a-\sqrt{-1})^2}\)
Solution:

Question 11.
Find the value of (- i)⁴ⁿ⁺³; when n is positive.
Solution:
(- i)⁴ⁿ⁺³
= (-i⁴ⁿ) (-i)³ = 1(- i³) = – (-i) = i

Question 12.
Find the square root of (a + 40i) + \(\sqrt{9-40 \sqrt{-i}}\)
Solution:

Question 13.
Express in the form of a + ib:
(i) \(\frac{3+5 i}{2-3 i}\)
Solution:

(ii) \(\frac{\sqrt{3}-i \sqrt{2}}{2 \sqrt{3}-i \sqrt{3}}\)
Solution

(iii) \(\frac{(\mathrm{I}+\mathrm{i})^2}{3-\mathrm{i}}\)
Solution:

(iv) \(\frac{(a+i b)^3}{a-i b}-\frac{(a-i b)^3}{a+i b}\)
Solution:

(v) \(\frac{1+i}{1-i}\)
Solution:
\(\frac{1+i}{1-i}\) = \(\frac{(1+i)^2}{2}=\frac{1-1+2 i}{2}\)
= i = 0 + i

Question 14.
Express the following points geometrically in the Argand plane.
(i) 1
Solution:
1 = 1 + i0 = (1, 0)

(ii) 3i
Solution:
3i = 0 + 3i = (0, 3)

(iii) – 2
Solution:
– 2 = – 2 + i0 = (- 2, 0)

(iv) 3 + 2i
Solution:
3 + 2i = (3, 2)

(v) – 3 + i
Solution:
– 3 + i = (- 3, 1)

(vi) 1-i
Solution:
1 – i = (1, – 1)

Question 15.
Show that the following numbers are equidistant from the origin:
√2 +i, 1 + i√2, i√3
Solution:
|√2 + i| = \(\sqrt{(\sqrt{2})^2+1^2}\) = √3
|1 + i√2| = \(\sqrt{1^2+(\sqrt{2})^2}\) = √3
and |i√3| = √3
∴ The points are equidistant from the origin.

Question 16.
Express each of the above complex numbers in the polar form.
Solution:

Question 17.
If 1, ω, ω² are the three cube roots of unity, prove that (1 + ω²)⁴ = ω
Solution:
L.H.S. = (1 + ω²)⁴ = (- ω)⁴ = ω⁴
= ω³ .ω = 1. ω = ω
= R.H.S. (Proved)

Question 18.
(1 – ω+ ω² ) (1 + ω – ω² ) = 4
Solution:
L.H.S. = (1 – ω+ ω² ) (1 + ω – ω² )
= (- ω – ω )(- ω² – ω² ) (∴ 1 + ω + ω² = 0)
= (- 2ω² – 2ω²) = 4ω³ = 4 = R.H.S.

Question 19.
(1 – ω) (1 – ω)² (1 – ω⁴) (1 – ω⁵) = 9
Solution:
L.H.S. =
(1 – ω) (1 – ω)² (1 – ω⁴) (1 – ω⁵)
= (1 – ω) (1 – ω²) (1 – ω) (1 – ω²)
= (1 – ω)² (1 – ω²)²
= {(1 – ω) (1 – ω²)}²
= (1 – ω² – ω + ω³)²
= {3 – (ω² + ω + 1)}²
= (3)² = 9 = R.H.S.

Question 20.
(2 + 5ω + 2ω² )⁶ = (2 + 2ω + 5ω² )⁶ =729
Solution:
L.H.S. = (2 + 5ω + 2ω² )⁶
(2 + 2ω² + 5ω)⁶ = {2(1 + ω² ) + 5ω}⁶
(- 2ω + 5ω)⁶ = (3ω)⁶ = 729ω⁶ = 729
Again, (2 + 2ω + 5ω² )⁶
= {2(1 + ω) + 5ω² )⁶
= (- 2ω² + 5ω² )⁶ = (3ω²)⁶
= 729ω¹² =729
∴ (2 + 5ω + 2ω² )⁶ = (2 + 2ω + 5ω² )⁶ =729

Question 21.
(1 – ω + ω² ) (1 – ω² + ω⁴) (1 – ω⁴ +ω²)  ….to 2n factors = 2²ⁿ
Solution:
L.H.S. = (1 – ω + ω² ) (1 – ω² + ω⁴) (1 – ω⁴ +ω²)  ….to 2n factors = 2²ⁿ
= (- ω – ω) (1 – ω² + ω) (1 – ω + ω² )  ….to 2n factors = 2²ⁿ
= (- 2ω) (- ω² – ω² ) (- ω – ω)  ….to 2n factors = 2²ⁿ
= [(- 2ω)(- 2ω) … to n factors] × [(- 2ω²)(- 2ω²) …. to n factors]
= (- 2ω)ⁿ × (- 2ω²)ⁿ = (4ω³)ⁿ = 4ⁿ = 2²ⁿ
R.H.S. (Proved)

Question 22.
Prove that x³ + y³ + z³– 3xyz
= (x + y + z) (x + ωy + ω²z) (x + yω² + zω)
Solution:
R.H.S. = (x + y + z) (x + ωy + ω²z) (x + yω² + zω)
= (x + y + z) (x + xyω² + zxω + xyω + y²ω³+ yz ω2 + zxω² + yzω⁴ + z²ω³)
= (x + y + z) [x² + y² + z² + xy (ω² + ω) +yz (ω² + ω) + zx(ω² + ω)]
= (x + y + z) (x² + y² + z² – xy – yz – zx)
= x³ + y³ + z³ – 3xyz = L.H.S. (Proved)

Question 23.
If x = a + b, y = aω + b ω², z = aω² + bω show that
(1) xyz = a³ + b³
Solution:
L.H.S. = xyz
= (a + b) (a ω + b ω²) (a ω² + b ω)
= (a + b) (a² ω³ + ab ω² + ab ω⁴ + b² ω³)
= (a + b) {a² + b² + ab(ω² + ω)}
= (a + b) (a² – ab + b²) = a³ + b³ = R.H.S. (Proved)

(2) x² + y² + z² = 6ab
Solution:
L.H.S. = x² + y² + z²
= (a + b)² + (a ω + b ω²)² (a ω² + b ω)²
= a² + b² + 2ab + a² ω² + b² ω⁴ + 2ab ω³ + a² ω⁴ + b² ω² + 2ab ω³
= a² + a² ω² + a² ω + b² + b² ω + b² ω² + 2ab + 2ab + 2ab
= a²(1 + ω² + ω) + b² (1 + ω + ω²) + 6ab
= 0 + 0 + 6ab = 6ab = R.H.S. (Proved)

(3) x³ + y³ + z³ = 3(a³ + b³)
Solution:
L.H.S. = x³ + y³ + z³
= (a + b)³ + (a ω + b ω²)³ + (aω² + b²)³
= a³ + 3a²b + 3ab² + b³ + a³ ω³ + 3a²b² bω² + 3a ωb² ω⁴ + b³ ω⁶ + a³ ω⁶ + 3a² ω⁴bω + 3a ω²b² ω² + b³ ω³
= a³ + a³ + a³ + b³ + b³ + b³ + 3a²b(1 + ω⁴ + ω⁵) + 3ab² (1 + ω⁵ + ω⁴)
= 3a³ + 3b³ + 0 + 0 = 3 (a³ + b³) = R.H.S. (Proved)

Question 24.
If ax + by + cz = X, cx + by + az = Y, bx + ay + cz = Z
show that (a² + b² + c² – ab – bc- ca) (x² +y² + z² – xy – yz – zx) = X² + Y² + Z² – YZ – ZX – XY
Solution:
L.H.S.
= (a² + b² + c² – ab – bc – ca) (x² + y² + z² – xy – yz – zx)
= (a + b ω + cω²) (a + bω² + cω) (x + yω + zω)² (x + yω² + z ω) (Refer Q.No.22)
= {(a + bω + cω²) (x + yω + zω²)} {(a + bω² + cω) (x + yω² + zω)}
= (ax + ayω + azω² + bx ω + byω² + bzω³ + cxω² + cyω³ + czω⁴) × (ax + ay ω² + azω + bxω² + byω⁴ + bzω³ + cxω + cyω³ + czω²)
= {(ax + cy + bz) + (cx + by + az) ω + (bx + ay + cz) ω²} x (ax + cy + bz) + (cx + by + az) ω + (bx + ay + cz) ω²}
= (X + Yω² + Zω) (X + Yω + Zω²)
= X² + Y² + Z² – XY – YZ – ZX (Refer Q. No. 22)
(Where X = ax + cy + bz.
Y = cx + by + az.
Z = bx + ay + cz).

Odisha State Board Elements of Mathematics Class 11 Solutions CHSE Odisha Chapter 5 Principles Of Mathematical Induction Ex 5 Textbook Exercise Questions and Answers.

CHSE Odisha Class 11 Math Solutions Chapter 5 Principles Of Mathematical Induction Exercise 5

Prove the following by induction.

Question 1.
1 + 2 + 3 + …… + n = \(\frac{n(n+1)}{2}\)
Solution:
Let p_n be the given statement

Question 2.
1² + 2² + …… + n² = \(\frac{n(n+1)(2 n+1)}{6}\)
Solution:
Let p_n be the given statement
when n = 1
1² =1 = \(\frac{1(1+1)(2 \times 1+1)}{6}\)
P₁ is true
Let P_k be true.
i.e. 1² + 2² + … + k² = \(\frac{k(k+1)(2 k+1)}{6}\)
we shall prove P_k + 1 is true i.e., 1² + 2² + … + k² + (k + 1)²
\(=\frac{(k+1)(k+2)(2 k+3)}{6}\)

∴ P_k+1 is true
∴ P_n is true of all values of n∈N.

Question 3.
1 + r + r²+ …. + rⁿ = \(\frac{r^{n+1}-1}{r-1}\)
Solution:
Let P_n be the given statement

Question 4.
5ⁿ – 1 is divisible by 4.
Solution:
Let P_n = 5ⁿ – 1
When n = 1,
5¹ – 1 = 4 is divisible by 4.
∴ P₁, is true.
Let P_k be true i.e.,
5^k – 1 is divisible by 4.
Let 5^k+1 – 1 = 4m, me Z
Now 5k + 1 – 1 = 5^k. 5 – 5 + 4
= 5 (5^k – 1) + 4
= 5 × 4m + 4 = 4 (5m + 1)
which is divisible by 4.
∴ P_k+1 is true.
∴ P_n is true for all values of n ∈ N

Question 5.
7²ⁿ + 2^3n-3  3^n-1 is divisible by 25 for any natural number n > 1.
Solution:
Let 7²ⁿ + 2^3n-3 . 3^n-1
when n = 1, 7¹ + 2⁰ . 3⁰ ⇒ 49 + 1 = 50
which is divisible by 25.
∴ P₁ is true. Let P_k be true.
i.e., 72^k + 2^3k-3 3^k-1 is divisible by 35
Now \(7^{2 \overline{k+1}}+2^{3 \overline{k+1}-3} 3^{\overline{k+1}-1}\)
=7^2k+2 + 2^3k. 3^k
= 7^2k. 7² + 2^3k-3 2³. 3^k-1. 3¹
= 7^2k. 49 + 2^3k-3. 3^k-1. 24
= 7^2k (25 + 24) + 24. 2^3k-3. 3^k-1
= 7^2k. 25 + 24 (7^2k + 2^3k-3. 3^k-1)
= 7^2k. 25 + 24 × 25m
Which is divisible by 25 (∵ P_k is true)
∴ P_k+1 is true
∴ P_n is true for all values of n > 1.

Question 6.
7. 5^2n-1 + 23ⁿ⁺¹ is divisible by 17 for every natural number n ≥ 1.
Solution:
Let P_n = 7. 5²ⁿ – 1 + 2³ⁿ⁺¹.
When n = 1, 7.5 + 2⁴ = 35 + 16 = 51
Which is divisible by 17.
P₁, is true.
Let P_k be true i.e., 7.5^2k-1 + 2^3k+1  is divisible by 17.
Let 7.5^2k-1 + 2^3k+1 = 17 m, m ∈ Z
Now, \(7.5^{2 \overline{k+1}-1}+2^{3 \overline{k+1}+1}\)
= 7.5^2k-1 + 2^3k+4
= 7.5^2k-1 . 5² + 2^3k+1 . 2³
= 25. 7. 5^2k-1 + 8. 2^3k+1
= (17 + 8) 7.5^2k-1 + 8. 2^3k+1
= 17. 7. 5^2k-1 + 8 (7. 5^2k-1 + 2^3k+1)
= 17 × 7 × 5^2k-1 + 8 × 17m
Which is divisible by 17.
Hence P_k+1. is true.
∴ P_n is true for all values of n ≥ 1.

Question 7.
4ⁿ⁺¹ + 15n + 14 is divisible by 9 for every natural number n ≥ 0.
Solution:
Let P_n = 4ⁿ⁺¹+ 15 n + 14
when n = 1, 4² + 15 + 14 = 45 is divisible by 9.
∴ P₁ is true. Let P_k be true.
i.e., 4^k+1 + 15k + 14 is divisible by 9.
Now, 4^k+1+1 + 15 (k + 1) + 14
= 4^k+2 + 15k + 29
= 4^k+1. 4 + 60k + 56 – 45k – 27
= 4 (4^k+1 + 15k + 14) – 9 (5k + 3)
Which is divisible by 9.
∴ P_k+1, is true.
∴ P_n is true for all values of n ≥ 0.

Question 8.
3^(2n-1) + 7 is divisible by 9 for every natural number n ≥ 2.
Solution:
Let P_n = 3^2(n-1) + 7
When n = 2. 3² + 7 = 16 is divisible by 8.
∴ P₂ is true.
Let P_k be true.
i.e., 5^2(k-1) + 7 is divisible by 8.
Let 3^2k-2 + 7 = 8m. m ∈ Z.
Now \(3^{2(\overline{k+1}-1)}\) + 7 = 3^2k + 7
= 3^2k-2. 3² + 63 – 56
= 9(3^2k-2 + 7) – 56
= 9 × 8m – 56 = 8 (9m – 7)
Which is divisible by 8.
P_k+1 is true.
P_n is true for all values on n ≥ 2.

Question 9.
5^(2n-4)  – 6n + 32 is divisible by 9 for every natural number n ≥ 5.
Solution:
Let P = 5^2(n-4) – 6n + 32
For n = 5, P₅ = 5² – 6. 5 + 32
= 25 – 30 + 32 = 27
Which is divisible by 9.
Hence P₅ is true.
Let P_k is true.
Let P_k is divisible by 9.
Let P_k = 5^2(k-4) – 6k + 32 = 9., m ≥ Z
5^2k+2 = 576 m + 24k  + 25 … (1)
we shall prove that P_k+1 is true.
Now 5^2(K+1)+2 – 24(k+1) – 25
= 5² (5^2k+2) – 24k – 24 – 25
= 5²[576m + 24k + 25] – 24k – 24 – 25
= 25 × 576 m + 25 × 24k + 25 × 25 – 24k – 24 – 25
= 25 × 576 m + 576 k + 576
= 576 [25 m + k + 1]
which is divisible by 576
∴ P_k+1 is true.
So by the method of induction P_n is true for all n.
i.e., 5²ⁿ⁺²  – 24n – 25 is divisible by 576 for all n ∈ N.
Hence P_k+1 is true.
So by methods of induction P_n is true.
i.e., 5²ⁿ⁺² – 24n – 25 is divisible by 576 for all n.

Question 10.
\(\frac{1}{1.2}+\frac{1}{2.3}+\ldots+\frac{1}{n(n+1)}=\frac{n}{n+1}\)
Solution:
when n = 1,

Question 11.
1.3 + 2.4 + 3.5 + …….. + n(n + 2) = \(\frac{n(n+1)(2 n+7)}{6}\)
Solution:
when n = 1,
we have 1.3 = 3 = \(\frac{3 \times 6}{6}\)
\(=\frac{1 \times 2 \times 9}{6}=\frac{1(1+1)(2 \times 1+7)}{6}\)

Question 12.
xⁿ – yⁿ = (x – y)(x^n-1 + x^n-2 y + … + xy^n-2 + y^n-1); x, y ∈ R [Hint : Write xⁿ⁺¹ – yⁿ⁺¹ = x(xⁿ – yⁿ) + yⁿ(x – y)]
Solution:
Let p(n) is
xⁿ – yⁿ = (x – y)(x^n-1 + x^n-2 y + … + xy^n-2 + y^n-1); x, y ∈ R
Step – 1:
For n = 2
x² – y² = (x – y) (x + y) (True)
∴ P(1) is true.
Step – 2:
Let P(k) is true.
i.e., x^k – y^k = (x – y)(x^k-1 + x^k-2y + … +xy^k-2 + y^k-1)
Step – 3:
Let us prove P_k+1 is true.
i.e., x^k+1 – y^k+1 = (x – y) (x^k + x^k-1y + … (xy^k-1 + y^k)
L.H.S. = x^k+1 – y^k+1
= x^k+1 – xy^k + xy^k – y^k+1
= x(x^k – y^k) + y^k (x – y)
= x(x – y)(x^k-1 + x^k-2 y + … + xy^k-2 + y^k-1) + y^k(x – y) [by (1)]
= (x – y) [x^k + x^k-1 y + … + xy^k-2 + xy^k-1 + y^k]
= R.H.S.
∴ P_(k+1) is true.
Step – 4:
By Principle Of Mathematical Induction P(n) is true for all n ∈ N.

Question 13.
1 + 3 + 5 + ……. +(2n – 1) = n²
Solution:
Let P(n) is : 1 + 3 + 5 + ……. +(2n – 1) = n².
Step – 1:
For n = 2
L.H.S. = 1 + 3 = 4 = 2² (R.H.S)
∴ P(1) is true.
Step – 2:
Let P(k) is true.
i.e., 1 + 3 + 5 … + (2k – 1) = k² …(1)
Step – 3:
We will prove that P(k + 1) is true
i.e., we want to prove.
1+ 3 + 5 + … + (2k – 1) + (2k + 1) = (k + 1)²
L.H.S. = 1 + 3 + 5 + … + (2k – 1) + (2k + 1)
= k² + 2k + 1          [By – (1)]
= (k + 1)² = R.H.S.
Step – 4:
By the Principle of Mathematical Induction P(n) is true for all n.
i.e., 1 + 3 + 5 ….+ (2n – 1) = n²

Question 14.
n > n; n is a natural number.
Solution:
Let P(n) is 2ⁿ > n
Step – 1:
2¹ > 1 (True)
∴ P(1) is true.
Step – 2:
Let P(k) is true.
⇒ 2^k > k
Step – 3:
We shall prove that P(k + 1) is true
i.e., 2^k+1 > k + 1
Now 2^k+1 = 2.2k > 2k ≥ k + 1 for k ∈ N.
∴ 2^k+1 > k + 1
⇒ P(k + 1) is true.
Step – 4:
By the Principle of Mathematical Induction P(n) is true for all n.
i.e., 2ⁿ > n for n ∈ N

Question 15.
(1, 2, 3 … n)³  > 8 (1³ + 2³ + 3³ + … + n³), for n > 3.
Solution:
Let P(n) is
(1, 2, 3 … n)³  > 8 (1³ + 2³ + 3³ + … + n³), for n > 3.
Step – 1:
For n = 4
(1. 2. 3. 4)³ = 24³ = 13824
8(1³+ 2³ + 3³ + 4³) = 808
∴ (1. 2 . 3 . 4)³ > 8(1³ + 2³ + 3³ + 4³)
∴ P(4) is true.
Step- 2:
Let P(k) is true.
(1. 2. 3…….k)³ > 8(1³ + 2³ + 3³ + …… + k³)
Step – 3:
We shall prove that P_(k+1) is true.
i.e., (1. 2. 3. …….. k(k+1))³ > 8(1³ + 2³ + … + k³ + (k + 1)³)
Now (1. 2. 3. …….. k(k + 1)³)
= (1. 2. 3 … k)³ (k + 1)³
> 8 (1³ + 2³ + … k³) (k + 1)³
> 8 (1³ + 2³ + … k³) + 8(k + 1)³
= 8 (1³ + 2³ + … + k³ + (k + 1)³)
P_(k+1) is true.
Step – 4:
By the Principle of Mathematical Induction P(n) is true for all n ∈ N and n > 3.

Question 16.
\(\frac{1}{n+1}+\frac{1}{n+2}+\ldots+\frac{1}{3 n+1}>\) for every positive integer n.
Solution:
Let P(n) is
\(\frac{1}{n+1}+\frac{1}{n+2}+\ldots+\frac{1}{3 n+1}>\)



Step-4:
By the Principle of Mathematical Induction P(n) is true for all n ∈ N.

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

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

Question 1.
What is the total number of functions that can be defined from the set {1, 2} to the set {1, 2, 3}?
Solution:
The total number of functions that can be defined from the set {1, 2} to the set {1, 2, 3} is 3² = 9.

Question 2.
A die of six faces marked with the integers 1, 2, 3, 4, 5, and 6 one on each face is thrown twice in succession, what is the total number of outcomes thus obtained?
Solution:
A die of six faces marked with the integers 1, 2, 3, 4, 5, and 6 one on each face, is thrown twice in succession.
∴ The total number of outcomes is 6² = 36.

Question 3.
Five cities A, B, C, D, and E are connected to each other by straight roads. What is the total number of such roads?
Solution:
Five cities A, B, C, D, and E are connected to each other by straight roads.
∴ The number of such roads is \(\frac{{ }^5 \mathrm{P}_2}{2 !}\) = 10

Question 4.
What is the total number of different diagonals of a given pentagon?
Solution:
The total number of different diagonals of a given pentagon is
\(\frac{{ }^5 \mathrm{P}_2}{2 !}\) – 5 = 10 – 5 = 5

Question 5.
There are two routes joining city A to city B and three routes joining B to city C. In how many ways can a person perform a journey from A to C?
Solution:
There are two routes joining city A to city B and three routes joining B to city C.
∴ By the fundamental principle of counting, the total number of journeys in which a person can perform from A to C is 2 × 3 = 6.

Question 6.
How many different four-lettered words can be formed by using the four letters a, b, c, and d, while the letters can be repeated?
Solution:
The number of different words that can be formed by using the four letters a, b, c, and d, while the letters can be repeated is 4⁴ = 256.

Question 7.
What is the sum of all three-digit numbers formed by using the digits 1, 2, and 3?
Solution:
The 3-digit numbers that can be formed by using the digits 1, 2, and 3 are 123, 132, 213, 231, 312, and 321.
∴ Their sum = 1332.

Question 8.
How many different words with two letters can be formed by using the letters of the word JUNGLE, each containing one vowel and one consonant?
Solution:
The word ‘Jungle’ contains 2 vowels and 4 consonants. Each word contains one vowel and one consonant.
The number of different words formed.
2 × ⁴P₁ × ²P₁ = 2 × 4 × 2 = 16

Question 9.
There are four doors leading to the inside of a cinema hall. In how many ways can a person enter into it and come out?
Solution:
There are four doors leading to the inside of a cinema hall. A person can enter into it and come out in 4² = 16 different ways. (By the principle of counting.)

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Odisha State Board Elements of Mathematics Class 11 CHSE Odisha Solutions Chapter 10 Sequences and Series Ex 10(a) Textbook Exercise Questions and Answers.

CHSE Odisha Class 11 Math Solutions Chapter 10 Sequences and Series Exercise 10(a)

Question 1.
Which of the following in a sequence?
(i) f(x) = [x], x ∈ R
(ii) f(x) = |x|, x ∈ R
(iii) f(x) = \(\sqrt[n]{\pi}\) ,n∈ N
Solution:
(iii) f(x) = \(\sqrt[n]{\pi}\) ,n∈ N is a sequence f(n) : N → X, X ⊂ R.

Question 2.
Determine if (t_n) is an arithmetic sequence if :
(i) t_n = an² + bn
Solution:
t_n = an² + bn
⇒ t_n+1 = a(n + 1)² + b(n – 1)
⇒ t_n+1 – t_n = a{(n + 1)² – n²} + b{n + 1 – n}
= a(2n + 1) + b
Which is not independent of n.
∴ (t_n) is not an A.P.

(ii) t_n = an + b
Solution:
t_n = an + b
⇒ t_n+1 = a(n + 1) + b
Now t_n+1 – t_n
= {a(n + 1) + b} – {an + b}
= a (constant)
∴ (t_n) is an arithmetic sequence.

(iii) t_n = an² + b
Solution:
t_n = an² + b
⇒ t_n+1 = a(n + 1)² + b
∴ t_n+1 – t_n = a[(n + 1)² – n²] + b – b
= a(2n + 1)
(does not independent of n)
∴ (t_n) is not an arithmetic sequence.

Question 3.
If a geometric series converges which of the following is true about its common ratio r?
(i) r > 1
(ii) -1 < r < 1
(iii) r > 0
Solution:
(ii) -1 < r < 1

Question 4.
If an arithmetic series ∑tn converges, which of the following is true about t_n?
(i) t_n < 1
(ii) |t_n| < 1
(iii) t_n = 0
(iv) t_n → 0
Solution:
(iii) t_n = 0

Question 5.
Which of the following is an arithmetic-geometric series?
(i) 1 + 3x + 7x² + 15x³+ ….
(ii) x + \(\frac{1}{2}\)x + \(\frac{1}{3}\)x² + ….
(iii) x + (1 + 2)x² + (1 + 2 + 3)x³ +…
(iv) x + 3x² + 5x³ + 7x⁴ + …
Solution:
(iv) x + 3x² + 5x³ + 7x⁴ + … is an arithmetic geometric series with a = 1, d = 2, r = x.

Question 6.
For an arithmetic sequence (t_n) t_p = q, t_q = p, (p ≠ q), find t_n.
Solution:
t_p = q ⇒ a + (p – 1)d = q    ……(1)
t_q = p ⇒ a + (q – 1)d = p    ……(2)
From (1) and (2) we have (p – q)d = q – p
⇒ d = (-1)
Putting d = (-1) in (1)
we have a = p + q – 1
∴ t_n = a + (n – 1)d
= (P + q – 1) + (n – 1) (-1)
= p + q – n

Question 7.
For an arithmetic series, ∑a_n S_p = q and S_q = p (p ≠ q) find S_p+q
Solution:
S_p = q and S_q = p

Question 8.
The sum of a geometric series is 3. The series of squares of its terms have a sum of 18. Find the series.
Solution:

Question 9.
The sum of a geometric series is 14, and the series of cubes of its terms have a sum of 392 find the series.
Solution:


∴ The series is \(\sum_{n=1}^{\infty} \frac{7}{2^{n-1}}\)

Question 10.
Find the sum as directed
(i) 1 + 2a + 3a² + 4a³ + …..(first n terms(a ≠ 1))
Solution:

(ii) 1 + (1 + x)y + (1 + x + x²)y² + …..(to infinity)
Solution:
Let S_∞ = 1 + (1 + x)y + (1x + x²)y² + …
⇒ S_n = 1 + (1 + 1 + x)y + (1 + x + x²)y² + ……+(1 + x + …. x^n-1)y^n-1

(iii) 1 + \(\frac{3}{5}\) + \(\frac{7}{25}\) + \(\frac{15}{125}\) + \(\frac{31}{625}\) + …..(to infinity)
Solution:

(iv) 1 + 4x + 8x² + 13x³ + 19x⁴ + …..(to infinity). Assuming that the series has a sum for |x| < 1.
Solution:

(v) 3.2 + 5.2² + 7.2³ + …..(first n terms)
Solution:
Sn = 3.2 + 5.22 + 7.3 + ….n terms = 2[3 + 5.2 + 7.22 + ….n terms]

Question 11.
Find the sum of the infinite series.
(i) \(\frac{1}{1 \cdot 2}+\frac{1}{2 \cdot 3}+\frac{1}{3 \cdot 4}+\ldots\)
Solution:

= 1 – \(\frac{1}{n+1}\)
∴ \(S_{\infty}=\lim _{n \rightarrow \infty} S_n=1\)

(ii) \(\frac{1}{1 \cdot 2 \cdot 3}+\frac{1}{2 \cdot 3 \cdot 4}+\frac{1}{3 \cdot 4 \cdot 5}+\ldots\)
Solution:
\(\frac{1}{1 \cdot 2 \cdot 3}+\frac{1}{2 \cdot 3 \cdot 4}+\frac{1}{3 \cdot 4 \cdot 5}+\ldots\)

(iii) \(\frac{1}{2 \cdot 5 \cdot 8}+\frac{1}{5 \cdot 8 \cdot 11}+\frac{1}{8 \cdot 11 \cdot 14}+\ldots\)
Solution:
Here t_n = \(\frac{1}{(3 n-1)(3 n+2)(3 n+5)}\)
The denominator of t_n is the product of 3 consecutive terms of A.P. Now multiplying and dividing by (3n + 5) – (3n – 1) we have
\(t_n=\frac{(3 n+5)-(3 n-1)}{6(3 n-1)(3 n+2)(3 n+5)}\)

(iv) \(\frac{3}{1^2 \cdot 2^2}+\frac{5}{2^2 \cdot 3^2}+\frac{7}{3^2 \cdot 4^2}+\ldots\) [Hint : take t_n = \(\frac{2 n+1}{n^2(n+1)^2}=\frac{(n+1)^2-n^2}{n^2(n+1)^2}\)]
Solution:

(v) \(\frac{1}{1 \cdot 5}+\frac{1}{3 \cdot 7}+\frac{1}{5 \cdot 9}+\ldots .\)
Solution:

Question 12.
Find S_n for the series.
(i) 1.2 + 2.3 + 3.4 + ….
Solution:

(ii) 1.2.3 + 2.3.4 + 3.4.5 + …
Solution:

(iii) 2.5.8 + 5.8.11 + 8.11.14 +…
Solution:

(iv) 1.2.3.4 + 2.3.4.5 + 3.4.5.6 + …
[Hint : t_n = (3n – 1) (3n + 2)(3n + 5)]
Solution:

(v) 1.5 + 2.6 + 3.7 + …
[Hint: t_n = n(n + 4) is not a product of two successive terms of an A.P. for the term following n should be n+1, not n+4. So the method of previous exercises is not applicable. Instead, write t_n = n² + 4n and find S_n = \(\sum_{k=1}^n k^2+4 \sum_{k=1}^n k\) applying formulae]
Solution:

(vi) 2.3 + 3.6 + 4.11 + …
[Hint : Take t_n = (n + 1) (n² + 2)]
Solution:

(vii) 1.3² + 2.5² + 3.7² + ….
Solution:

Question 13.
Find the sum of the first n terms of the series:
(i) 5 + 6 + 8 + 12 + 20 + …
Solution:

(ii) 4 + 5 + 8 + 13 + 20 + …
Solution:

Question 14.
(i) Find the sum of the product of 1,2,3….20 taken two at a time. [Hint: Required sum = \(\frac{1}{2}\left\{\left(\sum_{k=1}^{20} k\right)^2-\sum_{k=1}^{20} k^2\right\}\)]
Solution:
We know that
(x₁ + x₂ + x₃ + …. + x_n)²
= (x²₁ + xⁿ₂ + … + x²_n) + 2 (Sum of all possible Products taken two at a time)
∴ 2 (Sum of products of 1. 2, 3,…… 20 taken two at a time)
= (1 + 2 + 3 + … 20)² – (1² + 2² + … + 20²)
\(=\left(\frac{20 \times 21}{2}\right)^2-\frac{20(20+1)(40+1)}{6}\)
= (210)² – 70 × 41
= 44100 – 2870 = 41230
∴ The required sum = \(\frac{41230}{2}\) = 20615

(ii) Do the same for 1, 3, 5, 7,….19.
Solution:
Here the required sum

Question 15.
If a = 1 + x + x² + ….. and b = 1 + y + y² + ….|x| <  1 and |y| <  1, then prove that 1 – xy + x²y² + x³y³ + ….. =  \(\frac{a b}{a+b-1}\)
Solution:

Question 16.
If a, b, c are respectively the pth, qth, rth terms of an A.P., then prove that a(q – r) + b(r – p) + c(p – q) = 0
Solution:
Let the first term of an A.P. = A and the common difference = D.
According to the question
t_P = a, t_q = b, t_r = c
⇒ A + (p – q)D = a      …..(1)
A + (q – 1)D = b           …..(2)
A + (r – 1)D = c            …..(3)
L.H.S = a(q – r) + b (r – p) + c (p – q)
= (A + (p – 1)D) (q – r) + (A + (q – 1)D)
(r- p) + (A + (r – 1)D) (p – q)
= A (q – r + r – p + p – q) + D [(p – 1)
(q – r) + (q – 1) (r – p) + (r – 1) (p – q)]
= D (pq – pr + qr – pq + pr – qr) – D (q – r + r – p + p – q) = 0

Question 17.
If \(\frac{1}{a}, \frac{1}{b}, \frac{1}{c}\) are in A.P. and a + b + c ≠ 0, prove that \(\frac{\mathbf{b}+\mathbf{c}}{\mathbf{a}}, \frac{\mathbf{c}+\mathbf{a}}{\mathbf{b}}, \frac{\mathbf{a}+\mathbf{b}}{\mathbf{c}}\) are in A.P.
Solution:

Question 18.
If a², b², c² are in A.P. Prove that \(\frac{1}{b+c}, \frac{1}{c+a}, \frac{1}{a+b}\) are in A.P.
Solution:

Question 19.
If \(\frac{b+c}{a}, \frac{c+a}{b}, \frac{\mathbf{a}+\mathbf{b}}{c}\) are in A.P.,prove that \(\frac{\mathbf{1}}{\mathbf{a}}, \frac{\mathbf{1}}{\mathbf{b}}, \frac{\mathbf{1}}{\mathbf{c}}\) are inA.P.given a + b + c ≠ 0
Solution:

Question 20.
If (b – c)², (c – a)², (a – b)² are in A.P., prove that \(\frac{1}{b-c}, \frac{1}{c-a}, \frac{1}{a-b}\) are in A.P.
Solution:

Question 21.
If a, b, c are respectively the sum of p, q, r terms of an A.P., prove that \(\frac{a}{p}(q-r)+\frac{b}{q}(r-p)+\frac{c}{r}(p-q)\) = 0
Solution:

Question 22.
If a, b, c,d are in G.P., prove that (a² + b² + c²)(b² + c² + d²) = (ab + bc + cd)².
Solution:
Let a, b, c, and d are in G.P.
Let the common ratio = r
⇒ b = ar, c = ar², d = ar³
LHS = (a² + b² + c²) (b² + c² + d²)
= (a² + a²r² + a²r⁴) (a²r² + a²r⁴ + a²r⁶)
= a⁴r² (1 + r² + r⁴)²
= (a²r + a²r³ + a²r⁵)²
= (a.ar + ar.ar² + ar².ar³)²
= (ab + bc + cd)² = R.H.S. (proved)

 

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

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

Question 1.
Find the distance between the following pairs of points.
(i) (3, 4), (-2, 1);
Solution:
Distance between points (3, 4) and (-2, 1) is
\(\sqrt{(3+2)^2+(4-1)^2}=\sqrt{25+9}=\sqrt{34}\)

(ii) (-1, 0), (5, 3)
Solution:
The distance between the points (-1, 0) and (5, 3) is
\(\sqrt{(-1-5)^2+(0-3)^2}\)
= \(\sqrt{36+9}=\sqrt{45}=3 \sqrt{5}\)

Question 2.
If the distance between points (3, a) and (6, 1) is 5, find the value of a.
Solution:
Distance between the points. (3, a) and (6, 1) is
\(\sqrt{(3-6)^2+(a-1)^2}=\sqrt{9+(a-1)^2}\)
∴ \(\sqrt{9+(a-1)^2}=5\)
or, 9 + (a – 1)² = 25
or, (a – 1)² = 16
or, a – 1 = ± 4
a = 1 ± 4 = 5 or, – 3

Question 3.
Find the coordinate of the points which divides the line segment joining the points A (4, 6), B (-3, 1) in the ratio 2: 3 internally. Find also the coordinates of the point which divides \(\overline{\mathbf{A B}}\) in the same ratio externally.
Solution:

Question 4.
Find the coordinates of the mid-point of the following pairs of points.
(i) (-7, 3), (8, -4);
Solution:
Mid-point of the line segment joining the points (-7, 3) and (8, -4) are \(\left(\frac{-7+8}{2}, \frac{3-4}{2}\right)=\left(\frac{1}{2},-\frac{1}{2}\right)\)

(ii) (\(\frac{3}{4}\), -2), (\(\frac{-5}{2}\), 1)
Solution:
Mid-point of the line segment joining the points. (\(\frac{3}{4}\), -2) and (\(\frac{-5}{2}\), 1) is,
\(\left(\frac{\frac{3}{4}-\frac{5}{2}}{2}, \frac{-2+1}{2}\right)=\left(\frac{-7}{8}, \frac{-1}{2}\right)\)

Question 5.
Find the area of the triangle whose vertices are (1, 2), (3, 4) (\(\frac{1}{2}\), \(\frac{1}{4}\))
Solution:
Area of the triangle whose vertices are (1, 2), (3, 4) and (\(\frac{1}{2}\), \(\frac{1}{4}\)) is

Question 6.
If the area of the triangle with vertices (0, 0), (1, 0), (0, a) is 10 units, find the value of a.
Solution:
Area of the triangle with vertices (0, 0),(1,0), (0, a), is \(\frac{1}{2}\) × 1 × a = \(\frac{a}{2}\)
∴ \(\frac{a}{2}\) = 10 or a = 20

Question 7.
Find the value of a so that the points (1, 4), (2, 7), (3, a) are collinear.
Solution:
As points (1, 4), (2, 7), (3, a) are collinear, we have the area of the triangle with vertices (1, 4), (2, 7), and (3, a) is zero.
∴ \(\frac{1}{2}\) {1(7 – a) + 2(a – 4) + 3 (4 – 7)} = 0
or, 7 – a + 2a – 8 + 12 – 21 =0
⇒ a = 10

Question 8.
Find the slope of the lines whose inclinations are given.
(i) 30°
Solution:
The slope of the line whose inclination is 30°.
tan 30° = \(\frac{1}{\sqrt{3}}\)

(ii) 45°
Solution:
Slope = tan 45° = +1

(iii) 60°
Solution:
Slope = tan 60° = √3

(iv) 135°
Solution:
Slope = tan 135° = – 1

Question 9.
Find the inclination of the lines whose slopes are given below.
(i) \(\frac{1}{\sqrt{3}}\)
Solution:
The slope of the line is \(\frac{1}{\sqrt{3}}\)
∴ tan θ = \(\frac{1}{\sqrt{3}}\) or, θ = 30°
∴ The inclination of the line is 30°

(ii) 1
Solution:
Slope = 1 = tan 45°
∴ The inclination of the line is 45°.

(iii) √3
Solution:
Slope = √3 = tan 60°  ∴ θ = 60°
∴ Inclination = 60°

(iv) – 1
Solution:
Slope = – 1 = tan 135°
∴ Inclination = 135°

Question 10.
Find the angle between the pair of lines whose slopes are ;
(i) \(\frac{1}{\sqrt{3}}\), 1
Solution:

(ii) √3, -1
Solution:

Question 11.
(a) Show that the points (0, -1), (-2, 3), (6, 7), and (8, 3) are vertices of a rectangle.
Solution:


∴ The opposite sides are equal and two consecutive sides are perpendicular. So it is a rectangle.

(b) Show that the points (1, 1), (-1, -1), and (-√3, √3) are the vertices of an equilateral triangle.
Solution:

Question 12.
Find the coordinates of the point P(x, y) which is equidistant from (0, 0), (32, 10), and (42, 0).
Solution:

Question 13.
If the points (x, y) are equidistant from the points (a + b, b – a) and (a – b, a + b), prove that bx = ay.

Question 14.
The coordinate of the vertices of a triangle are (α₁, β₁), (α₂, β₂), and (α₃, β₃). Prove that the coordinates of its centroid is \(\left(\frac{\alpha_1+\alpha_2+\alpha_3}{3}, \frac{\beta_1+\beta_2+\beta_3}{3}\right)\)
Solution:

Question 15.
Two vertices of a triangle are (0, -4) and  (6, 0). If the medians meet at the point (2, 0), find the coordinates of the third vertex.
Solution:

∴ \(\frac{6+x}{3}\) = 2, \(\frac{-4+y}{3}\) = 0
⇒ x = 0, y = 4
∴ The coordinates of the 3rd vertex are (0, 4).

Question 16.
If the point (0, 4) divides the line segment joining(-4, 10) and (2, 1) internally, find the point which divides it externally in these same ratios.
Solution:

Question 17.
Find the ratios in which the line segment joining (-2, -3) arid (5, 4) is divided by the coordinate axes and hence find the coordinates of these points.
Solution:

Question 18.
In a triangle, one of the vertices is at (2, 5) and the centroid of the triangle is at (-1, 1). Find the coordinates of the midpoint of the side opposite to the given angular point.
Solution:

Question 19.
Find the coordinates of the vertices of a triangle whose sides have midpoints at (2, 1), (-1, 3), and (-2, 5).
Solution:

∴ x₂ + x₃ – 4 or, x₂ – 4 – 3 = 1
∴ x₁ = – 4 – x₂ = -4 – 1 = -5
Similarly y₁ + y₂ + y₃ = 5 + 1 + 3 = 9
As y₁ + y₂ = 10
we have y₃ = 9 – 10 = – 1
Again y₁ + y₃ = 6
or, y₁ = 6 – y₃ = 6 + 1 = 7
and y₂ = 10 – y₁ = 10 – 7 = 3
∴ The coordinates of A, B, and C are (-5, 7), (1, 3), and (3, -1).

Question 20.
If the vertices of a triangle have their coordinates given by rational numbers, prove that the triangle cannot be equilateral.
Solution:
Let us choose the contradiction method. Let the triangle is equilateral if the co¬ ordinate of the vertices is rational numbers.
Let ABC be an equilateral triangle with vertices A (a, 0), B (a, 0), and C (b, c) where a, b, c are rational.

⇒ a² = b² + c² = \(\frac{a^2}{4}\) + c²
⇒ c² =  a² – \(\frac{a^2}{4}\) = \(\frac{3a^2}{4}\) ⇒ c = \(\frac{\sqrt{3}}{2}\) a     ….(2)
Now b = \(\frac{a}{2}\), c = \(\frac{\sqrt{3}}{2}\) a
If a is rational then b is rational but c is irrational, i.e., the coordinates of the vertices are not rational, which contradicts the assumption.
Hence assumption is wrong.
So the triangle cannot be equilateral if the coordinate of the vertices is rational numbers.

Question 21.
Prove that the area of any triangle is equal to four times the area of the triangle formed by joining the midpoints of its sides.
Solution:

∴ The area of triangle ABC is four times the area of triangle DEF. (Proved)

Question 22.
Find the condition that the point (x, y) may lie on the line joining (1, 2) and (5, -3).
Solution:

∴ As points A, B, and C are collinear, we have the area of the triangle ABC as 0.
∴ \(\frac{1}{2}\) {1(-3 – y) + 5(y – 2) + x(2 + 3)} = 0
or, – 3 – y + 5y – 10 + 5x = 0
or, 5x + 4y = 13

Question 23.
Show that the three distinct points (a², a), (b², b), and (c², c) can never be collinear.
Solution:
Area of the triangle with vertices (a², a), (b², b) , and (c², c) is
\(\frac{1}{2}\) {a²(b – c) + b²(c – a) c²(a – b)}
= (a – b)(b – c)(a – c)
which is never equal to zero except when a = b = c, hence the points are not collinear.

Question 24.
If A, B, and C are points (-1, 2), (3, 1), and (-2, -3) respectively, then show that the points which divide BC, CA, and AB in the ratios (1: 3), (4: 3) and (-9: 4) respectively are collinear.
Solution:
Let the points P, Q, and R divides \(\overline{\mathrm{BC}}\), \(\overline{\mathrm{CA}}\), in \(\overline{\mathrm{AB}}\) the ratio 1: 3, 4: 3 and -9: 4

Question 25.
Prove analytically :
(a) The line segment joining the midpoints of two sides of a triangle is parallel to the third and half of its length.

Solution:
Let the coordinates of the triangle ABC be (x₁, y₁), (x₂, y₂) and (x₃, y₃)
The points D and E are the midpoints of the sides \(\overline{\mathrm{AB}}\) and \(\overline{\mathrm{AC}}\)

(b) The altitudes of a triangle are concurrent.
Solution:

(c) The perpendicular bisectors of the sides of a triangle are concurrent.
Solution:

(d) An angle in a semicircle is a right angle.
Solution:

∴ The angle in a semicircle is a right angle. (Proved)

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 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 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