Class 12

Punjab State Board PSEB 12th Class Maths Book Solutions Chapter 7 Integrals Ex 7.7 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 12 Maths Chapter 7 Integrals Ex 7.7

Direction (1 – 9): Integrate the function.

Question 1.
\(\sqrt{4-x^{2}}\)
Solution.

Question 2.
\(\sqrt{1-4 x^{2}}\)
Solution.
Let I = ∫ \(\sqrt{1-4 x^{2}}\) dx

= ∫ \(\sqrt{(1)^{2}-(2 x)^{2}}\) dx

Let 2x = t
⇒ 2 dx = dt

Question 3.
\(\sqrt{x^{2}+4 x+6}\)
Solution.

Question 4.
\(\sqrt{x^{2}+4 x+1}\)
Solution.

Question 5.
\(\sqrt{1-4 x-x^{2}}\)
Solution.

Question 6.
\(\sqrt{x^{2}+4 x-5}\)
Solution.

Question 7.
\(\sqrt{1+3 x-x^{2}}\)
Solution.

Question 8.
\(\sqrt{x^{2}+3 x}\)
Solution.

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

Question 10.
∫ \(\sqrt{1+x^{2}}\) dx is equal to
(A) \(\frac{x}{2} \sqrt{1+x^{2}}+\frac{1}{2} \log \left|x+\sqrt{1+x^{2}}\right|\) + C

(B) \(\frac{2}{3}\left(1+x^{2}\right)^{\frac{3}{2}}\) + C

(C) \(\frac{2}{3} x\left(1+x^{2}\right)^{\frac{3}{2}}\) + C

(D) \(\frac{x^{2}}{2} \sqrt{1+x^{2}}+\frac{1}{2} x^{2} \log \left|x+\sqrt{1+x^{2}}\right|\) + C
Solution.
We know that, ∫ \(\sqrt{a^{2}+x^{2}}\) dx = \(\frac{x}{2} \sqrt{a^{2}+x^{2}}+\frac{a^{2}}{2} \log \left|x+\sqrt{x^{2}+a^{2}}\right|\) + C

Here, a² = 1

∴ ∫ \(\sqrt{1+x^{2}}\) dx = \(\frac{x}{2} \sqrt{1+x^{2}}+\frac{1}{2} \log \left|x+\sqrt{1+x^{2}}\right|\) + C

Question 11.
∫ \(\sqrt{x^{2}-8 x+7}\) is equal to
(A) \(\frac{1}{2}(x-4) \sqrt{x^{2}-8 x+7}+9 \log \left|x-4+\sqrt{x^{2}-8 x+7}\right|\) + C

(B) \(\frac{1}{2}(x+4) \sqrt{x^{2}-8 x+7}+9 \log \left|x+4+\sqrt{x^{2}-8 x+7}\right|\) + C

(C) \(\frac{1}{2}(x-4) \sqrt{x^{2}-8 x+7}-3 \sqrt{2} \log \left|x-4+\sqrt{x^{2}-8 x+7}\right|\) + C

(D) \(\frac{1}{2}(x-4) \sqrt{x^{2}-8 x+7}+\frac{9}{2} \log \left|x-4+\sqrt{x^{2}-8 x+7}\right|\) + C
Solution.

Hence, the correct answer is (D).

Punjab State Board PSEB 12th Class Maths Book Solutions Chapter 7 Integrals Ex 7.4 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 12 Maths Chapter 7 Integrals Ex 7.4

Question 1.
\(\frac{3 x^{2}}{x^{6}+1}\)
Solution.
∫ \(\frac{3 x^{2}}{x^{6}+1}\) dx = ∫ \(\frac{3 x^{2}}{\left(x^{3}\right)^{2}+1}\) dx
Let x³ = t
⇒ 3x² dx = dt
⇒ dx = \(\frac{d t}{3 x^{2}}\)
∫ \(\frac{3 x^{2}}{\left(x^{3}\right)^{2}+1}\) dx = ∫ \(\frac{d t}{t^{2}+1}\)
= tan^-1 t + C
= tan^-1 (x³) + C

Question 2.
\(\frac{1}{\sqrt{1+4 x^{2}}}\)
Solution.
∫ \(\frac{1}{\sqrt{1+4 x^{2}}}\) dx = ∫ \(\) dx
Let 2x = t
⇒ 2 dx = dt
⇒ dx = \(\frac{1}{2}\)
∴ \(\int \frac{1}{\sqrt{1+4 x^{2}}} d x=\frac{1}{2} \int \frac{d t}{\sqrt{1+t^{2}}}\)

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

Question 4.
\(\frac{1}{\sqrt{9-25 x^{2}}}\)
Solution.

Question 5.
\(\frac{3 x}{1+2 x^{4}}\)
Solution.
∫ \(\frac{3 x}{1+2 x^{4}}\) dx = \(\frac{3}{2} \int \frac{x d x}{\frac{1}{2}+x^{4}}=\frac{3}{2} \int \frac{x d x}{\frac{1}{2}+\left(x^{2}\right)^{2}}\)

Let x² = t
⇒ 2x = \(\frac{d t}{d x}\)

⇒ dx = \(\frac{d t}{2 x}\)

Question 6.
\(\frac{x^{2}}{1-x^{6}}\)
Solution.

Question 7.
\(\frac{x-1}{\sqrt{x^{2}-1}}\)
Solution.

Question 8.
\(\frac{x^{2}}{\sqrt{x^{6}+a^{6}}}\)
Solution.

Question 9.
\(\frac{\sec ^{2} x}{\sqrt{\tan ^{2} x+4}}\)
Solution.

Question 10.
\(\frac{1}{\sqrt{x^{2}+2 x+2}}\)
Solution.

Question 11.
\(\frac{1}{9 x^{2}+6 x+5}\)
Solution.

Question 12.
\(\frac{1}{\sqrt{7-6 x-x^{2}}}\)
Solution.
7 – 6x – x² can be written as 7 – (x² + 6x + 9 – 9).
Therefore, 7 – (x² + 6x + 9 – 9) = 16 – (x² + 6x + 9)
= 16 – (x + 3)²
= (4)² – (x + 3)²

Question 13.
\(\frac{1}{\sqrt{(x-1)(x-2)}}\)
Solution.
(x – 1) (x – 2) can be written as x² – 3x + 2

Question 14.
\(\frac{1}{\sqrt{8+3 x-x^{2}}}\)
Solution.
8 + 3x – x² can be written as – (x² – 3x + \(\frac{9}{4}\) – \(\frac{9}{4}\))

Therefore, 8 – (x² – 3x + \(\frac{9}{4}\) – \(\frac{9}{4}\)) = \(\frac{41}{4}-\left(x-\frac{3}{2}\right)^{2}\)

Question 15.
\(\frac{1}{\sqrt{(x-a)(x-b)}}\)
Solution.
(x – a) (x – b) can be written as x² – (a + b) x + ab
Therefore,

Question 16.
\(\frac{4 x+1}{\sqrt{2 x^{2}+x-3}}\)
Solution.
∫ \(\frac{4 x+1}{\sqrt{2 x^{2}+x-3}}\) dx
Let 2x² + x – 3 = t
⇒ (4x + 1) dx = dt
⇒ dx = \(\frac{d t}{4 x+1}\)

= \(\int \frac{d t}{\sqrt{t}}=\int t^{-\frac{1}{2}} d t=\frac{t^{-\frac{1}{2}+1}}{\frac{1}{2}}+C=2 \sqrt{t}+C\)

= 2 \(\sqrt{2 x^{2}+x-3}\) + C

Question 17.
\(\frac{x+2}{\sqrt{x^{2}-1}}\)

Solution.

Question 18.
\(\frac{5 x-2}{1+2 x+3 x^{2}}\)
Solution.
Let 5x – 2 = A \(\frac{d}{d x}\) (1 +2x + 3x²) + B
⇒ 5x – 2 = A (2 + 6x) + B
Equating the coefficient of x and constant term on both sides, we get
5 = 6A
⇒ A = \(\frac{5}{6}\) ;
2A + B = – 2
⇒ B = – \(\frac{11}{3}\)

Question 19.
\(\frac{6 x+7}{\sqrt{(x-5)(x-4)}}\)
Solution.
\(\frac{6 x+7}{\sqrt{(x-5)(x-4)}}=\frac{6 x+7}{\sqrt{x^{2}-9 x+20}}\)

Let 6x + 7 = A \(\frac{d}{d x}\) (x² – 9x + 20) + B
⇒ 6x + 7 = A (2x – 9) + B
Equating the coefficients of x and constant term, we get
2A = 6
⇒ A = 3;
– 9A + B = 7
⇒ B = 34
∴ 6x + 7 = 3 (2x – 9) + 34

Question 20.
\(\frac{x+2}{\sqrt{4 x-x^{2}}}\)
Solution.
∫ \(\frac{x+2}{\sqrt{4 x-x^{2}}}\) dx

Question 21.
\(\frac{x+2}{\sqrt{x^{2}+2 x+3}}\)
Solution.

Substituting equations (ii) and (iii) in equation (i), we get
∫ \(\frac{x+2}{\sqrt{x^{2}+2 x+3}}\) dx = \(\frac{1}{2}\left[2 \sqrt{x^{2}+2 x+3}\right]+\log \mid(x+1)+\sqrt{x^{2}+2 x+3 \mid}\)

= \(\sqrt{x^{2}+2 x+3}+\log \mid(x+1)+\sqrt{x^{2}+2 x+3 \mid}\) + C

Question 22.
\(\frac{x+3}{x^{2}-2 x-5}\)
Solution.
Let (x + 3) = A \(\frac{d}{d x}\) (x² – 2x – 5) + B
⇒ (x + 3) = A (2x – 2) + B
⇒ x + 3 = 2Ax – 2A + B
On equating the coefficients of x and constant term on both sides, we get
2A = 1
⇒ A = \(\frac{1}{2}\);
– 2A + B = 3
⇒ B = 4
⇒ (x + 3) = \(\frac{1}{2}\) (2x – 2) + 4

Question 23.
\(\frac{5 x+3}{\sqrt{x^{2}+4 x+10}}\)
Solution.
Let 5x + 3 = A \(\frac{d}{d x}\) (x² + 4x + 10) + B
⇒ 5x + 3 = A (2x + 4) + B
⇒ 5x + 3 = 2Ax + 4A + B
On equating the coefficients of x and constant term, we get
2A = 5
A = \(\frac{5}{2}\);
4A + B = 3
⇒ B = – 7
⇒ 5x + 3 = (2x + 4) – 7

Direction (24 – 25): Choose the correct answer:

Question 24.
∫ \(\int \frac{d x}{x^{2}+2 x+2}\) equals
(A) x tan^-1 (x + 1) + C
(B) tan^-1 (x + 1) + C
(C) (x + 1) tan^-1 x + C
(D) tan^-1 x + C
Solution.
∫ \(\int \frac{d x}{x^{2}+2 x+2}\) dx = \(\int \frac{d x}{\left(x^{2}+2 x+1\right)+1}=\int \frac{1}{(x+1)^{2}+(1)^{2}} d x\)
Let x + 1 = t
⇒ dx = dt
∴ ∫ \(\frac{1}{t^{2}+1^{2}}\) dt = \(\frac{1}{1} \tan ^{-1}\left(\frac{t}{1}\right)\) + C

= tan \(\left(\frac{x+1}{1}\right)\) + C

= tan^-1 (x + 1) + C
Hence, the correct answer is (B).

Question 25.
∫ \(\frac{d x}{\sqrt{9 x-4 x^{2}}}\) equals
(A) \(\frac{1}{9} \sin ^{-1}\left(\frac{9 x-8}{8}\right)\) + C

(B) \(\frac{1}{2} \sin ^{-1}\left(\frac{8 x-9}{9}\right)\)

(C) \(\frac{1}{3} \sin ^{-1}\left(\frac{9 x-8}{8}\right)\)

(D) \(\frac{1}{2} \sin ^{-1}\left(\frac{9 x-8}{9}\right)\)
Solution.

Hence, the correct answer is (B).

Punjab State Board PSEB 12th Class Maths Book Solutions Chapter 7 Integrals Ex 7.3 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 12 Maths Chapter 7 Integrals Ex 7.3

Direction (1 – 22): Find the integrals of the functions:

Question 1.
sin² (2x + 5)
Solution.
= ∫ sin² (2x + 5) dx = ∫ \(\frac{1-\cos (4 x+10)}{2}\) dx
= ∫ 1 dx + \(\frac{1}{2}\) ∫ cos (4x + 10) dx
= \(\frac{1}{2}\) x – \(\frac{1}{2}\) (\(\left(\frac{\sin (4 x+10)}{4}\right)\)) + C
= \(\frac{x}{2}\) – \(\frac{1}{8}\) sin(4x + 10) + C

Question 2.
sin 3x cos 4x
Solution.
∫ sin 3x cos 4x dx = \(\frac{1}{2}\) ∫ {sin(3x + 4x) + sin(3x – 4x)} dx
[∵ 2 sin A cos B = sin (A + B) + sin(A – B)]
= \(\frac{1}{2}\) ∫ {sin 7x + sin (- x)}dx

= \(\frac{1}{2}\) ∫ {sin 7x – sin x} dx

= \(\frac{1}{2}\) ∫ sin 7x dx – \(\frac{1}{2}\) ∫ sin x dx

= \(\frac{1}{2}\) \(\left(\frac{-\cos 7 x}{7}\right)\) – \(\frac{1}{2}\) (- cos x) + C

= \(\frac{-\cos 7 x}{14}+\frac{\cos x}{2}\) + C

Question 3.
cos 2x cos 4x cos 6x
Solution.
∫ cos 2x (cos 4x cos 6x) dx = ∫ cos 2x [\(\frac{1}{2}\) {cos (4x +6x) + cos (4x – 6x)}] dx
[∵ 2 cos A cos B = cos (A + B) + cos (A – B)]
= \(\frac{1}{2}\) ∫ {cos 2x cos 10x + cos 2x cos (- 2x)}dx 2J

= \(\frac{1}{2}\) ∫ {cos 2x cos 10x + cos² 2x}dx

= \(\frac{1}{2}\) ∫ (cos 12x + cos 8x + 1 + cos 4x)dx

= \(\frac{1}{4}\left[\frac{\sin 12 x}{12}+\frac{\sin 8 x}{8}+x+\frac{\sin 4 x}{4}\right]\) + C

Question 4.
sin³ (2x + 1)
Solution.
Let I = ∫ sin³ (2x + 1) dx
⇒ = ∫ sin² (2x + 1) . sin (2x + 1) dx
= ∫ (1 – cos² (2x + 1)) sin(2x +1) dx
Let cos (2x + 1) = t
⇒ – 2 sin (2x + 1) dx = dt
⇒ sin (2x + 1) dx = \(\frac{-d t}{2}\)
∴ I = – \(\frac{1}{2}\) ∫ (1 – t²) dt
= \(\frac{-1}{2}\left\{t-\frac{t^{3}}{3}\right\}\) + C

Question 5.
sin³ x cos³ x
Solution.
Let I = ∫ sin³ x cos³ x dx
= ∫ cos³ x . sin² x . sin x dx
= ∫ cos³ x (1 – cos² x) sin x dx
Let cos x = t
⇒ – sin x dx = dt
⇒ dx = \(\frac{d t}{-\sin x}\)
⇒ I = – ∫ t³ (1 – t²) dt
= – ∫ (t³ – t⁵) dt
= – \(\left\{\frac{t^{4}}{4}-\frac{t^{6}}{6}\right\}\) + C

= – \(\left\{\frac{\cos ^{4} x}{4}-\frac{\cos ^{6} x}{6}\right\}\) + C

= \(\frac{\cos ^{6} x}{6}-\frac{\cos ^{4} x}{4}\) + C

Question 6.
sin x sin 2x sin 3x
Solution.
∫ sin x sin 2x sin 3x dx = ∫ [sin x . \(\frac{1}{2}\) {cos(2x – 3x) – cos(2x + 3x)}] dx
[∵ 2 sin A sin B = cos(A – B) – cos(A + B)]
= \(\frac{1}{2}\) ∫ (sin x cos (- x) – sin x cos 5x) dx
= \(\frac{1}{2}\) ∫ (sin x cos x – sin x cos 5x) dx

Question 7.
sin 4x sin 8x
Solution.
∫ sin 4x sin 8x dx = ∫ {\(\frac{1}{2}\) cos(4x – 8x) – cos(4x + 8x)} dx
= \(\frac{1}{2}\) ∫ (cos (- 4x) – cos 12x) dx
= \(\frac{1}{2}\) ∫(cos 4x – cos 12x) dx
= \(\frac{1}{2}\) \(\) + C
= \(\) [sin 4x – \([latex]\)[/latex] sin 12x] + C

Question 8.
\(\frac{1-\cos x}{1+\cos x}\)
Solution.
∫ \(\frac{1-\cos x}{1+\cos x}\) dx = ∫ \(\frac{2 \sin ^{2} \frac{x}{2}}{2 \cos ^{2} \frac{x}{2}}\) dx

[∵ 1 – cos x = 2 sin² \(\frac{x}{2}\) and
1 + cos x = 2 cos² \(\frac{x}{2}\)]

= ∫ tan² \(\frac{x}{2}\) dx

= (sec² \(\frac{x}{2}\) – 1) dx

= ∫ sec² \(\frac{x}{2}\) dx – ∫ 1 dx

= \(\left[\frac{\tan \frac{x}{2}}{\frac{1}{2}}-x\right]\) + C

= 2 tan \(\frac{x}{2}\) – x + C

Question 9.
\(\frac{\cos x}{1+\cos x}\)
Solution.
∫ \(\frac{\cos x}{1+\cos x}\) dx = ∫ \(\frac{\cos ^{2} \frac{x}{2}-\sin ^{2} \frac{x}{2}}{2 \cos ^{2} \frac{x}{2}}\) dx
[∵ cos x = cos² \(\frac{x}{2}\) – sin² \(\frac{x}{2}\) and
cos x = 2 cos² \(\frac{x}{x}\) – 1]

= \(\frac{1}{2}\) ∫ (1 – tan² \(\frac{x}{2}\)) dx

= \(\frac{1}{2}\) ∫ (1 – sec² \(\frac{x}{2}\) + 1 ) dx

= \(\frac{1}{2}\) ∫ (2 – sec² \(\frac{x}{2}\)) dx

= \(\frac{1}{2}\) ∫ 2 dx – \(\frac{1}{2}\) ∫ sec² \(\frac{x}{2}\) dx

= \(\frac{1}{2}\) [2x – \(\frac{\tan \frac{x}{2}}{\frac{1}{2}}\)] + C

= x – tan \(\frac{x}{2}\) + C

Question 10.
sin⁴ x
Solution.
∫ sin⁴ x dx = ∫ sin² x sin² x dx

Question 11.
cos⁴ 2x
Solution.
∫ cos⁴ 2x dx = ∫ (cos² 2x)² dx

Question 12.
\(\frac{\sin ^{2} x}{1+\cos x}\)
Solution.
∫ \(\frac{\sin ^{2} x}{1+\cos x}\) dx = ∫ \(\frac{\left(1-\cos ^{2} x\right)}{1+\cos x}\) dx

= ∫ \(\frac{(1-\cos x)(1+\cos x)}{1+\cos x}\) dx

[∵ a² – ² = (a – b) (a + b)]

= ∫ (1 – cos x) dx = x – sin x + C

Question 13.
\(\frac{\cos 2 x-\cos 2 \alpha}{\cos x-\cos \alpha}\)
Solution.
∫ \(\frac{\cos 2 x-\cos 2 \alpha}{\cos x-\cos \alpha}\) dx = ∫ \(\frac{\left(2 \cos ^{2} x-1\right)-\left(2 \cos ^{2} \alpha-1\right)}{\cos x-\cos \alpha}\) dx

= ∫ \(\frac{2\left(\cos ^{2} x-\cos ^{2} \alpha\right)}{\cos x-\cos \alpha}\) dx

= ∫ \(\frac{2(\cos x-\cos \alpha)(\cos x+\cos \alpha)}{\cos x-\cos \alpha}\) dx

= 2 ∫(cos x + cos α) dx
= 2 sin x + 2 cos α x + C
= 2 (sin x + x cos α) + C

Question 14.
\(\frac{\cos x-\sin x}{1+\sin 2 x}\)
Solution.
∫ \(\frac{\cos x-\sin x}{1+\sin 2 x}\) dx = ∫ \(\frac{\left(2 \cos ^{2} x-1\right)-\left(2 \cos ^{2} \alpha-1\right)}{\cos x-\cos \alpha}\) dx

Question 15.
tan³ 2x sec 2x
Solution.
∫ tan³ 2x sec 2x dx = ∫ tan² 2x tan 2x sec 2x dx
= ∫ (sec² 2x – 1) tan 2x sec 2x dx
= ∫ (sec² 2x tan 2x sec 2x – tan 2x sec 2x) dx
= ∫ sec² 2x tan 2x sec 2x dx – ∫ tan 2xsec 2x dx
= ∫ sec² 2x tan 2x sec 2x dx – \(\frac{\sec 2 x}{2}\) + C
Let sec 2x = t
∴ 2 sec 2x tan 2x dx = dt
∴ ∫ tan³ 2x sec 2x dx = \(\frac{1}{2}\) ∫ t² dt – \(\frac{\sec 2 x}{2}\) + C
= \(\frac{t^{3}}{6}-\frac{\sec 2 x}{2}\) + C

= \(\frac{(\sec 2 x)^{3}}{6}-\frac{\sec 2 x}{2}\) + C

Question 16.
tan4 x
Sol.
tan⁴ x dx = ∫ tan² x . tan² x dx
= ∫ [(sec² x – 1) tan² x] dx
= ∫ [sec² x tan² x – tan² x] dx
= ∫ sec² x tan² x dx – ∫ (sec² x – 1)dx
= ∫ sec² x tan² x dx – ∫ sec² x dx + ∫ 1 dx
= ∫ sec² x tan² x dx – tan x + x + C ……………….(i)
= ∫ sec² x tan² x dx
Let tan x = t
⇒ sec² x dx = dt
∫ sec² x tan² x dx = ∫ t² dt
= \(\frac{\tan ^{3} x}{3}\)
From Eq. (i), we get
∫ tan⁴ x dx = \(\frac{1}{3}\) tan³ x – tan x + x + C

Question 17.
\(\frac{\sin ^{3} x+\cos ^{3} x}{\sin ^{2} x \cos ^{2} x}\)
Solution.
∫ \(\frac{\sin ^{3} x+\cos ^{3} x}{\sin ^{2} x \cos ^{2} x}\) dx = \(\int \frac{\sin ^{3} x}{\sin ^{2} x \cos ^{2} x} d x+\int \frac{\cos ^{3} x}{\sin ^{2} x \cos ^{2} x} d x\)

= \(\int \frac{\sin x}{\cos ^{2} x} d x+\int \frac{\cos x}{\sin ^{2} x} d x\)

= ∫ tan x sec x dx + ∫ cot x cosec x dx

= sec x – cosec x + C

Question 18.
\(\frac{\cos 2 x+2 \sin ^{2} x}{\cos ^{2} x}\)
Solution.
∫ \(\frac{\cos 2 x+2 \sin ^{2} x}{\cos ^{2} x}\) dx = ∫ \(\frac{\cos 2 x+(1-\cos 2 x)}{\cos ^{2} x}\) dx
[∵ cos 2x = 1 – 2 sin² x]

∫ \(\frac{1}{\cos ^{2} x}\) dx
= ∫ sec² x dx = tan x + C

Question 19.
\(\frac{1}{\sin x \cos ^{3} x}\)
Solution.
∫ \(\frac{1}{\sin x \cos ^{3} x}\) dx

= ∫ tan x sec² x dx + ∫ \(\frac{\sec ^{2} x}{\tan x}\) dx
Let tan x = t
⇒ sec² x dx = dt
= ∫ t dt + ∫ \(\frac{1}{t}\) dt
= \(\frac{t^{2}}{2}\) + log |t| + C
= \(\frac{1}{2}\) tan² x + log |tan x| + C

Question 20.
\(\frac{\cos 2 x}{(\cos x+\sin x)^{2}}\)
Solution.
∫ \(\frac{\cos 2 x}{(\cos x+\sin x)^{2}}\) dx = ∫ \(\frac{\cos 2 x}{\cos ^{2} x+\sin ^{2} x+2 \sin x \cos x}\) dx
= ∫ \(\frac{\cos 2 x}{(1+\sin 2 x)}\) dx
Let 1 + sin 2x = t
⇒ 2 cos 2x dx = dt
∴ ∫ \(\frac{\cos 2 x}{(\cos x+\sin x)^{2}}\) dx = \(\frac{1}{2}\) ∫ \(\frac{1}{t}\) dt

= \(\frac{1}{2}\) log |t| + C

= \(\frac{1}{2}\) log|1 + sin 2x| + C

= \(\frac{1}{2}\) log |(sin x + cos x)²| + C

= log |sin x + cos x| + C

Question 21.
sin^-1 x (cos x)
Solution.
∫ sin^-1 x (cos x) dx = ∫ sin^-1 x [sin (\(\frac{\pi}{2}\) – x)] dx
= ∫ (\(\frac{\pi}{2}\) – x) = \(\frac{\pi}{2}\) ∫ dx – ∫ x dx
= \(\frac{\pi x}{2}-\frac{x^{2}}{2}\) + C

Question 22.
\(\frac{1}{\cos (x-a) \cos (x-b)}\)
Solution.

Question 23.
∫ \(\frac{\sin ^{2} x-\cos ^{2} x}{\sin ^{2} x \cos ^{2} x}\) dx is equal to
(A) tan x + cot x + C
(B) tan x + cosec x + C
(C) tan x + cot x + C
(D) tan x + sec x + C
Solution.
∫ \(\frac{\sin ^{2} x-\cos ^{2} x}{\sin ^{2} x \cos ^{2} x}\) dx = ∫ \(\left(\frac{\sin ^{2} x}{\sin ^{2} x \cos ^{2} x}-\frac{\cos ^{2} x}{\sin ^{2} x \cos ^{2} x}\right)\) dx
= ∫ (sec² x – cosec² x) dx
= tan x + cot x + C
Hence, the correct answer is (A).

Question 25.
∫ \(\frac{e^{x}(1+x)}{\cos ^{2}\left(e^{x} x\right)}\) dx equals
(A) – cot (e x^x) + C
(B) tan (x e^x) + C
(C) tan (e^x) + C
(D) cot (e^x) + C
Solution.
∫ \(\frac{e^{x}(1+x)}{\cos ^{2}\left(e^{x} x\right)}\) dx
Let x e^x = t
⇒ (e^x . x + e^x . 1) dx = dt
⇒ e^x (x + 1) dx = dt
∴ \(\int \frac{e^{x}(1+x)}{\cos ^{2}\left(e^{x} x\right)} d x=\int \frac{d t}{\cos ^{2} t}\) = ∫ sec² t dt
= tan t + C
= tan (e^x x) + C
Hence, the correct answer is (B).

Punjab State Board PSEB 12th Class Maths Book Solutions Chapter 7 Integrals Ex 7.2 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 12 Maths Chapter 7 Integrals Ex 7.2

Direction (1 – 37): Integrate the following functions.

Question 1.
\(\frac{2 x}{1+x^{2}}\)
Solution.
∫ \(\frac{2 x}{1+x^{2}}\) dx
Let 1 + x² = t
On differentiating w.rt. x, we get
2x dx = dt
⇒ x dx = \(\frac{d t}{2}\)
⇒ ∫ \(\frac{2 x}{1+x^{2}}\) dx = ∫ \(\frac{1}{t}\) dt
= log |t| + C
= log |1 + x²| + C

Question 2.
\(\frac{(\log x)^{2}}{x}\)
Sol.
∫ \(\frac{(\log x)^{2}}{x}\) dx
Let log x = t
On differentiating w.r.t. x, we get
\(\frac{1}{x}\) dx = dt
⇒ dx = x dt
⇒ ∫ \(\frac{(\log x)^{2}}{x}\) dx = ∫ t² dt
= \(\frac{t^{3}}{3}\) + C
= \(\frac{(\log x)^{3}}{3}\) + C

Question 3.
\(\frac{1}{x+x \log x}\)
Solution.
∫ \(\frac{1}{x+x \log x}\) dx = ∫ \(\frac{1}{x(1+\log x)}\) dx
Let 1 + log x = t
On differentiating w.r.t. x, we get
\(\frac{1}{x}\) dx = dt
⇒ dx = x dt
⇒ ∫ \(\frac{1}{x+x \log x}\) dx = ∫ \(\frac{1}{t}\) dt
= log |t| + C
= log |1 + log x| + C

Question 4.
sin x . sin (cos x)
Solution.
∫ sin x . sin(cos x) dx
Let cos x = t
On differentiating w.r.t. x, we get
– sin x dx = dt
⇒ ∫ sin x . sin (cos x) dx = – ∫ sin t dt
= – [- cos t] + C
= cos t + C = cos (cos x) + C

Question 5.
sin(ax + b) cos(ax + b)
Solution.
∫ sin(ax + b) cos(ax + b) dx
= – \(\frac{1}{2}\) ∫ 2 sin (ax + b) cos(ax + b) dx
= – \(\frac{1}{2}\) ∫ sin (2ax + 2b) dx
Let 2ax + 2b = t
On differentiating w.r.t. x, we get
2a dx = dt
∴ I = \(\frac{1}{2}\) ∫ sin t . \(\frac{d t}{2 a}\)

= \(\frac{1}{4 a}\) ∫ sin t dt

= – \(\frac{1}{4 a}\) cos t + C

= – \(\frac{1}{4 a}\) cos (2ax + 2) + C

Question 6.
\(\sqrt{a x+b}\)
Solution.
∫ \(\sqrt{a x+b}\) dx
Let ax + b = t
On differentiating w.r.t. x, we get
a dx = dt
∴ dx = \(\frac{1}{a}\) dt
⇒ ∫ \((a x+b)^{\frac{1}{2}} d x=\frac{1}{a} \int t^{\frac{1}{2}} d t\)

= \(\frac{1}{a}\left(\frac{t^{\frac{3}{2}}}{\frac{3}{2}}\right)\)

= \(\frac{2}{3 a}(a x+b)^{\frac{3}{2}}\) + C

Question 7.
x \(\sqrt{x+2}\)
Solution.
∫ x \(\sqrt{x+2}\) dx
Let (x + 2) = t
On differentiating w.r.t. x, we get
dx = dt

Question 8.
x \(\sqrt{1+2 x^{2}}\)
Solution.
∫ x \(\sqrt{1+2 x^{2}}\) dx
Let 1 + 2x² = t
On differentiating w.r.t. x, we get
4x dx = dt
∫ x \(\sqrt{1+2 x^{2}}\) dx = \(\int \frac{\sqrt{t} d t}{4}=\frac{1}{4} \int t^{\frac{1}{2}} d t\)

= \(\frac{1}{4}\left(\frac{t^{\frac{3}{2}}}{\frac{3}{2}}\right)+C=\frac{1}{6}\left(1+2 x^{2}\right)^{\frac{3}{2}}+C\)

Question 9.
(4x + 2) \(\sqrt{x^{2}+x+1}\)
Solution.
∫ (4x + 2) \(\sqrt{x^{2}+x+1}\) dx
Let x² + 2x + 1 = t
On differentiating w.r.t. x, we get
(2x + 1) dx = dt
∫ (4x + 2) \(\sqrt{x^{2}+x+1}\) dx = ∫ 2√t dt = 2 ∫ √t dt
= 2 \(\left(\frac{t^{\frac{3}{2}}}{\frac{3}{2}}\right)\) + C

= \(\frac{4}{3}\left(x^{2}+x+1\right)^{\frac{3}{2}}\) + C

Question 10.
\(\frac{1}{x-\sqrt{x}}\)
Solution.
∫ \(\frac{1}{x-\sqrt{x}}\) dx = ∫ \(\frac{1}{\sqrt{x}(\sqrt{x}-1)}\) dx
Let (√x – 1) = t
On differentiating w.r.t. x, we get
∴ \(\frac{1}{2 \sqrt{x}}\) dx = dt
⇒ dx = 2√x dt
⇒ ∫ \(\frac{1}{2 \sqrt{x}}\) dx = ∫ \(\frac{1}{\sqrt{x} t} 2 \sqrt{x} d t=\int \frac{2}{t} d t\)
= 2 log |t| + C
= 2 log |\(\sqrt{x}-1\)| + C

Question 11.
\(\frac{x}{\sqrt{x+4}}\), x > 0
Solution.
∫ \(\frac{x}{\sqrt{x+4}}\) dx
Let x + 4 = t
On differentiating w.r.t. x, we get
dx = dt

Question 11.
\(\frac{x}{\sqrt{x+4}}\), x > 0
Solution.
∫ \(\frac{x}{\sqrt{x+4}}[latex] dx
Let x + 4 = t
On differentiating w.r.t. x, we get
dx = dt

= [latex]\frac{2}{3}(x+4)^{\frac{1}{2}}\) (x + 4 – 12) + C

= \(\frac{2}{3} \sqrt{x+4}\) (x – 8) + C

Question 12.
\(\left(x^{3}-1\right)^{\frac{1}{3}} x^{5}\)
Solution.
∫ \(\left(x^{3}-1\right)^{\frac{1}{3}} x^{5}\) dx
Let x³ – 1 = t
On differentiating w.r.t. x, we get
3x² dx = dt
⇒ dx = \(\frac{d t}{3 x^{2}}\)
⇒ ∫ \(\left(x^{3}-1\right)^{\frac{1}{3}} x^{5}\) dx = ∫ (x³ – 1)^{\(\frac{4}{4}\)} x³ x² dx

Question 13.
\(\frac{x^{2}}{\left(2+3 x^{3}\right)^{3}}\)
Solution.
∫ \(\frac{x^{2}}{\left(2+3 x^{3}\right)^{3}}\) dx
On differentiating w.r.t. x, we get
9x² dx = dt
⇒ dx = \(\frac{d t}{9 x^{2}}\)
∫ \(\frac{x^{2}}{\left(2+3 x^{3}\right)^{3}}\) dx = \(\frac{1}{9} \int \frac{d t}{(t)^{3}}=\frac{1}{9}\left[\frac{t^{-3+1}}{-3+1}\right]+C\)

= – \(\frac{1}{18}\left(\frac{1}{t^{2}}\right)\) + C

= – \(\frac{1}{18\left(2+3 x^{3}\right)^{2}}\) + C

Question 14.
\(\frac{1}{x(\log x)^{m}}\), x > 0, m ≠ 1
Solution.
∫ \(\frac{1}{x(\log x)^{m}}\) dx
Let log x = t
On differentiating w.r.t. x, we get
\(\frac{1}{x}\) dx = dt
⇒ dx = x dt
∴ ∫ \(\frac{1}{x(\log x)^{m}}\) dx = ∫ \(\frac{d t}{(t)^{m}}\)

= \(\left(\frac{t^{-m+1}}{-m+1}\right)\) + C

= \(\frac{(\log x)^{1-m}}{(1-m)}\) + C

Question 15.
\(\frac{x}{9-4 x^{2}}\)
Solution.
∫ \(\frac{x}{9-4 x^{2}}\) dx
Let 9 – 4x² = t
On differentiating w.r.t. x, we get
– 8x dx = dt
⇒ dx = \(\frac{d t}{-8 x}\)
∫ \(\frac{x}{9-4 x^{2}}\) dx = \(\frac{-1}{8}\) ∫ \(\frac{1}{t}\) dt
= \(\frac{-1}{8}\) log |t| + C
= \(\frac{-1}{8}\) log |9 – 4x²| + C

Question 16.
e^{2x +3}
Solution.
∫ e^{2x + 3} dx
Let 2x + 3 = t
On differentiating w.r.t. x, we get
2 dx = dt
⇒ dx = \(\frac{1}{2}\) dt
∴ ∫ e^{2x + 3} dx = \(\frac{1}{2}\) ∫ e^t dt
= \(\frac{1}{2}\) (e^t) + C
= \(\frac{1}{2}\) e^{(2x + 3)} + C

Question 18.
\(\frac{e^{\tan ^{-1} x}}{1+x^{2}}\)
Solution.
∫ \(\frac{e^{\tan ^{-1} x}}{1+x^{2}}\) dx
Let tan^-1 x = t
⇒ \(\frac{1}{1+x^{2}}\) dx = dt
⇒ dx = (1 + x²) dt
∴ ∫ \(\frac{e^{\tan ^{-1} x}}{1+x^{2}}\) dx = ∫ e^t dt
= e^t + C
= e^tan-1 x + C

Question 19.
\(\frac{e^{2 x}-1}{e^{2 x}+1}\)
Solution.

= log |t| + C
= log |e^x + e^-x| + C.

Question 20.
\(\frac{e^{2 x}-e^{-2 x}}{e^{2 x}+e^{-2 x}}\)
Solution.
∫ \(\frac{e^{2 x}-e^{-2 x}}{e^{2 x}+e^{-2 x}}\) dx
Let e^2x + e^-2x = t
⇒ (2 e^2x – 2 e^-2x) dx = dt
⇒ dx = \(\frac{d t}{2\left(e^{2 x}-e^{-2 x}\right)}\)

∫ \(\frac{e^{2 x}-e^{-2 x}}{e^{2 x}+e^{-2 x}}\) dx = ∫ \(\frac{d t}{2 t}\)

= \(\frac{1}{2}\) ∫ \(\frac{1}{t}\) dt

= \(\frac{1}{2}\) log |t| + C

= \(\frac{1}{2}\) log |e^2x + e^-2x| + C.

Question 21.
tan² (2x – 3)
Solution.
∫ tan² (2x – 3) dx = ∫ sec² (2x – 3) dx – ∫ 1 dx
Let 2x – 3 = t
⇒ 2 dx = dt
⇒ dx = \(\frac{1}{2}\) dt
⇒ ∫ sec² (2x – 3) dx – ∫ 1 dx = \(\frac{1}{2}\) ∫ (sec² t dt – ∫ 1 dx
= \(\frac{1}{2}\) tan t – x + C
= \(\frac{1}{2}\) tan (2x – 3) – x + C

Question 22.
sec² (7 – 4x)
Solution.
∫ sec² (7 – 4x) dx
Let 7 – 4x = t
⇒ – 4 dx = dt
⇒ dx = \(\frac{d t}{-4}\)

∴ ∫ sec² (7 – 4x) dx = \(\frac{-1}{4}\) ∫ sec² t dt
= \(\frac{-1}{4}\) (tan t) + C
= \(\frac{-1}{4}\) tan (7 – 4x) + C

Question 23.
\(\frac{\sin ^{-1} x}{\sqrt{1-x^{2}}}\)
Solution.
∫ \(\frac{\sin ^{-1} x}{\sqrt{1-x^{2}}}\) dx
Let sin^-1 x = t

Question 24.
\(\frac{2 \cos x-3 \sin x}{6 \cos x+4 \sin x}\)
Solution.
∫ \(\frac{2 \cos x-3 \sin x}{6 \cos x+4 \sin x}\) dx = ∫ \(\frac{2 \cos x-3 \sin x}{2(3 \cos x+2 \sin x)}\) dx

Let 3 cos x + 2 sin x = t

⇒ (- 3 sin x + 2 cos x) dx = dt

⇒ dx = \(\frac{d t}{2 \cos x-3 \sin x}\)

∴ ∫ \(\frac{2 \cos x-3 \sin x}{6 \cos x+4 \sin x}\) dx = ∫ \(\frac{d t}{2 t}\) + \(\frac{1}{2}\) ∫ \(\frac{1}{t}\) dt

= \(\frac{1}{2}\) log |t| + C

= \(\frac{1}{2}\) log |2 sin x + 3 cos x| + C

Question 25.
\(\frac{1}{\cos ^{2} x(1-\tan x)^{2}}\)
Solution.
∫ \(\frac{1}{\cos ^{2} x(1-\tan x)^{2}}\) dx = ∫ \(\frac{\sec ^{2} x}{(1-\tan x)^{2}}\) dx
Let (1 – tan x) = t
⇒ – sec² x dx = dt
⇒ dx = \(\frac{d t}{-\sec ^{2} x}\)

∫ \(\frac{\sec ^{2} x}{(1-\tan x)^{2}}\) dx = ∫ \(\frac{-d t}{t^{2}}\)
= – ∫ t² dt

= – \(\left(\frac{-t^{-2+1}}{-2+1}\right)\) + C

= \(\frac{1}{t}+C=\frac{1}{(1-\tan x)}\) + C

Question 26.
\(\frac{\cos \sqrt{x}}{\sqrt{x}}\)
Solution.
∫ \(\frac{\cos \sqrt{x}}{\sqrt{x}}\) dx
Let √x = t
⇒ \(\frac{1}{2 \sqrt{x}}\) dx = dt
⇒ dx = 2 √x dt
∫ \(\frac{\cos \sqrt{x}}{\sqrt{x}}\) dx = 2 ∫ cos t dt
= 2 sin t + C
= 2 sin √x + C

Question 27.
\(\sqrt{\sin 2 x}\) cos 2x
Solution.
∫ \(\sqrt{\sin 2 x}\) cos 2x dx
Let sin 2x = t
⇒ 2 cos 2x dx = dt
⇒ dx = \(\frac{d t}{2 \cos 2 x}\)
∴ ∫ \(\sqrt{\sin 2 x}\) cos 2x dx = \(\frac{1}{2}\) ∫ (t)^{\(\frac{1}{2}\)} dt

= \(\frac{1}{2}\left(\frac{t^{\frac{3}{2}}}{\frac{3}{2}}\right)\)

= \(\frac{1}{3} t^{\frac{3}{2}}\) + C

= \(\frac{1}{3}(\sin 2 x)^{\frac{3}{2}}\) + C

Question 28.
\(\frac{\cos x}{\sqrt{1+\sin x}}\)
Solution.
∫ \(\frac{\cos x}{\sqrt{1+\sin x}}\) dx
Let 1 + sin x = t
⇒ cos x dx = dt
⇒ dx = \(\frac{d t}{\cos x}\)

∫ \(\frac{\cos x}{\sqrt{1+\sin x}}\) dx = \(\frac{d t}{\sqrt{t}}\)

= \(\frac{t^{\frac{1}{2}}}{\frac{1}{2}}\) + C

= 2√t + C

= 2 \({\sqrt{1+\sin x}}\) + C

Question 29.
cot x log sin x
Solution.
∫ cot x log sin x dx
Let log sin x = t .
⇒ \(\frac{1}{\sin x}\) . cos x dx = dt
⇒ cot x dx = dt
⇒ dx = \(\frac{d t}{\cot x}\)
∴ ∫ cot x log sin x dx = ∫ t dt
= \(\frac{t^{2}}{2}\) + C
= \(\frac{1}{2}\) (log sin x)² + C

Question 30.
\(\frac{\sin x}{1+\cos x}\)
Solution.
∫ \(\frac{\sin x}{1+\cos x}\) dx
Let 1 + cos x = t
⇒ – sin x dx = dt
⇒ dx = \(\frac{d t}{-\sin x}\)
∴ ∫ \(\frac{\sin x}{1+\cos x}\) dx = ∫ – \(\frac{d t}{t}\)
= – log |t| + C
= – log |1 + cos x| + C
= log \(\frac{1}{|1+\cos x|}\) + C

Question 31.
\(\frac{\sin x}{(1+\cos x)^{2}}\)
Solution.
∫ \(\frac{\sin x}{(1+\cos x)^{2}}\) dx
Let 1 + cos x = t
⇒ – sin x dx = dt
⇒ dx = \(=\frac{d t}{-\sin x}\)
∴ ∫ \(\frac{\sin x}{(1+\cos x)^{2}}\) dx = ∫ – \(\frac{d t}{t^{2}}\)

= ∫ t^-2 dt

= \(\frac{-t^{-2+1}}{-2+1}\) + C

= \(\frac{1}{t}\) + C

= \(\frac{1}{1+\cos x}\) + C

Question 32.
\(\frac{1}{1+\cot x}\)
Solution.

Question 33.
\(\frac{1}{1-\tan x}\)
Solution.

Question 34.
\(\frac{\sqrt{\tan x}}{\sin x \cos x}\)
Solution.
Let I = \(\frac{\sqrt{\tan x}}{\sin x \cos x}\) dx

= ∫ \(\frac{\sqrt{\tan x} \times \cos x}{\sin x \cos x \times \cos x}\) dx

= ∫ \(\frac{\sqrt{\tan x}}{\tan x \cos ^{2} x}\) dx

= ∫ \(\frac{\sec ^{2} x}{\sqrt{\tan x}}\) dx

Let x = tan t
⇒ sec² x dx = dt
⇒ dx = \(\frac{d t}{\sec ^{2} x}\)

∴ I = ∫ \(\frac{d t}{\sqrt{t}}\)

= ∫ t^{\(-\frac{1}{2}\)}

= 2√t + C

= 2√tan x + C

Question 35.
\(\frac{(1+\log x)^{2}}{x}\)
Solution.
∫ \(\frac{(1+\log x)^{2}}{x}\) dx
Let 1 + log x = t
⇒ \(\frac{1}{x}\)dx = dt
⇒ dx = x dt
∴ ∫ \(\frac{(1+\log x)^{2}}{x}\) dx = ∫ t² dt
= \(\frac{t^{3}}{3}\) + C
= \(\frac{(1+\log x)^{3}}{3}\) + C

Question 36.
\(\frac{(x+1)(x+\log x)^{2}}{x}\)
Solution.
∫ \(\frac{(x+1)(x+\log x)^{2}}{x}\) dx = ∫ \(\frac{x+1}{x}\) (x + log x)² dx

= ∫ (1 + \(\frac{1}{x}\)) (x + log x)² dx

Let x + log x = t
⇒ (1 + \(\frac{1}{x}\)) dx = dt

∴ ∫ (1 + \(\frac{1}{x}\)) (x + log x)² dx = ∫ t² dt

= \(\frac{t^{3}}{3}\) + C

= \(\frac{1}{3}\) (x + log x)³ + C

Question 37.
\(\frac{x^{3} \sin \left(\tan ^{-1} x^{4}\right)}{1+x^{8}}\)
Solution.
Let I = ∫ \(\frac{x^{3} \sin \left(\tan ^{-1} x^{4}\right)}{1+x^{8}}\) dx

Let tan^-1 x⁴ = t
⇒ \(\frac{1}{1+x^{8}}\) . 4x³ = \(\frac{d t}{d x}=\)

⇒ dx = \(\frac{\left(1+x^{8}\right)}{4 x^{3}}\) dt

∴ I = ∫ \(\frac{x^{3} \sin t}{\left(1+x^{8}\right)} \cdot \frac{1+x^{8} d t}{4 x^{3}}\)

= \(\frac{1}{4}\) ∫ sin t dt

= – \(\frac{1}{4}\) cos t + C

Direction (38 – 39) : Choose the correct answer.

Question 38.
∫ \(\frac{10 x^{9}+10^{x} \log _{e} 10}{x^{10}+10^{x}}\) dx equals
(A) 10^x – x¹⁰ + C

(B) 10^x + x¹⁰ + C

(C) (10^x – x¹⁰)^-1 + C

(D) log (10^x + x¹⁰) + C
Solution.
∫ \(\frac{10 x^{9}+10^{x} \log _{e} 10}{x^{10}+10^{x}}\) dx
Let x¹⁰ + 10^x = t
⇒ dx = \(\frac{d t}{10 x^{9}+10^{x} \log _{e} 10}\)
∴ ∫ \(\frac{10 x^{9}+10^{x} \log _{e} 10}{x^{10}+10^{x}}\) dx = ∫ \(\frac{dt}{t}\)
= log t + C
= log (10^x + x¹⁰) + C
Hence, the correct answer is (D).

Question 39.
∫ \(\frac{d x}{\sin ^{2} x \cos ^{2} x}\) ,
(A) tan x + cot x + C
(B) tan x – cot x + C
(C) tan x cot x + C
(D) tan x – cot 2x + C
Solution.

Hence, the correct answer is (B).

Punjab State Board PSEB 12th Class Maths Book Solutions Chapter 7 Integrals Ex 7.1 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 12 Maths Chapter 7 Integrals Ex 7.1

Direcflon (1 – 5):
Find the anti-derivative (or integral) of the following by the method of inspection.

Question 1.
sin 2x
Solution.
The anti-derivative of sin 2x is a function of x whose derivative is sin 2x.
It is known that,
\(\frac{d}{d x}\) (cos 2x) = – 2 sin 2x
⇒ sin 2x = – \(\frac{1}{2} \frac{d}{d x}\) (cos 2x)
sin 2x = \(\frac{d}{d x}\) (- \(\frac{1}{2}\) cos 2x)
Therefore, the anti-derivative of sin 2x is – \(\frac{1}{2}\) cos 2x + C.

Question 2.
cos 3x
Solution.
The anti-derivative of cos 3x is a function of x whose derivative is cos 3x.
It is known that,
\(\frac{d}{d x}\) (sin 3x) = 3 cos 3x
⇒ cos 3x = \(\frac{1}{3}\) \(\frac{d}{d x}\) (sin 3x)

∴ cos 3x = \(\frac{d}{d x}\) (\(\frac{1}{3}\) sin 3x)

Therefore, the anti-derivative of cos 3x is \(\frac{1}{3}\) sin 3x + C.

Question 3.
e^2x
Solution.
The anti-derivative of e^2x is the function of x whose derivative is e^2x.
It is known that,
\(\frac{d}{d x}\) (e^2x) = 2e^2x
⇒ e^2x = \(\frac{1}{2}\) \(\frac{d}{d x}\) (e^2x)

∴ e^2x = \(\frac{d}{d x}\) (\(\frac{1}{2}\) e^2x)

Therefore, the anti-derivative of e^2x is \(\frac{1}{2}\) e^2x + C.

Question 4.
(ax + b)²
Solution.
The anti-derivative of (ax + b)² is the function of x whose derivative is (ax + b)².
It is known that,
\(\frac{d}{d x}\) (ax + b)² = 3a (ax + b)²

⇒ (ax + b)² = \(\frac{1}{3 a}\) \(\frac{d}{d x}\) (ax + b)³

∴ (ax + b)² = \(\frac{d}{d x}\) (\(\frac{1}{3 a}\) (ax + b)³))

Therefore, the anti-derivative of (ax + b)² is \(\frac{1}{3 a}\) (ax + b)³) + C.

Question 5.
sin 2x – 4 e^3x
Solution.
The anti-derivative of (sin 2x – 4 e^3x) is the function of x whose derivative is (sin 2x – 4 e^3x).
It is known that,
\(\frac{d}{d x}\) (- \(\frac{1}{2}\) cos 2x – \(\frac{4}{3}\) e^3x) = sin 2x – 4 e^3x
Therefore, the anti-derivative of (sin 2x – 4 e^3x) is (- \(\frac{1}{2}\) cos 2x – \(\frac{4}{3}\) e^3x) + C.

Direction (6 – 20): Find the following integrals:

Question 6.
∫ (4 e^3x + 1) dx
Solution.
∫ (4 e^3x + 1) dx = 4 ∫ e^3x dx + ∫ 1 dx

= 4 \(\left(\frac{e^{3 x}}{3}\right)\) + x + C

= \(\frac{4}{3}\) e^3x + x + C {∵ ∫ e^nx dx = \(\frac{e^{n x}}{x}\)}

Question 7.
∫ x² (1 – \(\frac{1}{x^{2}}\)) dx
Solution.
∫ x² (1 – \(\frac{1}{x^{2}}\)) dx = ∫ (x² – 1) dx
= ∫ x² dx – ∫ 1 dx
= \(\frac{x^{3}}{3}\) – x + C.

Q.8.
∫ (ax² + bx + c) dx
Solution.
∫ (ax² + bx + c) dx = a ∫ x² dx + b ∫ x dx + c ∫ 1 dx

= \(\frac{a x^{2+1}}{2+1}+\frac{b x^{1+1}}{1+1}\) + cx + C

= \(a\left(\frac{x^{3}}{3}\right)+b\left(\frac{x^{2}}{2}\right)\) + cx + C

= \(\frac{a x^{3}}{3}+\frac{b x^{2}}{2}\) + cx + C

Question 9.
∫ (2x² + e^x) dx
Solution.
∫ (2x² + e^x) dx = 2 ∫ x² dx + ∫ e^x dx

= 2 \(\left(\frac{x^{2+1}}{2+1}\right)\) + e^x + C

= 2 \(\left(\frac{x^{3}}{3}\right)\) + e^x + C

= \(\frac{2}{3}\) x³ + e^x + C

Question 10.
∫ (√x – \(\frac{1}{\sqrt{x}}\))² dx
Solution.
∫ (√x – \(\frac{1}{\sqrt{x}}\))² dx = ∫ [(√x)² – 2 √x × \(\frac{1}{\sqrt{x}}\)] dx

= ∫ (x + \(\frac{1}{x}\) – 2) dx = ∫ x dx + ∫ \(\frac{1}{x}\) dx – 2 ∫ 1 dx

= \(\frac{x^{2}}{2}\) + log |x| – 2x + C.

Question 11.
∫ \(\frac{x^{3}+5 x^{2}-4}{x^{2}}\) dx
Solution.
∫ \(\frac{x^{3}+5 x^{2}-4}{x^{2}}\) dx = ∫ \(\left(\frac{x^{3}}{x^{2}}+5 \frac{x^{2}}{x^{2}}-\frac{4}{x^{2}}\right)\) dx

= ∫ (x + 5 – 4 x^-2) dx
= ∫ x dx + 5 ∫ 1 dx – 4 ∫ x^-2 dx
= \(\frac{x^{2}}{2}\) + 5x – 4 \(\left(\frac{x^{-2+1}}{-2+1}\right)\) + C

= \(\frac{x^{2}}{2}\) + 5x – 4 \(\left(\frac{x^{-1}}{-1}\right)\) + C

= \(\frac{x^{2}}{2}\) + 5x + \(\frac{4}{4}\) + C

Question 12.
∫ \(\frac{x^{3}+3 x+4}{\sqrt{x}}\) dx
Solution.
∫ \(\frac{x^{3}+3 x+4}{\sqrt{x}}\) dx

=

Question 13.
∫ \(\frac{x^{3}-x^{2}+x-1}{x-1}\) dx
Solution.
∫ \(\frac{x^{3}-x^{2}+x-1}{x-1}\) dx = ∫ \(\frac{x^{2}(x-1)+1(x-1)}{x-1}\) dx

= ∫ \(\frac{\left(x^{2}+1\right)(x-1)}{x-1}\) dx

= ∫ (x² + 1) dx

= ∫ x² dx + 1 dx

= \(\frac{x^{3}}{3}\) + x + C.

Question 14.
∫ (1 – x) √x dx
Solution.
∫ (1 – x) √x dx = ∫ \(\left(x^{\frac{1}{2}}-x^{\frac{3}{2}}\right)\) dx

=

Question 15.
∫ √x (3x² + 2x + 3) dx
Solution.
∫ √x (3x² + 2x + 3) dx

=

Question 16.
∫ (2x – 3 cos x + e^x) dx
Solution.
∫ (2x – 3 cos x + e^x) dx = 2 ∫ x dx – 3 ∫ cos x dx + ∫ e^x dx
= \(\frac{2 x^{2}}{2}\) – 3 (sin x) + e^x + C
= x² – 3 sin x + e^x + C

Question 17.
∫ (2x² – 3 sin x + 5√x) dx
Solution.
∫ (2x² – 3 sin x + 5√x) dx = 2 ∫ x² dx – 3 ∫ sin x dx + 5 ∫ x^{\(\frac{1}{2}\)} dx

= \(\frac{2 x^{3}}{3}\) – 3 (- cos x) + 5 \(\left(\frac{x^{3 / 2}}{\frac{3}{2}}\right)\) + C

= \(\frac{2}{3}\) x³ + 3 cos x + \(\frac{10}{3}\) x^{\(\frac{3}{2}\)} + C.

Question 18.
∫ sec x (sec x + tan x) dx
Solution.
∫ sec x (sec x + tan x) dx = ∫ (sec² x + sec x . tan x dx
= ∫ sec² x dx + ∫ sec x . tan x dx
= tan x + sec x + C

Question 19.
∫ \(\frac{\sec ^{2} x}{\ {cosec}^{2} x}\) dx
Solution.
∫ \(\frac{\sec ^{2} x}{\ {cosec}^{2} x}\) dx = ∫ \(\frac{\frac{1}{\cos ^{2} x}}{\frac{1}{\sin ^{2} x}}\) dx

= ∫ \(\frac{\sin ^{2} x}{\cos ^{2} x}\) dx

= ∫ tan² x dx

= ∫ (sec² x – 1) dx

= ∫ sec² x dx – ∫ 1 dx = tan x – x + C

Question 20.
∫ \(\int \frac{2-3 \sin x}{\cos ^{2} x} d x\)
Solution.
∫ \(\int \frac{2-3 \sin x}{\cos ^{2} x} d x\) = ∫ \(\left(\frac{2}{\cos ^{2} x}-\frac{3 \sin x}{\cos ^{2} x}\right)\) dx

= ∫ (2 sec² x – 3 sec x . tan x) dx
= 2 ∫ sec² x dx – ∫ tan x . sec x dx
= 2 tan x – 3 sec x + C.

Direction (21 – 22) : Choose the correct answer.

Question 21.
The anti-derivative of (√x + \(\frac{1}{\sqrt{x}}\)) equals.
(A) \(\frac{1}{3} x^{\frac{1}{3}}+2 x^{\frac{1}{2}}+C\)

(B) \(\frac{2}{3} x^{\frac{2}{3}}+\frac{1}{2} x^{2}+C\)

(C) \(\frac{2}{3} x^{\frac{3}{2}}+2 x^{\frac{1}{2}}+C\)

(D) \(\frac{3}{2} x^{\frac{3}{2}}+\frac{1}{2} x^{\frac{1}{2}}+C\)
Solution.
∫ (√x + \(\frac{1}{\sqrt{x}}\)) dx = ∫ \(x^{\frac{1}{2}}\) dx + ∫ \(x^{-\frac{1}{2}}\) dx

= \(\frac{x^{\frac{3}{2}}}{\frac{3}{2}}+\frac{x^{\frac{1}{2}}}{\frac{1}{2}}\) + C

= \(\frac{2}{3} x^{\frac{3}{2}}+2 x^{\frac{1}{2}}\) + C

Hence the correct answer is (C).

Question 22.
If \(\frac{d}{d x}\) f(x) = 4x³ – \(\frac{3}{x^{4}}\) such that f(2) = 0. Then, f(x) is
(A) x⁴ + \(\frac{1}{x^{3}}-\frac{129}{8}\)

(B) x⁴ + \(\frac{1}{x^{4}}+\frac{129}{8}\)

(C) x³ + \(\frac{1}{x^{3}}+\frac{129}{8}\)

(D) x³ + \(\frac{1}{x^{4}}-\frac{129}{8}\)
Solution.
It is given that,
\(\frac{d}{d x}\) f(x) = 4x³ – \(\frac{3}{x^{4}}\)

∴ Anti – derivative of 4x³ – \(\frac{3}{x^{4}}\) = f(x)

∴ f(x) = ∫ (4x³ – \(\frac{3}{x^{4}}\)) dx

⇒ f(x) = 4 ∫ x³ dx – 3 ∫ x^-4 dx

f(x) = \(4\left(\frac{x^{4}}{4}\right)-3\left(\frac{x^{-3}}{-3}\right)+C\)

⇒ f(x) = x⁴ + \(\frac{1}{x^{3}}\) + C
Also, f(2) = 0
∴ f(2) = (2)⁴ + \(\frac{1}{(2)^{3}}\) + C

⇒ 16 + \(\frac{1}{8}\) + C = 0

⇒ C = – (16 + \(\frac{1}{8}\))

⇒ C = – \(\frac{129}{8}\)

⇒ f(x) = x⁴ + \(\frac{1}{x^{3}}-\frac{129}{8}\)

Hence, the correct answer is (A).

Punjab State Board PSEB 12th Class Maths Book Solutions Chapter 6 Application of Derivatives Miscellaneous Exercise Questions and Answers.

PSEB Solutions for Class 12 Maths Chapter 6 Application of Derivatives Miscellaneous Exercise

Question 1.
Using differentials, find the approximate value of each of the following:
Solution.
(a) \(\left(\frac{17}{81}\right)^{\frac{1}{4}}\)
Solution.
Consider y = \(x^{\frac{1}{4}}\).

Let x = \(\frac{16}{81}\) and ∆x = \(\frac{1}{81}\).

Then, ∆y = (x + ∆x)^{\(\frac{1}{4}\)} – x^{\(\frac{1}{4}\)}

= \(\left(\frac{17}{81}\right)^{\frac{1}{4}}-\left(\frac{16}{81}\right)^{\frac{1}{4}}\)

= \(\left(\frac{17}{81}\right)^{\frac{1}{4}}-\frac{2}{3}\)

∴ \(\left(\frac{17}{81}\right)^{\frac{1}{4}}\) = \(\frac{2}{3}\) + ∆y

Now, dy is approximately equal to ∆y and is given by

dy = \(\left(\frac{d y}{d x}\right) \Delta x=\frac{1}{4(x)^{\frac{3}{4}}}(\Delta x)\)

= \(\frac{1}{4\left(\frac{16}{81}\right)^{\frac{3}{4}}}\left(\frac{1}{81}\right)=\frac{27}{4 \times 8} \times \frac{1}{81}=\frac{1}{32 \times 3}=\frac{1}{96}\) = 0.010

Hence, the approximate value of \(\left(\frac{17}{81}\right)^{\frac{1}{4}}\) is \(\frac{2}{3}\) + 0.010
= 0.667 + 0.010 = 0.677.

(b) Consider y = \(x^{-\frac{1}{5}}\).
Let x = 32 and ∆x = 1.
Then, ∆y = (x + ∆x)^{\(-\frac{1}{5}\)} – x^{\(-\frac{1}{5}\)}

= (33)^{\(-\frac{1}{5}\)} – (32)^{\(-\frac{1}{5}\)}

= (33)^{\(-\frac{1}{5}\)} – \(\frac{1}{2}\)
∴ (33)^{\(-\frac{1}{5}\)} = \(\frac{1}{2}\) + ∆y

Now, dy is approximately equal to ∆y and is given by
dy = \(\left(\frac{d y}{d x}\right) \Delta x=\frac{-1}{5(x)^{\frac{6}{5}}}(\Delta x)\) (As y = x^{\(-\frac{1}{5}\)})

= \(\frac{-1}{5(2)^{6}}(1)=-\frac{1}{320}\) = – 0.003.

Hence, the approximate value of (33)^{\(-\frac{1}{5}\)}
= \(\frac{1}{2}\) + (- 0.003) = 0.5 – 0.003 = 0.497.

Question 2.
Show that the function given by f(x) = \(\frac{\log x}{x}\) maximum at x = e.
Solution.
The given function is f(x) = \(\frac{\log x}{x}\)

Therefore, by second derivative test, f is the maximum at x = e.

Question 3.
The two equal sides of an isosceles triangle with fixed base b are decreasing at the rate of 3 cm per second. How fast s the area decreasing when two equal sides are equal to the base?
Solution.

Let ∆ ABC be isosceles, where BC is the base of fixed length b.
Let the length of the two equal sides of ∆ABC be a.
Draw AD ⊥ BC

Now, in ∆ADC, by applying the Pythagoras theorem, we have

AD = \(\sqrt{a^{2}-\frac{b^{2}}{4}}\)

:. Area of triangle (A) = \(\frac{1}{2} b \sqrt{a^{2}-\frac{b^{2}}{4}}\)

The rate of change of the area with respect to time (t) is given by

\(\frac{d A}{d t}=\frac{1}{2} b \cdot \frac{2 a}{2 \sqrt{a^{2}-\frac{b^{2}}{4}}} \cdot \frac{d a}{d t}=\frac{a b}{\sqrt{4 a^{2}-b^{2}}} \cdot \frac{d a}{d t}\)

It is given that the two equal sides of the triangle are decreasing at the rate of 3 cm per second.

∴ \(\frac{d a}{d t}\) = – 3 cm/s (Negative sign shows for decreasing)

∴\(\frac{d A}{d t}=\frac{-3 b}{\sqrt{4 a^{2}-b^{2}}} \mathrm{~cm}^{2} / \mathrm{s}\) cm² / s

When a = b, we have \(\frac{d A}{d t}=\frac{-3 b^{2}}{\sqrt{4 b^{2}-b^{2}}}=\frac{-3 b^{2}}{\sqrt{3 b^{2}}}\) = √3b cm² / s
Hence, if the two equal sides are equal to the base, then the area of the triangle ¡s decreasing at the rate of √3 db cm²/s.

Question 4.
Find the equation of the normal to curve y² = 4x at the point (1, 2).
Solution.
The equation of the given curve is y² = 4x.
On differentiating w.r.t. x, we get
2y \(\frac{d y}{d x}\) = 4

⇒ \(\frac{d y}{d x}\) = \(\frac{4}{2 y}=\frac{2}{y}\)

Slope of the tangent at (1, 2) is \(\left(\frac{d y}{d x}\right)_{(1,2)}=\frac{2}{2}\) = 1
and slope of the normal at the point (1, 2) is = \(\frac{-1}{1}\) = – 1.
∴ Equation of the normal at (1, 2) is
y – 2 = – 1(x – 1)
⇒ y – 2 = – x + 1
⇒ x + y – 3 = 0.

Question 5.
Show that the normal at any point 0 to the curve x = a cos0 + a0 sin0, y = a sin0 – a0 cos0 is at a constant distance from the origin.
Solution.
The given curve is x = a cos θ + a θ sin θ, y = a sin θ – a θ cos θ
On differentiating w.r.t. θ, we get dx
\(\frac{d x}{d \theta}\) = – a sin θ + a sin θ + a θ cos θ
= a θ cos θ

and \(\frac{d y}{d \theta}\) = a cos θ – a cos θ + a θ sin θ
= a θ sin θ

Slope of the tangent at θ,

\(\frac{d y}{d x}=\frac{d y}{d \theta} \cdot \frac{d \theta}{d x}=\frac{a \theta \sin \theta}{a \theta \cos \theta}\) = tan θ

∴ Slope of the normal at θ is \(\frac{1}{\tan \theta}\) = – cot θ.

The equation of the normal at a given point (x, y) is given by
y – a sin θ + a θ cos θ = – \(\frac{1}{\tan \theta}\) (x – a cos θ – a θ sin θ)

⇒ y sin θ – a sin2 θ + a θ sin θ cos θ = – x cos θ + a cos2 θ + a θ sin θ cos θ
⇒ x cos θ + y sin θ – a (sin² θ + cos² θ) = 0
⇒ x cos θ + y sin θ – a = 0
Now, the perpendicular distance of the normal from the origin is \(\frac{|-a|}{\sqrt{\cos ^{2} \theta+\sin ^{2} \theta}}=\frac{|-a|}{\sqrt{1}}=|-a|\), which is independent of θ.
Hence, the perpendicular distance of the normal from the origin is constant.

Question 6.
Find the intervals in which the function f given by f(x) = \(\frac{4 \sin x-2 x-x \cos x}{2+\cos x}\) is
(i) strictly increasing
(ii) strictly decreasing
Solution.
Given, f(x) = \(\frac{4 \sin x-2 x-x \cos x}{2+\cos x}\)

Now, f'(x) = 0
⇒ cos x = 0 or cos x = 4
But, cos x ≠ 4
∴ cos x = 0
⇒ x = \(\frac{\pi}{2}, \frac{3 \pi}{2}\)
Now, x = \(\frac{\pi}{2}\) and x = \(\frac{3 \pi}{2}\) divides (0, 2π) into three disjoint intervals i.e.,
(0, \(\frac{\pi}{2}\)), (\(\frac{\pi}{2}, \frac{3 \pi}{2}\)), and (\(\frac{3 \pi}{2}\), 2π)

In intervals (0, \(\frac{\pi}{2}\)) and (\(\frac{3 \pi}{2}\), 2π), f'(x) > 0.

Thus, f(x) is increasing for 0 < x < \(\frac{x}{2}\) and \(\frac{3 \pi}{2}\) < x < 2π.
In the interval \(\frac{\pi}{2}, \frac{3 \pi}{2}\), f'(x) < 0.
Thus, f(x) is decreasing for \(\frac{\pi}{2}\) < x < \(\frac{3 \pi}{2}\).

Question 7.
Find the intervals in which the function f given by f(x) = x³ + \(\frac{1}{x^{3}}\), x ≠ 0 is
(i) increasing
(ii) decreasing
Solution.
f(x) = x³ + \(\frac{1}{x^{3}}\)

∴ f'(x) = 3x² – \(\frac{3}{x^{4}}=\frac{3 x^{6}-3}{x^{4}}\)

Then, f’(x) = 0 = 3x⁶ – 3 = 0
⇒ x⁶ = 1
⇒ x = ± 1
Now, the points x = 1 and x = – 1 divide the real line into three disornt intervals i.e., (- ∞, – 1), (- 1, 1), and (1, ∞).
In intervals (- ∞, – 1) and (1, ∞) i.e., when x < – 1 and x > 1, f’(x) > 0.
Thus, when x < – 1 and x > 1, f is increasing.
In interval (- 1, 1) i.e., when – 1 < x < 1, f’(x) < 0.
Thus, when – 1 < x < 1, f is decreasing.

Question 8.
Find the maximum area of an isosceles triangle inscribed in the ellipse \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}\) = 1 with Its vertex at one end of the major
Solution.

A = area of isosceles ∆ APP’
= \(\frac{1}{2}\) . PP’ . AM

= \(\frac{1}{2}\) ab (2b sin θ) (a – a cos θ)

= ab (sin θ – \(\frac{1}{2}\) sin 2θ)

Differentiating w.r.t. θ, we get
\(\frac{d A}{d \theta}\) = ab (cos θ – cos 2θ)

For maxima and minima,
\(\frac{d A}{d \theta}\) = 0

∴ ab(cos θ – cos 2θ) = 0
⇒ cos 2θ = cos θ
⇒ 2θ = 2π – θ
⇒ θ = \(\frac{2 \pi}{3}\)

Now, \(\frac{d^{2} A}{d \theta^{2}}\) = ab(- sin θ + 2 sin 2θ)

At θ = \(\frac{2 \pi}{3}\),

\(\frac{d^{2} A}{d \theta^{2}}\) = ab \(\left(-\sin \frac{2 \pi}{3}+2 \sin \frac{4 \pi}{3}\right)\)

= ab \(\left[-\left(\frac{\sqrt{3}}{2}\right)+2\left(-\frac{\sqrt{3}}{2}\right)\right]\)

= ab \(\left(-\frac{\sqrt{3}}{2}-\frac{2 \sqrt{3}}{2}\right)=\frac{-3 \sqrt{3}}{2}\) ab < 0

⇒ A is maximum, when θ = \(\frac{2 \pi}{3}\) = 120°

Maximum value of A = ab(sin 120° – \(\frac{1}{2}\) sin 240°)
= ab \(\left[\frac{\sqrt{3}}{2}-\frac{1}{2}\left(-\frac{\sqrt{3}}{2}\right)\right]=\frac{3 \sqrt{3}}{4} a b\)

Thus, rnaximum area o the isosceles triangle is \(\frac{3 \sqrt{3}}{4} a b\) ab sq. unit.

Question 9.
A tank with rectangular base and rectangular sides, open at the top is to be constructed so that its depth in 2 m and volume is 8 m³. If building of tank costs ₹ 70 per sq m for the base and ₹ 45 per sq metre for sides. What is the cost of least expensive tank?
Solution.
Let l, b and h represent the length, breadth and height of the tank, respectively.
Then, we have height (ft) = 2 m
∴ Volume of the tank = 8 m³
⇒ Volume of the tank = l × b × h
⇒ 8 = l × b × 2
⇒ lb = 4
⇒ b = \(\frac{4}{l}\)
Now, area of the base = lb = 4
Area of the 4 walls (A) = 2h (l + b)
∴ A = 4 (l + \(\frac{4}{l}\))
⇒ \(\frac{d A}{d l}=4\left(1-\frac{4}{l^{2}}\right)\)
However, the length cannot be negative.
Therefore, we have l = 4
∴ b = \(\frac{4}{l}=\frac{4}{2}\) = 2

Now, \(\frac{d^{2} A}{d l^{2}}=\frac{32}{l^{3}}\)

when l = 2, then \(\frac{d^{2} A}{d l^{2}}=\frac{32}{8}\) = 4 > 0

Thus, by second derivative test, the area is the minimum when l = 2.
We have l = b = h = 2
∴ Cost of building the base = ₹ 70 × (lb)
= ₹ 70 (4) = ₹ 280
Cost of building the walls = ₹ 2h (l + b) × 45
= ₹ 90 (2) (2 + 2)
= ₹ 8 (90) = ₹ 720
Required total cost = ₹ 280 + 720 = ₹ 1000
Hence, the total cost of the tank will be ₹ 1000.

Question 10.
The sum of the perimeter of a circle and square is k, where k is some constant. Prove that the sum of their areas is least when the side of square is double the radius of the circle.
Solution.
Let x be the radius of the circle and y be the side of the square
Circumference of the circle = 2πx
Perimeter of square = 4y
Sum of perimeters of circle and square = 2πx + 4y = k ……………(i)
Area of circle = πx²
Area of square = y²
Sum of areas of circle and square = πx² + y² …………..(ii)
From Eq. (i),
y = \(\frac{k-2 \pi x}{4}\) ……………..(iii)

Question 11.
A window is in the form of a rectangle surmounted by a semicircular opening. The total perimeter of the window is 10 m. Find the dimensions of the window to admit maximum light through the whole opening.
Solution.

Let x and y be the length and breadth of the rectangular window.
Then, radius of the semicircular opening = \(\frac{x}{2}\)
It is given that the perimeter of the window is 10 m.
∴ x + 2y + \(\frac{\pi x}{2}\) = 10

⇒ (x + \(\frac{\pi}{2}\)) + 2y = 10

⇒ 2y = 10 – x (1 + \(\frac{\pi}{2}\))

⇒ y = 5 – x (\(\frac{1}{2}+\frac{\pi}{4}\))

∴ Area of the window (A) is given by

Therefore, by second derivative test, the area is the maximum when length x = \(\frac{20}{\pi+4}\) m.

y = \(5-\frac{20}{\pi+4}\left(\frac{2+\pi}{4}\right)=5-\frac{5(2+\pi)}{\pi+4}=\frac{10}{\pi+4} \mathrm{~m}\)

Hence, the required dimensions of the window to admit maximum light is given by length = \(\frac{20}{\pi+4}\) m and breadth = \(\frac{10}{\pi+4}\) m.

Question 12.
A point on the hypotenuse of a triangle is at distance a and b from the sides of the triangle. Show that the maximum length of the hypotenuse is \(\left(a^{\frac{2}{3}}+b^{\frac{2}{3}}\right)^{\frac{3}{2}}\).
Solution.
Let ∆ ABC be right angled at B.
Let AB = x and BC = y.
Let P be a point on the hypotenuse of the triangle such that P is at a distance of a and b from the sides AB and BC respectively.

Let ∠C = θ.
We have, AC = \(\sqrt{x^{2}+y^{2}}\)
Now, PC = b cosec θ and, AP = a sec θ
∴ AC = AP + PC
⇒ AC = b cosec θ + a sec θ ……………. (i)

∴ \(\frac{d(A C)}{d \theta}\) = – cosec θ cot θ + a sec θ tan θ

∴ \(\frac{d(A C)}{d \theta}\) = 0

⇒ a sec θ tan θ = cosec θ cot θ

It can be clearly shown that \(\frac{d^{2}(A C)}{d \theta^{2}}\) < 0 when tan θ = \(\left(\frac{b}{a}\right)^{\frac{1}{3}}\).

Therefore, by second derivative test, the length of the hypotenuse is the tan θ = \(\left(\frac{b}{a}\right)^{\frac{1}{3}}\).

Now, when tan θ = \(\left(\frac{b}{a}\right)^{\frac{1}{3}}\), then we have

AC = \(\frac{b \sqrt{a^{\frac{2}{3}}}+b^{\frac{2}{3}}}{b^{\frac{1}{3}}}+\frac{a \sqrt{a^{\frac{2}{3}}}+b^{\frac{2}{3}}}{a^{\frac{1}{3}}}\) [Using Eqs. (i) and (ii)]

= \(\sqrt{a^{\frac{2}{3}}+b^{\frac{2}{3}}}\left(b^{\frac{2}{3}}+a^{\frac{2}{3}}\right)=\left(a^{\frac{2}{3}}+b^{\frac{2}{3}}\right)^{\frac{3}{2}}\)

Hence, the maximum length of the hypotenuse is \(\left(a^{\frac{2}{3}}+b^{\frac{2}{3}}\right)^{\frac{3}{2}}\).

Question 13.
Find the points at which the function f given by f(x) = (x – 2)⁴ (x + 1)³ has –
(i) local maxima
(ii) local minima
(iii) point of inflexion
Solution.
The given function is f(x) = (x – 2)⁴ (x + 1)³
∴ f'(x) = 4 (x – 2)³ (x + 1)³ + 3(x + 1)² (x – 2)⁴
= (x – 2)³ (x +1)² [4(x + 1) + 3(x – 2)]
= (x – 2)³ (x + 1)² (7x – 2)
f'(x) = 0
⇒ x = – 1 and x = \(\frac{2}{7}\) or x = 2
Now, for values of x close to \(\frac{2}{7}\) and to the left of \(\frac{2}{7}\), f'(x) > 0.

Also, for values of x close to \(\frac{2}{7}\) and to the right of \(\frac{2}{7}\), f'(x) < 0.

Thus, x = \(\frac{2}{7}\) is the point of local maxima.
Now, for values of x close to 2 and to the left of 2, f'(x) < 0. Also, for values of x close to 2 and to the right of 2, f'(x) > 0.
Thus, x = 2 is the point of local minima.
Now, as the value of x varies through – 1, f (x) does not change its sign.
Thus, x = – 1 is the point of inflexion.

Question 14.
Find the absolute maximum and minimum values of the function f’given by f(x) = cos² x + sin x, x ∈ [0, π].
Solution.
Given, f(x) = cos² x + sin x
∴ f'(x) = 2 cos x (- sin x) + cos x = – 2 sin x cos x + cos x
Now, f’ (x) = 0
⇒ 2 sin x cos x = cos x
⇒ 2 sin x cos x – cos x = 0
⇒ cos x (2 sin x – 1) = 0
⇒ sin x = – or cos x = 0
⇒ x = \(\frac{\pi}{6}\) or \(\frac{\pi}{2}\) as x ∈ [0, π]
Now, evaluating the value of f at critical points x = \(\frac{\pi}{2}\) and x = \(\frac{\pi}{6}\) and at
the end points of the interval [0, π] (i.e., at x = 0 and x = π), we have

f(\(\frac{\pi}{6}\)) = cos² \(\frac{\pi}{6}\) + sin \(\frac{\pi}{6}\)
= \(\left(\frac{\sqrt{3}}{2}\right)^{2}+\frac{1}{2}=\frac{5}{4}\)

f(0) = cos² 0 + sin 0
= 1 + 0 = 1

f(π) = cos² π + sin π
= (- 1)² + 0 = 1

f(\(\frac{\pi}{2}\)) = cos² \(\frac{\pi}{2}\) + sin \(\frac{\pi}{2}\)
= 0 + 1 = 1

Hence, the absolute maximum value of f is \(\frac{5}{4}\) occurring at x = \(\frac{\pi}{6}\) and the absolute minimum value of f is 1 occurring at x = 0, \(\frac{\pi}{2}\) and π.

Question 15.
Show that the altitude of the right circular cone of maximum volume that can be inscribed in a sphere of radius r, is -.
Solution.

Let R be the radius and h be the height of cone.
∴OA = h – r
In ∆OAB, r² = R² + (h – r)²
⇒ r² = R² + h² + r² – 2rh
⇒ R² = 2rh – h²
The volume V of the cone is given by V = \(\frac{1}{3}\) πR²h
= \(\frac{1}{3}\) πh (2rh – h²)

= \(\frac{1}{3}\) π (2rh² – h³)

On differentiating w.r.t. h, we get
\(\frac{d V}{d h}\) = \(\frac{1}{3}\) π (4rh – 3h²)

For maximum or minimum put \(\frac{d V}{d h}\) = 0
⇒ 4rh = 3h²
⇒ 4r = 3h
∴ h = \(\frac{4 r}{3}\) (h ≠ 0)

Now, \(\frac{d^{2} V}{d h^{2}}\) = \(\frac{1}{3}\) π (4r – 6h)

At h = \(\frac{4 r}{3}\),

\(\left(\frac{d^{2} V}{d h^{2}}\right)_{h=\frac{4 r}{3}}\) = \(\frac{1}{3}\) π (4r – 6 × \(\frac{4 r}{3}\))

= \(\frac{pi}{3}\) (4r – 8r) = \(\frac{-4 r \pi}{3}\) < 0 ⇒ V is maximum when h = \(\frac{4 r}{3}\) Hence, volume of the cone is maximum when h = \(\frac{4 r}{3}\), which is the altitude of cone. Question 16. Let f be a function defined on [a, b] such that f'(x) > 0, for all x ∈ (a, b). Then prove that f is an increasing function on (a, b).
Solution.
Given, f'(x) > 0 on (a, b)
f is a differentiable function on (a, b).
Also, every differentiable function is continuous, therefore, f is continuous on (a, b).
Let x₁, x₂ ∈ [a, b] and x₂ > x₁; then by LMV theorem, there exist c ∈ [a, b] such that

f'(c) = \(\frac{f\left(x_{2}\right)-f\left(x_{1}\right)}{x_{2}-x_{1}}\)
⇒ f(x₂) – f(x₁) = (x₂ – x₁) f'(c)
⇒ f(x₂) – f(x₁) > 0 as x₂ > x₁ and f'(x) > 0
⇒ f(x₂) > f(x₁)
For x₁ < x₂
⇒ f(x₁) < f(x₂)
Hence, f is an increasing function on (a, b).

Question 17.
Show that the height of the cylinder of maximum volume that can be inscribed in a sphere of radius R, is . Also, find the maximum volume.
Solution.

Radius of the sphere = R
Let h be the height and x be the diameter of the base of the inscribed cylinder Then.
h² + x² = (2R)²
⇒ h² + x² = 4R² …………(i)
Volume of the cylinder = π (radius)² × height

⇒ V = π (\(\frac{x}{2}\))² . h
= \(\frac{1}{4}\) πx²h
⇒ V = \(\frac{1}{4}\) πh(4R² – h²) …………(ii)
[From Eq. (i), x² = 4R² – h²]

⇒ V = πR²h – \(\frac{1}{4}\) πh³
On differentiating w.r.t h, we get
\(\frac{d V}{d h}\) = πR²h – \(\frac{3}{4}\) πh²
= π (R² – \(\frac{3}{4}\) h²)

Put \(\frac{d V}{d h}\) = 0
⇒ R² = \(\frac{3}{4}\) h²
⇒ h = \(\frac{2 R}{\sqrt{3}}\)

Thus, volume of the cylinder is maximum when h = \(\frac{2 R}{\sqrt{3}}\).

Question 18.
Show that height of the cylinder of greatest volume which can be inscribed in a circular cone of height h and having
semi-vertical angle a is one-third that of one cone and the
greatest volume of cylinder Is ida3 tan2 a.
Solution.

Let VAR be the cone of height h, semi-vertical angle a and let x be the radius of the base of the cylinder A’ B’ DC which is inscribed in the cone VAB.
Then, OO’ is the height of the cylinder = VO – VO’ = h – x cot α
Volume of the cylinder,
V = πx² (h – x cot α) ………………(i)
On differentiating w.r.t. x, we get
\(\frac{d V}{d x}\) = 2πrh – 3πx² cot α
For maxima or minima put \(\frac{d V}{d x}=\)= 0
⇒ 2πxh – 3πx² cot α = 0

⇒ x = \(\frac{2 h}{3}\) tan α (∵ x ≠ 0)

Now, \(\frac{d^{2} V}{d x^{2}}\) = 2πh – 6π x cot α

At x= \(\frac{2 h}{3}\) tan α,

\(\frac{d^{2} V}{d x^{2}}\) = π (2h – 4h) = – 2πh < 0

Now, OO’ = h – x cot α = h – \(\frac{2 h}{3}\) = \(\frac{h}{3}\)

∴ The maximum volume of the cylinder is V = π (\(\frac{2 h}{3}\) tan α)² (h – \(\frac{2 h}{3}\))

= \(\frac{4}{27}\) πh³ tan² α.

Direction (19 – 24):
Choose the correct answer.

Question 19.
A cylindrical tank of radius 10 m is being filled with wheat at the rate of 314 cubic meter per hour. Then the depth of the wheat is increasing at the rate of
(A) 1 m/h
(B) 0.1 m/h
(C) 1.1 m/h
(D) 0.5 m/h
Solution.
Let r be the radius of the cylinder.
Then, volume (V) of the cylinder is given by,
V = π (radius)² × height
= π (10)² h (radius = 10 m)
= 100 πh
On differentiating w.r.t. t, we get
\(\frac{d V}{d t}\) = 100 π \(\frac{d h}{d t}\)
The tank is being filled with wheat at the rate of 314 cubic metres per hour.
\(\frac{d V}{d t}\) = 314 m³/h

Thus, we have
314 = 100π \(\frac{d h}{d t}\)

⇒ \(\frac{d h}{d t}\) = \(\frac{314}{100(3.14)}=\frac{314}{314}\) = 1

Hence, the depth of wheat is increasing at the rate of 1 m / h.
The correct answer is (A).

Question 20.
The slope of the tangent to the curve x = t² + 3t – 8, y = 2t² – 2t – 5 at the point (2, – 1) is
(A) \(\frac{22}{7}\)

(B) \(\frac{6}{7}\)

(C) \(\frac{7}{6}\)

(D) \(\frac{-6}{7}\)
Solution.
The given curve is x = t² + 3t – 8 and y = 2t² – 2t – 5
∴ \(\frac{d x}{d t}\) = 2t + 3 and

\(\frac{d y}{d t}\) =4t – 2

∴ \(\frac{d y}{d x}=\frac{d y}{d t} \cdot \frac{d t}{d x}=\frac{4 t-2}{2 t+3}\)

The given point is (2, – 1)
At x = 2, we have r2+3r-8=2
⇒ t² + 3t – 10 = 0
⇒ (t – 2) (t + 5) = 0
⇒ t = 2 or t = – 5
At y = – 1, we have
⇒ 2t² – 2t – 4 = 0
⇒ 2 (t² – t – 2) = 0
⇒ t = 2 or t = – 1
The common value oft is 2.
Hence, the slope of the tangent to the given curve at point (2, – 1) is

\(\left[\frac{d y}{d x}\right]_{t=2}=\frac{4(2)-2}{2(2)+3}=\frac{8-2}{4+3}=\frac{6}{7}\)
The correct answer is (B).

Question 21.
The line y = mx + 1 is a tangent to the curve y2 = 4# if the value of m is
(A) 1
(B) 2
(C) 3
(D) \(\frac{1}{2}\)
Solution.
The equation of the tangent to the given curve is y = mx + 1.
Now, substituting y = mx + 1 in y² = 4x, we get
⇒ (mx + 1)² = 4x
⇒ m²x² +1 + 2mx – 4x = 0
⇒ m²x² + x (2m – 4) + 1 = 0 …………….(i)
Since a tangent touches the curve at one point, the roots of equation (i) must be equal.
Therefore, we have
Discriminant = 0
⇒ (2m – 4)² – 4(m²) (1) = 0
⇒ 4m² + 16 – 16m – 4m² = 0
⇒ 16 – 16m = 0
⇒ m = 1
Hence, the required value of m is 1.
The correct answer is A.

Question 22.
The normal at the point (1, 1) on the curve 2y + x² = 3 is
(A) x + y = 0
(B) x – y = 0
(C) x + y + 1 = 0
(D) x – y = 0
Solution.
The equation of the given curve is 2y + x² = 3

\(\frac{2 d y}{d x}\) + 2x = 0

\(\frac{d y}{d x}\) = – x

Slope of the normal to the given curve at point (1, 1) is \(\frac{-1}{\left[\frac{d y}{d x}\right]_{(1,1)}}=\frac{-1}{-1}\) = 1
Hence, the equation of the normal to the given curve at (1, 1) is given as
⇒ y – 1 = 1 (x – 1)
⇒ y – 1 = x – 1
⇒ x – y = 0
The correct answer is (B).

Question 23.
The normal to the curve x² = 4y passing (1, 2) is
(A) x + y = 3
(B) x – y = 3
(C) x + y = 1
(D) x – y = 1
Solution.
The equation of the given curve is x = 4y
Differentiating w.r.t x, we get
2x = 4 \(\frac{d y}{d x}\)

∴ \(\frac{d y}{d x}=\frac{2 x}{4}=\frac{x}{2}\) = slope of the tangent
∴Slope of the normal = (- 1) / slope of tangent
∴ Equation of normal at (x₁, y₁) is
y – y₁ = (x – x₁) ……………..(i)
It passes through (1, 2), therefore
2 – y₁ = – \(\frac{2}{x_{1}}\) (1 – x₁)
⇒ 2x₁ – x₁y₁ = – 2 + 2x₁
x₁y₁ = 2
or y₁ = \(\frac{2}{x_{1}}\) …………(ii)
The point (x₁, y₁) lies on x² = 4y
x₁² = 4y₁ ……………(iii)
From Eqs. (i) and (iii), we get
x₁² = 4 . \(\frac{2}{x_{1}}\)
∴ x₁³ = 8
x₁ = 2
From Eq. (iii), 4 = 4y₁
∴ y₁ = 1
Putting these values in Eq. (i), we get
y – 1 = – \(\frac{2}{24}\) (x – 2)
y – 1 = – x + 2
⇒ x + y = 3
The correct answer is (A).

Question 24.
The points on the curve 9y² = x³, where the normal to the curve makes equal intercepts with the axes are
(A) (4, ± \(\frac{8}{3}\))

(B) (4, – \(\frac{8}{3}\))

(C) (4, ± \(\frac{3}{8}\))

(D)(± 4, \(\frac{3}{8}\))
Solution.
The equation of the given curve is 9y² = x³
Differentiating w.r.t. x, we get
18 y \(\text { dy }\) = 3x²

⇒ \(\frac{d y}{d x}=\frac{x^{2}}{6 y}\)

Let P(x₁, y₁) be the point where normal is drawn
Slope of tangent = \(\frac{x_{1}^{2}}{6 y_{1}}\)

∴ Slope of normal = \(\frac{6 y_{1}}{x_{1}^{2}}\)

Normal make equal intercepts on the curve
∴ Its slope = ± 1
– \(\frac{6 y_{1}}{x_{1}^{2}}\) = ± 1
⇒ 6y₁ = ± x₁² ……………..(i)
(x₁, y₁)lies on the curve 9y² = x³
⇒ 9y₁² = x₁³ …………..(ii)
Taking +ve sign, eliminating y₁ from Eq. (1) and Eq. (ii),
9 \(\left(\frac{x_{1}^{2}}{6}\right)^{2}\) = x₁³
⇒ \(\frac{9 x_{1}^{4}}{36}\) = x₁³
⇒ x₁ = 4
From Eq. (i),
y₁ = ± \(\frac{x_{1}^{2}}{6}=\pm \frac{16}{6}\)[Putting x = 4]
= ± \(\frac{8}{3}\)
∴ The point P is (4, ± \(\frac{8}{3}\))
The correct answer is (A).

Punjab State Board Syllabus PSEB 12th Class Hindi Book Solutions Guide Pdf is part of PSEB Solutions for Class 12.

PSEB 12th Class Hindi Guide | Hindi Guide for Class 12 PSEB

Hindi Guide for Class 12 PSEB | PSEB 12th Class Hindi Book Solutions

प्राचीन काव्य

• Chapter 1 सूरदास
• Chapter 2 मीराबाई
• Chapter 3 बिहारी
• Chapter 4 गुरु गोबिन्द सिंह

आधुनिक काव्य

• Chapter 5 मैथिलीशरण गुप्त
• Chapter 6 जयशंकर प्रसाद
• Chapter 7 सुमित्रानन्दन पंत
• Chapter 8 हरिवंशराय बच्चन
• Chapter 9 सच्चिदानन्द हीरानन्द, वात्स्यायन ‘अज्ञेय’
• Chapter 10 शिवमंगल सिंह ‘सुमन’
• Chapter 11 गिरिजा कुमार माथुर
• Chapter 12 धर्मवीर भारती
• Chapter 13 डॉ० चन्द्र त्रिखा
• Chapter 14 गीता डोगरा

निबन्ध भाग

• Chapter 15 सच्ची वीरता
• Chapter 16 क्या निराश हुआ जाए?
• Chapter 17 अगर ये बोल पाते : जलियाँवाला बाग़
• Chapter 18 समय नहीं मिला
• Chapter 19 शार्टकट सब ओर
• Chapter 20 गुरु गोबिन्द सिंह

कहानी भाग

• Chapter 21 मधुआ
• Chapter 22 तत्सत्
• Chapter 23 ठेस
• Chapter 24 उपेक्षिता

एकांकी भाग

• Chapter 25 वापसी
• Chapter 26 रीढ़ की हड्डी

हिन्दी साहित्य का इतिहास

• रीतिकाल
• आधुनिक काल

PSEB 12th Class Hindi Book Vyakaran व्याकरण

व्यावहारिक व्याकरण

• समास
• पद-परिचय
• छन्द
• अलंकार

PSEB 12th Class Hindi Book Rachana रचना-भाग

• निबंध-लेखन
• विज्ञापन लेखन
• सूचना लेखन
• पंजाबी से हिन्दी में अनुवाद
• पारिभाषिक शब्दावली

PSEB 12th Class Hindi Structure of Question Paper

कक्षा – बारहवीं (पंजाब)
विषय – हिंदी

समय : 3 घंटे

पूर्णांक लिखित : 80

नोट – (i) 05 अंक सुंदर लिखाई के लिए निर्धारित किए गए हैं। अक्षरों व शब्दों के सामान्य आकार, अक्षरों की सुस्पष्टता, अक्षरों व शब्दों के बीच की निश्चित दूरी, लिखने में एकसारता व प्रवाहयुक्त लेखन आदि के आधार पर अध्यापक परीक्षार्थी को लिखाई का मूल्यांकन करेगा।

• प्रश्न-पत्र में कुल 15 प्रश्न होंगे।
• सभी प्रश्न हल करने अनिवार्य होंगे।
• प्रश्न-पत्र के छह भाग (क से च तक) होंगे।

भाग – क : अति लघूत्तर प्रश्न (वस्तुनिष्ठ प्रश्न) (20 Marks)

प्रश्न 1. में (i) से (x) तक वस्तुनिष्ठ प्रश्न पूछे जायेंगे। प्रत्येक प्रश्न एक अंक का होगा। ये प्रश्न एक शब्द से एक वाक्य तक के उत्तर वाले अथवा हां/नहीं अथवा रिक्त स्थानों की पूर्ति करो अथवा सही/गलत अथवा बहुवैकल्पिक उत्तरों वाले, किसी भी प्रकार के हो सकते हैं।

• (i – iv) तक समास (अव्ययीभाव, तत्पुरुष, बहुब्रीहि तथा द्वंद्व) से संबंधित चार वस्तुनिष्ठ प्रश्न पूछे जायेंगे। (4 × 1 = 4)
• (v – vi) पद परिचय से संबंधित दो वस्तुनिष्ठ प्रश्न पूछे जाएंगे। (2 × 1 = 2)
• (vii – xi) तक पाठ्य-पुस्तक (हिंदी पुस्तक-12) में से पाँच वस्तुनिष्ठ प्रश्न पूछे जायेंगे। (5 × 1 = 5)
• (xii – xvi) तक हिंदी साहित्य का इतिहास (रीतिकाल एवं आधुनिक काल) में से पाँच वस्तुनिष्ठ प्रश्न पूछे जायेंगे। (5 × 1 = 5)
• (xvii – xviii) छंद से संबंधित दो वस्तुनिष्ठ प्रश्न पूछे जाएंगे। (2 × 1 = 2)
• (xvii – xviii) अलंकार से संबंधित दो वस्तुनिष्ठ प्रश्न पूछे जाएंगे। (2 × 1 = 2)

भाग – ख (पाठ्य-पुस्तक) (23 Marks)

प्रश्न 2. (i) हिंदी पुस्तक-12 में संकलित ‘प्राचीन काव्य’ में से दो पद्यांश दिये जायेंगे जिनमें से एक पद्यांश की सप्रसंग व्याख्या लिखने के लिये कहा जायेगा। प्रसंग के लिये 1 अंक तथा व्याख्या के लिये 3 अंक निर्धारित हैं। (1 + 3 = 4)

(ii) हिंदी पुस्तक-12 में संकलित ‘आधुनिक काव्य’ में से दो पद्यांश दिये जायेंगे जिनमें से एक पद्यांश की सप्रसंग व्याख्या लिखने के लिये कहा जायेगा। प्रसंग के लिये 1 अंक तथा व्याख्या के लिये 3 अंक निर्धारित हैं। (1 + 3 = 4)

प्रश्न 3. ‘प्राचीन काव्य’ तथा आधुनिक काव्य की विषय वस्तु से संबंधित दो लघूत्तर प्रश्न पूछे जायेंगे जिनमें से एक प्रश्न का उत्तर लगभग 50 शब्दों में लिखने के लिए कहा जायेगा। (2½)

प्रश्न 4. पाठ्य-पुस्तक में संकलित गद्य भाग की विषय वस्तु से संबंधित तीन निबंधात्मक प्रश्न पूछे जायेंगे जिनमें से एक प्रश्न का उत्तर लगभग 80 शब्दों में लिखने के लिये कहा जायेगा। (5)

नोट : प्रश्न-पत्र निर्माता पाठ्य-पुस्तक में संकलित गद्य भाग (निबंध, कहानी एवं एकाँकी) की सर्भ विधाओं को पूर्ण प्रतिनिधित्व दे।

प्रश्न 5. पाठ्य-पुस्तक में संकलित ‘निबंध’ भाग में से दो लघूत्तर प्रश्न पूछे जायेंगे जिनमें से एक प्रश्न का उत्तर लगभग 50 शब्दों में लिखने के लिये कहा जायेगा। (2½)

प्रश्न 6. पाठ्य-पुस्तक में संकलित ‘कहानी’ भाग में से दो लघूत्तर प्रश्न पूछे जायेंगे जिनमें से एक का उत्तर लगभग 50 शब्दों में लिखने के लिये कहा जायेगा। (2½)

प्रश्न 7. पाठ्य-पुस्तक में संकलित ‘एकाँकी’ भाग में से दो लघूत्तर प्रश्न पूछे जायेंगे जिनमें से एक प्रश्न का उत्तर लगभग 50 शब्दों में लिखने के लिये कहा जायेगा। (2½)

भाग – ग : हिंदी साहित्य का इतिहास (रीतिकाल एवं आधुनिक काल) (8 अंक)

प्रश्न 8. इस प्रश्न में हिंदी साहित्य के ‘रीतिकाल’ की प्रमुख परिस्थितियों, प्रमुख प्रवृत्तियों एवं प्रमुख कवियों से संबंधित दो निबंधात्मक प्रश्न पूछे जायेंगे जिनमें से एक प्रश्न का उत्तर लगभग 70-80 शब्दों में लिखने के लिये कहा जायेगा। (4)

प्रश्न 9. इस प्रश्न में हिंदी साहित्य के ‘आधुनिक काल’ की प्रमुख परिस्थितियों, प्रमुख प्रवृत्तियों एवं प्रमुख कवियों से संबंधित दो निबंधात्मक प्रश्न पूछे जायेंगे जिनमें से एक प्रश्न का उत्तर लगभग 70-80 शब्दों में लिखने के लिये कहा जायेगा। (4)

भाग – घ (रचनात्मक लेखन) (7 अंक)

प्रश्न 10. यह प्रश्न निबंध रचना से संबंधित होगा। कोई चार विषय देकर उनमें से किसी एक विषय पर लगभग 230-250 शब्दों में निबंध लिखने के लिये कहा जायेगा। भूमिका के 2 अंक, विषय वस्तु के 4 अंक और उपसंहार के 1 अंक निर्धारित हैं। (2 + 4 + 1 = 7)

भाग – ङ (व्यावहारिक ज्ञान) (9 अंक)

प्रश्न 11. इस प्रश्न में लगभग 40-50 शब्दों का पंजाबी में एक गद्यांश दिया जायेगा जिसका अनुवाद हिंदी में लिखना होगा। (3)

प्रश्न 12. अंग्रेजी के पाँच पारिभाषिक शब्द दिए जायेंगे जिनमें से किन्हीं तीन शब्दों के हिंदी रूप लिखकर वाक्यों में प्रयोग करने के लिए कहा जायेगा। (3)

प्रश्न 13. विज्ञापन और सूचना से संबंधित दो प्रश्न पूछे जायेंगे जिनमें से एक प्रश्न का उत्तर लिखने के लिये कहा जायेगा। (3)

भाग – च (छंद एवं अलंकार) (8 अंक)

प्रश्न 14. कोई दो छंद देकर किसी एक छंद का लक्षण एवं उदाहरण लिखने के लिये कहा जायेगा। (1 + 3 = 4)

प्रश्न 15. कोई दो अलंकार देकर किसी एक अलंकार की परिभाषा एवं उदाहरण लिखने के लिये कहा जायेगा। (1 + 3 = 4)

आंतरिक मूल्यांकन (20 अंक)

आंतरिक मूल्यांकन की रूपरेखा

1. भाषायी कौशलों का मूल्यांकन (12 अंक)

• श्रवण कौशल (3)
• वाचन कौशल (3)
• पठन कौशल (3)
• लेखन कौशल (3)

2. पुस्तक बैंक (2 अंक)
विद्यार्थी द्वारा स्कूल के पुस्तकालय में हिंदी विषय की पुस्तकों के संग्रह में योगदान देने के आधार पर मूल्यांकन किया जाए। स्कूल में बने पुस्तक बैंक में समय पर पुस्तकें जमा करवाने व पुस्तकों का रखरखाव करने आदि के आधार पर मूल्यांकन किया जाए।

3. परियोजना कार्य : (6 अंक)
इसके अंतर्गत निर्धारित पाठ्यक्रम के आधार पर अध्यापक द्वारा विद्यार्थियों को परियोजना तैयार करने को कहा जाएगा, जिसका मार्गदर्शन अध्यापक करेगा। विद्यार्थी परियोजना लिखते समय उसे रुचिपूर्ण बनाने के लिए चित्रों का प्रयोग कर सकता है। इसके अंक निम्नलिखित प्रकार से निर्धारित होंगे। विषय वस्तु की समझ एवं प्रस्तुतिकरण (2 + 4 = 6)

PSEB 12th Class Hindi Syllabus

कक्षा – बारहवीं (पंजाब)
विषय – हिंदी
समय : 3 घंटे

पूर्णांक (लिखित) = 75 + 5 (सुंदर लिखाई) = 80
आंतरिक मूल्यांकन : 20

विषय-वस्तु	अंक
भाग – क : अति लघूत्तर प्रश्न (वस्तुनिष्ठ प्रश्न)	20
समास (अव्ययीभाव, तत्पुरुष, बहुब्रीहि तथा द्वंद्व)
पाठ्य-पुस्तक
हिंदी साहित्य का इतिहास (रीतिकाल एवं आधुनिक काल)
छंद
अलंकार
भाग – ख : पाठ्य-पुस्तक (हिंदी पुस्तक-12)	23
भाग – ग : हिंदी साहित्य का इतिहास (रीतिकाल एवं आधुनिक काल)	8
भाग – घ : रचनात्मक लेखन : निबंध लेखन	7
भाग – ङ : व्यावहारिक ज्ञान	9
1. पंजाबी गद्यांश का हिंदी अनुवाद	3
2. पारिभाषिक शब्दावली (J से लेकर Z तक)	3
3. विज्ञापन लेखन, सूचना लेखन	3
भाग-च : छन्द एवं अलंकार	8
1. छंद (दोहा, सोरठा, सवैया, कवित्त, चौपाई)	4
2. अलंकार (अनुप्रास, उपमा, रूपक, यमक, श्लेष)।	4

Punjab State Board Syllabus PSEB 12th Class Environmental Education Book Solutions Guide Pdf in English Medium & Punjabi Medium & Hindi Medium are part of PSEB Solutions for Class 12.

PSEB 12th Class Environmental Education Guide | Environmental Education Guide for Class 12 PSEB in English Medium

PSEB 12th Class Environmental Education Book Solutions in Punjabi Medium

• Chapter 1 ਜੈਵਿਕ ਵਿਭਿੰਨਤਾ / ਜੀਵ ਅਨੇਕਰੂਪਤਾ (ਭਾਗ-1) [Biodiversity (Part-I)]
• Chapter 2 ਜੈਵਿਕ ਵਿਭਿੰਨਤਾ / ਜੀਵ ਅਨੇਕਰੂਪਤਾ (ਭਾਗ-2) [Biodiversity (Part-II)]
• Chapter 3 ਜੈਵਿਕ ਵਿਭਿੰਨਤਾ / ਜੀਵ ਅਨੇਕਰੂਪਤਾ (ਭਾਗ-3) [Biodiversity (Part-III)]
• Chapter 4 ਵਾਤਾਵਰਣੀ ਪ੍ਰਬੰਧਣ (ਭਾਗ-1) [Environmental Management (Part-1)]
• Chapter 5 ਵਾਤਾਵਰਣੀ ਪ੍ਰਬੰਧਣ (ਭਾਗ-2) [Environmental Management (Part-II)]
• Chapter 6 ਵਾਤਾਵਰਣੀ ਪ੍ਰਬੰਧਣ (ਭਾਗ-3) [Environmental Management (Part-III)]
• Chapter 7 ਸਟੱਬਲ ਬਰਨਿੰਗ ਦੀਆਂ ਸਮੱਸਿਆਵਾਂ (Problems of Stubble Burning)
• Chapter 8 ਕਾਇਮ ਰਹਿਣਯੋਗ / ਝੱਲਣਯੋਗ / ਟਿਕਾਊ ਵਿਕਾਸ (ਭਾਗ-1) [Sustainable Development (Part-I)]
• Chapter 9 ਕਾਇਮ ਰਹਿਣਯੋਗ / ਝੱਲਣਯੋਗ / ਟਿਕਾਊ ਵਿਕਾਸ (ਭਾਗ-2) [Sustainable Development (Part-II)
• Chapter 10 ਕਾਇਮ ਰਹਿਣਯੋਗ / ਝੱਲਣਯੋਗ / ਟਿਕਾਊ ਵਿਕਾਸ (ਭਾਗ-3) [Sustainable Development (Part-III)]
• Chapter 11 ਕਾਇਮ ਰਹਿਣਯੋਗ / ਝੱਲਣਯੋਗ / ਟਿਕਾਊ ਖੇਤੀਬਾੜੀ (ਭਾਗ-1) [Sustainable Agriculture (Part-I)]
• Chapter 12 ਕਾਇਮ ਰਹਿਣਯੋਗ / ਝੱਲਣਯੋਗ / ਟਿਕਾਊ ਖੇਤੀਬਾੜੀ (ਭਾਗ-2) [Sustainable Agriculture (Part-II)]
• Chapter 13 ਕਾਇਮ ਰਹਿਣਯੋਗ / ਝੱਲਣਯੋਗ / ਟਿਕਾਊ ਖੇਤੀਬਾੜੀ (ਭਾਗ-3) [Sustainable Agriculture (Part-III)]
• Chapter 14 ਵਾਤਾਵਰਣੀ ਕਿਰਿਆ-(ਭਾਗ-1) [Environmental Action (Part-I)]
  ਮਨੁੱਖ ਜਾਤੀ ਦੀਆਂ ਮੁੱਢਲੀਆਂ ਜ਼ਰੂਰਤਾਂ-ਖਾਧ ਪਦਾਰਥ, ਪਾਣੀ, ਪਨਾਹਗਾਹ ਅਤੇ ਈਂਧਨ ਦੀ ਸਾਰਿਆਂ ਲਈ ਪੂਰਤੀ ਕਰਨਾ [Meeting Basic Human Needs (Food, Water, Shelter and Fuel For All)]
• Chapter 15 ਵਾਤਾਵਰਣੀ ਕਿਰਿਆ (ਭਾਗ-2) [Environmental Action (Part-II)]
  ਵਸੋਂ ਤੇ ਕੰਟਰੋਲ (Population Control)
• Chapter 16 ਵਾਤਾਵਰਣੀ ਕਿਰਿਆ (ਭਾਗ-3) [Environmental Action (Part-III)]
  ਖਪਤ ਦੇ ਬਦਲਦੇ ਨਮੂਨੇ (Changing Consumption Patterns)
• Chapter 17 ਵਾਤਾਵਰਣੀ ਕਿਰਿਆ (ਭਾਗ-4) [Environmental Action (Part-IV)]
  ਵਾਤਾਵਰਣੀ ਪ੍ਰਦੂਸ਼ਣ ਦੀ ਰੋਕਥਾਮ ਅਤੇ ਕਾਬੂ (Prevention and Control of Environmental Pollution)
• Chapter 18 ਵਾਤਾਵਰਣੀ ਕਿਰਿਆ (ਭਾਗ-5) [Environmental Action (Part-V)]
  ਵਿਅਰਥ ਪਦਾਰਥਾਂ ਦਾ ਪ੍ਰਬੰਧਣ (Waste Management)
• Chapter 19 ਵਾਤਾਵਰਣੀ ਕਿਰਿਆ (ਭਾਗ-6) [Environmental Action (Part-VI)]
• Chapter 20 ਨਸ਼ਾ-ਮਾੜੇ ਪ੍ਰਭਾਵ-II [Drugs-Ill Effects-II]

Punjab State Board PSEB 12th Class Maths Book Solutions Chapter 6 Application of Derivatives Ex 6.5 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 12 Maths Chapter 6 Application of Derivatives Ex 6.5

Question 1.
Find the maximum and minimum values, if any, of the following functions given by
(i) = (2x – 1)² + 3
(ii) f(x) = 9x² + 12x + 2
(iii) f(x) = – (x – 1)² + 10
(iv) g(x) = x³ + 1
Solution.
(i) The given function is f(x) = (2x – 1)² + 3.
It can be observed that (2x – 1)² ≥ 0 for every x ∈ R.

Therefore, f(x) = (2x – 1)² + 3 ≥ 3 for every x ∈ R.
The minimum value of f is attained when 2x – 1 = 0
⇒ x = \(\frac{1}{2}\)
∴ Minimum value of f = f(\(\frac{1}{2}\))
= (2 . \(\frac{1}{2}\) – 1)² + 3 = 3
Hence, function f does not have a maximum value.

(ii) The given function is f(x) = 9x² + 12x + 2
= (3x + 2)² – 2
It can be observed that (3x + 2)² ≥ 0 for every x ∈ R.
Therefore, f(x) = (3x + 2)² – 2 ≥ – 2 for every x ∈ R.
The minimum value of f is attained when 3x +2 = 0
⇒ x = \(\frac{-2}{3}\)
∴ Minimum value of f = f(\(\frac{-2}{3}\))= (3 . (\(\frac{- 2}{3}\)) + 2)² – 2 = – 2.
Hence, function f does not have a maximum value.

(iii) The given function is f(x) = -(x – 1)² + 10.
It can be observed that (x – 1)² ≥ 0 for every x ∈ R.
Therefore, f(x) = – (x – 1)² + 10 ≤ 10 for every x ∈ R.
The minimum value off is attained when (x – 1) = 0
⇒ x = 1
∴ Minimum value of f = f(1) = -(1 – 1)² + 10 = 10
Hence, function f does not have a minimum value.

(iv) The given function is g(x) = x³ + 1.
We observe that the value of f(x) increases when the value of x increases and f(x) can be made as large as we please by giving large value to x. So, f(x) does not have the maximum value.
Similarly, f(x) can be made as small as we please by giving smaller values of x.
So, f(x) does not have the minimum value.

Question 2.
Find the maximum and minimum values, if any, of the following functions given by
(i) f(x) = |x + 2| – 1
(ii) g(x) = – |x + 1| + 3
(iii) h(x) = sin (2x) + 5
(iv) f(x) = |sin 4x| + 3|
(v) h(x) = x + 1, x e (-1, 1)
Solution.
(i) Given, f(x) = |x + 2 | – 1
We know that, | x + 2 | ≥ 0 for every x ∈ R.
Therefore, f(x) = |x + 2| – 1 ≥ – 1 for every x ∈ R.
The minimum value of / is attained when | x + 2| = 0
⇒ x = – 2
∴ Minimum value of f = f(- 2) = = |- 2 + 2| – 1 = – 1
Hence, function f does not have a maximum value.

(ii) Given, g(x) = – | x + 1| + 3
We know that -| x + 1| < 0 for every x ∈ R.
Therefore, g(x) = -|x + 1| + 3 ≤ 3 for every x ∈ R.
The maximum value of g is attained when |x + 1| = 0
⇒ x = – 1
∴ Maximum value of g = g(- 1) = -|- 1 + 1| + 3 = 3
Hence, function g does not have a minimum value.

(iii) Given, h(x) = sin 2x + 5
We know that, – 1 ≤ sin 2x ≤ 1
⇒ – 1 + 5 ≤ sin 2x + 5 ≤ 1 + 5
⇒ 4 ≤ sin 2x + 5 ≤ 6
Hence, the maximum and minimum values of h are 6 and 4, respectively.

(iv) Given,f(x) = |sin 4x + 3|
We know that, – 1 < sin 4x < 1
⇒ 2 ≤ sin 4x + 3 ≤ 4
2 ≤ |sin 4x + 3| ≤ 4
Hence, the maximum and minimum values of f are 4 and 2, respectively.

(v) h(x) = x + 1, x ∈ (- 1, 1)
Here, if a point x₀ is closest to – 1, then we find \(\frac{x_{0}}{2}\) + 1 < x₀ ∈ (- 1, 1).
Also, if x₁ is closest to 1, then
x₁ + 1 < \(\frac{x_{1}+1}{2}\) + 1 for all x₁ ∈ (- 1, 1).
Hence, function h(x) has neither maximum nor minimum value in (- 1, 1).

Question 3.
Find the local maxima and local minima, if any, of the following functions, Find also the local maximum and the local minimum values, as the case may be :
(i) f(x) = x²
(ii) g(x) = x³ – 3x
(iii) h(x) = sin x + cos x, 0 < x < \(\frac{\pi}{2}\)
(iv) f(x) = sin x – cos x, 0 < x < 2π
(v) f(x) = x³ – 6x² + 9x + 15
(vi) g(x) = \(\frac{x}{2}+\frac{2}{x}\), x > 0
(vii) g(x) = \(\frac{1}{x^{2}+2}\)
(viii) f(x) = \(x \sqrt{1-x}\), x > 0
Solution.
(i) Given, f(x) = x²
⇒ f'(x) = 2x
Now, f'(x) = 0
⇒ x = 0
Thus, x = 0 is the only critical point which could possibly be the point of local maxima or local minima of f.
We have f”(0) = 2, which is positive.
Therefore, by second derivative test, x = 0 is a point of local minima and local minimum value of f at x = 0 is f(0) = 0.

(ii) Given, g(x) = x³ – 3x
⇒ g'(x) = 3x² – 3
Now, g'(x) = 0
⇒ 3x² = 3
⇒ x = ± 1
∴ g'(x) = 6x
⇒ g'(1) = 6 > 0
⇒ g'(- 1) = – 6 > 0
By second derivative test x = 1 is a point of local minima and local minimum value of g at x = 1 is
g(1) = 1³ – 3
= 1 – 3 = – 2.
However, x = – 1 is a point of local maxima and local maximum value of g at x = – 1 is
g(- 1) = (- 1)³ – 3(- 1)
= – 1 + 3 = 2.

(iii) Given, h(x) = sinx + cos x, 0 < x < \(\frac{\pi}{2}\)
∴ h'(x) = – sin x – cos x = -(sin x + cos x)
h'(x) = 0
⇒ sin x = cos x
⇒ tan x = 1
⇒ x = \(\frac{\pi}{4}\) ∈ (0, \(\frac{\pi}{2}\))
h”(x) = – sin x – cos x
= – (sin x + cos x)
h(\(\frac{\pi}{4}\)) = – \(\left(\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}\right)=-\frac{2}{\sqrt{2}}=-\sqrt{2}<0\)
Therefore, by second derivative test, x = \(\frac{\pi}{4}\) is a point of local maxima and
and the local maximum value of h at x = \(\frac{\pi}{4}\) is h(\(\frac{\pi}{4}\)) = sin \(\frac{\pi}{4}\) + cos \(\frac{\pi}{4}\)
= \(\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}\) = √2.

(iv) f(x) = sin x – cos x, 0 < x < 2π
∴ f'(x) = cos x + sin x f'(x) = 0
⇒ cos x = – sin x ⇒ tan x = 1
⇒ x = \(\frac{3 \pi}{4}\), \(\frac{7 \pi}{4}\) ∈ (0, 2π)
f”(x) = – sin x + cos x f”(\(\frac{3 \pi}{4}\)) = \(-\sin \frac{3 \pi}{4}+\cos \frac{3 \pi}{4}=-\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}}=-\sqrt{2}>0\)

f”(\(\frac{7 \pi}{4}\)) = \(-\sin \frac{7 \pi}{4}+\cos \frac{7 \pi}{4}=\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}=\sqrt{2}>0\)

Therefore, by second derivative test, x = \(\frac{3 \pi}{4}\) is a point of local maxima and the local maximum value of f at x = \(\frac{3 \pi}{4}\) is
f(\(\frac{3 \pi}{4}\)) = sin \(\frac{3 \pi}{4}\) – cos \(\frac{3 \pi}{4}\)
= \(\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}\) = √2.

However, x = \(\frac{7 \pi}{4}\) is a point of local minima and the local minimum value of f at x = \(\frac{7 \pi}{4}\) is
f(\(\frac{7 \pi}{4}\)) = sin \(\frac{7 \pi}{4}\) – cos \(\frac{7 \pi}{4}\)
= \(-\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}}=-\sqrt{2}\)

(v) Given, f(x) = x³ – 6x² + 9x + 15
f'(x) = 3x² – 12x + 9
f'(x) = 0
⇒ 3(x² – 4x + 3) = 0
⇒ 3(x – 1) (x – 3) = 0
⇒ x = 1, 3
Now, f'(x) = 6x – 12 = 6 (x – 2)
f'(1) = 6(1 – 2) = – 6 < 0 ⇒ f'(3) = 6(3 – 2) = 6 > 0
Therefore, by second derivative test, x = 1 is a point of local maxima and the local maximum value of f at x = 1 is
f(1) = 1 – 6 + 9 + 15 = 19.
However, x = 3 is a point of local minima and the local minimum value of f at x = 3 is
f(3) = 27 – 54 + 27 + 15 = 15

(vi) Given, g(x) = \(\frac{x}{2}+\frac{2}{x}\), x > 0

∴ g'(x) = \(\frac{1}{2}-\frac{2}{x^{2}}\)

Now, g'(x) = 0 gives \(\frac{2}{x^{2}}=\frac{1}{2}\)
⇒ x² = 4
⇒ x = ± 2
Therefore, by second derivative test, x = 2 is a point of local minima and the local minimum value of g at x = 2 is
g(2) = \(\frac{2}{2}+\frac{2}{2}\) = 1 + 1 = 2.

(vii) Given, g(x) = \(\frac{1}{x^{2}+2}\)

∴ g'(x) = \(\frac{-(2 x)}{\left(x^{2}+2\right)^{2}}\)
Now, g'(x) = 0
⇒ \(\frac{-2 x}{\left(x^{2}+2\right)^{2}}\) = 0
⇒ x = 0
Now, for values close to x = 0 and to the left of 0, g’(x) > 0.
Also, for values close to x = 0 and to the right of 0, g’(x) < 0.
Therefore, by first derivative test, x = 0 is a point of local maxima and the local maximum value of g(0) is \(\frac{1}{0+2}=\frac{1}{2}\).

(viii) Given, f(x) = x\(\sqrt{1-x}\)

Therefore, by second derivative test, x = \(\frac{2}{3}\) is a point of local maxima and the local maximum value of f at x = \(\frac{2}{3}\) is
f(\(\frac{2}{3}\)) = \(\frac{2}{3} \sqrt{1-\frac{2}{3}}=\frac{2}{3} \sqrt{\frac{1}{3}}=\frac{2}{3 \sqrt{3}}=\frac{2 \sqrt{3}}{9}\).

Question 4.
Prove that the following functions do not have maxima or minima:
(i) f(x) = e^x
(ii) g(x) = log x
(iii) h(x) = x³ + x² + x + 1
Solution.
(i) Given function is f(x) = e^x
⇒ f'(x) = e^x
Now, if f'(x) = 0, then e^x = 0.
But, the exponential function can never assume 0 for any value of x. ,
Therefore, there does not exist c ∈ R such that f'(c) = 0.
Hence, function f does not have maxima or minima.

(ii) Given function is g(x) = log x
⇒ g'(x) = \(\frac{1}{x}\)
Since, log x is defined for a positive number x, g’ (x) > 0 for any x.
Therefore, there does not exist ce R such that g’ (c) = 0.
Hence, function g does not have maxima or minima.

(iii) Given function is h(x) = x³ + x² + x + 1
⇒ h'(x) = 3x + 2x + 1
Now, h(x) = 0
⇒ 3x² + 2x +1 = 0
⇒ x = \(\frac{-2 \pm 2 \sqrt{2} i}{6}=\frac{-1 \pm \sqrt{2} i}{3}\) ∉ R
Therefore, there does not exist c ∈ R such that h'(c) = 0.
Hence, function h does not have maxima or minima.

Question 5.
Find the absolute maximum value and the absolute minimum value of the following functions in the given intervals :
(i) f(x) = x³, x ∈ [- 2, 2]
(ii) f(x) = sin x + cos x, x ∈ [0, π]
(iii) f(x) = 4x – \(\frac{1}{2}\) x², x ∈ [- 2, \(\frac{9}{2}\)]
(iv) f(x) = (x – 1)² + 3, x ∈ [- 3, 1]
Solution.
(i) The given function is f(x) = x³
∴ f'(x) = 3x²
Now, f(x) = 0
⇒ x = 0
Then, we evaluate the value of f at critical point x = 0 and at end points of the interval [- 2, 2].
∴ f(0) = 0; f(- 2) = (- 2) 3 = – 8; 012f(2) = (2)³ – 8
Hence, we can conclude that the absolute maximum value of f on [- 2, 2] is 8 occurring at x = 2.
Also, the absolute minimum value of f on [- 2, 2] is – 8 occurring at x = – 2.

(ii) The given function is f(x) = sin x + cos x.
f'(x) = cos x – sin x
Now, f'(x) = 0
⇒ sin x = cos x
⇒ tan x = 1
⇒ x = \(\frac{\pi}{4}\)
Then, we evaluate the value of f at critical point x = \(\frac{\pi}{4}\) and at the end points of the interval [0, π].

∴ f(\(\frac{\pi}{4}\)) = sin \(\frac{\pi}{4}\) + cos \(\frac{\pi}{4}\)
= \(\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}+\frac{2}{\sqrt{2}}=\sqrt{2}\)

f(0) = sin 0 + cos 0 = 0 + 1 = 1
f(π) = sin π + cos π = 0 – 1 = – 1
Hence, we can conclude that the absolute maximum value of f on [0, π] is 2 occurring at x = \(\frac{\pi}{4}\) and the absolute minimum value of f on [0, π] is – 1 occurring at x = π.

(iii) The given function is f(x) = 4x – \(\frac{1}{2}\) x²
f(x) = 4 – \(\frac{1}{2}\) (2x) = 4 – x
Now, f'(x) = 0
⇒ x = 4
Then, we evaluate the value of f at critical point x = 4 and at the end [- 2, \(\frac{9}{2}\)]
∴ f(4) = 16 – \(\frac{1}{2}\) (16) = 16 – 8 = 8

f(- 2) = – 8 – \(\frac{1}{2}\) (4) = – 8 – 2 = – 10

f(\(\frac{1}{2}\)) = \(4\left(\frac{9}{2}\right)-\frac{1}{2}\left(\frac{9}{2}\right)^{2}=18-\frac{81}{8}\)

= 18 – 10.125 = 7.875
Hence, we can conclude that the absolute maximum value of f on is 8 occurring at x = 4 and the absolute minimum value of f on – 10 occurring at x = – 2.

(iv) The given function is f(x) = (x – 1)² + 3
∴ f'(x) = 2(x – 1)
Now, f'(x) = 0
⇒2 (x – 1) = 0
⇒ x = 1
Then, we evaluate the value of f at critical point x = 1 and at the end points of the interval [- 3, 1].
∴ f(1) = (1 – 1)² + 3 = 0 + 3 = 3;
f(- 3) = (- 3 – 1)² + 3 = 16 + 3 = 19
Hence, we can conclude that the absolute maximum value of / on [- 3, 1] is 19 occurring at x = – 3 and the minimum value of f on [- 3, 1] is 3 occurring at x = 1.

Question 6.
Find the maximum profit that a company can make, if the profit function is given by p(x) = 41 – 72x – 18x²
Solution.
The profit function is given as p(x) = 41 – 72x – 18x²
On differentiating both sides w.r.t. x, we get
p'(x) = – 72 – 36x = – 36 (2 + x)
Again, differentiating both sides w.r.t. x, wt ,et
p”(x) = – 36
For maxima or minima, put p(x) = 0
⇒ – 36 (2 + x) = 0
⇒ x + 2 = 0
⇒x = – 2
At x = – 2, p'(-2) = – 36 < 0
∴ x = – 2 is a point of maxima.
Maximum profit, p(- 2) = 41 – 72 (- 2) – 81 (- 2)²
= 41 + 144 – 72 = 113
Hence, the maximum profit that a company can make, is 113 units.

Question 7.
Find both the maximum value and the minimum value of 3x⁴ – 8x³ + 12x² – 48x + 25 on the interval [0, 3].
Solution.
Let f(x) = 3x⁴ – 8x³ + 12x² – 48x + 25
⇒ f'(x) = 12x³ – 24x² + 24x – 48
= 12 (x³ – 2x² + 2x – 4)
= 12 (x² (x – 2) + 2 (x – 2))
= 12(x – 2) (x² + 2)
For maxima or minima put f'(x) = 0
12(x – 2) (x² + 2) = 0
⇒ if x – 2 = 0 ⇒ x = 2 e [0, 3]
And if, x² + 2 = 0
⇒ x² = – 2
⇒x = √- 2
Hence, the only real root is x = 2 which is considered as critical point.
Now, we evaluate the value of f at critical point x = 2 and at the end point of the interval [0, 3].
At x = 2, f(2) = 3 × 2⁴ – 8 × 2³ + 12 × 2² – 48 × 2 + 25
= 48 – 64 + 48 – 96 + 25 = – 39

At x = 0, f(0) = 0 – 0 + 0 – 0 + 25 = 25
At x = 3, f(3) = 3 × 3⁴ – 8 × 3³ + 12 × 3² – 48 × 3 + 25
= 243 – 216 + 108 – 144 + 25 = 16
Hence, we can conclude that the absolute maximum value of f is 25 at x = 0 and the absolute minimum value of f is – 39 at x = 2.

Question 8.
At what points in the interval [0, 2π], does the function sin2x attain its maximum value?
Solution.
Let f(x) = sin 2x
⇒ f'(x) = 2 cos 2x
Now, f'(x) = 0
⇒ cos 2x = 0
⇒ 2x = \(\frac{\pi}{2}, \frac{3 \pi}{2}, \frac{5 \pi}{2}, \frac{7 \pi}{2}\)

⇒ x = \(\frac{\pi}{4}, \frac{3 \pi}{4}, \frac{5 \pi}{4}, \frac{7 \pi}{4}\)

Then, we evaluate the value off at critical points x = \(\frac{\pi}{4}, \frac{3 \pi}{4}, \frac{5 \pi}{4}, \frac{7 \pi}{4}\) and at the end points of the interval [0, 2π].
f(\(\frac{\pi}{4}\)) = sin \(\frac{\pi}{2}\) = 1;

f(\(\frac{3 \pi}{4}\)) = sin \(\frac{3 \pi}{2}\) = – 1;

f(\(\frac{5 \pi}{4}\)) = sin \(\frac{5 \pi}{2}\) = 1;

f(\(\frac{7 \pi}{4}\)) = sin \(\frac{7 \pi}{2}\) = – 1;

f(0) = sin 0 = 0;
f(2π) = sin 2π = 0
Hence, we can conclude that the absolute maximum value of f on [0, 2π] is occurring at x = \(\frac{\pi}{4}\) and x = \(\frac{5 \pi}{4}\).

Question 9.
What is the maximum value of the function sin x + cos x?
Solution.
Let f(x) = sin x + cos x
f”(x) = – sin x – cos x = – (sin x + cos x)
Now, f”(x) will be negative when (sin x + cos x) is positive i.e., when sin x and cos x are both positive.
Also, we know that, sin x and cos x both are positive in the first quadrant.
Then, f”(x) will be negative when x ∈ (0, \(\frac{\pi}{2}\))
Thus, we consider x = \(\frac{\pi}{4}\)

f”(\(\frac{\pi}{4}\)) = – (sin \(\frac{\pi}{4}\) + cos \(\frac{\pi}{4}\))

= –\(\left(\frac{2}{\sqrt{2}}\right)\) = – √2
∴ By second derivative test, f will be the maximum at x = \(\frac{\pi}{4}\) and the maximum value of f is
f(\(\frac{\pi}{4}\)) = sin \(\frac{\pi}{4}\) + cos \(\frac{\pi}{4}\)
= \(\frac{1}{\sqrt{2}} \times \frac{1}{\sqrt{2}}=\frac{2}{\sqrt{2}}=\sqrt{2}\).

Question 10.
Find the maximum value of 2x³ – 24x + 107in the interval [1, 3]. Find the maximum value of the same function in [- 3, – 1].
Solution.
Let f(x) = 2x³ – 24x + 107
⇒ f(x) = 6x² – 24
= 6(x² – 4)
Now, f'(x) = 0
⇒ 6 (x² – 4) = 0
⇒ x² = 4
⇒ x = ± 2
We first consider the interval [1, 3]
Then, we evaluate the value of f at the critical point x = 2 ∈ [1, 3] and at the end points of the interval [1, 3].
f(2) = 2(8) – 24(2) +107 = 16 – 48 +107 = 75
f(1) = 2(1)-24(1)+ 107 = 2 – 24 +107 = 85
f(3) = 2(27) – 24(3) +107 = 54 – 72 +107 = 89
Hence, the absolute maximum value of f(x) in the interval [1, 3] is 89 occurring at x = 3.
Next, we consider the interval [- 3, – 1],
Evaluate the value of f at the critical point x = – 2 ∈ [- 3, – 1] and at the end points of the interval [1, 3].
f(- 3) = 2(- 27) – 24(- 3) +107 = – 54 + 72 +107 = 125
f(- 1) = 2(- 1) – 24(- 1) + 107 = – 2 + 24 +107 = 129
f(- 2) = 2(- 8) – 24(- 2) + 107 = – 16 + 48 +107 = 139
Hence, the absolute maximum value of f(x) in the interval [- 3, – 1] is 139 occurring at x = – 2.

Question 11.
It is given that at x = 1, the function x⁴ – 62x² + ax + 9 attains its maximum value, on the interval [0, 2]. Find the value of a.
Solution.
Let f(x) = x⁴ – 62x² + ax + 9
f'(x) = 4x³ – 124x + a
It is given that function f attains its maximum value on the interval [0, 2] at x = 1.
∴ f'(1) = 0
⇒ 4 – 124 + a = 0
⇒ a = 120
Hence, the value of a is 120.

Question 12.
Find the maximum and minimum values of x + sin 2x on [0, 2π].
Solution.
Let f(x) = x + sin 2x
⇒ f'(x) = 1 + 2 cos 2x
Now, f'(x) = 0
⇒ cos 2x = – \(\frac{1}{2}\) = – cos \(\frac{\pi}{3}\)

= cos (π – \(\frac{\pi}{3}\))

= cos \(\frac{2 \pi}{3}\)

2x = 2n ± \(\frac{2 \pi}{3}\), n ∈ Z

x = nπ ± \(\frac{\pi}{3}\), n ∈ Z
⇒ x = \(\frac{\pi}{3}, \frac{2 \pi}{3}, \frac{4 \pi}{3}, \frac{5 \pi}{3}\) ∈[0, 2π]

Then, we evaluate the value of f at critical points x = \(\frac{\pi}{3}, \frac{2 \pi}{3}, \frac{4 \pi}{3}, \frac{5 \pi}{3}\) and at the end points of the interval [0, 2π],
∴ f(\(\frac{\pi}{3}\)) = \(\frac{\pi}{3}+\sin \frac{2 \pi}{3}=\frac{\pi}{3}+\frac{\sqrt{3}}{2}\);

f(\(\frac{2 \pi}{3}\)) = \(\frac{2 \pi}{3}+\sin \frac{4 \pi}{3}=\frac{2 \pi}{3}-\frac{\sqrt{3}}{2}\);

f(\(\frac{4 \pi}{3}\)) = \(\frac{4 \pi}{3}+\sin \frac{8 \pi}{3}=\frac{4 \pi}{3}+\frac{\sqrt{3}}{2}\);

f(\(\frac{5 \pi}{3}\)) = \(\frac{5 \pi}{3}+\sin \frac{10 \pi}{3}=\frac{5 \pi}{3}-\frac{\sqrt{3}}{2}\);
f(0) = 0 + sin 0 = 0;
f(2π) = 2π + sin 4π = 2π + 0 = 2π
Thus, maximum value is 2π at x = 2π and minimum value is 0 at x = 0.

Question 13.
Find two numbers whose sum is 24 and whose product is as large as possible.
Solution.
Let one number be x.
Then, the other number is (24 – x).
Let f(x) denote the product of the two numbers.
Thus, we have P(x) = x(24 – x) = 24x – x²
P'(x) = 24 – 2x;
P”(x) = – 2
Now, P’ (x) = 0
⇒ x =12
Also, P”(12) = – 2 < 0
By second derivative test, x = 12 is the point of local maxima of P.
Hence, the product of the number is the maximum when the numbers are 12 and 24 – 12 = 12.

Question14.
Find two positive numbers x and y such that x + y = 60 and xy³ is maximum.
Solution.
The two numbers are x and y such that x + y – 60
⇒ y = 60 – x
Let f(x) = xy³
⇒ f(x) = x(60 – x)³
⇒ f'(x) = (60 – x)³ – 3x (60 – x)²
= (60 – x)² [60 – x – 3x]
= (60 – x)² (60 – 4x)
Also, f(x) = – 2 (60 – x) (60 – 4x) – 4 (60 – x)²
= – 2 (60 – x) [60 – 4x + 2(60 – x)]
= – 2 (60 – x) (180 – 6x)
= -12(60 – x) (30 – x)
Now, f'(x) = 0
⇒ x = 60 or x = 15
When, x = 60 then, f”(x) = 0
When x = 15, then f'(x) – 12(60 – 15) (30 – 15) = – 12 × 45 × 15 < 0 .
∴ By second derivative test, x = 15 is a point of local maxima of f.
Thus, function xy³ is maximum when x = 15 and y = 60 -15 = 45.
Hence, the required numbers are 15 and 45.

Question 15.
Find two positive numbers x and y such that their sum is 35 and the product x²zy⁵ is a maximum.
Solution.
Let one number be x. Then, the other number is y = (35 – x).
Let P(x) = x²y⁵, then
P(x) = x² (35 – x)⁵
∴ P'(x) = 2x (35 – x)⁵ – 5x² (35 – x)⁴
= x (35 – x)⁴ [2(35 – x) – 5x]
= x (35 – x)⁴ (70 – 7x)
= 7x (35 – x)4^{4/sup> (10 – x)}

And p”(x) = 7(35 – x)⁴ (10 – x) + 7x [- (35 – x)⁴ – 4(35 – x)³ (10 – x)]
= 7 (35 – x)⁴ (10 – x) – 7x (35 – x)⁴ – 28x (35 – x)³ (10 – x)
= 7 (35 – x)³ [(35 – x)(10 – x) – x(35 – x) – 4x(10 – x)]
= 7 (35-x)³ [350 – 45x + x² – 35x + x² – 40x + 4x²]
= 7 (35 – x)³ (6x² – 120x + 350)
Now, P'(x) = 0
⇒ x = 0, x = 35, x = 10
When, x = 35, f'(x) = f(x) = 0 and y = 35 – 35 = 0.
This will make the product x²y⁵ equal to 0.
When x = 0, then y = 35 -0 = 35 and the product x2y2 will be 0.
∴ x = 0 and x = 35 cannot be the possible values of x.
When x = 10 then, we have
P'(x) = 7(35 – 10)³ (6 × 100 – 120 × 10 + 350)
= 7(25)³ (- 250) < 0
∴ By second derivative test, P(x) will be the maximum when x = 10 and y = 35 – 10 = 25.
Hence, the required numbers are 10 and 25.

Question 16.
Find two positive numbers whose sum is 16 and the sum of whose cubes is minimum.
Solution.
Let one number be x.
Then, the other number is (16 – x).
Let the sum of the cubes of these numbers be denoted by S(x). Then, S(x) = x3 + (16 -x)3
∴ S'(x) = 3x² – 3(16 – x)²;
S'(x) = 6x + 6(16 – x)
Now, S'(x) = 0
⇒ 3x² – 3(16 – x)² = 0
⇒ x² – (16 – x)² = 0
⇒ x² – 256 – x² + 32x = 0
⇒ x =\(\frac{256}{32}\) = 8
Now, S'(8) = 6(8) + 6(16 -8) = 48 + 48 = 96 > 0.
By second derivatives test, x = 8 is the point of local minima of S.
Hence, the sum of the cubes of the numbers is the minimum when the numbers are 8 and 16 – 8 = 8.

Question 17.
A square piece of tin of side 18 cm is to be made into a box without top, by cutting a square from each corner and folding up the flaps to form the box. What should be the side of the square to be cut off so that the volume of the box is the maximum possible?
Solution.

Let the side of the square to be cut-off be x cm (0 < x < 9).
Then, the length and the breadth of the box will be (18 – 2x) cm each and the height of the box is x cm.
Let V the volume of the open box formed by folding up the flaps, then
V = x (18 – 2x) (18 – 2x)
= 4x (9 – x)²
= 4x (81 + x² – 18x)
= 4(x³ – 18x² + 81x)
On differentiating twice w.r.t. x, we get
\(\frac{d V}{d x}\) = 4(3x² – 36x + 81)
= 12(x² – 12x+27)
and \(\frac{d^{2} V}{d x^{2}}\) = 12 (2x – 12) = 24 (x – 6)
For maxima put \(\frac{d V}{d x}\) = 0
⇒ 12(x – 12x + 27) = 0
⇒ x² – 12x + 27 = 0
⇒ (x – 3) (x – 9) = 0
⇒ x = 3, 9
But x = 9 is not possible,
∵ 2x = 2 × 9 = 18
Which is equal to side of square piece.
At x = 3,
\(\left(\frac{d^{2} V}{d x^{2}}\right)_{x=3}\) = 24 (3 – 6) = – 72 < 0

∴ By second derivative test, x = 5 is the point of maxima.
Hence, the side of the square to be cut-off to make the volume of the box maximum possible is 5 cm.

Question 18.
A rectangular sheet of tin 45cm by 24cm is to be made into a box without top, by cutting off square from each comer and folding up the flaps. What should be the side of the square to be cut off so that the volume of the box is the maximum possible?
Solution.

Let the side of the square to be cut off be x cm.
Then, the height of the box is x, the length is 45 – 2x and the breadth is 24 f'(x) 2x.
Let V be the corresponding volume of the box then,
V = x (24 – 2x) (45 – 2x)
V = x (4x² – 138x + 1080)
=4x³ – 138 x² + 1080x
On differentiating twice w.r.t. x,

\(\frac{d V}{d x}\) = 12x² – 276x + 1080

\(\frac{d^{2} V}{d x^{2}}\) = 24x – 276
For maxima put \(\frac{d V}{d x}\) = o
⇒ 12x² – 276x + 1080 = 0
⇒ x² – 23x + 90 = 0
= (x – 18) (x – 5) = 0
⇒ x = 5, 18
It is not possible to cut-off a square of side 18 cm from each corner of the rectangular sheet.
Thus, x cannot be equal to 18.
At x = 5,
\(\left(\frac{d^{2} V}{d x^{2}}\right)_{x=5}\) = 24 × 5 – 276
= 120 – 276
= – 156 < 0
∴ By second derivative test, x = 5 is the point of maxima.
Hence, the side of the square to be cut-off to make the volume of the box maximum possible is 5 cm.

Question 19.
Show that of all the rectangles inscribed in a given fixed circle, the square has the maximum area.
Solution.
Let a rectangle of length l and breadth b be inscribed in the given circle of radius a.
Then, the diagonal passes through the centre and is of length 2a cm.

Now, by applying the Pythagoras theorem, we have (2 a)² = l² + b²
⇒ b² = 4 a² – l²
⇒ b = \(\sqrt{4 a^{2}-l^{2}}\)
∴ Area of rectangle, A = l \(\sqrt{4 a^{2}-l^{2}}\)

∴ By the second derivative test, when l = √2a then the area of the rectangle is the maximum.
Since, l = b = √2a, therefore the rectangle is a square.
Hence, it has been proved that of all the rectangles inscribed in the given fixed circle the square has the maximum area.

Question 20.
Show that th right circular cylinder of given surface and maximum volume Is such that its heigh is equal to the diameter of the base.
Solution.

Let r and h be the radius and height of the cylinder, respectively.
Then, the surface area (S) of the cylinder is given by
S = 2πr² + 2πrh
⇒ h = \(\frac{S-2 \pi r^{2}}{2 \pi r}=\frac{S}{2 \pi}\left(\frac{1}{r}\right)-r\)

Let V be the volume of the cylinder. Then,

V = πr²h
= πr² \(\left[\frac{S}{2 \pi}\left(\frac{1}{r}\right)-r\right]=\frac{S r}{2}-\pi r^{3}\)

On differentiating w.r.t. x, we get
\(\frac{d V}{d r}=\frac{S}{2}-3 \pi r^{2}\)

For maxima or mmima put \(\frac{d V}{d r}\) = 0
\(\frac{S}{2}\) = 3πr²
r² = \(\frac{S}{6 \pi}\)

Now, \(\frac{d^{2} V}{d r^{2}}\) = – 6π \(\left(\sqrt{\frac{S}{6 \pi}}\right)\) < 0

∴ By second derivative test, the volume is the maximum when r² = \(\frac{S}{6 \pi}\).

Now, when r² = \(\frac{S}{6 \pi}\), then h = \(\frac{6 \pi r^{2}}{2 \pi}\left(\frac{1}{r}\right)\) – \(\frac{1}{r}\) = 3r – r = 2r.
Hence, the volume is the maximum when the height is twice the radius si.e., when the height is equal to the diameter.

Question 21.
Of all the closed cylindrical cans (right circular), of a given volume of 100 cubic cm, find the dimensions of the can which has the minimum surface area?
Solution.

 

Let r and h be the radius and height of the cylinder, respectively.
Then, volume (V) of the cylinder is given by
V = πr²h = 100
∴ h = \(\frac{100}{\pi r^{2}}\)
Surface area (S) of the cylinder is given by S = 2πr² + 2πrh = 2πr² + \(\frac{200}{r}\)
On differentiating w.r.t. x, we get h
∴ \(\frac{d S}{d r}\) = 4πr – \(\frac{200}{r^{2}}\)
Now, for maxima or minima put \(\frac{d S}{d r}\) = 0

⇒ 4πr = \(\frac{200}{r^{2}}\)

⇒ r³ = \(\frac{200}{4 \pi}=\frac{50}{\pi}\)

⇒ r = \(\left(\frac{50}{\pi}\right)^{\frac{1}{3}}\)

Again, differentiating w.r.t. x, we get.
\(\frac{d^{2} S}{d r^{2}}\) = 4π + \(\frac{400}{r^{3}}\)

Now, it is observed that when r = latex]\left(\frac{50}{\pi}\right)^{\frac{1}{3}}[/latex], \(\frac{d^{2} S}{d r^{2}}\) > 0.

∴ By second derivative test, the surface area is the minimum when the radius of the cylinder is \(\left(\frac{50}{\pi}\right)^{\frac{1}{3}}\) cm.

when r = \(\left(\frac{50}{\pi}\right)^{\frac{1}{3}}\), then h = \(\frac{100}{\pi\left(\frac{50}{\pi}\right)^{\frac{2}{3}}}=2\left(\frac{50}{\pi}\right)^{\frac{1}{3}}\) cm

Hence, the required dimensions of the can which has the minimum surface area, is given by radius = \(\left(\frac{50}{\pi}\right)^{\frac{1}{3}}\) cm and height = 2 \(\left(\frac{50}{\pi}\right)^{\frac{1}{3}}\) cm.

Question 22.
A wire of length 28 m is to be cut into two pieces. One of the pieces is to be made into a square and the other into a circle. What should be the length of the two pieces so that the combined area of the square and the circle is minimum?
Solution.
Let a piece of length l be cut from the given wire to make a square.
Then, the other piece of wire to be made into a circle is of length (28 – l) m.
Now, side of square = \(\frac{l}{4}\)
Let r be the radius of the circle. Then, 2πr = 28 – l
The combined area of the square and the circle (A) is given by A = (side of the square)² + πr²

∴ By second derivative test, the area (A) is the minimum when l = \(\frac{112}{\pi+4}\)

Hence, the combined area is the minimum when the length of the wire in making the square is \(\frac{112}{\pi+4}\) cm while the length of the wire in making the circle is 28 – \(\frac{112}{\pi+4}=\frac{28 \pi}{\pi+4}\) cm.

Question 23.
Prove that the volume of the largest cone that can be inscribed in a sphere of radius R, is \(\frac{8}{27}\) of the volume of the sphere.
Solution.
Let r and h be the radius and height of the cone respectively inscribed in a sphere of radius R.
Let V be the volume of the cone.
Then, V = \(\frac{1}{3}\) πr²h
Height of the cone is given by h = R + AB
= R + \(\sqrt{R^{2}-r^{2}}\) [∵ ABC is right angled triangle]

For maxima put
\(\frac{d V}{d r}\) = 0

⇒ \(\frac{2}{3}\) πrR = \(\frac{3 \pi r^{3}-2 \pi r R^{2}}{3 \sqrt{R^{2}-r^{2}}}\)

⇒ 2R = \(\frac{3 r^{2}-2 R^{2}}{\sqrt{R^{2}-r^{2}}}\)

⇒ 2R \(\sqrt{R^{2}-r^{2}}\) = (3r² – 2R²)²
⇒ 4R² (R² – r²)
= (3r² – 2R²)²
⇒ 4R⁴ – 4R²r² = 9r⁴ + 4R⁴ – 12r²R²
⇒ 9r⁴ = 8R²r²
⇒ r² = \(\frac{8}{9}\) R²

∴ By second derivative test, the volume of the cone is the maximum when r² = \(\frac{8}{9}\) R².

When r² = \(\frac{8}{9}\) R², then
h = R + \(\sqrt{R^{2}-\frac{8}{9} R^{2}}=R+\sqrt{\frac{1}{9} R^{2}}=R+\frac{R}{3}=\frac{4}{3} R\)

Therefore, V = \(\frac{1}{3} \pi\left(\frac{8}{9} R^{2}\right)\left(\frac{4}{3} R\right)=\frac{8}{27}\left(\frac{4}{3} \pi R^{3}\right)\)

= \(\frac{8}{27}\) × (volume of the sphere)
Hence, the volume of the largest cone that can be inscribed in the sphere, is \(\frac{8}{27}\) of the volume of the sphere.

Question 24.
Show that the right circular cone of least curved surface and given volume has an altitude equal toJ time the radius of the base.
Solution.
Let r be the radius of the base, h be the height, V be the volume and S be the curved surface area of the cone.
Then, V = \(\frac{1}{3}\) πr²h
⇒ 3V = πr²h
⇒ 9V² = π² r⁴ h²
⇒ h² = \(\frac{9 V^{2}}{\pi^{2} r^{4}}\) ……………..(i)
And S = πrl
⇒ S = πr \(\sqrt{r^{2}+h^{2}}\) (∵ l = \(\sqrt{h^{2}+r^{2}}\))
⇒ S² = π² r² (r² + h²)
= π² r² (\(\frac{9 V^{2}}{\pi^{2} r^{4}}\) + r²) [Using Eq. (i)]

⇒ S² = \(\frac{9 V^{2}}{r^{2}}\) + π² r⁴ …………(ii)

Hence, S² and therefore S is minimum when 9V² = 2π²r⁶
On putting 9 V² = 2π²r⁶ in Eq. (i) we get
2π²r⁶ = π²r⁴h²
⇒ 2r² = h²
⇒ h = √2r
Hence, altitude of right circular cone is √2 times the radius of the base.

Question 25.
Show that the semi-vertical angle of the cone of the maximum volume and of given slant height is tan^-1 √2.
Solution.

Let θ be the semi-vertical angle of the cone.
It is clear that θ ∈ [0, \(\frac{\pi}{2}\)]
Let r, h and l be the radius, height and the slant height of the cone, respectively.
The slant height of the cone is given as constant.
Now, r = l sin θ and h = l cos θ
The volume (V) of the cone is given by V = \(\frac{1}{3}\) πr² h
⇒ \(\frac{1}{3}\) π (l² sin² θ) (l cos θ)
= \(\frac{1}{3}\) π l² sin² θ cos θ
On differentiating w.r.t. θ, we get
∴ \(\frac{d V}{d \theta}\) = \(\frac{l^{2} \pi}{3}\) [sin² θ (- sin θ) + cos θ (2 sin θ cos θ)]

= \(\frac{l^{3} \pi}{3}[\) [- sin³ θ + 2sin θ cos² θ]

Again, differentiating w.r.t. θ, we get
\(\frac{d^{2} V}{d \theta^{2}}\) = \(\frac{l^{3} \pi}{3}\) [- 3 sin² θ cos θ + 2 cos³ θ – 4 sin² θ cos θ]

= \(\frac{l^{3} \pi}{3}\) [2 cos³ θ – 7 sin² θ cos θ]
For maxima put \(\frac{d V}{d \theta}\) = 0
⇒ sin³ θ = 2 sin θ cos θ v d0
⇒ tan² θ = 2
⇒ tan θ = √2
⇒ θ = tan^-1 √2
Now, when θ = tan^-1 √2, then tan² θ = 2 or sin² θ = 2 cos² θ
Then, we have

\(\frac{d^{2} V}{d \theta^{2}}\) = \(\frac{l^{3} \pi}{3}\) [2 cos³ θ – 14 cos³ θ]
= – 4πl³ cos³ θ < 0 for θ ∈ [0, \(\frac{\pi}{2}\)]

∴ By second derivative test, the volume (V) is the maximum when θ = tan^-1 √2.
Hence, for a given slant height, the semi-vertical angle of the cone of the maximum volume is tan^-1 √2.

Question 26.
Show that the semi-vertical angle of right circular cone of given surface area and maximum volume is sin^-1 (\(\frac{1}{3}\)).
Solution.
With usual notation, given that total surface area S = πrl + πr²

Direction (27 – 29): Choose the correct answer.

Question27.
The point on the curve x² = 2y which is nearest to the point (0, 5) is
(A) (2√2, 4)
(B) (2√2, 0)
(C) (0, 0)
(D) (2, 2)
Solution.
Let d be the distance of the point (x, y) on x² = 2y from the point (0, 5),
then
d = \(\sqrt{(x-0)^{2}+(y-5)^{2}}=\sqrt{x^{2}+(y-5)^{2}}\) ……………(i)

= \(\sqrt{2 y+(y-5)^{2}}\) [Putting x² = 2y]

= \(\sqrt{y^{2}-8 y+25}\)

= \(\sqrt{y^{2}-8 y+4^{2}+9}=\sqrt{(y-4)^{2}+9}\)

d is least when (y – 4)² = 0 i.e., when y = 4
when y = 4 then x² = 2 × 4
⇒ x = ± √8 = ± 2√2
∴ The points (2√2, 4) and (- 2√2, 4) on the given curve are nearest to the point (0, 5).
The correct answer is (A).

Question 28.
For all real values of x, the minimum value of \(\frac{1-x+x^{2}}{1+x+x^{2}}\) is
(A) 0
(B) 1
(C) 3
(D) \(\frac{1}{3}\)
Solution.

The correct answer is (D).

Question 29.
The maximum value of [x (x – 1) + 1]^{\(\frac{1}{3}\)},0 < x < 1 is
(A) \(\left(\frac{1}{3}\right)^{\frac{1}{3}}\)

(B) \(\frac{1}{2}\)

(C) 1
(D) 0
Solution.
Let f(x) = [x (x – 1) + 1]^{\(\frac{1}{3}\)}

∴ f'(x) = \(\frac{2 x-1}{3[x(x-1)+1]^{\frac{2}{3}}}\)

Now, f'(x) = 0
⇒ x = \(\frac{1}{2}\)
Then, we evaluate the value of f at critical point x = \(\frac{1}{2}\) and at the point of the interval [0, 1]
{i.e., at x = 0 and x = 1}.
f(0) = [0 (0 – 1) + 1]^{\(\frac{1}{3}\)} = 1;

f(1) = [1 (1 – 1) + 1]^{\(\frac{1}{3}\)} = 1

f(\(\frac{1}{2}\)) = \(\left[\frac{1}{2}\left(\frac{-1}{2}\right)+1\right]^{\frac{1}{3}}=\left(\frac{3}{4}\right)^{\frac{1}{3}}\)

Hence, we can conclude that the maximum value off in the interval [0, 1] is 1.
The correct answer is (C).