Category: ML Aggarwal Solutions

Well-structured Understanding ISC Mathematics Class 12 Solutions Chapter 8 Integrals Ex 8.8 facilitate a deeper understanding of mathematical principles.

ML Aggarwal Class 12 Maths Solutions Section A Chapter 8 Integrals Ex 8.8

Evaluate the following (1 to 12) integrals:

Question 1.
(i) ∫ \(\frac{d x}{x^2-4 x+8}\)
(ii) ∫ \(\frac{d x}{x^2+2 x+2}\)
Solution:
(i) ∫ \(\frac{d x}{x^2-4 x+8}\)
= ∫ \(\frac{d x}{x^2-4 x+4+4}\)
= ∫ \(\frac{d x}{(x-2)^2+2^2}\)
[put x – 2 = t
⇒ dx = dt]
= ∫ \(\frac{d t}{t^2+a^2}\)
= \(\frac{1}{2} \tan ^{-1} \frac{t}{2}\) + C
[∵ ∫ \(\frac{d x}{x^2+a^2}=\frac{1}{a} \tan ^{-1} \frac{x}{a}\)]
= \(\frac{1}{2} \tan ^{-1}\left(\frac{x-2}{2}\right)\) + C

(ii) ∫ \(\frac{d x}{x^2+2 x+2}\)
Solution:
Let I = ∫ \(\frac{d x}{x^2+2 x+2}\)
= ∫ \(\frac{d x}{x^2+2 x+1+1}\)
= ∫ \(\frac{d x}{(x+1)^2+1^2}\)
= ∫ \(\frac{d t}{t^2+1^2}\) ;
when x + 1 = t
⇒ dx = dt
= \(\frac{1}{1} \tan ^{-1}\left(\frac{t}{1}\right)\) + C
= tan^-1 (x + 1) + C

Question 1 (old).
(ii) ∫ \(\frac{d x}{5-8 x-x^2}\) (NCERT)
Solution:
Let I = ∫ \(\frac{d x}{5-8 x-x^2}\)

Question 2.
(i) ∫ \(\frac{d x}{x^2-6 x+13}\) (NCERT)
(ii) ∫ \(\frac{x+3}{x^2-2 x-5}\) dx (NCERT)
Solution:
(i) ∫ \(\frac{d x}{x^2-6 x+13}\)
= ∫ \(\frac{d x}{x^2-6 x+9+4}\)
= ∫ \(\frac{d x}{(x-3)^2+2^2}\)
put x – 3 = t
dx = dt
= ∫ \(\frac{d t}{t^2+2^2}\)
= \(\frac{1}{2} \tan ^{-1} \frac{t}{2}\) + C
= \(\frac{1}{2} \tan ^{-1}\left(\frac{x-3}{2}\right)\) + C

(ii) Let I = ∫ \(\frac{x+3}{x^2-2 x-5}\) dx

Question 3.
(i) ∫ \(\frac{x+2}{2 x^2+6 x+5}\) dx
(ii) ∫ \(\frac{2 x+1}{18-4 x-x^2}\) dx
Solution:
(i) Let I = ∫ \(\frac{x+2}{2 x^2+6 x+5}\) dx

(ii) Let I = ∫ \(\frac{2 x+1}{18-4 x-x^2}\) dx
Let 2x + 1 = A \(\frac{d}{d x}\) (18 – 4x – x²) + B
= A (- 4 – 2x) + B
On comparing the coefficient of x and constant terms on both sides ; we get
2 = – 2 A
⇒ A = – 1 ;
1 = – 4A + B
B = – 3

Question 4.
(i) ∫ \(\frac{e^x}{2 e^{2 x}+3 e^x+5}\) dx
(ii) ∫ \(\frac{\cos x}{6+4 \sin x-\cos ^2 x}\) dx
Solution:
(i) ∫ \(\frac{e^x}{2 e^{2 x}+3 e^x+5}\) dx ;
put e^x = t
e^x dx = dt
∴ I = ∫ \(\frac{d t}{2 t^2+3 t+5}\)
= \(\frac{1}{2}\) ∫ \(\frac{d t}{t^2+\frac{3 t}{2}+\frac{5}{2}}\)
= \(\frac{1}{2}\) ∫ \(\frac{d t}{t^2+\frac{3 t}{2}+\frac{9}{16}-\frac{9}{16}+\frac{5}{2}}\)

(ii) Let I = ∫ \(\frac{\cos x}{6+4 \sin x-\cos ^2 x}\) dx

= tan^-1 (t + 2) + C
= tan^-1 (2 + sin x) + C

Question 5.
(i) ∫ \(\frac{d x}{\sqrt{2 x-x^2}}\) (NCERT)
(ii) ∫ \(\frac{d x}{\sqrt{7-6 x-x^2}}\)
Solution:
(i) ∫ \(\frac{d x}{\sqrt{2 x-x^2}}\)
= ∫ \(\frac{d x}{\sqrt{-\left(x^2-2 x+1-1\right)}}\)
= ∫ \(\frac{d x}{\sqrt{1-(x-1)^2}}\)
put x – 1 = t
⇒ dx = dt
= ∫ \(\frac{d t}{\sqrt{1-t^2}}\)
= sin^-1 t + C
= sin^-1 (x – 1) + C
[∵ ∫ \(\frac{d x}{\sqrt{a^2-x^2}}\) = sin^-1 \(\frac{x}{a}\) + C]

(ii) Let I = ∫ \(\frac{d x}{\sqrt{7-6 x-x^2}}\)

Question 5 (old).
(ii) ∫ \(\frac{d x}{\sqrt{7-6 x-x^2}}\)
Solution:
Let I = ∫ \(\frac{d x}{\sqrt{7-6 x-x^2}}\)
= ∫ \(\frac{d x}{\sqrt{x^2-3 x+\frac{9}{4}-\frac{9}{4}+2}}\)
= ∫ \(\frac{d x}{\sqrt{\left(x-\frac{3}{2}\right)^2-\left(\frac{1}{2}\right)^2}}\)
put x – \(\frac{3}{2}\) = t
⇒ dx = dt
= ∫ \(\frac{d t}{\sqrt{t^2-\left(\frac{1}{2}\right)^2}}\)
= log |t + \(\sqrt{t^2-\frac{1}{4}}\)| + C
= log |x – \(\frac{3}{2}+\sqrt{x^2-3 x+2}\)| + C
[∵∫ \(\frac{d x}{\sqrt{x^2-a^2}}\) = log |x + \(\sqrt{x^2-a^2}\)| + C|]

Question 6.
(i) ∫ \(\frac{d x}{\sqrt{7-6 x-x^2}}\) (NCERT)
(ii) ∫ \(\frac{d x}{\sqrt{(x-1)(x-2)}}\) (NCERT)
Solution:
(i) Let I = ∫ \(\frac{d x}{\sqrt{7-6 x-x^2}}\)

(ii) Let I = ∫ \(\frac{d x}{\sqrt{(x-1)(x-2)}}\)

Question 6 (old).
(ii) ∫ \(\frac{d x}{\sqrt{x^2+2 x+2}}\) (NCERT)
Solution:
Let I = ∫ \(\frac{d x}{\sqrt{x^2+2 x+2}}\)
= ∫ \(\frac{d x}{\sqrt{x^2+2 x+1+1}}\)
= ∫ \(\frac{d x}{\sqrt{(x+1)^2+1^2}}\)
put x + 1 = t
⇒ dx = dt
= ∫ \(\frac{d t}{\sqrt{t^2+1^2}}\)
= log |t + \(\sqrt{t^2+1}\)| + C
[∵∫ \(\frac{d x}{\sqrt{x^2-a^2}}\) = log |x + \(\sqrt{x^2-a^2}\)| + C|]
= log |x + 1 + \(\sqrt{x^2+2 x+2}\)| + C

Question 7.
(i) ∫ \(\frac{d x}{\sqrt{5-4 x-2 x^2}}\)
(ii) ∫ \(\frac{d x}{\sqrt{5 x-4 x^2}}\) (ISC 2020)
Solution:
(i) Let I = ∫ \(\frac{d x}{\sqrt{5-4 x-2 x^2}}\)
= ∫ \(\frac{d x}{\sqrt{-2\left(x^2+2 x-\frac{5}{2}\right)}}\)
= ∫ \(\frac{d x}{\sqrt{2} \sqrt{-\left(x^2+2 x+1-1-\frac{5}{2}\right)}}\)
= \(\frac{1}{\sqrt{2}} \int \frac{d x}{\sqrt{-(x+1)^2+\frac{7}{2}}}\)
= \(\frac{1}{\sqrt{2}} \int \frac{d x}{\sqrt{\left(\sqrt{\frac{7}{2}}\right)^2-(x+1)^2}}\)
= \(\frac{1}{\sqrt{2}} \sin ^{-1}\left(\frac{x+1}{\sqrt{\frac{7}{2}}}\right)\) + c
= \(\frac{1}{\sqrt{2}} \sin ^{-1}\left[\sqrt{\frac{2}{7}}(x+1)\right]\) + c

(ii) Let I = ∫ \(\frac{d x}{\sqrt{5 x-4 x^2}}\)

Question 7 (old).
(ii) ∫ \(\frac{d x}{\sqrt{8+3 x-x^2}}\)
Solution:
Let I = ∫ \(\frac{d x}{\sqrt{8+3 x-x^2}}\)
= ∫ \(\frac{d x}{\sqrt{-\left(x^2-3 x-8\right)}}\)
= ∫ \(\frac{d x}{\sqrt{-\left(x^2-3 x+\frac{9}{4}-\frac{9}{4}-8\right)}}\)
= ∫ \(\frac{d x}{\sqrt{\frac{41}{4}-\left(x-\frac{3}{2}\right)^2}}\)
= ∫ \(\frac{d x}{\sqrt{\left(\frac{\sqrt{41}}{2}\right)^2-\left(x-\frac{3}{2}\right)^2}}\)
= sin^-1 \(\left(\frac{x-\frac{3}{2}}{\frac{\sqrt{41}}{2}}\right)\) + c
= sin^-1 \(\left(\frac{2 x-3}{\sqrt{41}}\right)\) + c
[∵∫ \(\frac{d x}{\sqrt{a^2-x^2}}\) = sin^-1 \(\frac{x}{a}\) + c]

Question 8.
(i) ∫ \(\frac{x-1}{\sqrt{x^2-x}}\) dx (NCERT)
(ii) ∫ \(\frac{5 x+3}{\sqrt{x^2+4 x+10}}\) dx
Solution:
(i) Let x – 1 = A \(\frac{d}{d x}\) (x² – x) + B
= A (2x – 1) + B
⇒ x – 1 = 2Ax – A + B
∴ 2A = 1
A = \(\frac{1}{2}\)
and – A + B = – 1
⇒ B = – 1 + \(\frac{1}{2}\) = – \(\frac{1}{2}\)
∴ I = ∫ \(\frac{x-1}{\sqrt{x^2-x}}\) dx
= ∫ \(\frac{\frac{1}{2}(2 x-1)-\frac{1}{2}}{\sqrt{x^2-x}}\) dx
= \(\frac{1}{2} \int \frac{(2 x-1) d x}{\sqrt{x^2-x}}-\frac{1}{2} \int \frac{d x}{\sqrt{x^2-x}}\)
= \(\frac{1}{2}\) I₁ – \(\frac{1}{2}\) I₂
where I₁ = \(\frac{(2 x-1) d x}{\sqrt{x^2-x}}\)
put x² – x = t
⇒ (2x – 1) dx = dt

(ii) Let I = ∫ \(\frac{5 x+3}{\sqrt{x^2+4 x+10}}\) dx

Question 8 (old).
(i) ∫ \(\frac{d x}{\sqrt{5 x^2-2 x}}\) (NCERT)
(ii) ∫ \(\frac{d x}{\sqrt{9 x-4 x^2}}\) (NCERT)
Solution:
(i) Let I = ∫ \(\frac{d x}{\sqrt{5 x^2-2 x}}\)
= ∫ \(\frac{d x}{\sqrt{5\left(x^2-\frac{2}{5} x+\frac{1}{25}-\frac{1}{25}\right)}}\)
= \(\frac{1}{\sqrt{5}} \int \frac{d x}{\sqrt{\left(x-\frac{1}{5}\right)^2-\left(\frac{1}{5}\right)^2}}\)
= \(\frac{1}{\sqrt{5}} \log \left|x-\frac{1}{5}+\sqrt{\left(x-\frac{1}{5}\right)^2-\left(\frac{1}{5}\right)^2}\right|\) + c
= \(\frac{1}{\sqrt{5}} \log \left|\frac{5 x-1}{5}+\frac{\sqrt{5 x^2-2 x}}{\sqrt{5}}\right|\) + c
[∵∫ \(\frac{d x}{\sqrt{a^2-x^2}}\) = log |x + \(\sqrt{x^2-a^2}\)| + c]

(ii) Let I = ∫ \(\frac{d x}{\sqrt{9 x-4 x^2}}\)

Question 9.
(i) ∫ \(\frac{3 x+5}{\sqrt{x^2-8 x+7}}\) dx
(ii) ∫ \(\frac{x^2}{\sqrt{x^6-2 x^2-3}}\) dx
Solution:
(i) Let I = ∫ \(\frac{3 x+5}{\sqrt{x^2-8 x+7}}\) dx

∴ from (1) ; we have
I = 3 \(\sqrt{x^2-8 x+7}\) + 17 log |x – 4 + \(\sqrt{x^2-8 x+7}\)| + C

(ii) Let I = ∫ \(\frac{x^2}{\sqrt{x^6-2 x^2-3}}\) dx ;
put x³ = t
⇒ 3x² dx = dt
∴ I = ∫ \(\frac{d t}{3 \sqrt{t^2-2 t-3}}\)
= \(\frac{1}{3} \int \frac{d t}{\sqrt{t^2-2 t+1-4}}\)
= \(\frac{1}{3} \int \frac{d t}{\sqrt{(t-1)^2-2^2}}\)
put t – 1 = u
⇒ dt = du
= \(\frac{1}{3} \int \frac{d u}{\sqrt{u^2-2^2}}\)
= \(\frac{1}{3}\) log |u + \(\sqrt{u^2-4}\)| + C
[∵∫ \(\frac{d x}{\sqrt{x^2+^2}}\) = log |x + \(\sqrt{x^2-a^2}\)| + c]
= \(\frac{1}{3}\) log |t – 1 + \(\sqrt{t^2-2 t-3}\)|
= \(\frac{1}{3}\) log |x – 1 + \(\sqrt{x^6-2 x^3-3}\)| + C

Question 10 (old).
(i) ∫ \(\frac{x+3}{\sqrt{5-4 x-x^2}}\) dx
Solution:
Let I = \(\frac{(x+3) d x}{\sqrt{5-4 x-x^2}}\)

Question 11 (old).
(ii) ∫ \(\frac{x+2}{\sqrt{(x-2)(x-3)}}\) dx
Solution:
Let I = ∫ \(\frac{x+2}{\sqrt{(x-2)(x-3)}}\) dx

Question 12 (old).
(i) ∫ \(\frac{x+2}{\sqrt{x^2+2 x+3}}\) dx
Solution:
Let I = ∫ \(\frac{x+2}{\sqrt{x^2+2 x+3}}\) dx

Students can track their progress and improvement through regular use of ML Aggarwal Maths for Class 12 Solutions Chapter 8 Integrals Ex 8.7.

ML Aggarwal Class 12 Maths Solutions Section A Chapter 8 Integrals Ex 8.7

Very Short answer type questions (1 to 3):

Evaluate the following (1 to 16) integrals:

Question 1.
(i) ∫ \(\frac{1}{x^2-16}\) dx (NCERT)
(ii) ∫ \(\frac{1}{9-x^2}\) dx
(iii) ∫ \(\frac{d x}{\sqrt{1-x^2}}\)
(iv) ∫ \(\frac{d x}{x^2+16}\)
Solution:
(i) ∫ \(\frac{1}{x^2-16}\) dx
= (i) ∫ \(\frac{d x}{x^2-16}\)
= \(\frac{1}{2 \times 4} \log \left|\frac{x-4}{x+4}\right|\) + C
= \(\frac{1}{8} \log \left|\frac{x-4}{x+4}\right|\) + C
[∵ ∫ \(\frac{d x}{x^2-a^2}\) = \(\frac{1}{2 a} \log \left|\frac{x-a}{x+a}\right|\) + C]

(ii) ∫ \(\frac{1}{9-x^2}\) dx
= ∫ \(\frac{d x}{3^2-x^2}\)
= \(\frac{1}{2 \times 3} \log \left|\frac{3+x}{3-x}\right|\) + C
= \(\frac{1}{6} \log \left|\frac{3+x}{3-x}\right|\) + C
[∵ ∫ \(\frac{d x}{x^2-a^2}\) = \(\frac{1}{2 a} \log \left|\frac{x-a}{x+a}\right|\) + C]

(iii) ∫ \(\frac{d x}{\sqrt{1-x^2}}\)
= ∫ \(\frac{d x}{\sqrt{1^2-x^2}}\)
= sin^-1 \(\frac{x}{1}\) + C
= sin^-1 x + C
[∵ ∫ \(\frac{d x}{a^2-x^2}\) = sin^-1 \(\frac{x}{a}\) + C]

(iv) ∫ \(\frac{d x}{x^2+16}\)
= ∫ \(\frac{d x}{x^2+4^2}\)
=\(\frac{1}{4}\) tan^-1 \(\frac{x}{4}\) + C
[∵ ∫ \(\frac{d x}{x^2+a^2}\) = \(\frac{1}{a} \tan ^{-1} \frac{x}{a}\) + C]

Question 2.
(i) ∫ \(\frac{d x}{\sqrt{9-x^2}}\)
(ii) ∫ \(\frac{d x}{\sqrt{25+9 x^2}}\)
Solution:
(i) Let I = ∫ \(\frac{d x}{\sqrt{9-x^2}}\)
= \(\frac{1}{2} \int \frac{d x}{\sqrt{\left(\frac{3}{2}\right)^2-x^2}}\)
= \(\frac{1}{2} \sin ^{-1}\left(\frac{x}{\frac{3}{2}}\right)\) + C
[∵ ∫ \(\frac{d x}{x^2-a^2}\) = sin^-1 \(\frac{x}{a}\) + C]
= \(\frac{1}{2} \sin ^{-1}\left(\frac{2 x}{3}\right)\) + C

(ii) ∫ \(\frac{d x}{\sqrt{25+9 x^2}}\)
= \(\frac{1}{9} \int \frac{d x}{x^2+\left(\frac{5}{3}\right)^2}\)
= \(\frac{1}{9} \times \frac{1}{\frac{5}{3}} \tan ^{-1}\left(\frac{x}{\frac{5}{3}}\right)\) + C
= \(\frac{1}{15} \tan ^{-1}\left(\frac{3 x}{5}\right)\) + C
[∵ ∫ \(\frac{d x}{x^2+a^2}\) = \(\frac{1}{a} \tan ^{-1} \frac{x}{a}\) + C]

Question 2 (old).
(i) ∫ \(\frac{d x}{\sqrt{16-9 x^2}}\) (NCERT Exemplar)
Solution:
(i) ∫ \(\frac{d x}{\sqrt{16-9 x^2}}\)
= \(\frac{1}{3} \int \frac{d x}{\sqrt{\left(\frac{4}{3}\right)^2-x^2}}\)
= \(\frac{1}{3} \sin ^{-1}\left(\frac{x}{\frac{4}{3}}\right)\) + C
[∵ ∫ \(\frac{d x}{a^2-x^2}\) = sin^-1 \(\frac{x}{a}\) + C]
= \(\frac{1}{3} \sin ^{-1}\left(\frac{3 x}{4}\right)\) + C

Question 3.
(i) ∫ \(\frac{d x}{x \sqrt{9 x^2-16}}\)
(ii) ∫ \(\frac{d x}{\sqrt{x^2+5}}\)
Solution:
(i) ∫ \(\frac{d x}{x \sqrt{9 x^2-16}}\)
= \(\frac{1}{3} \int \frac{d x}{x \sqrt{x^2-\left(\frac{4}{3}\right)^2}}\)
= \(\frac{1}{3} \times \frac{1}{\frac{4}{3}} \sec ^{-1}\left(\frac{x}{\frac{4}{3}}\right)\) + C
= \(\frac{1}{4} \sec ^{-1}\left(\frac{3 x}{4}\right)\) + C
[∵ ∫ \(\frac{d x}{x \sqrt{x^2-a^2}}=\frac{1}{a} \sec ^{-1} \frac{x}{a}\) + C]

(ii) Let I = ∫ \(\frac{d x}{\sqrt{x^2+5}}\)
= ∫ \(\frac{d x}{\sqrt{x^2+(\sqrt{5})^2}}\)
= log |x + \(\sqrt{x^2+5}\)| + C
[∵ ∫ \(\frac{d x}{\sqrt{x^2+a^2}}\) = log |x + \(\sqrt{x^2+a^2}\)| + C

Question 3 (old).
(ii) ∫ \(\frac{d x}{\sqrt{x^2+4}}\)
Solution:
∫ \(\frac{d x}{\sqrt{x^2+4}}\)
= ∫ \(\frac{d x}{\sqrt{x^2+2^2}}\)
= log |x + \(\sqrt{x^2+4}\)| + C
[∵ ∫ \(\frac{d x}{\sqrt{x^2+a^2}}\) = log |x + \(\sqrt{x^2+a^2}\)| + C

Question 4.
(i) ∫ \(\frac{1}{9 x^2-1}\) dx
(ii) ∫ \(\frac{d x}{32-2 x^2}\)
Solution:
(i) ∫ \(\frac{1}{9 x^2-1}\) dx
= \(\frac{1}{9} \int \frac{d x}{x^2-\left(\frac{1}{3}\right)^2}\)
= \(\frac{1}{9} \times \frac{1}{2 \times \frac{1}{3}} \log \left|\frac{x-\frac{1}{3}}{x+\frac{1}{3}}\right|\) + C
= \(\frac{1}{6} \log \left|\frac{3 x-1}{3 x+1}\right|\) + C

(ii) ∫ \(\frac{d x}{32-2 x^2}\)
= \(\frac{1}{2} \int \frac{d x}{16-x^2}\)
= \(\frac{1}{2} \int \frac{d x}{4^2-x^2}\)
= \(\frac{1}{2} \times \frac{1}{2 \times 4} \log \left|\frac{4+x}{4-x}\right|\) + C
= \(\frac{1}{16} \log \left|\frac{4+x}{4-x}\right|\) + C
[∵ ∫ \(\frac{d x}{a^2-x^2}=\frac{1}{2 a} \log \left|\frac{a+x}{a-x}\right|\) + C]

Question 5.
(i) ∫ \(\frac{d x}{\sqrt{15-8 x^2}}\)
(ii) ∫ \(\frac{d x}{\sqrt{1+4 x^2}}\) (NCERT)
Solution:
(i) ∫ \(\frac{d x}{\sqrt{15-8 x^2}}\)

(ii) ∫ \(\frac{d x}{\sqrt{1+4 x^2}}\)
= \(\frac{1}{2} \int \frac{d x}{\sqrt{\left(\frac{1}{2}\right)^2+x^2}}\)
= \(\frac{1}{2} \log \left|x+\sqrt{x^2+\frac{1}{4}}\right|\) + C
[∵ ∫ \(\frac{d x}{\sqrt{x^2+a^2}}\) = log |x + \(\sqrt{x^2+a^2}\)| + C

Question 6.
(i) ∫ \(\frac{2 y^2}{y^2+4}\) dy (ISC 2015)
(ii) ∫ \(\frac{x^4+1}{x^2+1}\) dx
Solution:
(i) ∫ \(\frac{2 y^2}{y^2+4}\) dy
= 2 ∫ \(\frac{y^2+4-4}{y^2+4}\) dy
= 2 ∫ [1 – \(\frac{4}{y^2+4}\)] dy
= 2 ∫ dy – 8 ∫ \(\frac{d y}{y^2+2^2}\)
= 2y – \(\frac{8}{2} \tan ^{-1}\left(\frac{y}{2}\right)\) + C
= 2y – 4 tan^-1 (\(\frac{y}{2}\)) + C

(ii) Let I = ∫ \(\frac{x^4+1}{x^2+1}\) dx
= ∫ [x² – 1 + \(\frac{2}{x^2+1}\)] dx
= \(\frac{x^{3}}{3}\) – x + 2 tan^-1 \(\frac{x}{1}\) + c
= \(\frac{x^{3}}{3}\) – x + 2 tan^-1 x + c
[∵ ∫ \(\frac{d x}{\sqrt{x^2+a^2}}=\frac{1}{a} \tan ^{-1} \frac{x}{a}\) + c]

Question 7.
(i) ∫ \(\frac{x}{x^4-1}\) dx (NCERT Exemplar)
(ii) ∫ \(\frac{3 x^2}{1+x^6}\) dx (NCERT)
Solution:
(i) I = ∫ \(\frac{x}{x^4-1}\) dx
put x² = t
⇒ 2x dx = dt
∴ I = ∫ \(\int \frac{d t}{2\left(t^2-1\right)}\)
= \(\frac{1}{2} \int \frac{d t}{t^2-1^2}\)
= \(\frac{1}{2} \times \frac{1}{2 \times 1} \log \left|\frac{t-1}{t+1}\right|\) + C
= \(\frac{1}{4} \log \left|\frac{x^2-1}{x^2+1}\right|\) + C
[∵ ∫ \(\frac{d x}{x^2-a^2}=\frac{1}{2 a} \log \mid \frac{x-a}{x+a}\) + C]

(ii) Let I = \(\frac{3 x^2}{1+x^6}\) dx
∴ I = ∫ \(\frac{d t}{t^2+1}\)
= tan^-1 t + C
= tan^-1 (x³) + C

Question 8.
(i) ∫ \(\frac{x^2}{\sqrt{x^6+a^6}}\) dx (NCERT)
(ii) ∫ \(\frac{x^3}{\sqrt{1-x^8}}\) dx (NCERT)
Solution:
(i) Let I = ∫ \(\frac{x^2}{\sqrt{x^6+a^6}}\) dx
put x³ = t
⇒ 3x² dx = dt
∴ I = ∫ \(\frac{d t}{3 \sqrt{t^2+\left(a^3\right)^2}}\)
= \(\frac{1}{3}\) log |t + \(\sqrt{t^2+a^6}\)| + C
= \(\frac{1}{3}\) log |x³ + \(\sqrt{x^6+a^6}\)| + C

(ii) I = ∫ \(\frac{x^3}{\sqrt{1-x^8}}\) dx
put x⁴ = t
⇒ 4x³ dx = dt
∴ I = \(\int \frac{d t}{4 \sqrt{1-t^2}}\)
= \(\frac{1}{4} \sin ^{-1} \frac{t}{1}\) + C
= \(\frac{1}{4}\) sin^-1 (x⁴) + C
[∵ ∫ \(\frac{d x}{\sqrt{a^2-x^2}}\) = sin^-1 \(\frac{x}{a}\) + C]

Question 9.
(i) ∫ \(\frac{\sqrt{x}}{\sqrt{a^3-x^3}}\) dx (NCERT Exemplar)
(ii) ∫ \(\frac{3 x}{1+2 x^4}\) dx (NCERT)
Solution:
(i) Let I = ∫ \(\frac{\sqrt{x} d x}{\sqrt{a^3-x^3}}\)

(ii) Let I = ∫ \(\frac{3 x}{1+2 x^4}\) dx
= ∫ \(\frac{3 x}{1+2\left(x^2\right)^2}\) dx
put x² = t
∴ 2x dx = dt
∴ I = 3 ∫ \(\frac{d t}{2\left[1+2 t^2\right]}\)
= \(\frac{3}{4} \int \frac{d t}{t^2+\frac{1}{2}}\)
= \(\frac{3}{4} \int \frac{d t}{t^2+\left(\frac{1}{\sqrt{2}}\right)^2}\)
= \(\frac{3}{4} \cdot \frac{1}{\frac{1}{\sqrt{2}}} \tan ^{-1}\left(\frac{t}{\frac{1}{\sqrt{2}}}\right)\) + C
= \(\frac{3 \sqrt{2}}{4}\) tan^-1 (√2 x²) + C

Question 10.
(i) ∫ \(\frac{\cos x}{\sqrt{4-\sin ^2 x}}\) dx (NCERT)
(ii) ∫ \(\frac{\sec ^2 x}{\sqrt{\tan ^2 x+4}}\) dx (NCERT)
Solution:
(i) Let I = ∫ \(\frac{\cos x}{\sqrt{4-\sin ^2 x}}\) dx
put sin x = t
⇒ cos x dx = dt
= ∫ \(\frac{d t}{\sqrt{4-t^2}}\)
= ∫ \(\frac{d t}{\sqrt{2^2-t^2}}\)
= sin^-1 (\(\frac{t}{2}\)) + c
= sin^-1 (\(\left(\frac{\sin x}{2}\right)\)) + c

(ii) Let I = ∫ \(\frac{\sec ^2 x}{\sqrt{\tan ^2 x+4}}\) dx
put tan x = t
⇒ ∫ sec² x dx = dt
= ∫ \(\frac{d t}{\sqrt{4+t^2}}\)
= ∫ \(\frac{d t}{\sqrt{t^2+2^2}}\)
= log |t + \(\sqrt{t^2+4}\)| + c
[∵ ∫ \(\frac{d x}{\sqrt{x^2+a^2}}\) = log |x + \(\sqrt{x^2+a^2}\)| + c
= log |tan x + \(\sqrt{\tan ^2 x+4}\)| + c

Question 11.
(i) ∫ \(\frac{\sin x}{16-9 \cos ^2 x}\) dx
(ii) ∫ \(\frac{\tan ^2 x \sec ^2 x}{1-\tan ^6 x}\) dx
Solution:
(i) Let I = ∫ \(\frac{\sin x}{16-9 \cos ^2 x}\) dx
put cos x = t
⇒ – sin x dx = dt

(ii) Let I = ∫ \(\frac{\tan ^2 x \sec ^2 x}{1-\tan ^6 x}\) dx
put tan³ x = t
⇒ 3 tan² x sec² x dx = dt
∴ I = ∫ \(\frac{d t}{3\left(1-t^2\right)}\)
= \(\frac{1}{3} \times \frac{1}{2 \times 1} \log \left|\frac{1+t}{1-t}\right|\) + C
[∵ \(\frac{d x}{a^2+x^2}=\frac{1}{2 a} \log \left|\frac{a+x}{a-x}\right|\) + C]
= \(\frac{1}{6} \log \left|\frac{1+\tan ^3 x}{1-\tan ^3 x}\right|\) + C

Question 11 (old).
(ii) ∫ latex]\frac{\ {cosec}^2 x}{1-\cot ^2 x}[/latex] dx
Solution:
Let I = ∫ latex]\frac{\ {cosec}^2 x}{1-\cot ^2 x}[/latex] dx
put cos x = t
⇒ – cosec² x dx = dt
∴ I = ∫ \(\frac{-d t}{1-t^2}\)
= – \(\frac{1}{2} \log \left|\frac{1+t}{1-t}\right|\)
= – \(\frac{1}{2} \log \left|\frac{1+\cot x}{1-\cot x}\right|\) + C
[∵ \(\frac{d x}{a^2-x^2}=\frac{1}{2 a} \log \left|\frac{a+x}{a-x}\right|\) + C]

Question 12.
(i) ∫ \(\frac{\sin x+\cos x}{\sqrt{\sin 2 x}}\) dx
(ii) ∫ \(\frac{d x}{2 \sin ^2 x+5 \cos ^2 x}\)
Solution:
(i) Let I = ∫ \(\frac{\sin x+\cos x}{\sqrt{\sin 2 x}}\) dx
put sin x – cos x = t ……….(1)
⇒ (cos x + sin x) dx = dt
On squaring (1) ; we have
sin² x + cos² x – 2 sin x cos x = t²
⇒ 1 – sin 2x = t²
⇒ sin 2x = 1 – t²
∴ I = ∫ \(\frac{d t}{\sqrt{1-t^2}}\)
= sin^-1 (t) + C
= sin^-1 (sin x – cos x) + C

(ii) Let I = ∫ \(\frac{d x}{2 \sin ^2 x+5 \cos ^2 x}\)
Divide Numerator and denominator by cos² x ; we get
put tan x = t
⇒ sec² x dx = dt
∴ I = ∫ \(\frac{d t}{2 t^2+5}\)
= \(\frac{1}{2}\) ∫ \(\frac{d t}{t^2+\left(\sqrt{\frac{5}{2}}\right)^2}\)
= \(\frac{1}{2} \times \frac{1}{\sqrt{\frac{5}{2}}} \tan ^{-1}\left(\frac{t}{\sqrt{\frac{5}{2}}}\right)\) + C
[∵ ∫ \(\frac{d x}{x^2+a^2}=\frac{1}{a} \tan ^{-1} \frac{x}{a}\) + C]
= \(\frac{1}{\sqrt{10}} \tan ^{-1}\left(\sqrt{\frac{2}{5}} \tan x\right)\) + C

Question 12 (old).
(ii) ∫ \(\frac{\sin x+\cos x}{9+16 \sin 2 x}\) dx
Solution:
Let I = ∫ \(\frac{\sin x+\cos x}{9+16 \sin 2 x}\) dx

Question 13.
(i) ∫ \(\frac{\cos x}{\cos 3 x}\) dx
(ii) ∫ \(\frac{d x}{2-3 \cos 2 x}\)
Solution:
(i) Let I = ∫ \(\frac{\cos x}{\cos 3 x}\) dx

(ii) Let I = ∫ \(\frac{d x}{2-3 \cos 2 x}\)
= ∫ \(\frac{d x}{2-3\left(1-2 \sin ^2 x\right)}\)
= ∫ \(\frac{d x}{-1+6 \sin ^2 x}\)
Divide numerator and denominator by cos² x ; we have
= ∫ \(\frac{\sec ^2 x d x}{-\sec ^2 x+6 \tan ^2 x}\)
=∫ \(\frac{\sec ^2 x d x}{-1-\tan ^2 x+6 \tan ^2 x}\)

Question 13 (old).
(ii) ∫ \(\frac{d x}{11 \sin ^2 x+4 \cos ^2 x+5}\)
Solution:
Let I = ∫ \(\frac{d x}{11 \sin ^2 x+4 \cos ^2 x+5}\)
Divide numerator and denominator by cos² x ; we have

Question 14.
(i) ∫ \(\frac{d x}{e^x-e^{-x}}\)
(ii) ∫ \(\sqrt{\frac{1+x}{1-x}}\) dx (NCERT Exemplar)
Solution:
(i) Let I = ∫ \(\frac{d x}{e^x-e^{-x}}\)
= ∫ \(\frac{d x}{e^x-\frac{1}{e^x}}\)
= ∫ \(\frac{e^x d x}{e^{2 x}-1}\)
put e^x = t
⇒ e^x dx = dt
= ∫ \(\frac{d t}{t^2-1^2}\)
= \(\frac{1}{2} \log \left|\frac{t-1}{t+1}\right|\) + C
= \(\frac{1}{2} \log \left|\frac{e^x-1}{e^x+1}\right|\) + C

(ii) Let I = ∫ \(\sqrt{\frac{1+x}{1-x}}\) dx
put x = cos θ
⇒ dx = – sin θ dθ

= – [θ + sin θ] + C
= – cos^-1 x – \(\sqrt{1-x^{2}}\) + C
[∵ cos θ = x
⇒ θ = cos^-1 x
and sin θ = \(\sqrt{1-\cos ^2 \theta}\)
= \(\sqrt{1-x^2}\)]

Question 15.
(i) ∫ \(\frac{x-1}{\sqrt{x^2-1}}\) dx (NCERT)
(ii) ∫ \(\frac{x+2}{\sqrt{x^2-1}}\) dx (NCERT)
Solution:
(i) Let I = ∫ \(\frac{x-1}{\sqrt{x^2-1}}\) dx
= \(\int \frac{x d x}{\sqrt{x^2-1}}-\int \frac{d x}{\sqrt{x^2-1}}\)
= I₁ – \(\frac{d x}{\sqrt{x^2-1}}\) …..(1)
where I₁ = ∫ \(\frac{x d x}{\sqrt{x^2-1}}\) ……….(1)
put x² = t
⇒ 2x dx = dt
∴ I₁ = ∫ \(\frac{d t}{2 \sqrt{t-1}}\)
= \(\frac{1}{2}\) (t – 1)^-1/2 dt
= \(\frac{1}{2} \frac{(t-1)^{1 / 2}}{1 / 2}\)
= \(\sqrt{x^{2}-1}\)
I = \(\sqrt{x^2-1}-\log \left|x+\sqrt{x^2-1}\right|\) + C

(ii) Let I = ∫ \(\frac{x+2}{\sqrt{x^2-1}}\) dx
= \(\int \frac{x d x}{\sqrt{x^2-1}}+2 \int \frac{d x}{\sqrt{x^2-1}}\)
= \(\frac{1}{2}\) ∫ (x^{2} – 1)^{\(-\frac{1}{2}\)} (2x dx) + 2 ∫ \(\frac{d x}{\sqrt{x^2-1^2}}\)
= \(\frac{1}{2} \frac{\left(x^2-1\right)^{\frac{-1}{2}+1}}{\left(\frac{-1}{2}+1\right)}\) + 2 log |x + \(\sqrt{x^2-1}\)| + C
[∵ ∫ [f(x)]ⁿ f'(x) dx = \(\frac{(f(x))^{n+1}}{n+1}\) + C]
= \(\sqrt{x^2-1}\) + 2 log |x + \(\sqrt{x^2-1}\)| + C

Question 16.
(i) ∫ \(\sqrt{\frac{a+x}{a-x}}\) dx (NCERT Exemplar)
(ii) ∫ \(\frac{\sqrt{x^2-a^2}}{x}\) dx
Solution:
(i) Let I = ∫ \(\sqrt{\frac{a+x}{a-x}}\) dx
∴ I = ∫ \(\sqrt{\frac{(a+x)}{a-x} \times \frac{a+x}{a+x}}\) dx
= ∫ \(\frac{a+x}{\sqrt{a^2-x^2}}\) dx
= ∫ \(a \int \frac{d x}{\sqrt{a^2-x^2}}+\int \frac{x d x}{\sqrt{a^2-x^2}}\)
∴ I = a sin^-1 \(\frac{x}{a}\) + I₁
where I₁ = ∫ \(\frac{d t}{2 \sqrt{a^2-t}}\)
= \(\frac{1}{2}\) ∫ (a² – t)^-1/2 dt
= – \(\frac{1}{2} \frac{\left(a^2-t\right)^{1 / 2}}{1 / 2}\)
= – \(\sqrt{a^2-x^2}\)
∴ From (1) ; we have
∴ I = a sin^-1 \(\frac{x}{a}\) – \(\sqrt{a^2-x^2}\) + C

(ii) Let I =∫ \(\frac{\sqrt{x^2-a^2}}{x}\) dx

Question 17.
If ∫ \(\frac{2^x}{\sqrt{1-4^x}}\) dx = k sin^-1 (2^x) + C, then what is the value of k?
Solution:
Let I = ∫ \(\frac{2^x}{\sqrt{1-4^x}}\) dx
= ∫ \(\frac{2^x d x}{\sqrt{1-\left(2^x\right)^2}}\)
put 2^x = t
⇒ 2^x log 2 dx = dt
⇒ I = ∫ \(\frac{d t}{\log 2 \sqrt{1-t^2}}\)
= \(\frac{1}{\log 2} \int \frac{d t}{\sqrt{1-t^2}}\)
I = \(\frac{1}{\log 2}\) sin^-1 t + C
[∵ ∫ \(\frac{d x}{\sqrt{a^2-x^2}}\) = sin^-1 \(\frac{x}{a}\) + C
∴ I = \(\frac{1}{\log 2}\) sin^-1 (2^x) + C ………….(1)
Also given
∫ \(\frac{2^x d x}{1-4 x}\) = k sin^-1 (2^x) + C ………..(2)
From (1) and (2) ; we have
k = \(\frac{1}{\log 2}\)

Parents can use ML Aggarwal Class 12 ISC Solutions Chapter 8 Integrals Ex 8.6 to provide additional support to their children.

ML Aggarwal Class 12 Maths Solutions Section A Chapter 8 Integrals Ex 8.6

Very Short answer type questions (1 to 3):

Evaluate the following (1 to 8) integrals:

Question 1.
(i) ∫ tan (2 – 3x) dx
(ii) ∫ cot (7 – 4x) dx
Solution:
(i) Let I = ∫ \(\frac{\sin (2-3 x)}{\cos (2-3 x)}\) dx
put cos (2 – 3x) = t
⇒ – sin (2 – 3x) (- 3) dx = dt
= ∫ \(\frac{d t}{3}\)
= \(\frac{1}{3}\) log |t| + C
= \(\frac{1}{3}\) log |cos (2 – 3x)| + C

(ii) Let I = ∫ cot (7 – 4x) dx
= ∫ \(\frac{\cos (7 x-4)}{\sin (7 x-4)}\) dx
put sin (7x – 4) = t
⇒ cos (7x – 4) . 7 dx = dt
= ∫ \(\frac{d t}{7 t}\)
= \(\frac{1}{7}\) log |t| + C
= \(\frac{1}{7}\) log |sin (7x – 4)| + C

Question 2.
(i) ∫ \(\frac{1+\tan ^2 x}{2 \tan x}\) dx
(ii)∫ \(\frac{1}{\sin x \cos x}\) dx
Solution:
(i) Let I = ∫ \(\frac{1+\tan ^2 x}{2 \tan x}\) dx
= \(\frac{1}{2} \int \frac{\sec ^2 x d x}{\tan x}\)
= \(\frac{1}{2}\) log |tan x| + C
[∵ ∫ \(\frac{f^{\prime}(x)}{f(x)}\) dx = log |f(x)| + C]

Aliter:
Let I = ∫ \(\frac{\sec ^2 x d x}{2 \tan x}\)
= ∫ \(\frac{d x}{2 \sin x \cos x}\)
= ∫ \(\frac{d x}{\sin 2 x}\)
= ∫ cosec 2x dx
= \(\frac{\log |\ {cosec} 2 x-\cot 2 x|}{2}\) + C

(ii) Let I = ∫ \(\frac{d x}{\sin x \cos x}\)
= ∫ \(\frac{2 d x}{\sin 2 x}\)
= 2 ∫ cosec 2x dx
= \(\frac{2 \log |\ {cosec} 2 x-\cot 2 x|}{2}\) + C
= log |cosec 2x – cot 2x| + C

Question 3.
(i) ∫ \(\frac{\cos 2 x}{\sin x}\) dx
(ii) ∫ \(\frac{\sin 2 x}{\sin 4 x}\) dx
(iii) ∫ \(\frac{\sin ^2 x-\cos ^2 x}{\sin x \cos x}\) dx
Solution:
(i) Let I = ∫ \(\frac{\cos 2 x}{\sin x}\) dx
= ∫ \(\frac{\left(1-2 \sin ^2 x\right)}{\sin x}\) dx
= ∫ cosec x dx – 2 ∫ sin x dx + C
= log |cosec x – cot x| + 2 cos x + C

(ii) Let I = ∫ \(\frac{\sin 2 x}{\sin 4 x}\) dx
= ∫ \(\frac{\sin 2 x d x}{2 \sin 2 x \cos 2 x}\) dx
= \(\frac{1}{2}\) ∫ sec 2x dx
= \(\frac{1}{2} \frac{\log |\sec 2 x+\tan 2 x|}{2}\) + C
= \(\frac{1}{4}\) log |sec 2x + tan 2x| + C

(iii) Let I = ∫ \(\frac{\sin ^2 x-\cos ^2 x}{\sin x \cos x}\) dx
= \(\int \frac{\sin x}{\cos x} d x-\int \frac{\cos x}{\sin x} d x\)
= – \(\int \frac{-\sin x}{\cos x} d x-\int \frac{\cos x d x}{\sin x}\)
= – log |cos x| – log |sin x| + C
[∵ ∫ \(\frac{f^{\prime}(x)}{f(x)}\) dx = log |f(x)| + C]
= log |sec x| – log |sin x| + C
[∵ a log b = log b^q]

Question 4.
(i) ∫ \(\frac{\sin x}{\sin (x-\alpha)}\) dx
(ii) ∫ \(\frac{\cos x}{\cos (x+\alpha)}\) dx
Solution:
(i) Let I = ∫ \(\frac{\sin x}{\sin (x-\alpha)}\) dx
= ∫ \(\frac{\sin (x-\alpha+\alpha) d x}{\sin (x-\alpha)}\)
= ∫ cos α dx + sin α ∫ \(\frac{\cos (x-\alpha)}{\sin (x-\alpha)}\) dx
= x cos α + sin α log |sin (x – α)| + C
[∵ ∫ \(\frac{f^{\prime}(x)}{f(x)}\) dx = log |f(x)| + C]

(ii) Let I = ∫ \(\frac{\cos x}{\cos (x+\alpha)}\) dx
= ∫ \(\frac{\cos x}{\cos (x+\alpha)}\) dx
= ∫ \(\frac{\cos (x+\alpha-\alpha)}{\cos (x+\alpha)}\) dx
= ∫ \(\left[\frac{\cos (x+\alpha) \cos \alpha-\sin (x+\alpha) \sin \alpha}{\cos (x+\alpha)}\right]\) dx
= ∫ cos α dx – sin α ∫ \(\frac{\sin (x+\alpha) d x}{\cos (x+\alpha)}\)
[∵ ∫ \(\frac{f^{\prime}(x)}{f(x)}\) dx = log |f(x)| + C]
= x cos α + sin α log |cos (x + α)| + C

Question 5.
(i) ∫ \(\frac{d x}{\sin (x-\alpha) \sin (x-\beta)}\)
(ii) ∫ \(\frac{d x}{\cos (x-a) \cos (x-b)}\)
Solution:
(i) Let I = ∫ \(\frac{d x}{\sin (x-\alpha) \sin (x-\beta)}\)

(ii) Let I = ∫ \(\frac{d x}{\cos (x-a) \cos (x-b)}\)

Question 6.
(i) ∫ \(\frac{d x}{\sin (x-a) \cos (x-b)}\)
(ii) ∫ tan³ x dx (ISC 2016)
Solution:
(i) Let I = ∫ \(\frac{d x}{\sin (x-a) \cos (x-b)}\)

(ii) Let I = ∫ tan³ x dx
= ∫ tan x (sec² – 1) dx
= ∫ tan x (sec² x dx) – ∫ tan x dx
= \(\frac{\tan ^2 x}{2}\) + log |cos x| + C
[∵ ∫ [f(x)]ⁿ f'(x) dx = \(\frac{[f(x)]^{n+1}}{n+1}\), n ≠ 1]

Question 7.
(i) ∫ \(\frac{\sin 2 x}{\sin 5 x \sin 3 x}\) dx
(ii) ∫ \(\frac{1}{\sin x \cos ^2 x}\) dx
Solution:
(i) Let I = ∫ \(\frac{\sin 2 x}{\sin 5 x \sin 3 x}\) dx

(ii) Let I = ∫ \(\frac{1}{\sin x \cos ^2 x}\) dx
= ∫ \(\frac{\left(\sin ^2 x+\cos ^2 x\right) d x}{\sin x \cos ^2 x}\)
= ∫ tan x sec x dx + ∫ cosec x dx
= sec x + log |cosec x – cot x| + C

Question 8.
(i) ∫ \(\frac{d x}{\sqrt{1-\sin x}}\)
(ii) ∫ \(\frac{\sin x+\cos x}{1+\sin 2 x}\) dx
Solution:
(i) Let I = ∫ \(\frac{d x}{\sqrt{1-\sin x}}\)
= ∫ \(\frac{d x}{\sqrt{1-\cos \left(\frac{\pi}{2}-x\right)}}\)
= ∫ \(\frac{d x}{\sqrt{2 \sin ^2\left(\frac{\pi}{4}-\frac{x}{2}\right)}}\)
= \(\frac{1}{\sqrt{2}} \int \frac{d x}{\sin \left(\frac{\pi}{4}-\frac{x}{2}\right)}\)
= \(\frac{1}{\sqrt{2}} \int \ {cosec}\left(\frac{\pi}{4}-\frac{x}{2}\right)\)
= \(\frac{1}{\sqrt{2}} \frac{\log \left|\ {cosec}\left(\frac{\pi}{4}-\frac{x}{2}\right)-\cot \left(\frac{\pi}{4}-\frac{x}{2}\right)\right|}{-\frac{1}{2}}\) + C
= – √2 log \(\left|\ {cosec}\left(\frac{\pi}{4}-\frac{x}{2}\right)-\cot \left(\frac{\pi}{4}-\frac{x}{2}\right)\right|\) + C

Aliter:

(ii) Let I = ∫ \(\frac{\sin x+\cos x}{1+\sin 2 x}\) dx

Question 9.
∫ \(\frac{\sqrt{2} \sin x}{\sin \left(x-\frac{\pi}{4}\right)}\) dx = ax + b log |sin (x – \(\frac{\pi}{4}\))| + C, find the values of a and b.
Solution:
Let I = ∫ \(\frac{\sqrt{2} \sin x}{\sin \left(x-\frac{\pi}{4}\right)}\) dx

Also given I = ∫ \(\frac{\sqrt{2} \sin x}{\sin \left(x-\frac{\pi}{4}\right)}\) dx
= ax + b log |sin (x – \(\frac{\pi}{4}\))| + C …………(2)
∴ From (1) and (2) ; we have
a = 1 ;
b = 1.

The availability of ML Aggarwal Class 12 Solutions ISC Chapter 8 Integrals Ex 8.5 encourages students to tackle difficult exercises.

ML Aggarwal Class 12 Maths Solutions Section A Chapter 8 Integrals Ex 8.5

Very Short answer type questions (1 to 3):

Evaluate the following (1 to 23) integrals:

Question 1.
(i) ∫ 2x sin (x² + 1) dx
(ii) ∫ x³ cos (x⁴) dt
Solution:
(i) Let I = ∫ 2x sin (x² + 1) dx
put x² + 1 = t
⇒ 2x dx = dt
∴ I = ∫ sin t dt
= – cos t + C
= – cos (x² + 1) + C

(ii) Let I = ∫ x³ cos (x⁴) dt
put x⁴ = t
⇒ d (x⁴) = dt
⇒ 4x³ dx = dt
⇒ x³ dx = \(\frac{1}{4}\) dt
∴ I = ∫ cos x⁴ (x³ dx)
= ∫ cos t . \(\frac{d t}{4}\)
= \(\frac{1}{4}\) sin t + c
= \(\frac{1}{4}\) sin x⁴ + c

Question 2.
(i) ∫ \(\frac{\cos \sqrt{x}}{\sqrt{x}}\) dx
(ii) ∫ \(\frac{\sin \sqrt{x}}{\sqrt{x}}\) dx
(iii) ∫ \(\frac{\sec ^2 \sqrt{x}}{\sqrt{x}}\) dx
(iv) ∫ \(\frac{\ {cosec}^2 \sqrt{x}}{\sqrt{x}}\) dx
Solution:
(i) Let I = ∫ \(\frac{\cos \sqrt{x}}{\sqrt{x}}\) dx
put √x = t
⇒ \(\frac{1}{2 \sqrt{x}}\) dx = dt
⇒ \(\frac{d x}{\sqrt{x}}\) = 2 dt
∴ I = ∫ cos t (2 dt) = 2 sin t + C

(ii) Let I = ∫ \(\frac{\sin \sqrt{x}}{\sqrt{x}}\) dx
put √x = t
⇒ \(\frac{1}{\sqrt{x}}\) dx = dt
∴ I = ∫ sin t dt
= – cos t + c
= – cos √x + c

(iii) Let I = ∫ \(\frac{\sec ^2 \sqrt{x}}{\sqrt{x}}\) dx
put √x = t
⇒ d (√x) = dt
⇒ \(\frac{1}{2 \sqrt{x}}\) dx = dt
⇒ \(\frac{d x}{\sqrt{x}}\) = 2 dt
∴ I = 2 ∫ sec² t dt
= 2 tan t + c
= 2 tan √x + c

(iv) Let I = ∫ \(\frac{\ {cosec}^2 \sqrt{x}}{\sqrt{x}}\) dx
put √x = t
⇒ \(\frac{1}{2 \sqrt{x}}\) dx = dt
∴ I = ∫ cosec² t (2 dt) = – 2 cot t + C
= – 2 cot √x + C.

Question 3.
(i) ∫ sin x . e^{cos x} dx
(ii) ∫ \(\frac{e^{2 x}}{1+e^x}\) dx
Solution:
(i) Let I = ∫ sin x . e^{cos x} dx
put cos x = t
⇒ – sin x dx = dt
= ∫ e^t (- dt)
= – e^t + C
= – e^{cos x} + C

(ii) Let I = ∫ \(\frac{e^{2 x}}{1+e^x}\) dx
= ∫ \(\frac{e^x \cdot e^x d x}{1+e^x}\) dx
put e^x = t
⇒ d (e^x) = dt
⇒ e^x dx = dt
= ∫ \(\frac{t d t}{1+t}\)
= ∫ \(\frac{1+t-1}{1+t}\)
= ∫ \(\left[1-\frac{1}{t+1}\right]\) dt
= t – log |1 + t| + c
= e^x – log |1 + e^x| + c

Question 4.
(i) ∫ sin x sin (cos x) dx (NCERT)
(ii) ∫ x³ sec² (x⁴ + 3) dx
Solution:
(i) Let I = ∫ sin x sin (cos x) dx
put cos x = t
⇒ – sin x dx = dt
∴ I = ∫ sin t (- dt)
= cos t + C
= cos (cos x) + C

(ii) Let I = ∫ x³ sec² (x⁴ + 3) dx
put x⁴ + 3 = t
⇒ 4x³ dx = dt
= ∫ sec² t (\(\frac{d t}{4}\)) = \(\frac{1}{4}\) tan t + C
= \(\frac{1}{4}\) tan (x⁴ + 4) + C

Question 5.
(i) ∫ \(\frac{1}{x^2} \tan ^2\left(\frac{1}{x}\right)\) dx
(ii) ∫ \(\frac{1}{x^2} \sin ^2\left(\frac{1}{x}\right)\) dx (ISC 2017)
(iii) ∫ 2x sec³ (x² + 5) tan (x² + 5) dx
Solution:
(i) Let I = ∫ \(\frac{1}{x^2} \tan ^2\left(\frac{1}{x}\right)\) dx
put \(\frac{1}{x}\) = t
⇒ – \(\frac{1}{x^2}\) dx = dt
∴ I = ∫ tan² t (- dt)
= ∫ – (sec² t dt + ∫ dt + C
= – tan t + t + C
= – tan (\(\frac{1}{x}\)) + \(\frac{1}{x}\) + C

(ii) Let I = ∫ \(\frac{1}{x^2} \sin ^2\left(\frac{1}{x}\right)\) dx

(iii) Let I = ∫ 2x sec³ (x² + 5) tan (x² + 5) dx
put x² + 5 = t
⇒ 2x dx = dt
= ∫ sec³ t tan t dt
= ∫ sec² t (sec t tan t dt)
= \(\frac{\sec ^3 t}{3}\) + C
[∵ [f(x)]ⁿ f'(x) dx = \(\frac{[f(x)]^{n+1}}{n+1}\) + C, where n ≠ – 1]
= \(\frac{1}{3}\) sec³ (x² + 5) + C

Question 6.
(i) ∫ \(\frac{\sin (\log x)}{x}\) dx
(ii) ∫ \(\frac{\ {cosec}^2(\log x)}{x}\) dx
Solution:
(i) Let I = ∫ \(\frac{\sin (\log x)}{x}\) dx
put log x = t
⇒ d (log x) dt
⇒ \(\frac{1}{x}\) dx = dt
∴ I = ∫ sin t dt
= – cos t + c
= – cos (log x) + c

(ii) Let I = ∫ \(\frac{\ {cosec}^2(\log x)}{x}\) dx
put log x = t
⇒ \(\frac{1}{x}\) dx = dt
∴ I = ∫ cosec² t dt
= – cot t + C
= – cot (log x) + C

Question 7.
(i) ∫ \(\frac{e^{\tan ^{-1} x}}{1+x^2}\) dx
(ii) ∫ \(\frac{\sin \left(2 \tan ^{-1} x\right)}{1+x^2}\) dx
Solution:
(i) Let I = ∫ \(\frac{e^{\tan ^{-1} x}}{1+x^2}\) dx
put tan^-1 x = t
⇒ \(\frac{1}{1+x^2}\) dx = dt
= ∫ e^t dt
= e^t + C
= e^{tan-1 x} + C

(ii) Let I = ∫ \(\frac{\sin \left(2 \tan ^{-1} x\right)}{1+x^2}\) dx
put tan^-1 x = t
⇒ \(\frac{1}{1+x^2}\) dx = dt
∴ I = ∫ sin 2t dt
= – \(\frac{\cos 2 t}{2}\) + C
= – \(\frac{1}{2}\) cos (2 tan^-1 x) + C

Question 8.
(i) ∫ \(\frac{e^{-1 / x}+1}{x^2}\) dx
(ii) ∫ \(\frac{e^{m \tan ^{-1} x}}{1+x^2}\) dx
Solution:
(i) Let I = ∫ \(\frac{e^{-1 / x}+1}{x^2}\) dx
= ∫ (e^-1/x + 1) \(\frac{1}{x^{2}}\) dx
put – \(\frac{1}{x}\) = t
⇒ \(\frac{1}{x^{2}}\) dx = dt
= ∫ (e^t + 1) dt
= e^t + t + C
= e^{– 1/x} – \(\frac{1}{x}\) + C

(ii) Let I = ∫ \(\frac{e^{m \tan ^{-1} x}}{1+x^2}\) dx
put tan^-1 x = t
⇒ \(\frac{1}{1+x^2}\) dx = dt
∴ I = ∫ e^mt dt
= \(\frac{e^{m t}}{m}\) + C
= \(\frac{e^{m \tan ^{-1} x}}{m}\) + C

Question 9.
(i) ∫ e^x cosec² (e^x) dt
(ii) ∫ \(\frac{(x+1) e^x}{\cot ^2\left(x e^x\right)}\) dx
Solution:
(i) Let I = ∫ e^x cosec² (e^x) dt
put e^x = t
⇒ e^x dx = dt
∴ I = ∫ cosec² t dt
= – cot t + C
= – cot (e^x) + C

(ii) Let I = ∫ \(\frac{(x+1) e^x}{\cot ^2\left(x e^x\right)}\) dx
put x e^x = t
d (x e^x) = dt
⇒ (x e^x + e^x) dx = dt
⇒ (x + 1) e^x dx = dt
∴ I = ∫ \(\int \frac{d t}{\cot ^2 t}\)
= ∫ tan² t dt
= ∫ (sec² t – 1) dt
= tan t – t + C
= tan (x e^x) – x e^x + C

Question 10.
(i) ∫ x² e^x3 (sin e^x3) dx
(ii) ∫ \(\frac{\sin ^2(\log x)}{x}\) dx
(iii) ∫ \(\frac{\sqrt{\tan x}}{\sin ^2 x}\) dx
Solution:
(i) Let I = ∫ x² e^x3 (sin e^x3) dx
put e^x3 = t
⇒ e^x3 . 3x² dx = dt
∴ I = \(\frac{1}{3}\) ∫ sin t dt
= – \(\frac{\cos t}{3}\) + C
= – \(\frac{1}{3}\) cos (e^x3) + C

(ii) Let I = ∫ \(\frac{\sin ^2(\log x)}{x}\) dx
put log x = t
⇒ \(\frac{1}{x}\) dx = dt
∴ I = ∫ sin² t dt
= ∫ \(\frac{1-\cos 2 t}{2}\) dt
= \(\frac{1}{2}\left[t-\frac{\sin 2 t}{2}\right]\) + C
= \(\frac{1}{2}\) [log x – \(\frac{1}{2}\) sin (2 log x)] + C

(iii) Let I = ∫ \(\frac{\sqrt{\tan x}}{\sin ^2 x}\) dx
= ∫ \(\frac{\sqrt{\tan x}}{\frac{\sin ^2 x}{\cos ^2 x} \cos ^2 x}\) dx
= ∫ \(\frac{\sqrt{\tan x}}{\tan ^2 x}\) sec² x dx
= ∫ (tan x)^-3/2 sec² x dx
put tan x = t
⇒ sec² x dx = dt
= ∫ t^-3/2 dt
= \(\frac{t^{-3 / 2}+1}{-\frac{3}{2}+1}\) + C
= – 2 t^-1/2 + C
= – \(\frac{2}{\sqrt{\tan x}}\) + C

Question 11.
(i) ∫ x sin³ (x²) cos (x²) dx
(ii) ∫ tan 2x sec 2x dx (NCERT)
Solution:
(i) Let I = ∫ x sin³ (x²) cos (x²) dx
put sin x² = t
⇒ cos (x²) . 2x dx = dt
= ∫ t³ \(\frac{d t}{2}\)
= \(\frac{1}{2}\) \(\frac{t^4}{4}\) + C
= \(\frac{t^4}{8}\) + C
= \(\frac{1}{8}\) sin⁴ (x² + C

(ii) Let I = ∫ tan³ 2x sec 2x dx
= ∫ tan² 2x (tan 2x sec 2x) dx
= ∫ (sec² 2x – 1) tan 2x sec 2x dx
put sec 2x = t
⇒ (sec 2x tan 2x) 2 dx = dt
= ∫ (t² – 1) \(\frac{d t}{2}\)
= \(\frac{1}{2}\left[\frac{t^3}{3}-t\right]\) + C
= \(\frac{t^3}{6}-\frac{1}{2} t\) + C
= \(\frac{1}{6}\) sec³ 2x – \(\frac{1}{2}\) sec 2x + C

Question 12.
(i) ∫ sin³ (2x + 1) dx (NCERT)
(ii) ∫ sin³ x cos² x dx (NCERT)
Solution:
(i) Let I = ∫ sin³ (2x + 1) dx
= ∫ sin² (2x + 1) sin (2x + 1) dx
= ∫ (1 – cos² (2x + 1)) sin (2x + 1) dx
put cos (2x + 1) = t
⇒ – 2 sin (2x + 1) dx = dt
∴ I = ∫ (1 – t²) \(\frac{d t}{- 2}\)
= \(\) + C
= – \(\frac{1}{2}\) cos (2x + 1) + \(\frac{1}{6}\) cos³(2x + 1) + C

(ii) Let I = ∫ sin³ x cos² x dx
= ∫ \(\frac{\cos ^3 x}{\sin ^2 x}\) dx
= ∫ \(\frac{\left(1-\sin ^2 x\right) \cos x}{\sin ^2 x}\) dx
put sin x = t
⇒ cos x dx = dt
∴ I = ∫ \(\frac{\left(1-t^2\right) d t}{t^2}\)
= ∫ \(\frac{1}{t^2}\) dt – ∫ dt
= – \(\frac{1}{t}\) – t + C
= – \(\frac{1}{sin x}\) – sin x + C
= – cosec x – sin x + C

Question 14.
(i) ∫ \(\frac{\cos ^3 x}{\sqrt{\sin x}}\) dx
(ii) ∫ \(\frac{\sin x \cos ^3 x}{1+\cos ^2 x}\) dx
Solution:
(i) Let I = ∫ \(\frac{\cos ^3 x}{\sqrt{\sin x}}\) dx
= ∫ \(\frac{\cos ^2 x \cos x d x}{\sqrt{\sin x}}\)
= ∫ \(\frac{\left(1-\sin ^2 x\right) \cos x d x}{\sqrt{\sin x}}\)
put sin x = t
⇒ cos x dx = dt
= ∫ \(\frac{1-t^2}{\sqrt{t}}\) dt
= ∫ \(\left[\frac{1}{\sqrt{t}}-t^{3 / 2}\right]\) dt
put sin x = t
⇒ cos x dx = dt
= ∫ \(\frac{1-t^2}{\sqrt{t}}\) dt
= ∫ \(\left[\frac{1}{\sqrt{t}}-t^{3 / 2}\right]\) dt
= \(\frac{t^{\frac{-1}{2}+1}}{\left(\frac{-1}{2}+1\right)}-\frac{t^{\frac{3}{2}+1}}{\left(\frac{3}{2}+1\right)}\) + c
= 2√t – \(\frac{2}{5}\) t^5/2 + c
= 2 \(\sqrt{sin x}\) – \(\frac{2}{5}\) t^5/2 + c

(ii) Let I = ∫ \(\frac{\sin x \cos ^3 x}{1+\cos ^2 x}\) dx
put cos x = t
⇒ – sin x dx = dt

Question 15.
(i) ∫ cosec x log (cosec x – cot x) dx
(ii) ∫ \(\frac{\log \left(\tan \frac{x}{2}\right)}{\sin x}\) dx
Solution:
(i) Let I = ∫ cosec x log (cosec x – cot x) dx
put log (cosec x – cot x) = t
⇒ d {log (cosec x – cot x)} = dt
[- cot x cosec x + cosec² x] dx = dt
⇒ cosec x dx = \(\frac{1}{\ {cosec} x-\cot x}\) dt
∴ I = ∫ t dt
= \(\frac{t^{2}}{2}\) + c
= \(\frac{1}{2}\) [log (cosec x – cot x)]² + c

(ii) Let I = ∫ \(\frac{\log \left(\tan \frac{x}{2}\right)}{\sin x}\) dx
put log (tan \(\frac{x}{2}\)) = t
⇒ \(\frac{1}{\tan \frac{x}{2}} \sec ^2 \frac{x}{2} \cdot \frac{1}{2}\) dx = dt
⇒ \(\frac{1}{2 \sin \frac{x}{2} \cos \frac{x}{2}}\) dx = dt
⇒ \(\frac{1}{\sin x}\) dx = dt
∴ I = ∫ t dt
= \(\frac{t^{2}}{2}\) + c
= \(\frac{\left[\log \left(\tan \frac{x}{2}\right)\right]^2}{2}\) + C

Question 16.
(i) ∫ sec⁴ x dx
(ii) ∫ tan² x sec⁴ x dx (NCERT Exemplar)
Solution:
(i) Let I = ∫ sec⁴ x dx
= ∫ sec² x (1 + tan² x) dx
put tan x = t
⇒ sec² x dx = dt
∴ I = ∫ (1 + t² dt
= t + \(\frac{t^{3}}{3}\) + C
= tan x + \(\frac{\tan ^3 x}{3}\) + C

(ii) Let I = ∫ tan² x sec⁴ x dx
= ∫ tan² x . sec² x . sec² x dx
= ∫ tan² x (1 + tan² x) . sec² x dx
put tan x = t
sec² x dx = dt
= ∫ t² (1 + t²) dt
= \(\frac{t^3}{3}+\frac{t^5}{5}\) + C
= \(\frac{1}{3}\) tan³ x + \(\frac{1}{5}\) tan⁵ x + C

Question 17.
(i) ∫ \(\frac{\tan ^5 \sqrt{x} \sec ^2 \sqrt{x}}{\sqrt{x}}\) dx
(ii) ∫ \(\frac{\cos x-\sin x}{\sqrt{1+\sin 2 x}}\) dx
Solution:
(i) Let I = ∫ \(\frac{\tan ^5 \sqrt{x} \sec ^2 \sqrt{x}}{\sqrt{x}}\) dx
put tan √x = t
sec² √x \(\left(\frac{1}{2 \sqrt{x}}\right)\) dx = dt
∴ I = ∫ t⁵ (2 dt)
= \(\frac{2 t^6}{6}\) + C
= \(\frac{1}{3}\) tan⁶ (√x) + C

(ii) Let I = ∫ \(\frac{\cos x-\sin x}{\sqrt{1+\sin 2 x}}\) dx
= ∫ \(\frac{(\cos x-\sin x) d x}{\sqrt{\cos ^2 x+\sin ^2 x+2 \sin x \cos x}}\)
= ∫ \(\frac{(\cos x-\sin x) d x}{\sqrt{(\cos x+\sin x)^2}}\)
put cos x + sin x = t
⇒ (- sin x + cos x) dx = dt
= ∫ \(\frac{d t}{t}\)
= log |t| + C
= log |cos x + sin x| + C

Question 18.
(i) ∫ \(\frac{x}{\sqrt{x+4}}\) dx (NCERT)
(ii) ∫ x \(\sqrt{x+2}\) dx (NCERT)
Solution:
(i) Let I = ∫ \(\frac{x d x}{\sqrt{x+4}}\)
= ∫ \(\frac{(x+4-4)}{\sqrt{x+4}}\) dx
= ∫ \(\sqrt{x+4}\) dx – 4 ∫ (x + 4)^{\(-\frac{1}{2}\)} dx
= \(\frac{2}{3}\) (x + 4)\(\frac{3}{2}\) – 4 \(\frac{(x+4)^{-\frac{1}{2}+1}}{\left(-\frac{1}{2}+1\right)}\) + C
= \(\frac{2}{3}\) (x + 4)\(\frac{3}{2}\) – 8 \(\sqrt{x+4}\) + C
∴ I = \(\frac{2}{3} \sqrt{x+4}\) (x + 4 – 12)
= \(\frac{2}{3}\) (x – 8) \(\sqrt{x+4}\) + C

(ii) Let I = ∫ x \(\sqrt{x+2}\) dx
= ∫ (x + 2 – 2) \(\sqrt{x+2}\) dx
= ∫ (x + 2)^{\(\frac{3}{2}\)} dx – 2 ∫ (x + 2)^{\(\frac{1}{2}\)} dx
= \(\frac{2}{5}(x+2)^{\frac{5}{2}}-2 \times \frac{2}{3}(x+2)^{\frac{3}{2}}\) + C
= \(\frac{2}{5}(x+2)^{\frac{5}{2}}-\frac{4}{3}(x+2)^{\frac{3}{2}}\) + C

Question 19.
(i) ∫ \(\frac{x}{\left(x^2+1\right)^2}\) dx
(ii) ∫ \(\frac{x^2}{(4+x)^{3 / 2}}\) dx
Solution:
(i) Let I = ∫ \(\frac{x}{\left(x^2+1\right)^2}\) dx
put x² = t
⇒ 2x dx = dt
∴ I = ∫ \(\frac{d t}{2(t+1)^2}\)
= \(\frac{1}{2} \frac{(t+1)^{-2+1}}{(-2+1)}\) + C
= – \(\frac{1}{2(t+1)}\) + C
= – \(\frac{1}{2\left(x^2+1\right)}\) + C

(ii) Let I = ∫ \(\frac{x^2}{(4+x)^{3 / 2}}\) dx
put 4 + x = t
⇒ dx = dt

Question 20.
(i) ∫ 4x³ \(\sqrt{5-x^2}\) dx
(ii) ∫ \(\frac{d x}{x-\sqrt{x}}\) (NCERT)
Solution:
(i) Let I = ∫ 4x³ \(\sqrt{5-x^2}\) dx
put 5 – x² = t
⇒ x² = 5 – t
⇒ 2x dx = – dt
∴ I = ∫ 2 (5 – t) √t (- dt)
= – 10 t^{\(\frac{1}{2}\) + 1} / (\(\frac{1}{2}\) + 1) + 2 t^{\(\frac{5}{2}\)} / \(\frac{5}{2}\) + C
= – \(\frac{20}{3}\) t^3/2 + \(\frac{4}{5}\) t^5/2 + C
= \(\frac{4}{5}\) (5 – x²)^5/2 – \(\frac{20}{3}\) (5 – x²)^5/2 + C

(ii) Let I = ∫ \(\frac{d x}{x-\sqrt{x}}\)
put √x = t
⇒ x = t²
⇒ dx = 2t dt
∴ I = ∫ \(\frac{2 t d t}{t^2-t}\)
= ∫ \(\frac{2 t d t}{t(t-1)}\)
= 2 ∫ \(\frac{d t}{t-1}\)
= 2 log |t – 1| + C
= 2 log |√x – 1| + C

Question 21.
(i) ∫ \(\frac{x}{1+\sqrt{x}}\) dx (NCERT Exemplar)
(ii) ∫ \(\frac{x^{\frac{1}{2}}}{1+x^{\frac{3}{4}}}\) dx
Solution:
(i) Let I = ∫ \(\frac{x}{1+\sqrt{x}}\) dx
put √x = t
⇒ x = t²
⇒ dx = 2t dt

(ii) Let I = ∫ \(\frac{x^{\frac{1}{2}}}{1+x^{\frac{3}{4}}}\) dx
put x^1/4 = t
⇒ x = t⁴
⇒ dx = 4t³

Question 22.
(i) ∫ \(\frac{d x}{\sqrt{x+1}+\sqrt[3]{x+1}}\)
(ii) ∫ \(\frac{d x}{\sqrt{1+\sqrt{x}}}\)
Solution:
(i) Let I = ∫ \(\frac{d x}{\sqrt{x+1}+\sqrt[3]{x+1}}\)

(ii) Let I = ∫ \(\frac{d x}{\sqrt{1+\sqrt{x}}}\)
put \(\sqrt{1+\sqrt{x}}\) = t
⇒ 1 + √x = t
⇒ √x = t² – 1
⇒ x = (t² – 1)²
⇒ dx = 4t (t² – 1)
∴ I = ∫ \(\frac{4 t\left(t^2-1\right)}{t}\)
= 4 \(\left[\frac{t^3}{3}-t\right]\) + C
= \(\frac{4}{3}\) (1 + √x)^3/2 – 4 \(\sqrt{1+\sqrt{x}}\) + C

Question 23.
(i) ∫ \(\frac{\sin 2 x}{\left(a^2+b^2 \sin ^2 x\right)^2}\) dx
(ii) ∫ \(\frac{\sqrt{1+x^2}}{x^4}\) dx (NCERT Exemplar)
(iii) ∫ \(\frac{1}{x^2 \sqrt{1+x^2}}\) dx
Solution:
(i) Let I = ∫ \(\frac{\sin 2 x}{\left(a^2+b^2 \sin ^2 x\right)^2}\) dx
put a² + b² sin² x = t
⇒ 2b² sin x cos x dx = dt
⇒ b² sin 2x dx = dt

(ii) Let I = ∫ \(\frac{\sqrt{1+x^2}}{x^4}\) dx
put x = tan θ
dx = sec² θ
∴ I = ∫ \(\frac{\sqrt{1+\tan ^2 \theta}}{\tan ^4 \theta}\) sec² θ dθ
= ∫ \(\frac{\sec ^3 \theta d \theta}{\tan ^4 \theta}\)
= ∫ \(\frac{\frac{1}{\cos ^3 \theta}}{\frac{\sin ^4 \theta}{\cos ^4 \theta}}\) dθ
= ∫ \(\frac{\cos \theta d \theta}{\sin ^4 \theta}\)
= ∫ (sin θ)^-4 cos θ dθ
= \(\frac{(\sin \theta)^{-4+1}}{(-4+1)}\) + C
[∵ ∫ [f(x)]ⁿ f'(x) dx = \(\frac{[f(x)]^{n+1}}{n+1}\) + C]

(iii) Let I = ∫ \(\frac{1}{x^2 \sqrt{1+x^2}}\) dx

Question 24.
(i) If ∫ x e^kx2 dx = \(\frac{1}{4}\) e^2x2 + C, then find the value of k.
(ii) If ∫ x⁶ sin (5x)⁷ dx = \(\frac{k}{5}\) cos (5x⁷) + C, then what is the value of k ?
Solution:
(i) Let I = ∫ x e^kx2 dx
put x² = t
2x dx = dt
= ∫ e^kt \(\frac{d t}{2}\)
= \(\frac{1}{2} \frac{e^{k t}}{k}\) + C
= \(\frac{1}{2 k}\) e^kx2 + C ……….(1)
Also given
I = ∫ x e^kx2 dx
= \(\frac{1}{4}\) e^2x2 + C …………..(2)
From (1) and (2) ; we have
\(\)
⇒ 2k = 4
⇒ k = 2

(ii) put x⁷ = t
⇒ 7x⁶ dx = dt
⇒ x⁶ dx = \(\frac{d t}{7}\)
∴ ∫ x⁶ sin (5x⁷) dx = ∫ sin 5t \(\frac{d t}{7}\)
= – \(\frac{\cos 5 x^7}{35}\) + C ……….(1)
Also given,
∫ x⁶ sin (5x⁷) dx = \(\frac{k}{5}\) cos (5x⁷) + C ………..(2)
∴ from (1) and (2) ; we have
\(\frac{k}{5}=-\frac{1}{35}\)
k = – \(\frac{1}{7}\)

Regular engagement with ML Aggarwal Class 12 Solutions Chapter 8 Integrals Ex 8.4 can boost students confidence in the subject.

ML Aggarwal Class 12 Maths Solutions Section A Chapter 8 Integrals Ex 8.4

Very Short answer type questions (1 to 9):

Evaluate the following (1 to 21) integrals:

Question 1.
(i) ∫ x³ (1 + x⁴)³ dx
(ii) ∫ (2x + 4) \(\sqrt{x²+4x+3}\) dx
Solution:
(i) Let I = ∫ x³ (1 + x⁴)³ dx
= \(\frac{1}{4}\) ∫ (1 + x⁴)x³ dx
= \(\frac{1}{4} \frac{\left(1+x^4\right)^4}{4}\) + C
[∵ [f(x)]ⁿ f'(x) dx = \(\frac{[f(x)]^{n+1}}{n+1}\) + C, where n ≠ – 1]
= \(\frac{\left(1+x^4\right)^4}{16}\) + C

(ii) Let I = ∫ (2x + 4) \(\sqrt{x²+4x+3}\) dx
= ∫ (x² + 4x + 3)^1/2 (2x + 4) dx
= \(\) + C
[∵ [f(x)]ⁿ f'(x) dx = \(\frac{[f(x)]^{n+1}}{n+1}\) + C, where n ≠ – 1]
= \(\frac{2}{3}\) (x² + 4x + 3)^3/2 + C

Question 2.
(i) ∫ x (x² + 3)^3/2 dx
(ii) ∫ \(\frac{4 x+1}{\sqrt{2 x^2+x-7}}\) dx
Solution:
(i) ∫ x (x² + 3)^3/2 dx
= \(\frac{1}{2}\) ∫ (x² + 3)^3/2 dx
= \(\frac{1}{2} \frac{\left(x^2+3\right)^{\frac{3}{2}+1}}{\left(\frac{3}{2}+1\right)}\) + C
= \(\frac{1}{5}\) (x² + 3)^5/2 + C

(ii) Let I = ∫ \(\frac{4 x+1}{\sqrt{2 x^2+x-7}}\) dx
= ∫ (2x² + x – 7)^{\(-\frac{1}{2}\)} (4x + 1) dx
[∵ [f(x)]ⁿ f'(x) dx = \(\frac{[f(x)]^{n+1}}{n+1}\) + C, where n ≠ – 1]
= \(\frac{\left(2 x^2+x-7\right)^{-\frac{1}{2}+1}}{-\frac{1}{2}+1}\) + C
= 2 \(\sqrt{2 x^2+x-7}\) + C

Question 3.
(i) ∫ \(\frac{3 x+1}{\left(3 x^2+2 x-5\right)^3}\) dx
(ii) ∫ \(\frac{2 x}{\sqrt[3]{x^2+1}}\) dx
Solution:
(i) Let I = ∫ \(\frac{3 x+1}{\left(3 x^2+2 x-5\right)^3}\) dx
= ∫ (3x² + 2x – 5)^{– 3} (3x + 1) dx
= \(\frac{1}{2}\) ∫ (3x² + 2x – 5)^{– 3} (6x + 2) dx
= \(\frac{1}{2} \frac{\left(3 x^2+2 x-5\right)^{-3+1}}{(-3+1)}\) + C
= – \(\frac{1}{4}\) ∫ (3x² + 2x – 5)^{– 2} + C
= – \(\frac{1}{4\left(3 x^2+2 x-5\right)^2}\) + C

(ii) Let I = ∫ \(\frac{2 x}{\sqrt[3]{x^2+1}}\) dx
put x² + 1 = t
⇒ 2x + dx = dt
= ∫ \(\frac{d t}{t^{1 / 3}}\)
= ∫ t^{\(-\frac{1}{3}\)} dt
= \(\frac{t^{-\frac{1}{3}+1}}{-\frac{1}{3}+1}\) + C
= \(\frac{3}{2}\) t^2/3 + C
= \(\frac{3}{2}\) (x² + 1)^2/3 + C

Question 3 (old).
(ii) ∫ (4x + 2) \(\sqrt{x^2+x-3}\) dx
Solution:
Let I = ∫ (4x + 2) \(\sqrt{x^2+x-3}\) dx
= ∫ (x² + x – 3)^{\(\frac{1}{2}\)} 2 (2x + 1) dx
= \(\frac{2\left(x^2+x-3\right)^{\frac{1}{2}+1}}{\frac{1}{2}+1}\) + C
[∵ [f(x)]ⁿ f'(x) dx = \(\frac{[f(x)]^{n+1}}{n+1}\) + C, where n ≠ – 1]
= \(\frac{2\left(x^2+x-3\right)^{\frac{1}{2}+1}}{\frac{1}{2}+1}\) + C
= \(\frac{4}{3}\) (x² + x – 3)^3/2 + C

Question 4.
(i) ∫ e^x (a + be^x)⁶ dx
(ii) ∫ \(\frac{(\log x)^2}{x}\) dx
Solution:
(i) Let I = ∫ e^x (a + be^x)⁶ dx
= \(\frac{1}{b}\) ∫ (a + be^x)⁶ be^x dx
= \(\frac{1}{b} \frac{\left(a+b e^x\right)^{6+1}}{6+1}\) + C
= \(\frac{1}{7 b}\) (a + be^x)⁷ + C

(ii) Let I = ∫ \(\frac{1}{x}\) (log x)² dx
put log x = t
⇒ d (log x) = dt
⇒ \(\frac{1}{x}\) dx = dt
∴ I = ∫ t² dt
= \(\frac{t^{3}}{3}\) + c
= \(\frac{(\log x)^3}{3}\) + c

Question 5.
(i) ∫ \(\frac{\sqrt{3+\log x}}{x}\) dx
(ii) ∫ \(\frac{3(x+1)(x+\log x)^2}{x}\) dx
Solution:
(i) Let I = ∫ \(\frac{\sqrt{3+\log x}}{x}\) dx
= ∫ (3 + log x)^1/2 \(\frac{1}{x}\) dx
[∵ \(\frac{d}{d x}\) (3 + log x) = 0 + \(\frac{1}{x}\) = \(\frac{1}{x}\)]
= \(\frac{(3+\log x)^{\frac{1}{2}+1}}{\frac{1}{2}+1}\) + C
[∵ [f(x)]ⁿ f'(x) dx = \(\frac{[f(x)]^{n+1}}{n+1}\) + C, where n ≠ – 1]
= \(\frac{2}{3}\) (3 + log x)^3/2 + C

(ii) Let I = ∫ \(\frac{3(x+1)(x+\log x)^2}{x}\) dx
= 3 ∫ (x + log x)² (1 + \(\frac{1}{x}\)) dx
[∵ \(\frac{d}{d x}\) (x + log x) = 1 + \(\frac{1}{x}\)]
= 3 \(\frac{(x+\log x)^3}{3}\) + C
[∵ [f(x)]ⁿ f'(x) dx = \(\frac{[f(x)]^{n+1}}{n+1}\) + C, where n ≠ – 1]
= (x + log x)³ + C

Question 6.
(i) ∫ \(\frac{\sqrt{\tan ^{-1} x}}{1+x^2}\) dx
(ii) ∫ \(\frac{\left(\sin ^{-1} 2 x\right)^3}{\sqrt{1-4 x^2}}\) dx
Solution:
(i) Let I = ∫ \(\frac{\sqrt{\tan ^{-1} x}}{1+x^2}\) dx
= ∫ \(\left(\tan ^{-1} x\right)^{\frac{1}{2}} \cdot \frac{1}{1+x^2}\) dx
= \(\frac{\left(\tan ^{-1} x\right)^{\frac{1}{2}+1}}{\frac{1}{2}+1}\) + C
[∵ \(\frac{d}{d x}\) tan^-1 x = \(\frac{1}{1+x^2}\)]
= \(\frac{2}{3}\left(\tan ^{-1} x\right)^{\frac{3}{2}}\) + C

(ii) As \(\frac{d}{d x}\) sin^-1 2x = \(\frac{1}{\sqrt{1-4 x^2}}\) × 2
∴ ∫ \(\frac{\left(\sin ^{-1} 2 x\right)^3}{\sqrt{1-4 x^2}}\) dx = \(\frac{1}{2}\) ∫ (sin^-1 2x)³ \(\frac{2}{\sqrt{1-4 x^2}}\) dx
[∵ [f(x)]ⁿ f'(x) dx = \(\frac{[f(x)]^{n+1}}{n+1}\) + C, where n ≠ – 1]
= \(\frac{1}{2} \frac{\left(\sin ^{-1} 2 x\right)^{3+1}}{(3+1)}\) + C
= \(\frac{1}{8}\) (sin^-1 2x)⁴ + C

Question 7.
(i) ∫ sin x cos⁵ x dx
(ii) ∫ cos³ x (ax + b) sin (ax + b) dx
Solution:
(i) Let I =∫ sin x cos⁵ x dx
= – ∫ (cos x)⁵ (- sin x) dx
= – \(\frac{(\cos x)^{5+1}}{5+1}\) + C
= – \(\frac{1}{6}\) cos⁶ x + C

(ii) Let I = ∫ cos³ x (ax + b) sin (ax + b) dx
= – \(\frac{1}{a}\) ∫ (cos (ax + b))³ (- a sin (ax + b)) dx
[∵ [f(x)]ⁿ f'(x) dx = \(\frac{[f(x)]^{n+1}}{n+1}\) + C, where n ≠ – 1]
= – \(\frac{1}{a} \frac{(\cos (a x+b))^4}{4}\) + C
= – \(\frac{1}{4 a}\) cos⁴ (ax + b) + C

Question 8.
(i) ∫ sec³ x tan x dx
(ii) ∫ cot³ x cosec² x dx
Solution:
(i) Let I = ∫ sec³ x tan x dx
= ∫ sec² x (sec x tan x) dx
= \(\frac{(\sec x)^{2+1}}{2+1}\) + C
= \(\frac{\sec ^3 x}{3}\) + C

(ii) Let I = ∫ cot³ x cosec² x dx
= ∫ (cot x)³ cosec² x dx
put cos x = t
⇒ d (cot x) = dt
⇒ – cosec² x dx = dt
⇒ cosec² x dx = – dt
∴ I = ∫ t³ (- dt)
= \(\frac{-t^4}{4}\) + c
= \(\frac{- 1}{4}\) cot⁴ x + c

Question 9.
(i) ∫ \(\sqrt{tan x}\) (1 + tan² x) dx
(ii) ∫ \(\frac{\sin x}{1+\cos x}\) dx
Solution:
(i) Let I = ∫ \(\sqrt{tan x}\) (1 + tan² x) dx
= ∫ (tan x)^{\(\frac{1}{2}\)} sec² x dx
= \(\frac{(\tan x)^{\frac{1}{2}+1}}{\frac{1}{2}+1}\) + C
= \(\frac{2}{3} \tan \frac{3}{2} x\) + C

(ii) Let I = ∫ \(\frac{\sin x}{1+\cos x}\) dx
= – ∫ \(\frac{-\sin x d x}{1+\cos x}\)
= – log (1 + cos x) + C
[∵ ∫ \(\frac{f^{\prime}(x) d x}{f(x)}\) = log |f(x) + C]

Question 10.
(i) ∫ \(\frac{\cot x}{\sqrt{\sin x}}\) dx
(ii) ∫ \(\frac{\tan x}{\sqrt{\cos x}}\) dx
Solution:
(i) Let I = ∫ \(\frac{\cot x}{\sqrt{\sin x}}\) dx
= ∫ \(\frac{\cos x}{(\sin x)^{3 / 2}}\) dx
put sin x = t
⇒ cos x dx = dt
∴ I = ∫ \(\frac{d t}{t^{3 / 2}}\)
= ∫ t^-3/2 dt
= \(\frac{t^{-3 / 2+1}}{\left(\frac{-3}{2}+1\right)}\) + c
= – 2t^-1/2 + c
= \(\frac{-2}{\sqrt{\sin x}}\) + c

(ii) Let I = ∫ \(\frac{\tan x}{\sqrt{\cos x}}\) dx
= ∫ \(\frac{\sin x}{(\cos x)^{3 / 2}}\) dx
put cos x = t
⇒ d (cos x) = dt
⇒ – sin x dx = dt
⇒ sin x dx = – dt
∴ I = ∫ \(\frac{-d t}{t^{3 / 2}}\)
= – \(\frac{t^{\frac{-3}{2}+1}}{\left(\frac{-3}{2}+1\right)}\) + c
= \(\frac{2}{t^{1 / 2}}\) + c
= \(\frac{2}{\sqrt{\cos x}}\) + c

Question 11.
(i) ∫ \(\frac{\sin x}{\sqrt{3+2 \cos x}}\) dx
(ii) ∫ \(\frac{1+\sin x}{\sqrt{x-\cos x}}\) dx.
Solution:
(i) Let I = ∫ \(\frac{\sin x}{\sqrt{3+2 \cos x}}\) dx
= – \(\frac{1}{2}\) ∫ (3 + 2 cos x)^{– \(\frac{1}{2}\)} (- 2 sin x) dx
= – \(\frac{1}{2} \frac{(3+2 \cos x)^{-\frac{1}{2}+1}}{-\frac{1}{2}+1}\) + C
= – \(\sqrt{3+2 cos x}\) + C

(ii) Let I = ∫ \(\frac{1+\sin x}{\sqrt{x-\cos x}}\) dx
put x – cos x = t
⇒ (1 + sin x) dx = dt
∴ I = ∫ \(\frac{d t}{\sqrt{t}}\)
= ∫ t^-1/2 dt
= \(\frac{t^{\frac{-1}{2}+1}}{\frac{-1}{2}+1}\) + c
= 2√t + c
= 2 \(\sqrt{x-cos x}\) + c

Question 12.
(i) ∫ \(\frac{\sin x}{(1+\cos x)^2}\) dx
(ii) ∫ \(\frac{\left(3 \tan ^2 x+2\right) \sec ^2 x}{\left(\tan ^3 x+2 \tan x+5\right)^2}\) dx
Solution:
(i) Let I = ∫ \(\frac{\sin x}{(1+\cos x)^2}\) dx
= ∫ (1 + cos x)^-2 sin x dx
= – ∫ (1 + cos x)^-2 (- sin x dx)
= – \(\frac{(1+\cos x)^{-2+1}}{-2+1}\) + C
= \(\frac{1}{1+\cos x}\) + C

(ii) Let I = ∫ \(\frac{\left(3 \tan ^2 x+2\right) \sec ^2 x}{\left(\tan ^3 x+2 \tan x+5\right)^2}\) dx
Since \(\frac{d}{d x}\) (tan³ x + 2 tan x + 5)
= 3 tan² x sec² x + 2 sec² x
= sec² x (3 tan² x + 2)
∴ I = ∫ (tan³ x + 2 tan x + 5)^{– 2} {(3 tan² x + 2) sec² x dx}
= \(\frac{\left(\tan ^3 x+2 \tan x+5\right)^{-2+1}}{-2+1}\) + C
= – \(\frac{1}{\tan ^3 x+2 \tan x+5}\) + C
[∵ [f(x)]ⁿ f'(x) dx = \(\frac{[f(x)]^{n+1}}{n+1}\) + C, where n ≠ – 1]

Question 13.
(i) ∫ (e^3x + 1)² e^3x dx
(ii) ∫ \(\frac{e^{4 x}}{\left(1+e^{4 x}\right)^2}\) dx
Solution:
(i) Let I = ∫ (e^3x + 1)² e^3x dx
[∵ \(\frac{d}{d x}\) (e^3x + 1) = 3e^3x]
= \(\frac{1}{3}\) ∫ (e^3x + 1)² (3e^3x dx)
= \(\frac{1}{3} \frac{\left(e^{3 x}+1\right)^3}{3}\) + C
= \(\frac{1}{9}\) (e^3x + 1)³ + C

(ii) Let I = ∫ \(\frac{e^{4 x}}{\left(1+e^{4 x}\right)^2}\) dx
= ∫ (1 + e^4x)^{– 3} e^4x dx
= \(\frac{1}{4}\) ∫ (1 + e^4x)^{– 2} 4 e^4x dx
= \(\frac{1}{4} \frac{\left(1+e^{4 x}\right)^{-2+1}}{-2+1}\) + C
= – \(\frac{1}{4} \frac{1}{\left(1+e^{4 x}\right)}\) + C

Question 14.
(i) ∫ tan⁴ x dx
(ii) ∫ cot⁴ x dx
Solution:
(i) Let I = ∫ tan⁴ x dx
= ∫ tan² x . tan² x dx
= ∫ tan² x (sec² x – 1)
= ∫ tan² x sec² x dx – ∫ tan² x dx
= ∫ (tan x)² sec² x dx – ∫ (sec² x – 1) dx
= \(\frac{(\tan x)^{2+1}}{(2+1)}\) – tan x + x + C
= \(\frac{\tan ^3 x}{3}\) – tan x + x + C

(ii) Let I = ∫ cot⁴ x dx
= ∫ cot² x . cot² x dx
= ∫ cot² x (cosec² x – 1) dx
= ∫ (cot x)² cosec² x dx – ∫ (cosec² x – 1) dx
= – ∫ (cot x)² (- cosec² x dx) – ∫ cosec² x dx + ∫ dx + C
= – \(\frac{\cot ^3 x}{3}\) – cot x + x + C

Question 15.
(i) ∫ \(\frac{1+\tan x}{x+\log \sec x}\) dx
(ii) ∫ \(\frac{1+\sin 2 x}{x+\sin ^2 x}\) dx
Solution:
(i) Since \(\frac{d}{d x}\) (x + log sec x) = 1 + \(\frac{1}{\sec x}\) × sec x tan x
= 1 + tan x
∴ ∫ \(\frac{1+\tan x}{x+\log \sec x}\) dx = log |x + log sec x| + C
[∵∫ \(\frac{f^{\prime}(x) d x}{f(x)}\) = log |f(x)| + C]

(ii) Since \(\frac{d}{d x}\) (x + sin² x) = 1 + 2 sin x cos x
= 1 + sin 2x
∴ I = ∫ \(\frac{1+\sin 2 x}{x+\sin ^2 x}\) dx
= log |x + sin² x| + C
[∵ ∫ \(\frac{f^{\prime}(x) d x}{f(x)}\) = log |f(x)| + C]

Question 16.
(i) ∫ \(\frac{\tan x \sec ^2 x}{1-\tan ^2 x}\) dx
(ii) ∫ \(\frac{x^5}{1+x^6}\) dx
Solution:
(i) Since \(\frac{d}{d x}\) (1 – tan² x) = – 2 tan x sec² x
Let I = ∫ \(\frac{\tan x \sec ^2 x}{1-\tan ^2 x}\) dx
= – \(\frac{1}{2} \int \frac{-2 \tan x \sec ^2 x d x}{1-\tan ^2 x}\)
= – \(\frac{1}{2}\) log |1 – tan² x| + C

(ii) Since \(\frac{d}{d x}\) (1 + x⁶) = 6x⁵
I = ∫ \(\frac{x^5}{1+x^6}\) dx
= ∫ \(\frac{6 x^5 d x}{1+x^6}\)
= \(\frac{1}{6}\) log |1 + x⁶| + C

Question 17.
(i) ∫ \(\frac{2 \cos x-3 \sin x}{6 \cos x+4 \sin x}\) dx
(ii) ∫ \(\frac{\sin 2 x-\sin x}{\sin ^2 x+\cos x+7}\) dx
Solution:
(i) Let I = ∫ \(\frac{2 \cos x-3 \sin x}{6 \cos x+4 \sin x}\) dx
= \(\frac{1}{2} \int \frac{2 \cos x-3 \sin x}{3 \cos x+2 \sin x}\) dx
put 3 cos x + 2 sin x = t
⇒ d (3 cos x + 2 sin x) = dt
⇒ (- 3 sin x + 2 cos x) dx = dt
⇒ (2 cos x – 3 sin x) dx = dt
∴ I = ∫ \(\frac{d t}{2 t}\)
= \(\frac{1}{2}\) log |t| + c
= \(\frac{1}{2}\) log |3 cos x + 2 sin x| + c

(ii) Since \(\frac{d}{d x}\) (sin² x + cos x + 7) = 2 sin x cos x – sin x
= sin 2x – sin x
∴ I = ∫ \(\frac{\sin 2 x-\sin x}{\sin ^2 x+\cos x+7}\) dx
= log |sin² x + cos x + 7| + C
[∵ ∫ \(\frac{f^{\prime}(x) d x}{f(x)}\) = log |f(x)| + C]

Question 18.
(i) ∫ \(\frac{x+\tan x}{x^2-2 \log \cos x}\) dx
(ii) ∫ \(\frac{1}{x \log x}\) dx
Solution:
(i) Since \(\frac{d}{d x}\) (x² – 2 lo0g cos x) = 2x – \(\frac{2}{\cos x}\) (- sin x)
= 2x + 2 tan x
Let I = ∫ \(\frac{(x+\tan x) d x}{x^2-2 \log \cos x}\)
= \(\frac{1}{2}\) ∫ \(\frac{2(x+\tan x) d x}{x^2-2 \log \cos x}\)
= \(\frac{1}{2}\) log |x² – 2 log cos x| + C

(ii) Let I = ∫ \(\frac{d x}{x \log x}\) ;
put log x = t
⇒ \(\frac{1}{x}\) dx = dt
∴ I = ∫ \(\frac{d t}{t}\)
= log |t| + c
= log |log x| + c

Question 19.
(i) ∫ \(\frac{2 x^2-1}{x\left(x^2-\log x\right)}\) dx
(ii) ∫ \(\frac{e^{2 x}+1}{e^{2 x}-1}\) dx
Solution:
(i) Let I = ∫ \(\frac{2 x^2-1}{x\left(x^2-\log x\right)}\) dx
= \(\frac{\frac{2 x^2-1}{x}}{x^2-\log x}\) dx
= \(\frac{\left(2 x-\frac{1}{x}\right) d x}{x^2-\log x}\) dx
As \(\frac{d}{d x}\) (x² – log x) = 2x – \(\frac{1}{x}\)
I = log |x² – log x| + C
[∵ ∫ \(\frac{f^{\prime}(x) d x}{f(x)}\) = log |f(x)| + C]

(ii) Let I = ∫ \(\frac{e^{2 x}+1}{e^{2 x}-1}\) dx
Divivde Num and Deno. by e^x ; we have
= ∫ \(\frac{e^x+e^{-x}}{e^x-e^{-x}}\) dx
put e^x – e^{– x} = t
⇒ (e^x + e^{– x}) dx = dt
∴ I = ∫ \(\frac{d t}{t}\)
= log |e^x – e^{– x}| + C.

Question 20.
(i) ∫ \(\frac{1}{1+e^{-x}}\) dx
(ii) ∫ \(\frac{\tan x}{a+b \tan ^2 x}\) dx
Solution:
(i) Let I = ∫ \(\frac{1}{1+e^{-x}}\) dx
= ∫ \(\frac{e^x d x}{1+e^x}\)
= log |1 + e^x| + C
[∵ ∫ \(\frac{f^{\prime}(x) d x}{f(x)}\) = log |f(x)| + C]

(ii) Let I = ∫ \(\frac{\tan x}{a+b \tan ^2 x}\) dx
= ∫ \(\frac{\frac{\sin x d x}{\cos x}}{a+b \frac{\sin ^2 x}{\cos ^2 x}}\)
= ∫ \(\frac{\sin x \cos x d x}{a \cos ^2 x+b \sin ^2 x}\)
Since, \(\frac{d}{d x}\) (a cos² x + b sin² x)
= 2a cos x (- sin x) + 2b sin x cos x
= 2 (b – a) cos x sin x
∴ I = \(\frac{1}{2(b-a)} \int \frac{2(b-a) \sin x \cos x d x}{a \cos ^2 x+b \sin ^2 x}\)
= \(\frac{1}{2(b-a)}\) log |a cos² x + b sin² x| + C
[∵ ∫ \(\frac{f^{\prime}(x) d x}{f(x)}\) = log |f(x)| + C]

Question 21.
(i) ∫ \(\frac{1-\cot x}{1+\cot x}\) dx
(ii) ∫ \(\frac{\cos 2 x}{(\cos x+\sin x)^2}\) dx
Solution:
(i) Let I = ∫ \(\frac{1-\cot x}{1+\cot x}\) dx
= ∫ \(\frac{1-\frac{\cos x}{\sin x}}{1-\frac{\cos x}{\sin x}}\) dx
= ∫ \(\frac{(\sin x-\cos x)}{\sin x+\cos x}\) dx
putting sin x + cos x = t
⇒ d (sin x + cos x) = dt
⇒ (cos x – sin x) dx = dt
⇒ (sin x – cos x) dx = – dt
∴ I = ∫ \(-\frac{d t}{t}\)
= – log |t| + c
= – log |cos x + sin x| + c

(ii) Let I = ∫ \(\frac{\cos 2 x}{(\cos x+\sin x)^2}\) dx
Thus I = ∫ \(\frac{\cos ^2 x-\sin ^2 x}{(\cos x+\sin x)^2}\) dx
= ∫ \(\frac{\cos x-\sin x}{\cos x+\sin x}\) dx
put cos x + sin x = t
⇒ d (cos x + sin x) = dt
⇒ (- sin x + cos x) dx = dt
∴ I = ∫ \(\frac{d t}{t}\)
= log |t| + c
= log |sin x + cos x| + c

After:
I = ∫ \(\frac{\cos 2 x d x}{\cos ^2 x+\sin ^2 x+2 \sin x \cos x}\)
= \(\frac{\cos 2 x d x}{1+\sin 2 x}\)
= \(\frac{2 \cos 2 x d x}{1+\sin 2 x}\)
= \(\frac{1}{2}\) log |1 + sin 2x| + c
[∵ ∫ \(\frac{f^{\prime}(x) d x}{f(x)}\) = log |f(x)| + C]

Peer review of Class 12 ISC Maths Solutions Chapter 8 Integrals Ex 8.3 can encourage collaborative learning.

ML Aggarwal Class 12 Maths Solutions Section A Chapter 8 Integrals Ex 8.3

Very short answer type questions (1 to 10):

Find all the anti-derivatives of the following (1 to 6) functions :

Question 1.
(i) (ax + b)³
(ii) 4e^3x + 1 (NCERT)
Solution:
(i) ∫ (ax + b)³ = \(\frac{(a x+b)^{3+1}}{(3+1) a}\) + C
= \(\frac{(a x+b)^4}{4 a}\) + C
[∵ ∫ (ax + b)ⁿ dx = \(\frac{(a x+b)^{n+1}}{(n+1) a}\) + C ; n ≠ – 1]

(ii) ∫ 4e^3x + 1 dx
= 4 ∫ e^3x dx + ∫ dx
= \(\frac{4 e^{3 x}}{3}\) + x + C

Question 1 (old).
(i) sin 2x (NCERT)
(ii) cos 3x (NCERT)
Solution:
(i) ∫ sin 2x dx = – \(\frac{\cos 2 x}{2}\) + C
[∵ ∫ sin mx dx = \(\frac{\cos m x}{m}\) + C]

(ii) ∫ cos 3x dx = \(\frac{\sin 3 x}{3}\) + C

Question 2.
(i) \(\frac{1}{\sqrt{7-2 x}}\)
(ii) \(\sqrt{a x+b}\) (NCERT)
Solution:
(i) ∫ \(\frac{1}{\sqrt{7-2 x}}\) dx
= ∫ (7 – 2x)^{– \(\frac{1}{2}\)} dx
= \(\frac{(7-2 x)^{-\frac{1}{2}+1}}{\left(-\frac{1}{2}+1\right)(-2)}\) + C
= – \(\sqrt{7-2x}\) + C

(ii) ∫ \(\sqrt{a x+b}\) dx
= ∫ (ax + b)^1/2 dx
= \(\frac{(a x+b)^{\frac{1}{2}+1}}{\left(\frac{1}{2}+1\right) a}\)
= \(\frac{2}{3 a}\) (ax + b)^3/2 + C

Question 2 (old).
(i) sin mx (NCERT)
(ii) e^2x (NCERT)
Solution:
(i) ∫ sin mx dx = – \(\frac{\cos m x}{m}\) + C

(ii) ∫ e^2x dx
= \(\frac{e^{2 x}}{2}\) + C
[∵ ∫ e^mx dx = \(\frac{e^{m x}}{m}\) + C]

Question 3.
(i) tan² (2x – 3) (NCERT)
(ii) 5 cosec² (2 – 7x)
Solution:
(i) ∫ tan² (2x – 3) dx
= ∫ [sec² (2x – 3) – 1] dx
= ∫ [sec² (2x – 3) dx – ∫ dx
= \(\frac{\tan (2 x-3)}{2}\) – x + c

(ii) ∫ 5 cosec² (2 – 7x) dx
= \(\frac{-5 \cot (2-7 x)}{-7}\) + C
= \(\frac{5}{7}\) cot (2 – 7x) + C

Question 4.
(i) \(\sqrt{e^x}\)
(ii) cosec (3x + 2) cot (3x + 2)
Solution:
(i) ∫ \(\sqrt{e^x}\) dx
= ∫ e^x/2 dx
= \(\frac{e^{x / 2}}{1 / 2}\) + C
= 2e^x/2 + C

(ii) ∫ cos (3x + 2) cot (3x + 2) dx
= – \(\frac{\ {cosec}(3 x+2)}{3}\) + C
[∵ ∫ cot θ cosec θ dθ = – cosec θ + C]

Evaluate the following (7 to 21) integrals :

Question 5.
(i) ∫ e^{2x + 3} dx
(ii) ∫ e^{(3a log x + e3x log a)} dx
Solution:
(i) ∫ e^{2x + 3} dx
= \(\frac{e^{2 x+3}}{2}\) + C
[∵ ∫ e^mx dx = \(\frac{e^{m x}}{m}\) + C]

(ii) ∫ e^{(3a log x + e3x log a)} dx
= ∫ (e^{log x3a} + e^{log a3x}) dx
[∵ a log b = log b^a
and e^{log x} = x]
= ∫ (x^3a + a) dx
= \(\frac{x^{3 a+1}}{3 a+1}+\frac{a^{3 x}}{3 \log a}\) + C

Question 6.
(i) ∫ \(\sqrt{1+cos x}\) dx
(ii) ∫ \(\sqrt{1+sin x}\) dx
Solution:
(i) ∫ \(\sqrt{1+cos x}\) dx
= ∫ \(\sqrt{2 \cos ^2 \frac{x}{2}}\) dx
= √2 ∫ cos \(\frac{x}{2}\) dx
= √2 \(\frac{\sin \frac{x}{2}}{\frac{1}{2}}\) + C
= 2√2 sin \(\frac{x}{2}\) + C

(ii) ∫ \(\sqrt{1+sin x}\) dx

Question 6 (old).
(ii) ∫ \(\sqrt{1-sin x}\) dx
Solution:

Question 7.
(i) ∫ sec² (7 – x) dx
(ii) ∫ a^{3x + 5} dx, a ≥ 0
Solution:
(i) ∫ sec² (7 – x) dx
= ∫ sec² t (- dt)
[put 7 – x = t
⇒ dx = – dt]
= – tan t + C
= – tan (7 – x) + C

(ii) ∫ a^{3x + 5} dx, a > 0
= \(\frac{a^{3 x+5}}{3(\log a)}\) + C
[∵ ∫ a^mx dx = \(\frac{a^{m x}}{m \log a}\) + C

Question 8.
(i) ∫ \(\frac{1-\tan ^2 x}{1+\tan ^2 x}\) dx
(ii) ∫ sec² x cosec² x dx
Solution:
(i) ∫ \(\frac{1-\tan ^2 x}{1+\tan ^2 x}\) dx
= ∫ cos 2x dx
= \(\frac{\sin 2 x}{2}\) + C

(ii) ∫ sec² x cosec² x dx
= ∫ \(\frac{\left(\sin ^2 x+\cos ^2 x\right) d x}{\sin ^2 x \cos ^2 x}\)
= ∫ \(\frac{1}{\cos ^2 x}\) dx + ∫ \(\frac{1}{\sin ^2 x}\) dx
= ∫ sec² x dx + ∫ cosec² x dx
= tan x – cot x + C

Question 9.
(i) ∫ \(\frac{x}{2 x-3}\) dx
(ii) ∫ \(\frac{x+2}{3 x^2+5 x-2}\) dx
Solution:
(i) Let I = ∫ \(\frac{x}{2 x-3}\) dx
= \(\frac{1}{2} \int\left[\frac{2 x-3+3}{2 x-3}\right]\) dx
= \(\frac{1}{2} \int\left[1+\frac{3}{2 x-3}\right]\) dx
= \(\frac{1}{2}\left[x-\frac{3 \log |2 x-3|}{2}\right]\) + C
= \(\frac{x}{2}\) – \(\frac{3}{4}\) log |2x – 3| + C

(ii) Let I = ∫ \(\frac{x+2}{3 x^2+5 x-2}\) dx
= ∫ \(\frac{(x+2) d x}{(x+2)(3 x-1)}\)
= ∫ \(\frac{d x}{3 x-1}\)
= \(\frac{\log |3 x-1|}{3}\) + C

Question 10.
(i) ∫ \(\frac{x}{(x+1)^2}\) dx (ISC 2010)
(ii) ∫ \(\frac{x}{(x+1)^2}\) dx (NCERT Exemplar)
Solution:
(i) ∫ \(\frac{x}{(x+1)^2}\) dx

(ii) ∫ \(\frac{x}{(x+1)^2}\) dx
= ∫ \(\left[\frac{x^2-1+3}{x+1}\right]\) dx
= ∫ [(x – 1) + \(\frac{3}{x+1}\)] dx
= \(\frac{x^{2}}{2}\) – x + 3 log |x + 1| + C

Question 11.
(i) ∫ \(\frac{x^3}{2 x+1}\) dx
(ii) ∫ \(\frac{x^3+4 x^2-3 x-2}{x+2}\) dx
Solution:
(i) ∫ \(\frac{x^3}{2 x+1}\) dx

(ii) ∫ \(\frac{x^3+4 x^2-3 x-2}{x+2}\) dx
= ∫ [(x² + 2x – 7) + \(\frac{12}{x+2}\)] dx
= \(\frac{x^3}{3}\) + x² – 7x + 12 log |x + 2| + C

Question 11 (old).
(i) ∫ \(\frac{1}{(2 x-3)^{3 / 2}}\) dx
(ii) ∫ \(\frac{x}{2 x-3}\) dx
Solution:
(i) ∫ \(\frac{1}{(2 x-3)^{3 / 2}}\) dx
= ∫ (2x – 3)^{– 3/2} dx
= \(\frac{(2 x-3)^{-\frac{3}{2}+1}}{\left(-\frac{3}{2}+1\right) \cdot 2}\)
= – \(\frac{1}{\sqrt{2 x-3}}\) + C

(ii) ∫ \(\frac{x}{2 x-3}\) dx
= \(\frac{1}{2} \int\left[\frac{2 x-3+3}{2 x-3}\right]\) dx
= \(\frac{1}{2} \int\left[1+\frac{3}{2 x-3}\right]\) dx
= \(\frac{1}{2}\left[x-\frac{3 \log |2 x-3|}{2}\right]\) + C
= \(\frac{x}{2}\) – \(\frac{3}{4}\) log |2x – 3| + C

Question 12.
(i) ∫ \(\frac{1}{\sqrt{x+1}+\sqrt{x+2}}\) dx
(ii) ∫ \(\frac{1}{\sqrt{2 x+3}-\sqrt{2 x}}\) dx
Solution:
(i) ∫ \(\frac{1}{\sqrt{x+1}+\sqrt{x+2}}\) dx
= ∫ \(\frac{\sqrt{x+1}-\sqrt{x+2}}{(x+1)-(x+2)}\) dx
= – \(\left[\frac{2}{3}(x+1)^{3 / 2}-\frac{2}{3}(x+2)^{3 / 2}\right]\) + C

(ii) ∫ \(\frac{1}{\sqrt{2 x+3}-\sqrt{2 x}}\) dx

Question 13.
(i) ∫ \(\frac{x+1}{\sqrt{2 x-1}}\) dx
(ii) ∫ x \(\sqrt{3 x-2}\) dx
Solution:
(i) ∫ \(\frac{x+1}{\sqrt{2 x-1}}\) dx

(ii) ∫ x \(\sqrt{3 x-2}\) dx
= \(\frac{1}{3}\) ∫ (3x – 2 + 2) \(\sqrt{3 x-2}\) dx
= \(\frac{1}{3}\) ∫ (3x – 2)^3/2 dx + \(\frac{2}{3}\) ∫ (3x – 2)^1/2 dx
= \(\frac{1}{3} \frac{(3 x-2)^{5 / 2}}{\frac{5}{2} \times 3}+\frac{2}{3} \frac{(3 x-2)^{3 / 2}}{\frac{3}{2} \times 3}\) + C
= \(\frac{2}{45}\) (3x – 2)^5/2 + \(\frac{4}{27}\) (3x – 2)^3/2 + C

Question 14.
(i) ∫ cos² x dx (NCERT)
(ii) ∫ sin⁴ dx
Solution:
(i) ∫ cos² x dx
= ∫ \(\frac{1+\cos 2 x}{2}\) dx
= \(\frac{1}{2}\left[x+\frac{\sin 2 x}{2}\right]\) + C

(ii) ∫ sin⁴ dx
= ∫ \(\left[\frac{1-\cos 2 x}{2}\right]^2\) dx
= \(\frac{1}{4}\) ∫ [1 + cos² 2x – 2 cos 2x] dx
= \(\frac{1}{4}\) ∫ [1 + \(\frac{1+\cos 4 x}{2}\) – 2 cos 2x] dx
= \(\frac{1}{8}\) ∫ [3 + cos 4x – 4 cos 2x] dx
= \(\frac{1}{8}\) [3x + \(\frac{\sin 4 x}{4}\) – 2 sin 2x] + C

Question 15.
(i) ∫ sin⁵ (2x + 5) dx (NCERT)
(ii) ∫ cos⁴ 2x dx (NCERT)
Solution:
(i) ∫ sin⁵ (2x + 5) dx
= ∫ \(\frac{1-\cos (4 x+10)}{2}\) dx
[∵ sin² θ = \(\frac{1-\cos 2 \theta}{2}\)]
= ∫ [x – \(\frac{\sin (4 x+10)}{4}\)] + C

(ii) ∫ cos⁴ 2x dx
= ∫ (cos² 2x)² dx
= ∫ \(\left[\frac{1+\cos 4 x}{2}\right]^2\) dx
= \(\frac{1}{4}\) ∫ [1 + 2 cos 4x + cos² 4x] dx
= \(\frac{1}{4}\) ∫ [1 + 2 cos 4x + \(\frac{1+\cos 8 x}{2}\)] dx
= \(\frac{1}{8}\) ∫ [3 + 4 cos 4x + cos 8x] dx
= \(\frac{1}{8}\) [3x + 4 \(\frac{\sin 4 x}{4}\) + \(\frac{\sin 8 x}{8}\)] + c
= \(\frac{3 x}{8}+\frac{\sin 4 x}{8}+\frac{\sin 8 x}{64}\) + c

Question 16.
(i) ∫ \(\frac{1}{1-\sin \frac{x}{2}}\) dx
(ii) ∫ sin x \(\sqrt{1-cos 2x}\) dx
Solution:
(i) ∫ \(\frac{1}{1-\sin \frac{x}{2}}\) dx

(ii) ∫ sin x \(\sqrt{1-cos 2x}\) dx
= ∫ sin x \(\sqrt{2 \sin ^2 x}\) dx
= √2 ∫ sin² x dx
= \(\sqrt{2} \int \frac{1-\cos 2 x}{2}\) dx
= \(\frac{1}{\sqrt{2}}\left[x-\frac{\sin 2 x}{2}\right]\) + C
= \(\frac{x}{\sqrt{2}}\) – \(\frac{1}{2 \sqrt{2}}\) sin 2x + C.

Question 17.
(i) ∫ cos 3x cos 5x dx
(ii) ∫ sin 4x sin 8x dx
(iii) ∫ sin 4x cos 3x dx
(iv) ∫ sin 2x cos 3x dx (NCERT)
Solution:
(i) ∫ cos 3x cos 5x dx
= \(\frac{1}{2}\) ∫ (2 cos 5x cos 3x) dx
= \(\frac{1}{2}\) ∫ [cos 8x + cos 2x] dx
= \(\frac{1}{2}\left[\frac{\sin 8 x}{8}+\frac{\sin 2 x}{2}\right]\) + C

(ii) ∫ sin 4x sin 8x dx
= \(\frac{1}{2}\) ∫ (2 sin 8x sin 4x) dx
= \(\frac{1}{2}\) ∫ [cos 4x – cos 12 x] dx
[∵ 2 sin A sin B = cos (A – B) – cos (A + B)]
= \(\frac{1}{2}\left[\frac{\sin 4 x}{7}-\frac{\sin 12 x}{12}\right]\) + C

(iii) ∫ sin 4x cos 3x dx
= \(\frac{1}{2}\) ∫ 2 sin 4x cos 3x dx
= \(\frac{1}{2}\) ∫ (sin 7x + sin x) dx
[∵ 2 sin A cos B = sin (A + B) + sin (A – B)]
= \(\frac{1}{2}\left[-\frac{\cos 7 x}{7}-\cos x\right]\) + C
= \(\frac{1}{14}\) [- cos 7x – 7 cos x] + C

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

Question 18.
(i) ∫ \(\frac{\sin 4 x}{\cos x}\) dx
(ii) ∫ sin x sin 2x sin 3x dx
Solution:
(i) ∫ \(\frac{\sin 4 x}{\cos x}\) dx
= ∫ \(\frac{2 \sin 2 x \cos 2 x}{\cos x}\) dx
= ∫ \(\frac{2 \times 2 \sin x \cos x \cos 2 x d x}{\cos x}\) dx
= 2 ∫ 2 cos 2x sin x dx
= 2 ∫ [sin 3x – sin x] dx
= 2 [- \(\frac{\cos 3 x}{3}\) + cos x]
= \(\frac{2}{3}\) (3 cos x – cos 3x)

(ii) ∫ sin x sin 2x sin 3x dx
= \(\frac{1}{2}\) ∫ (2 sin 3x sin 2x) sin x dx
= \(\frac{1}{2}\) ∫ [cos x – cos 5x] sin x dx
= \(\frac{1}{4}\) ∫ 2 sin x cos x dx – \(\frac{1}{4}\) ∫ 2 cos 5x sin x dx
= \(\frac{1}{4}\left(-\frac{\cos 2 x}{2}\right)\) – \(\frac{1}{4}\) ∫ [sin 6x – sin 4x] dx
= \(-\frac{\cos 2 x}{8}+\frac{\cos 6 x}{24}-\frac{\cos 4 x}{16}\) + C

Question 19.
(i) ∫ sin⁶ x dx
(ii) ∫ sin⁴ x cos² x dx.
Solution:
(i) ∫ sin⁶ x dx
= ∫ (sin² x)³ dx
= ∫ \(\left[\frac{1-\cos 2 x}{2}\right]^3\) dx
= \(\frac{1}{8}\) ∫ [1 – \(\frac{1}{4}\) (cos 6x + 3 cos 2x) – 3 cos 2x + 3 \(\left(\frac{1+\cos 4 x}{2}\right)\)]
[∵ cos³ θ = \(\frac{1}{4}\) [cos 3θ + 3 cos θ]
and cos² θ = \(\frac{1+\cos 2 \theta}{2}\)]
= \(\frac{1}{8 \times 4}\) ∫ [4 – cos 6x – 3 cos 2x – 12 cos 2x + 6 + 6 cos 4x] dx
= \(\frac{1}{32}\) ∫ [10 + 6 cos 4x – 15 cos 2x – cos 6x]
= \(\frac{1}{32}\left[10 x+\frac{6 \sin 4 x}{4}-\frac{15 \sin 2 x}{2}-\frac{\sin 6 x}{6}\right]\) + C
= \(\frac{1}{192}\) [60x + 9 sin 4x – 45 sin 2x – sin 6x] + C

(ii) ∫ sin⁴ x cos² x dx.
= \(\frac{1}{4}\) ∫ sin² x (4 sin² x cos² x) dx
= \(\frac{1}{4}\) ∫ \(\left[\frac{1-\cos 2 x}{2}\right]\) [2 sin x cos x]² dx
= \(\frac{1}{4} \int\left[\frac{1-\cos 2 x}{2}\right]\left[\frac{1-\cos 4 x}{2}\right]\) dx
= \(\frac{1}{16}\) ∫ [1 – cos 4x – cos 2x + (2 cos 4x cos 2x) \(\frac{1}{2}\)] dx
= \(\frac{1}{16}\) ∫ [1 – cos 4x – cos 2x – \(\frac{1}{2}\) (cos 6x + cos 2x)] dx
= \(\frac{1}{32}\) ∫ [2 – 2 cos 4x – cos 2x + cos 6x] dx
= \(\frac{1}{32}\left[2 x-\frac{2 \sin 4 x}{4}-\frac{\sin 2 x}{2}+\frac{\sin 6 x}{6}\right]\) + C
= \(\frac{1}{192}\) [12x – 3 sin 4x – 3 sin 2x + sin 6x] + C

Question 20.
Evaluate ∫ cos mx cos nx dx, where m, n are positive integers, m ≠ n. What happens if m n?
Solution:
Case – I:
When m ≠ n
∴ ∫ cos mx cos nx dx = ∫ 2 cos mx cos nx dx
= \(\frac{1}{2}\) ∫ [cos (m + n) x + cos (m – n) dx
= \(\frac{1}{2}\left[\frac{\sin (m+n) x}{m+n}+\frac{\sin (m-n) x}{m-n}\right]\) + C

Case-II:
When m = n
∫ cos mx cos nx dx = ∫ cos² mx dx
= ∫ \(\left[\frac{1+\cos 2 m x}{2}\right]\) dx
= \(\frac{1}{2}\left[x+\frac{\sin 2 m x}{2 m}\right]\) + C

Continuous practice using ISC Maths Class 12 Solutions Chapter 8 Integrals Ex 8.2 can lead to a stronger grasp of mathematical concepts.

ML Aggarwal Class 12 Maths Solutions Section A Chapter 8 Integrals Ex 8.2

Very short answer type questions (1 to 14) :

Evaluate the following (1 to 20) integrals :

Question 1.
(i) ∫ (ax² + bx + c) dx (NCERT)
(ii) ∫ (x^2/3 + 1) dx. (NCFRT)
Solution:
(i) ∫ (ax² + bx + c) dx
= a ∫ x² dx + b ∫ x dx + ∫ c dx
= \(\frac{a x^3}{3}+\frac{b x^2}{2}\) + cx + C
[∵ ∫ xⁿ dx = \(\frac{x^{n+1}}{n+1}\) + C; n ≠ 1]

(ii) ∫ (x^2/3 + 1) dx
= ∫ x^2/3 dx + ∫ 1 dx
= \(\frac{x^{\frac{2}{3}+1}}{\frac{2}{3}+1}\) + x + C
= \(\frac{3}{5}\) x^5/3 + x + C

Question 2.
(i) ∫ (\(\sqrt{3x}\) + \(\frac{1}{\sqrt{x}}\)) dx
(ii) ∫ \(\left(\frac{2 a}{\sqrt{x}}-\frac{b}{x^2}+3 c \sqrt[3]{x^2}\right)\) dx (NCERT Exampler)
Solution:
(i) ∫ (\(\sqrt{3x}\) + \(\frac{1}{\sqrt{x}}\)) dx
= ∫ \(\sqrt{3x}\) dx + ∫ \(\frac{1}{\sqrt{x}}\)) dx
= √3 ∫ x^{\(\frac{1}{2}\)} dx + ∫ x^{\(-\frac{1}{2}\)} dx
= √3 \(\frac{x^{\frac{1}{2}+1}}{\left(\frac{1}{2}+1\right)}+\frac{x^{-\frac{1}{2}+1}}{-\frac{1}{2}+1}\) + C
= \(\frac{2}{\sqrt{3}} x^{3 / 2}\) + 2√x + C

(ii) ∫ \(\left(\frac{2 a}{\sqrt{x}}-\frac{b}{x^2}+3 c \sqrt[3]{x^2}\right)\) dx
= ∫ \(\frac{2 a}{\sqrt{x}}\) dx – b ∫ x^{– 2} dx + 3c ∫ (x²)^1/3 dx
= 2a ∫ x^{\(-\frac{1}{2}\)} dx – b ∫ x^{– 2} dx + 3c ∫ x^2/3 dx
= 2a \(\frac{x^{-\frac{1}{2}+1}}{\left(-\frac{1}{2}+1\right)}-b \frac{x^{-2+1}}{(-2+1)}+3 c \frac{x^{\frac{2}{3}+1}}{\left(\frac{2}{3}+1\right)}\) + C
= 4a√x + \(\frac{b}{x}\) + \(\frac{9 c}{5}\) x^5/3 + C.

Question 3.
(i) ∫ \(\frac{2 \cos x}{3 \sin ^2 x}\) dx
(ii) ∫ (2x² 3 sin x + 5√x) dx (NCERT)
Solution:
(i) ∫ \(\frac{2 \cos x}{3 \sin ^2 x}\) dx
= \(\frac{2}{3}\) ∫ cot x cosec x dx
= – \(\frac{2}{3}\) cosec x + C

(ii) ∫ (2x² – 3 sin x + 5√x) dx
= 2 ∫ x² dx – 3 ∫ sin x dx + 5 ∫ √x dx
= 2 \(\frac{x^{3}}{3}\) – 3 (- cos x) + 5 \(\frac{x^{3 / 2}}{3 / 2}\) dx
= \(\frac{2}{3}\) x³ + 3 cos x + \(\frac{10}{3}\) x^3/2 + C

Question 4.
(i) ∫ (2x – 3 cos x + e^x) dx (NCERT)
(ii) ∫ x² (1 – \(\frac{1}{x^2}\)) dx (NCERT)
Solution:
(i) ∫ (2x – 3 cos x + e^x) dx
= ∫ 2x 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

(ii) ∫ x² (1 – \(\frac{1}{x^2}\)) dx
= ∫ x² dx – ∫ 1 dx
= \(\frac{x^{3}}{3}\) – x + C

Question 5.
(i) ∫ (x^3/2 + 2 e^x – \(\frac{1}{x}\)) dx (NCERT)
(ii) ∫ (√x + \(\frac{1}{\sqrt{x}}\)) dx (NCERT)
Solution:
(i) ∫ (x^3/2 + 2 e^x – \(\frac{1}{x}\)) dx
= \(\frac{x^{\frac{3}{2}+1}}{\left(\frac{3}{2}+1\right)}\) + 2 e^x – log |x| + C
= \(\frac{2}{5}\) x^5/2 + 2 e^x – log |x| + C
[∵ ∫ \(\frac{d x}{x}\) = log |x| + C ;
∫ xⁿ dx = \(\frac{x^{n+1}}{n+1}\) ; n ≠ – 1]

(ii) ∫ (√x + \(\frac{1}{\sqrt{x}}\)) dx
= ∫ √x dx + ∫ \(\frac{d x}{\sqrt{x}}\)
= \(\frac{x^{\frac{1}{2}+1}}{\left(\frac{1}{2}+1\right)}+\frac{x^{-\frac{1}{2}+1}}{-\frac{1}{2}+1}\) + C
= \(\frac{2}{3}\) x^3/2 + 2 √x + C

Question 6.
(i) ∫ (1 – x) √x dx
(ii) ∫ √x (3x² + 2x + 3) dx (NCERT)
Solution:
(i) ∫ (1 – x) √x dx
= ∫ √x dx + ∫ x^3/2 dx
= \(\frac{x^{\frac{1}{2}+1}}{\left(\frac{1}{2}+1\right)}-\frac{x^{\frac{3}{2}+1}}{\left(\frac{3}{2}+1\right)}\) + C
= \(\frac{2}{3} x^{3 / 2}-\frac{2}{5} x^{5 / 2}\) + C

(ii) ∫ √x (3x² + 2x + 3) dx
= ∫ 3x^5/2 dx + ∫ 2x^3/2 dx + ∫ 3x^1/2 dx
= \(\frac{3 x^{\frac{5}{2}+1}}{\frac{5}{2}+1}+\frac{2 x^{\frac{3}{2}+1}}{\frac{3}{2}+1}+\frac{3 x^{\frac{1}{2}+1}}{\frac{1}{2}+1}\) + C
= \(\frac{6}{7} x^{7 / 2}+\frac{4}{5} x^{5 / 2}\) + 2x^3/2 + C

Question 7.
(i) ∫ \(\frac{x^3+5 x^2+4 x+1}{x^2}\) dx (ISC 2018)
(ii) ∫ \(\frac{x^3+3 x+4}{\sqrt{x}}\) dx (NCERT)
Solution:
(i) ∫ \(\frac{x^3+5 x^2+4 x+1}{x^2}\) dx
= ∫ \(\left[\frac{x^3}{x^2}+\frac{5 x^2}{x^2}+\frac{4 x}{x^2}+\frac{1}{x^2}\right]\) dx
= ∫ x dx + ∫ 5 dx + 4 ∫ \(\frac{d x}{x}\) + ∫ x^-2 dx
= \(\frac{x^{2}}{2}\) + 5x + 4 log |x| + \(\frac{x^{-2+1}}{-2+1}\) + C
= \(\frac{x^{2}}{2}\) + 5x + 4 log |x| – \(\frac{1}{x}\) + C

(ii) ∫ \(\frac{x^3+3 x+4}{\sqrt{x}}\) dx
= ∫ x^5/2 dx + 3 ∫ √x dx + 4 ∫ x^{\(-\frac{1}{2}\)} dx
= \(\frac{x^{7 / 2}}{\frac{7}{2}}+\frac{3 x^{3 / 2}}{\frac{3}{2}}+\frac{4 x^{-\frac{1}{2}+1}}{\left(-\frac{1}{2}+1\right)}\) + C
= \(\frac{2}{7} x^{\frac{7}{2}}\) + 2x^3/2 + 8√x + C

Question 8.
(i) ∫ x² (2x – \(\frac{1}{x}\))² dx
(ii) ∫ (x + √x)³ dx
Solution:
(i) ∫ x² (2x – \(\frac{1}{x}\))² dx
= ∫ x² [4x² + \(\frac{1}{x^{2}}\) – 4] dx
= ∫ 4x⁴ dx + ∫ dx – 4 ∫ x² dx
= \(\frac{4 x^5}{5}+x-\frac{4 x^3}{3}\) + C

(ii) ∫ (x + √x)³ dx
= ∫ [x³ + x^3/2 + 3x^3/2 (x + √x)] dx
= ∫ x³ dx + ∫ x^3/2 dx + 3 ∫ x^5/2 dx + 3 ∫ x² dx
= \(\frac{x^4}{4}+\frac{2 x^{5 / 2}}{5}+\frac{3 x^{7 / 2}}{\frac{7}{2}}+\frac{3 x^3}{3}\) + C
= \(\frac{x^4}{4}+\frac{2}{5} x^{5 / 2}+\frac{6}{7} x^{7 / 2}\) + x³ + C

Question 9.
(i) ∫ \(\frac{(x+a)^2}{\sqrt{x}}\) dx
(ii) ∫ \(\frac{x^3-x^2+x-1}{x-1}\) dx
Solution:
(i) ∫ \(\frac{(x+a)^2}{\sqrt{x}}\) dx
= ∫ \(\frac{x^2+a^2+2 a x}{\sqrt{x}}\) dx
= ∫ x^3/2 dx + a² ∫ x^{– \(\frac{1}{2}\)} dx + 2a ∫ √x dx
= \(\frac{2}{5}\) x^5/2 + 2a²√x + \(\frac{4 a}{3}\) x^3/2 + C

(ii) ∫ \(\frac{x^3-x^2+x-1}{x-1}\) dx
= ∫ \(\frac{x^3-x^2+x-1}{x-1}\) dx
= ∫ \(\frac{(x-1)\left(x^2+1\right)}{(x-1)}\) dx
= ∫ (x² + 1) dx
= \(\frac{x^{3}}{3}\) + x + C

Question 10.
(i) ∫ \(\frac{1-\sin x}{\cos ^2 x}\) dx (NCERT)
(ii) ∫ \(\frac{e^{6 \log x}-e^{5 \log x}}{e^{4 \log x}-e^{3 \log x}}\) dx (NCERT Exemplar)
Solution:
(i) ∫ \(\frac{1-\sin x}{\cos ^2 x}\) dx
= \(\int \frac{1}{\cos ^2 x} d x-\int \frac{\sin x}{\cos ^2 x} d x\)
= ∫ sec² x dx – ∫ tan x sec x dx
= tan x – sec x + C

(ii) Let I = ∫ \(\frac{e^{6 \log x}-e^{5 \log x}}{e^{4 \log x}-e^{3 \log x}}\) dx
= ∫ \(\left(\frac{e^{\log x^6}-e^{\log x^5}}{e^{\log x^4}-e^{\log x^3}}\right)\) dx
= ∫ \(\frac{x^6-x^5}{x^4-x^3}\) dx
= = ∫ \(\frac{x^5(x-1)}{x^3(x-1)}\) dx
= \(\frac{x^{3}}{3}\) + C

Question 11.
(i) ∫ sec x (sec x + tan x) dx
(ii) ∫ \(\frac{\sec ^2 x}{\ {cosec}^2 x}\) dx
Solution:
(i) ∫ sec x (sec x + tan x) dx
= ∫ sec² x dx + ∫ sec x tan x dx
= tan x + sec x + C

(ii) ∫ \(\frac{\sec ^2 x}{\ {cosec}^2 x}\) dx
= ∫ \(\) × sin² x dx
= ∫ tan² x dx
= ∫ (sec² x – 1) dx
= tan x – x + C

Question 12.
(i) ∫ \(\left(\frac{\cot x}{\sin x}-\frac{\tan x}{\cos x}-3 \tan ^2 x\right)\) dx
(ii) ∫ \(\frac{2-3 \sin x}{\cos ^2 x}\) dx
Solution:
(i) ∫ \(\left(\frac{\cot x}{\sin x}-\frac{\tan x}{\cos x}-3 \tan ^2 x\right)\) dx
= ∫ cot x cosec x dx – ∫ tan x sec x dx – 3 ∫ (sec² x – 1) dx
= – cosec x – sec x – 3 tan x + 3x + C
[∵ ∫ cot x cosec x dx = – cosec x + C
and ∫ tan x sec x dx = sec x + C]

(ii) ∫ \(\frac{2-3 \sin x}{\cos ^2 x}\) dx
= ∫ 2 sec² x dx – 3 ∫ tan x sec x dx
= 2 tan x – 3 sec x + C

Question 13.
(i) ∫ \(\frac{1}{1+\cos x}\) dx
(ii) ∫ \(\frac{1}{1+\sin x}\) dx
Solution:
(i) ∫ \(\frac{1}{1+\cos x}\) dx
= ∫ \(\frac{1}{1+\cos x} \times \frac{1-\cos x}{1-\cos x}\)
= ∫ \(\frac{1-\cos x}{\sin ^2 x}\) dx
= ∫ cosec² dx – ∫ cot x cosec x dx
= – cot x + cosec x + C

(ii) ∫ \(\frac{1}{1+\sin x}\) dx
= ∫ \(\frac{(1+\sin x) d x}{1-\sin ^2 x}\)
= ∫ \(\frac{(1+\sin x)}{\cos ^2 x}\) dx
= ∫ sec^{2</sup x dx + ∫ tan x sec x dx
= tan x + sec x + C}

Question 14.
(i) ∫ \(\frac{1}{1+\ {cosec} x}\) dx
(ii) ∫ \(\frac{\sin ^2 x}{1+\cos x}\) dx (NCERT)
Solution:
(i) ∫ \(\frac{1}{1+\ {cosec} x}\) dx

(ii) ∫ \(\frac{\sin ^2 x}{1+\cos x}\) dx
= ∫ \(\frac{1-\cos ^2 x}{1+\cos x}\) dx
= ∫ \(\frac{(1-\cos x)(1+\cos x)}{1+\cos x}\) dx
= ∫ (1 – cos x) dx
= ∫ dx – ∫ cos x dx
= x – sin x + c

Question 15.
(i) ∫ (√x – sin \(\frac{x}{2}\) cos \(\frac{x}{2}\) + 5) dx
(ii) ∫ \(\frac{\cos 2 x}{\sin ^2 x \cos ^2 x}\) dx
Solution:
(i) ∫ (√x – sin \(\frac{x}{2}\) cos \(\frac{x}{2}\) + 5) dx
= ∫ √x dx – ∫ \(\frac{x}{2}\) cos \(\frac{x}{2}\) dx + ∫ 5 dx
= \(\frac{x^{\frac{1}{2}+1}}{\frac{1}{2}+1}\) – ∫ \(\frac{1}{2}\) sin x dx + 5x
[∵ sin 2θ = 2 sin θ cos θ]
= \(\frac{2}{3}\) x^3/2 + \(\frac{1}{2}\) cos x + 5x + C

(ii) ∫ \(\frac{\cos 2 x}{\sin ^2 x \cos ^2 x}\) dx
= \(\int \frac{\cos ^2 x-\sin ^2 x}{\sin ^2 x \cos ^2 x}\) dx
= \(\int \frac{1}{\sin ^2 x} d x-\int \frac{1}{\cos ^2 x} d x\)
= ∫ cosec² x dx – ∫ sec² x dx
= – cot x – tan x + C

Question 16.
(i) ∫ \(\frac{\sec x}{\sec x+\tan x}\) dx
(ii) ∫ \(\frac{\ {cosec} x}{\ {cosec} x-\cot x}\) dx
Solution:
(i) ∫ \(\frac{\sec x}{\sec x+\tan x}\) dx
= ∫ \(\frac{\sec x(\sec x-\tan x)}{\sec ^2 x-\tan ^2 x}\) dx
= ∫ sec x (sec x – tan x) dx
[∵ sec² θ – tan² θ = 1]
= ∫ sec² x dx – ∫ sec x tan x dx + C
= tan x – sec x + C

(ii) ∫ \(\frac{\ {cosec} x}{\ {cosec} x-\cot x}\) dx
= ∫ \(\frac{\ {cosec} x(\ {cosec} x+\cot x) d x}{(\ {cosec} x-\cot x)(\ {cosec} x+\cot x)}\)
= ∫ \(\frac{\ {cosec}^2 x+\cot x \ {cosec} x d x}{\ {cosec}^2 x-\cot ^2 x}\)
[∵ 1 + cot² θ = cosec² θ ]
= ∫ cosec² x dx + ∫ cot x cosec x dx
= – cot x – cosec x + C

Question 17.
(i) ∫ \(\frac{x^6+1}{x^2+1}\) dx
(ii) ∫ \(\frac{6 \sin ^3 x+5 \cos ^3 x}{\sin ^2 x \cos ^2 x}\) dx
Solution:
(i) ∫ \(\frac{x^6+1}{x^2+1}\) dx
= ∫ \(\frac{\left(x^2+1\right)\left(x^4-x^2+1\right)}{x^2+1}\) dx
= ∫ (x⁴ – x² + 1) dx
= ∫ x⁴ dx + ∫ x² dx + ∫ 1 dx
= \(\frac{x^5}{5}-\frac{x^3}{3}\) + x + c
[∵ ∫ xⁿ dx = \(\frac{x^{n+1}}{n+1}\) + c ; n ≠ – 1]

(ii) ∫ \(\frac{6 \sin ^3 x+5 \cos ^3 x}{\sin ^2 x \cos ^2 x}\) dx
= 6 ∫ \(\frac{\sin ^3 x}{\sin ^2 x \cos ^2 x}\) dx + 5 ∫ \(\frac{\cos ^3 x d x}{\cos ^2 x \sin ^2 x}\)
= 6 ∫ tan x sec x dx + 5 ∫ cot x cosec x dx
= 6 sec x – 5 cosec x + C

Question 18.
(i) ∫ \(\sqrt{1+\cos 2 x}\) dx
(ii) ∫ \(\sqrt{1-\cos 2 x}\) dx
Solution:
(i) ∫ \(\sqrt{1+\cos 2 x}\) dx
= ∫ \(\sqrt{2 \cos ^2 x}\) dx
= √2 ∫ cos x dx
= √2 sin x + C

(ii) ∫ \(\sqrt{1-\cos 2 x}\) dx
= ∫ \(\sqrt{2 \sin ^2 x}\) dx
= √2 ∫ sin x dx
= – √2 cos x + C

Question 19.
(i) ∫ cot^-1 \(\left(\frac{\sin 2 x}{1-\cos 2 x}\right)\) dx
(ii) ∫ tan^-1 \(\left(\sqrt{\frac{1-\cos 2 x}{1+\cos 2 x}}\right)\) dx
Solution:
(i) ∫ cot^-1 \(\left(\frac{\sin 2 x}{1-\cos 2 x}\right)\) dx
= ∫ cot^-1 \(\left(\frac{2 \sin x \cos x}{2 \sin ^2 x}\right)\) dx
= ∫ cot^-1 (cot x) dx
= ∫ x dx
= \(\frac{x^{2}}{2}\) + C

(ii) ∫ tan^-1 \(\left(\sqrt{\frac{1-\cos 2 x}{1+\cos 2 x}}\right)\) dx
= ∫ tan^-1 \(\sqrt{\frac{2 \sin ^2 x}{2 \cos ^2 x}}\) dx
= ∫ tan^-1 (tan x) dx
= ∫ x dx
= \(\frac{x^{2}}{2}\) + C

Question 20.
(i) ∫ tan^-1 (cosec x – cot x) dx
(ii) ∫ cos^-1 \(\left(\frac{1-\tan ^2 x}{1+\tan ^2 x}\right)\) dx
Solution:
(i) ∫ tan^-1 (cosec x – cot x) dx

(ii) ∫ cos^-1 \(\left(\frac{1-\tan ^2 x}{1+\tan ^2 x}\right)\) dx
= ∫ cos^-1 (cos 2x) dx
[∵ cos 2θ = \(\frac{1-\tan ^2 \theta}{1+\tan ^2 \theta}\)]
= ∫ 2x dx
= x² + c

Question 21.
If f’(x) = 4x³ – 6 and f (0) = 3, find f (x).
Solution:
Given f’ (x) = 4x³ – 6
On integrating both sides w.r.t. x, we get
∫ f’(x) dx = ∫ (4x³ – 6) dx
f(x) = x⁴ – 6x + C ………..(1)
given f(0) = 3
i.e. when x = 0 ;
f(x) = 3
∴ from (1) ;
3 = c
Thus, from (1) ;
f(x) = x⁴ – 6x + 3.

Effective ISC Mathematics Class 12 Solutions Chapter 8 Integrals Ex 8.1 can help bridge the gap between theory and application.

ML Aggarwal Class 12 Maths Solutions Section A Chapter 8 Integrals Ex 8.1

Very short answer type questions :

Evaluate the following integrals :

Question 1.
(i) ∫ x⁵ dx
(i) ∫ \(\frac{1}{x^3}\) dx
(iii) ∫ \(\frac{1}{\sqrt{y}}\) dy
Solution:
(i) ∫ x⁵ dx = \(\frac{x^{5+1}}{5+1}\) + C
[∵ ∫ x⁵ dx = \(\frac{x^{n+1}}{n+1}\) + C ; n ≠ 1]
= \(\frac{x^6}{6}\) + C

(ii) ∫ \(\frac{1}{x^3}\) dx
= ∫ x^{– 3} dx
= \(\frac{x^{-3+1}}{-3+1}\) + C
= \(\frac{x^{-2}}{-2}\) + C
= – \(\frac{1}{2 x^2}\) + C

(iii) ∫ \(\frac{1}{\sqrt{y}}\) dy
= ∫ y^{\(\frac{-1}{2}\)} dy
= \(\frac{y^{-\frac{1}{2}+1}}{-\frac{1}{2}+1}\) + C
= 2√y + C.

Question 2.
(i) ∫ \(\frac{1}{\sqrt[3]{x}}\) dx
(ii) ∫ sin² x cosec² x dx
(iii) ∫ sin x sec² x dx
Solution:
(i) ∫ \(\frac{1}{\sqrt[3]{x}}\) dx
= ∫ x^{– 1/3} dx
= \(\frac{x^{-\frac{1}{3}+1}}{-\frac{1}{3}+1}\) + C
= \(\frac{3}{2}\) x^2/3 + C

(ii) ∫ sin² x cosec² x dx
= ∫ sin² x × \(\frac{1}{\sin ^2 x}\) dx
= ∫ 1 dx = x + C

(iii) ∫ sin x sec² x dx
= ∫ \(\frac{\sin x}{\cos ^2 x}\) dx
= ∫ tan x sec x dx
= sec x + C
[∵ ∫ tan x sec x dx = sec x + C]

Question 3.
(i) ∫ 5^x dx
(ii) ∫ 7^x e^x dx
(iii) ∫ \(\frac{4^x}{9^x}\) dx
Solution:
(i) ∫ 5^x dx
= \(\frac{5^x}{\log _e 5}\) + C
[∵ ∫ a^x dx
= \(\frac{a^x}{\log _e a}\) + C

(ii) ∫ 7^x e^x dx
= ∫ (7e)^x dx
= \(\frac{(7 e)^x}{\log 7 e}\) + C

(iii) ∫ \(\frac{4^x}{9^x}\) dx
= ∫ \(\left(\frac{4}{9}\right)^x\) dx
= \(\frac{\left(\frac{4}{9}\right)^x}{\log \frac{4}{9}}\) dx + C

Question 4.
(i) ∫ \(\frac{\sin x}{1-\sin ^2 x}\) dx
(ii) ∫ \(\frac{\cos 2 x+2 \sin ^2 x}{\cos ^2 x}\) dx (NCERT)
(iii) ∫ \(\sqrt{\frac{1}{2}(1+\cos 2 x)}\) dx
Solution:
(i) ∫ \(\frac{\sin x}{1-\sin ^2 x}\) dx
= ∫ \(\frac{\sin x}{\cos ^2 x}\) dx
= ∫ tan x sec x dx
= sec x + C

(ii) ∫ \(\frac{\cos 2 x+2 \sin ^2 x}{\cos ^2 x}\) dx
= ∫ \(\frac{1-2 \sin ^2 x+2 \sin ^2 x}{\cos ^2 x}\) dx
[∵ cos 2θ = 1 – 2 sin² θ]
= ∫ \(\frac{1}{\cos ^2 x}\) dx
= ∫ sec² x dx = tan x
[∵ ∫ sec² x dx = tan x + C]

(iii) ∫ \(\sqrt{\frac{1}{2}(1+\cos 2 x)}\) dx
= ∫ \(\sqrt{\frac{1}{2} \times 2 \cos ^2 x}\) dx
= ∫ cos x dx
= sin x + C

Question 5.
(i) ∫ e^{2 log x} x^{– 3} dx
(ii) ∫ 5^{2 log₅ x} dx
(iii) ∫ 2^{log₄ x} dx
Solution:
(i) ∫ e^{2 log x} x^{– 3} dx
= ∫ e^{log x2} x^{– 3} dx
= ∫ x² . x^{– 3} dx
= ∫ x^{– 1} dx
= ∫ \(\frac{1}{x}\) dx
= log |x| + C
[∵ ∫ \(\frac{1}{x}\) dx = log |x| + C]

(ii) ∫ 5^{2 log₅ x} dx
= ∫ 5^{log₅ x2} dx
= ∫ x² dx
= \(\frac{x^{3}}{3}\) + C

(iii) ∫ 2^{log₄ x} dx
= ∫ 2^{\(\frac{\log x}{2 \log 2}\)} dx
= ∫ \(2^{\frac{1}{2} \log _2 x}\) dx
= ∫ 2^{log₂ x1/2} dx
= ∫ x^{\(\frac{1}{2}\)} dx
= \(\frac{x^{\frac{1}{2}+1}}{\left(\frac{1}{2}+1\right)}\) + C
= \(\frac{2}{3}\) x^{\(\frac{3}{2}\)} + C
[∵ ∫ xⁿ dx = \(\frac{x^{n+1}}{n+1}\) + C]

Students can cross-reference their work with Understanding ISC Mathematics Class 12 Solutions Chapter 7 Applications of Derivatives MCQs to ensure accuracy.

ML Aggarwal Class 12 Maths Solutions Section A Chapter 7 Applications of Derivatives MCQ’s

Choose the correct answer from the given four options in questions (1 to 47):

Question 1.
If the radius of a circle is increasing at the rate of 2 cm / sec, then the area of the circle when its radius is 20 cm is increasing at the rate of
(a) 80 π m² / sec
(b) 80 m² / sec
(c) 80 π cm² /sec
(d) 80 cm² /sec
Solution:
(c) 80 π cm² /sec

Let r be the radius of circle at any time t
Then A = area of circle = πr²
∴ \(\frac{d A}{d t}\) = 2 cm / sec
Given \(\frac{d r}{d t}\) = 2 cm/sec ;
r = 20 cm
∴ \(\frac{d A}{d t}\) = 2π × 20 × 2 = 80π cm² / sec

Question 2.
The sides of an equilateral triangle are increasing at the rate of 2 cm/sec. The rate at which its area increases, when side is 10 cm, is
(a) 10 cm² / sec
(b) 10√3 cm² /sec
(c) \(\frac{10}{3}\) cm² / sec
(d) √3 cm² / sec
Solution:
(b) 10√3 cm² /sec

Let x be the length of each side of equilateral triangle at any time t.
Then A = area of equilateral triangle = \(\frac{\sqrt{3}}{4}\) x²
∴ \(\frac{d \mathrm{~A}}{d t}=\frac{\sqrt{3}}{4} \times 2 x \frac{d x}{d t}\)
= \(\frac{\sqrt{3}}{2} x \frac{d x}{d t}\)
Given \(\frac{d x}{d t}\) = 2 cm / sec ;
x = 10 cm
∴ \(\frac{d A}{d t}\) = \(\frac{\sqrt{3}}{2}\) × 10 × 2 = 10√3 cm² / sec.

Question 3.
A spherical ice ball is melting at the rate of 100 π cm³ / min. The rate at which its radius is decreasing, when its radius is 15 cm, is
(a) \(\frac{1}{9}\) cm / min
(b) \(\frac{1}{9 \ pi}\) cm / min
(c) \(\frac{1}{18}\) cm / min
(d) \(\frac{1}{36}\) cm / min
Solution:
(a) \(\frac{1}{9}\) cm / min

Let r be the radius of spherical ice ball
Then \(\frac{d V}{d t}\) = – 100π cm³ / min.
⇒ \(\frac{d}{d t}\) (\(\frac{4}{3}\) πr³) = – 100π
⇒ \(\frac{4}{3}\) π × 3r² \(\frac{d r}{d t}\) = – 100π
⇒ 4π × 15² . \(\frac{d r}{d t}\) = – 100π [given r = 15]
⇒ \(\frac{d r}{d t}\) = – \(\frac{1}{9}\) cm / min.

Question 4.
The radius of a cylinder is increasing at the rate of 3 cm/sec and its height is decreasing at the rate of 4 cm/sec. The rate of change of its volume when radius is 4 cm and height is 6 cm, is
(a) 64 cm³ / sec
(b) 144 cm³/ sec
(c) 80 cm³/ sec
(d) 80 π cm³ / sec
Solution:
(d) 80 π cm³ / sec

Let r be the radius and h be the height of cylinder at any time.
Then V = volume of cylinder = πr²h
∴ \(\frac{d V}{d t}=\pi\left[r^2 \frac{d h}{d t}+2 r h \frac{d r}{d t}\right]\)
Given \(\frac{d r}{d t}\) = 3 cm/sec ;
\(\frac{d h}{d t}\) = – 4 cm/sec ;
r = 4 cm
and h = 6 cm
Thus \(\frac{d V}{d t}\) = π [4² × (- 4) + 2 × 4 × 6 × 3]
= π [- 64 + 144]
= 80 π cm³ / sec

Question 5.
A point on the curve y² = 18x at which the ordinate increases at twice the rate of abscissa is
(a) (2, 4)
(b) (2, – 4)
(c) \(\left(-\frac{9}{8}, \frac{9}{2}\right)\)
(d) \(\left(\frac{9}{8}, \frac{9}{2}\right)\)
Solution:
(d) \(\left(\frac{9}{8}, \frac{9}{2}\right)\)

Given eqn. of curve y² = 18x
⇒ 2y \(\frac{d y}{d x}\) = 18
⇒ \(\frac{d y}{d x}=\frac{9}{y}\)
⇒ 2y \(\frac{d y}{d t}\) = 18 \(\frac{d x}{d t}\)
⇒ \(\frac{d y}{d t}=\frac{9}{y} \frac{d x}{d t}\) ……..(1)
Also \(\frac{d y}{d t}\) = 2 \(\frac{d x}{d t}\)
from (1) and (2) ; we have
⇒ 2 \(\frac{d x}{d t}\) = \(\frac{9}{y}\) \(\frac{d x}{d t}\)
⇒ y = \(\frac{9}{2}\) \(
∴ from given curve, we have,
[latex]\left(\frac{9}{2}\right)^2\) = 18x
⇒ x = \(\frac{81}{4 \times 18}=\frac{9}{8}\)
Thus required point on given curve be \(\left(\frac{9}{8}, \frac{9}{2}\right)\).

Question 6.
A ladder, 5 metres long standing on a floor, leans against a vertical wall. If the top of the ladder slides downwards at the rate of 10 cm/sec, then the rate at which the angle between the floor and the ladder is decreasing when the lower end of the ladder is 2 metres away from the wall is
(a) \(\frac{1}{10}\) radian / sec
(b) \(\frac{1}{20}\) radian/sec
(c) 5 radian/sec
(d) 10 radian/sec
Solution:
(b) \(\frac{1}{20}\) radian/sec

Let the bottom of the ladder be at a distance x m from the wall and the top be at a height h from the ground.
Then x² + y² = 25 ………..(1)
and tan θ = \(\frac{y}{x}\) …………(2)
Diff. eqn. (1) both sides w.r.t. t, we have
2x \(\frac{d x}{d t}\) + 2y \(\frac{d y}{d t}\) = 0
Given \(\frac{d y}{d t}\) = – 10 cm / sec = – \(\frac{1}{10}\) m / sec
⇒ 2x \(\frac{d x}{d t}\) – \(\frac{2 y}{10}\) = 0
⇒ \(\frac{d x}{d t}=\frac{y}{10 x}\)
Diff. eqn. (2) both sides w.r.t. t, we have

Question 7.
The curve y = x^1/5 has at (0, 0)
(a) a vertical tangent (parallel to y-axis)
(b) a horizonlal tangent (parallel to x-axis)
(c) an oblique tangent
(d) no tangent
Solution:
(a) a vertical tangent (parallel to y-axis)

Equation of given curve be,
y = x^1/5
∴ slope of tangent to given curve at point
P(0, 0) = (\(\frac{d y}{d x}\))_{(0, 0)} → ∞
Thus tangent is parallel to y-axis.

Question 8.
The equation of the tangent to the curve y = (4 – x²)^2/3 at x = 2 is
(a) x = – 2
(b) x = 2
(c) y = 2
(d) y = – 2
Solution:
(b) x = 2

eqn. of given curve be y = (4 – x²)^2/3
When x = 2 ; y = 0
Thus any point on given curve be (2, 0)
∴ \(\frac{d y}{d x}\) = \(\frac{2}{3}\) (4 – x²)^-1/3 (- 2x)
∴ slope of tangent to given curve at x = 2 = (\(\frac{d y}{d x}\))_{x = 2} = ∞
Thus required eqn. of tangent to given curve at (2, 0) be given by
y – 0 = \(\frac{\frac{1}{d x}}{d y}\) (x – 2)
⇒ x – 2 = 0
⇒ x = 2.

Question 9.
The equation of the tangent to the curve y = e^2x at (0, 1) is
(a) y + 1 = 2x
(b) 1 – y = 2x
(c) y – 1 = 2x
(d) none of these
Solution:
(c) y – 1 = 2x

eqn. of given curve be
y = e^2x …………(1)
∴ \(\frac{d y}{d x}\) = 2e^2x
Thus slope of tangent to curve (1) at (0, 1) = (\(\frac{d y}{d x}\))_{(0, 1)}
=2 e^{2 × 0} = 2
Thus eqn. of tangent to given curve at point (0, 1) be given by
y – 1 = 2(x – 0)
⇒ y – 1 = 2x.

Question 10.
The tangent to the curve y = e at the point (0, 1) meets x-axis at:
(a) (0, 1)
(b) (- \(\frac{1}{2}\), o)
(c) (2, 0)
(d) (0, 2)
Solution:
(b) (- \(\frac{1}{2}\), o)

Equation of given curve be y = e^2x ………….(1)
Differentiate equation (1) both sides w.r.t. x; we get
\(\frac{d y}{d x}\) = 2e^2x
∴ slope of tangent to given curve (1) at (0, 1) = (\(\frac{d y}{d x}\))_{(0, 1)}
= 2 × e⁰ = 2
Thus required equation of tangent to given curve (1) at (0, 1) is given by
y – 1 = 2 (x – 0)
⇒ y – 1 = 2x …………(2)
Equation (2) meets x-axis at y = 0
Hence the coordinates of required point are (- \(\frac{1}{2}\), o).

Question 11.
The tangent to the curve x² = 2y at the point (1, \(\frac{1}{2}\)) makes with the x-axis an angle of
(a) 0
(b) \(\frac{\pi}{6}\)
(c) \(\frac{\pi}{4}\)
(d) \(\frac{\pi}{3}\)
Solution:
(c) \(\frac{\pi}{4}\)

Given eqn. of curve be x² = 2y
∴ 2x = 2 \(\frac{d y}{d x}\)
⇒ \(\frac{d y}{d x}\) = x
∴ slope of tangent to given curve at (1, \(\frac{1}{2}\)) = (\(\frac{d}{d x}\))_{(1, \(\frac{1}{2}\))} = 1
Let θ be the angle made by tangent with the +ve direction of x-axis.
∴ slope of tangent to given curve = tan θ
∴ tan θ = 1
⇒ θ = \(\frac{\pi}{4}\)

Question 12.
The tangents to the curve x² + y² = 2 at points (1, 1) and (- 1, 1) are
(a) parallel
(b) at right angles
(c) neither parallel nor at right angles
(d) none of these
Solution:
eqn. of given curve be
x² + y² = 2 …………(1)
Diff. eqn. (1) w.r.t. x; we have
2x + 2y \(\frac{d y}{d x}\) = 0
⇒ \(\frac{d y}{d x}\) = – \(\frac{x}{y}\)
∴ m₁ = slope of tangent to given curve at (1, 1) = (\(\frac{d y}{d x}\))_{(0, 1)} = – 1
m₂ = slope of tangent to given curve at (- 1, 1) = (\(\frac{d y}{d x}\))_{(- 1, 1)} = 1
∴ m₁ m₂ = – 1 × 1 = – 1
Hence, tangents to given curve at points (1, 1) and (- 1, 1) are at right angles.

Question 13.
The point on the curve y² = x, where tangent make an angle of with the x-axis, is
(a) \(\left(\frac{1}{2}, \frac{1}{4}\right)\)
(b) \(\left(\frac{1}{4}, \frac{1}{2}\right)\)
(c) (4, 2)
(d) (1, 1)
Solution:
(b) \(\left(\frac{1}{4}, \frac{1}{2}\right)\)

eqn. of given curve be
y²= x …………….(1)
diff, both sides of eqn. (1) w.r.t. x;
2y \(\frac{d y}{d x}\) = 1
⇒ \(\frac{d y}{d x}\) = \(\frac{1}{2 y}\)
∴ slope oftangentto curve (1) = \(\frac{d y}{d x}\) = \(\frac{1}{2 y}\)
Also, slope of tangent to given curve (1) = tan \(\frac{\pi}{4}\) = 1
∴ \(\frac{1}{2 y}\) = 1
⇒ y = \(\frac{1}{2}\)
∴ from (1) ;
x = \(\frac{1}{4}\)
Thus, required point on given curve be (\(\left(\frac{1}{4}, \frac{1}{2}\right)\))

Question 14.
The point on the curve y = 6x – x² where the tangent is parallel to the line 4x – 2y – 1 = 0 is
(a) (2, 8)
(b) (8, 2)
(c) (6, 1)
(d) (4, 2)
Solution:
(a) (2, 8)

Let P (x₁, y₁) be any point on curve
y = 6x – x² ……….(1)
∴ \(\frac{d y}{d x}\) = 6 – 2x
Thus slope of tangent to given curve at
P(x₁, y₁) = \(\left(\frac{d y}{d x}\right)_{\mathrm{P}\left(x_1, y_1\right)}\) = 6 – 2x₁
eqn.of given line be 4x – 2y – 1 = 0 ………….(2)
∴ slope of hne (2) = \(\frac{- 4}{- 2}\) = 2
Since it is given that tangent to given curve is parallel to line (2)
∴ 6 – 2x₁ = 2
⇒ 2x₁ = 4
⇒ x₁ = 2
Also P (x₁, y₁) lies on eqn. (1) ;
y₁ = 6x₁ – x₁
= 6 × 2 + 2² = 8
Hence coordinates of any point on given curve be (2, 8).

Question 15.
The points at which the tangents to the curve y = x³ – 12x + 18 are parallel to x-axis are :
(a) (2, – 2), (- 2, – 34)
(b) (2, 34), (- 2, 0)
(c) (0, 34), (- 2, 0)
(d) (2, 2), (- 2, 34)
Solution:
(d) (2, 2), (- 2, 34)

Equation of given curve be
y = x³ – 12x + 18 …………..(1)
Let the required points on curve (1) are (x₁, y₁)
Differentiate equation (1) w.r.t. x; we get
\(\frac{d y}{d x}\) = 3x² – 12
slope of tangent to given curve (1) at (x₁, y₁)
= \(\left(\frac{d y}{d x}\right)_{\left(x_1, y_1\right)}\)
= 3x₁² – 12
Also tangents are parallel to x – axis.
∴ \(\left(\frac{d y}{d x}\right)_{\left(x_1, y_1\right)}\) = 0
[since slope of the tangent is 0]
3x₁² – 12 = 0
⇒ x₁ = ± 2
Also, the point (x₁, y₁) lies on curve (1)
y₁= x₁ – 12x₁ + 18 …………….(2)
when x₁ = 2
∴ from (2) ; we have
y₁ = 8 – 24 + 18 = 2
when x₁ = 2
∴ from (2) ; we have
y₁ = – 8 + 24 + 18 = 34
Thus the required points on given curve (1) are (2, 2) and (- 2, 34).

Question 16.
The point at which the tangent to the curve y = 2x² – x + 1 is parallel to the line y = 3x + 9 is
(a) (2, 1)
(b) (1, 2)
(c) (3, 9)
(d) (- 2, 1)
Solution:
(b) (1, 2)

Let P (x₁, y₁) be any point on given curve
y = 2x² – x + 1
∴ y₁ = 2x₁² – x₁ + 1 ………..(1)
∴ slope of tangent to given curve at P (x₁, y₁)
= \(\left(\frac{d y}{d x}\right)_{\left(x_1, y_1\right)}\)
= 4x₁ – 1
Again eqn. of given line be 3x – y + 9 = 0 …………..(2)
∴ slope of line (2) = \(\frac{- 3}{- 1}\) = 3
Since tangent to curve (1) is parallel to line (2)
∴ 4x₁ – 1 = 3
⇒ x₁ = 1
∴ from (1);
y₁ = 2 – 1 + 1 = 2
Thus, required point on tangent to given curve be (1, 1).

Question 17.
If the tangent to the curve x = t² – 1, y = t² – t is parallel to x-axis, then
(a) t = 0
(b) t = 2
(c) t = \(\frac{1}{2}\)
(d) t = – \(\frac{1}{2}\)
Solution:
(c) t = \(\frac{1}{2}\)

Given eqn. of curve be,
x = t² – 1
and y = t² – t
∴ \(\frac{d x}{d t}\) = 2t ;
\(\frac{d y}{d t}\) = 2t – 1
Thus \(\frac{d y}{d x}\) = \(\frac{\frac{d y}{d t}}{\frac{d x}{d t}}\)
= \(\frac{2 t-1}{2 t}\)
Since tangent to given curve is parallel to x-axis
∴ \(\frac{d y}{d x}\) = 0
⇒ \(\frac{2 t-1}{2 t}\) = 0
⇒ t = \(\frac{1}{2}\).

Question 18.
The tangent to the curve x = e^t cos t, y = e^t sin t at t = \(\frac{\pi}{4}\) makes with x-axis an angle
(a) 0
(b) \(\frac{\pi}{4}\)
(c) \(\frac{\pi}{3}\)
(d) \(\frac{\pi}{2}\)
Solution:
(d) \(\frac{\pi}{2}\)

Given parametric eqns. of curve be,
x = e^t cos t ……………..(1)
y = e^t sin t …………..(2)
Diff. eqns. (1) and (2) w.r.t. t; we have
\(\frac{d x}{d t}\) = e^t (- sin t) + cos t . e^t
and \(\frac{d y}{d t}\) = e^t cos t + sin t e^t
∴ \(\frac{d y}{d x}\) = \(\frac{\frac{d y}{d t}}{\frac{d x}{d t}}\)
= \(\frac{e^t(\cos t+\sin t)}{e^t(\cos t-\sin t)}\)
= \(\frac{\cos t+\sin t}{\cos t-\sin t}\)
∴ slope of tangent to given curve at t = \(\frac{\pi}{4}\)
= \(\left(\frac{d y}{d x}\right)_{t=\frac{\pi}{4}}\)
= \(\frac{\cos \frac{\pi}{4}+\sin \frac{\pi}{4}}{\cos \frac{\pi}{4}-\sin \frac{\pi}{4}}\)
= \(\frac{\sqrt{2}}{0}\) → ∞
Let the tangent to given curve at t = \(\frac{\pi}{4}\) makes an angle θ with x – axis.
Then slope of tangent to given curve at t = \(\frac{\pi}{4}\) = tan θ
∴ tan θ = ∞
⇒ θ = \(\frac{\pi}{2}\).

Question 19.
The equation to the normal to the curve sin x at (0, 0) is
(a) x = 0
(b) y =
(c) x + y = 0
(d) x – y = 0
Solution:
(c) x + y = 0

Given eqn. of curve be y = sin x …………..(1)
∴ \(\frac{d y}{d x}\) = cos x
∴ \(\left(\frac{d y}{d x}\right)_{(0,0)}\) = cos 0 = 1
Thus required equation of normal at (0, 0) to given curve be
y – 0 = \(\frac{-1}{\left(\frac{d y}{d x}\right)_{(0,0)}}\) (x – 0)
⇒ y = – 1 (x)
⇒ x + y = 0.

Question 20.
The slope of the normal to the curve X2 + 3y +ý = 5 at the poInt (1, 1) ¡s
(a) – \(\frac{2}{5}\)
(b) \(\frac{5}{2}\)
(c) \(\frac{2}{5}\)
(d) – \(\frac{5}{2}\)
Solution:
(b) \(\frac{5}{2}\)

eqn. of given curve be
x² + 3y + y² = 5 …………(1)
duff. eqn. (1) both sides w.r.t. X; we have
2x + 3 \(\frac{d y}{d x}\) + 2y \(\frac{d y}{d x}\) = 0
⇒ (2y + 3) \(\frac{d y}{d x}\) = – 2x
⇒ \(\frac{d y}{d x}\) = \(\frac{-2 x}{2 y+3}\)
∴ slope of normal to given curve at (1, 1) = \(\frac{-1}{\left(\frac{d y}{d x}\right)_{(1,1)}}\)
= \(\frac{-1}{\frac{-2 \times 1}{2 \times 1+3}}\)
= \(\frac{5}{2}\).

Question 21.
The point on the curve 9y² = x³, where the normal to the curve makes equal intercepts with the axes is
(a) (4, ± 8/3)
(b) (- 4, 8/3)
(c) (- 4, – 8/3)
(d) (8/3, 4)
Solution:
(a) (4, ± 8/3)

Given eqn. of curve be
9y² = x³ ………….(1)
Let the required point on curve (1) be (x₁, y₁)
∴ 9y₁² = x₁³ ………….(2)
Diff. eqn. (1) both sides w.r.t. x; we get
18 y = \(\frac{d y}{d x}\) = 3x²
Since normal to curve makes an equal intercept with axes.
∴ slope of normal at (x₁, y₁) = ± 1
∴ \(\frac{-1}{\left(\frac{d y}{d x}\right)_{\left(x_1, y_1\right)}}\) = ± 1
⇒ \(\frac{-6 y_1}{x_1^2}\) = ± 1
⇒ \(\frac{6 y_1}{x_1^2}\) = ± 1
when \(\frac{6 y_1}{x_1^2}\) = 1
⇒ 6y₁ = x₁² ………….(3)
∴ from (2) ; we have
\(\frac{9 x_1^4}{36}\) = x₁³
⇒ x₁³ (x₁ – 4) = 0
⇒ x₁ = 0, 4
When x₁ = 0
∴ from (3) ;
y₁ = 0
When x₁ = 4
∴ from (2) ;
y₁ = ± 8/3
When \(\frac{6 y_1}{x_1^2}\) = – 1
⇒ 6y₁ = – x₁² …………(4)
∴ from (2) ;
x₁ = 0, 4
When x₁ = 4
∴ from (4) ;
y₁ = 0
When x₁ = 0
∴ from (4) ; y₁ = 0
Thus required points are (0. 0), (4, ± 8/3).

Question 22.
The equation of the normal to the curve 3x² – y² = 8 which is parallel to x + 3y = 8 is
(a) x – 3y = 8
(b) x – 3y + 8 = 0
(c) x + 3y ± 8 = 0
(d) x + 3y = 0
Solution:
Given eqn. of curve be 3x² – y² = 8 ………..(1)
Let (x₁, y₁) be point on curve (1) where normal to curve (1) || to given line
x+3y – 8 = 0
Diff. (1) w.r.t. x; we get
6x – 2y \(\frac{d y}{d x}\) = 0
⇒ \(\frac{d y}{d x}\) = \(\frac{d3 x}{y}\)
∴ \(\left(\frac{d y}{d x}\right)_{\left(x_1, y_1\right)}=\frac{3 x_1}{y_1}\)
Since the normal to curve (1) || to line x + 3y – 8 = 0
∴ slope of normal at (x₁, y₁) = slope of line
x + 3y – 8 = 0
⇒ \(\frac{-1}{\left(\frac{d y}{d x}\right)_{\left(x_1, y_1\right)}}=-\frac{1}{3}\)
⇒ \(-\frac{y_1}{3 x_1}=-\frac{1}{3}\)
⇒ x₁ = y₁ …………….(2)
also point (x₁, y₁) lies on eqn. (1)
∴ 3x₁² – y₁² = 8
⇒ 3x₁² – x₁² = 8
⇒ x₁² = 4
⇒ x₁ = ± 2
from (2) ;
y₁ = ± 2
Thus point of contact be (± 2, ± 2).
∴ required eqn. of normal at (2, 2) to given curve be
y – 2 = – (x – 2)
3y – 6 = – x + 2
x + 3y -8 = 0
and eqn. of normal at (- 2, – 2) to given curve be
y – 2 = – \(\frac{1}{3}\) (x – 2)
⇒ 3y – 6 = – x + 2
⇒ x + 3y – 8 = 0
and eqn. of normal at (- 2, – 2) to given curve be
y + 2 = – \(\frac{1}{3}\) (x + 2)
⇒ x + 3y + 8 = 0

Question 23.
The angle between the tangents to the curve y = x² – 5x + 6 at the point (2, 0) and (3, 0) is
(a) \(\frac{\pi}{2}\)
(b) \(\frac{\pi}{3}\)
(c) \(\frac{\pi}{4}\)
(d) \(\frac{\pi}{6}\)
Solution:
(a) \(\frac{\pi}{2}\)

eqn. of given curve be y = x² – 5x + 6
∴ \(\frac{d y}{d x}\) = 2x – 5
m₁ = slope of tangent to given curve at (2, 0)
= \(\left(\frac{d y}{d x}\right)_{(2,0)}\)
= 2 × 2 – 5 = – 1
and m₂ = slope of tangent to given curve at (3, 0)
= \(\left(\frac{d y}{d x}\right)_{(2,0)}\)
= 2 × 3 – 5 = 1
Here m₁m₂ = (- 1 ) × 1 = – 1
Thus both tangents to given curve at points (2, 0) and (3, 0) cut orthogonally
i.e. the angle between them is \(\frac{\pi}{2}\).

Question 24.
If the curve ay + x² = 7 and x³ = y cut orthogonally at (1, 1), then a is equal to
(a) 1
(b) – 6
(c) 6
(d) 0
Solution:
(c) 6

Given eqns. of curve be,
ay + x² = 7 ……….(1)
and x³ = y ………..(2)
Diff. (1) and (2) both sides w.r.t. x, we get
a \(\frac{d y}{d x}\) + 2x = 0
∴ \(\frac{d y}{d x}\) = – \(\frac{2 x}{a}\)
∴ m₁ = \(\left(\frac{d y}{d x}\right)_{(1,1)}\)
= \(-\frac{2}{a}\)
and 3x² = \(\frac{d y}{d x}\)
∴ m₂ = \(\left(\frac{d y}{d x}\right)_{(1,1)}\) = 3
Since both cuves cut orthogonally.
∴ m₁m₂ = – 1
⇒ – \(\frac{2}{3}\) × a = – 1
⇒ a = 6

Question 25.
The two curves x³ – 3xy² + 2 = 0 and 3x²y – y³ – 2 = 0 intersect at an angle of:
(a) \(\frac{\pi}{4}\)
(b) \(\frac{\pi}{3}\)
(c) \(\frac{\pi}{2}\)
(d) \(\frac{\pi}{6}\)
Solution:
(c) \(\frac{\pi}{2}\)

Given eqns. of curves be,
x³ – 3xy² = – 2 ……………(1)
and 3x²y – y³ = 2 …………..(2)
From (1) and (2) ; we get
x³ – 3xy² = – 3x²y + y³
⇒ x³ – 3xy² + 3x²y – y³ = 0
⇒ (x – y)³ = 0
⇒ x = y
putting x =y in eqn. (1); we get
x³ – 3x³ = – 2
⇒ x³ = 1
⇒ x = + 1
∴ from (3) ;
y = + 1
Hence the points of intersection of two curves be (1, 1).
Diff. (1) w.r.t. x; we get
3x² – 3 [2xy \(\frac{d y}{d x}\) + y²] = 0
⇒ \(\frac{d y}{d x}=\frac{x^2-y^2}{2 x y}\) ……..(3)
Diff. (2) w.r.t. x; we get
3 [x² \(\frac{d y}{d x}\) + y . 2x] – 3y² \(\frac{d y}{d x}\) = 0
\(\frac{d y}{d x}=-\frac{2 x y}{x^2-y^2}\) ………….(4)

at (1, 1) :
using (3);
m₁ = \(\left(\frac{d y}{d x}\right)_{(1,1)}\) = 0
m₂ = \(\left(\frac{d y}{d x}\right)_{(1,1)}\) = ∞
If θ be the angle of intersection between two curves.
Then tan θ = \(\left|\frac{m_1-m_2}{1+m_1 m_2}\right|\)
= \(\left|\frac{m_2\left(\frac{m_1}{m_2}-1\right)}{m_2\left(\frac{1}{m_2}+m_1\right)}\right|\) = ∞
⇒ θ = \(\frac{\pi}{2}\)
Hence both curves intersect orthogonally at (1, 1).

Question 26.
The interval on which the function f(x) = 2x³ + 9x² + 12x – 1 is decreasing in:
(a) [- 1, ∞)
(b) [- 2, – 1]
(c) (- ∞, – 2]
(d) [- 1, 1]
Solution:
(b) [- 2, – 1]

Given f (x) = 2x³ + 9x² + 12x + 20
∴ f’(x) = 6x² + 18x + 12
= 6 (x² + 3x + 2)
= 6 (x + 1) (x + 2)
Now f’(x) > 0
⇒ (x + 1) (x + 2) > 0
⇒ x > – 1 or x < – 2
∴ f(x) is strictly increasing in (- ∞, – 2) ∪ (- 1, ∞).
Now f’(x) < 0
⇒ (x + 1) (x + 2) < 0
⇒ – 2 < x < – 1
[If (x – a) (x – b) < 0 and a < b ⇒ a < x < b]
∴ f(x) is strictly decreasing in [- 2, – 1].
[If f’(x) < 0 ∀ x ∈ (a, b)Thenf(x) is decreasg in [a, b]]

Question 27.
The interval in which the function f(x) = 2x³ + 3x² – 12x + 1 is strictly increasing is
(a) [- 2, 1]
(b) (- ∞, – 2] ∪ [1, ∞)
(e) (- ∞, 1]
(f)(- ∞, – 1] ∪ [2, ∞)
Solution:
(b) (- ∞, – 2] ∪ [1, ∞)

Given f(x) = 2x³ + 3x² – 12x + 1
∴ f’(x) = 6x² + 6x – 12
= 6 (x² + x – 2)
= 6 (x – 1) (x + 2)
Now f’(x) > 0 iff 6 (x – 1) (x + 2) > 0
iff x > 1 or x< – 2 [if (x – a) (x – b) > 0 and a > b ⇒ x > a or x < b]
iff x ∈ (- ∞, – 2) ∪ (1, ∞)
Thus f(x) is strictly increasing in (- ∞, – 2] ∪ [1, x).

Question 28.
The function f (x) = x² e^{– x} is monotonic increasing when
(a) x ∈ R – [0, 2]
(b) 0 < x < 2
(c) 2 < x < ∞
(d) x < 0
Solution:
(b) 0 < x < 2

Given f(x) = x² e^{– x}
∴ f'(x) = x² e^{– x} (- 1) + e^{– x} 2x
= e^{– x} [- x² + 2x]
= e^{– x} (2 – x) x
For f(x) is to be monotomically decreasing we must have
f’(x) ≥ 0
⇒ e^{– x} (2 – x) x ≥ 0 [where e^{– x} ≥ 0]
⇒ – (x – 2)x ≤ 0
⇒ (x – 2) x ≤ 0
⇒ 0 < x < 2
⇒ x ∈ (0, 2)
[if (x – a) (x – b)< 0 and a < b then a < x < b].

Question 29.
The function f(x) = tan x – x
(a) always increases
(b) always decreases
(c) never increases
(d) sometimes increases and decreases.
Solution:
(a) always increases

Given f(x) = tan x – x
∴ f'(x) = sec² x – 1 ≥ 0
[sec² x ≥ 1
⇒ sec² x – 1 ≥ 0 ∀ x ∈ R]
Hence the function f(x) is always increases.

Question 30.
The function f(x) = x⁴ – 4x is strictly
(a) decreasing in [1, ∞)
(b) increasing in [1, ∞)
(c) Increasing in (- ∞, 1]
(d) increasing in [- 1, 1]
Solution:

Given f(x) = x⁴ – 4x
f’(x) = 4x³ – 4
= 4 (x³ – 1)
= 4 (x – 1) (x² + x + 1)
= 4 (x – 1) (x² + x + \(\frac{1}{4}\) + \(\frac{3}{4}\))
= 4 (x – 1) [(x + \(\frac{1}{2}\))² + \(\frac{3}{4}\)]
Now f’ (x) > 0
iff 4 (x – 1) [(x + \(\frac{1}{2}\))² + \(\frac{3}{4}\)] > 0
iff x – 1 > 0
[∵ (x + \(\frac{1}{2}\)) + \(\frac{3}{4}\) > 0 ∀ x ∈ R]
iff x > 1
iff x ∈ (1, ∞)
Thus f(x) is strictly increasing in [1, ∞)

Question 31.
The function f (x) = x² – 2x is strictly decreasing in the interval
(a) (- ∞, 1]
(b) [1, ∞)
(c) [- 1, ∞)
(d) none of these
Solution:
(a) (- ∞, 1]

Given f(x) = x² – 2x
∴ f’(x) = 2x – 2
Now f’(x) < 0 iff 2x – 2 < 0
iff x < 1 iff x ∈ (- ∞, 1)
Thus,f(x) is strictly decreasing in (- ∞, 1].

Question 32.
y = x (x – 3)² decreases for the values of x given by:
(a) 1 < x < 3
(b) x < 0 (c) x > 0
(d) 0 < x < \(\frac{3}{2}\)
Solution:
(a) 1 < x < 3

Given y = x (x – 3)² = f(x)
∴ \(\frac{d y}{d x}\) = (x – 3)² + 2x (x – 3)
= (x – 3) (x – 3 + 2x)
= 3 (x – 3) (x – 1)
since f(x) is decreases if f’ (x) < 0
i.e. \(\frac{d y}{d x}\) < 0
⇒ (x – 3) (x – 1) < 0
⇒ 1 < x < 3
[if a < b and (x – a) (x- b) < 0 ⇒ a < x < b]

Question 33.
Which of the following functions is decreasing on (o, \(\frac{\pi}{2}\))
(a) sin 2x
(b) tan x
(c) cos x
(d) cos 3x
Solution:
When 0 < x < \(\frac{\pi}{2}\)
⇒ 0 < 2x < π For option (A) ; f (x) = sin 2x ∴ f’(x) = 2 cos 2x may be > 0 or < 0.
Hence f(x) may be increasing or decreasing.
For option (B) ;
f (x) = tan x
∴ f'(x) = sec² x > 0 ∀ x ∈ (0)
f(x) is increasing on (o, \(\frac{\pi}{2}\)).
For option (C) ;
f (x) = cos x
∴ f’(x) = – sin x < 0 ∀ x ∈ (o, \(\frac{\pi}{2}\))
∴ f(x) is decreasing on (o, \(\frac{\pi}{2}\)).
For option (D) ;
f (x) = cos 3x
∴ f ‘(x) = 3 sin 3x
[∵ 0 < x < \(\frac{\pi}{2}\) ⇒ 0 < 3x < \(\frac{3 \pi}{2}\)]
it may be positive or negative.
Thus f(x) may be increasing or decreasing.

Question 34.
The function f (x) = x^x, x > 0, is increasing on the interval
(a) (0, e]
(b) (0, \(\frac{1}{e}\))
(c) [\(\frac{1}{e}\), ∞)
(d) none of these
Solution:
(c) [\(\frac{1}{e}\), ∞)

Given f(x) = x^x
∴ log f(x) = Iog x^x = x log x;
Diff, both sides w.r.t. x, we have
⇒ \(\frac{1}{f(x)}\) f’(x) = x × \(\frac{1}{x}\) + log x × 1
⇒ f’ (x) = x^x (1 + log x)
Now f’(x) ≥ 0 iff x^x (1 + log x) ≥ 0
iff 1 + log x ≥ 0 [∵ x^x > 0]
iff log x ≥ 1 iff x ≥ \(\frac{1}{e}\)
Thus f(x) is increasing in [\(\frac{1}{e}\), ∞).

Question 35.
The value of p so that the function f (x) = sin x – cos x – px + q decreases for all real values of x is given by
(a) p ≥ √2
(b) p < √2
(c) p ≥ 1
(d) p < 1
Solution:
(a) p ≥ √2

f’(x) = cos x + sin x – p
f(x) is decreasing iff f’cos ≤ 0
iff cosx + sinx – p ≤ 0
iff cos x + sin x ≤ p
iff √2 sin (x + \(\frac{\pi}{4}\)) ≤ p
Since |sin x| ≤ 1
∴ p ≥ √2

Question 36.
If x is real, x² – 8x + 17 is
(a) – 1
(b) 0
(c) 1
(d) 2
Solution:
(c) 1

Let f(x) = x² – 8x + 17
= x² – 8x + 16 + 1
= (x – 4)² + 1 ≥ 1 ∀ x ∈ R
Hence 1 be the minimum value of f (x) and is attains at x = 4.

Question 37.
The smallest value of the polynomial x³ – 18x² + 96x in [0, 9] is
(a) 126
(b) 135
(c) 160
(d) 0
Solution:
(d) 0

Given f(x) = x³ – 18x² + 96x
∴ f’(x) = 3x² – 36x+96
= 3 (x² – 12x + 32)
= 3 (x – 8) (x – 4)
For critical points, f’ (x) = 0
⇒ 3 (x – 8) (x – 4) = 0
⇒ x = 4, 8
Now we compute f (x) at critical points x = 4, 8 and at end points x = 0, 9.
Here, f(0) = 0;
f(4) = 4³ – 18 x 4² + 96 x 4
= 64 – 288 + 384
= 160
f(8) = 8³ – 18 × 64 + 96 × 8 = 128
f(9) = 9³ – 18 × 81 + 96 × 9 = 135
∴ smallest value of f(x) = f(0) = 0.

Question 38.
The minimum value of (x² + \(\frac{250}{x}\)) is
(a) 75
(b) 50
(c) 25
(d) 55
Solution:
(a) 75

Let f(x) = x² + \(\frac{250}{x}\)
∴ f’(x) = 2x – \(\frac{250}{x^{2}}\)
For maxima/minima, f’ (x) = 0
⇒ 2x – \(\frac{250}{x^{2}}\) = 0
⇒ x³ = 125 = 5³
[∵ x³ – 5³
⇒ (x – 5) (x² + 5x + 25) = 0
⇒ x = 5 and x² + 5x + 25 = 0 does not gives any real values of x]
⇒ f”(x) = 2 + \(\frac{500}{x^{3}}\)
∴ f”(5) = 2 + \(\frac{500}{125}\)
= 6 > 0
Hence x = 5 is a point of minima and min.
value of f(x) = f(5) = 25 + \(\frac{250}{5}\) = 75

Question 39.
The function f(x) = \(\frac{x}{2}+\frac{2}{x}\) has a local minimum at
(a) x = 1
(b) x = 2
(c) x = – 2
(d) x = – 1
Solution:
(b) x = 2

Given f(x) = \(\frac{x}{2}+\frac{2}{x}\)
∴ f'(x) = \(\frac{1}{2}-\frac{2}{x^2}\)
For local maxima/minima, f'(x) = 0
⇒ \(\frac{1}{2}-\frac{2}{x^2}\) = 0
⇒ \(\frac{1}{2}=\frac{2}{x^2}\)
⇒ x² = 4
⇒ x = ± 2
at x = 2;
f”(2) = \(\frac{4}{2^3}=\frac{1}{2}\) >0
Thus f(x) is minimise when x = 2.

Question 40.
Let the function f: R → R be defined by f(x) = 2 x+ cos x, then f(x)
(a) has a minimum at x = it
(b) has a maximum at x = 0
(c) is a decreasing function
(d) is an increasing function [NCERT Exemplar]
Solution:
(d) is an increasing function

Given f(x) = 2x + cos x
∴ f’(x) = 2 – sinx
Now – 1 ≤ sin x ≤ 1
⇒ 1 ≥ – sin x ≥ – 1
⇒ 3 ≥ 2 – sin x ≥ 1
⇒ 1 ≤ f’(x) ≤ 3
∴ f'(x) > 0
Thus f(x) is an increasing function.

Question 41.
At x = \(\frac{5 \pi}{6}\), f(x) = 2 sin 3x + 3 cos 3x is
(a) 0
(b) maximum
(c) minimum
(d) none of these
Solution:
(d) none of these

f(x) = 2 sin 3x + 3 cos 3x
f’(x) = 6 cos 3x – 9 sin 3x
f” (x) = – 18 sin 3x – 27 cos 3x
∴ f’(5 π / 6) = 6 cos \(\frac{5 \pi}{2}\) – 9 sin 5π /2
= 6 cos (2π+ π/2) – 9 sin (2π + π/2)
= 6 × 0 – 9 × 1
= – 9 ≠ 0
f(5 π / 6) = 2 sin 5 π / 2 + 3 cos 5 π / 2
= 2 × 1 + 3 × 0
=2 ≠ 0
Thus x = \(\frac{5 \pi}{2}\) is not an extreme point.

Question 42.
The f(x) = x^x has a stationary point at
(a) x = e
(b) x = \(\frac{1}{e}\)
(c) x = 1
(d) x = √e
Solution:
(b) x = \(\frac{1}{e}\)

Given f (x) = x^x;
Taking logarithm on both sides ; we have
∴ log f(x) = x log x
Differentiate both sides wr.t. x; we have
\(\frac{1}{f(x)}\) f’(x) = x . \(\frac{1}{x}\) + log x . 1
⇒ f ‘(x) = f(x) [1 + log x]
= x^x (1 + log x)
For stationary point, we put f ‘(x) = 0
⇒ x^x (1 + log x) = 0
⇒ 1 + log x = 0 [∵ x^x > o]
⇒ log x = – 1
⇒ x = e^-1
= \(\frac{1}{e}\)
When x slightly < \(\frac{1}{e}\)
⇒ log x < – 1
⇒ 1 + log x < 0
⇒ f'(x) < 0 When x slightly > \(\frac{1}{e}\)
⇒ log x > – 1
⇒ 1 + log x > 0
⇒ f’(x) > 0
Thus f'(x) changes its sign from -ve to +ve.
∴ x = \(\frac{1}{e}\) be a point of minima.

Question 43.
The maximum value of \(\frac{\log x}{x}\) is
(a) e
(b) 2e
(c) \(\frac{1}{e}\)
(d) \(\frac{2}{e}\)
Solution:
(c) \(\frac{1}{e}\)

Given y = \(\frac{\log x}{x}\)

Thus y is maximise at x = e
∴ Maximum value of f(x) = f(e) = \(\frac{\log e}{e}=\frac{1}{e}\).

Question 44.
The minimum value of \(\frac{x}{\log _e x}\) is
(a) e
(b) 1/e
(c) 1
(d) none of these
Solution:
(a) e

Given f(x) = \(\frac{x}{\log _e x}\) is defined for x > 0
∴ f'(x) = \(\frac{\log x \cdot 1-x \times \frac{1}{x}}{(\log x)^2}\)
= \(\frac{\log x-1}{(\log x)^2}\)
For maxima / minima,
f’ (x) = 0
⇒ log x = 1
⇒ log x = log e
⇒ x = e
when x slightly < e
⇒ log x< 1
⇒ log x – 1 < 0
⇒ f'(x) < 0 when x slightly > e
⇒ log x > log e
⇒ log x – 1 > 0
⇒ f'(x) > 0
Thusf’ (x) changes its sign from negative to positive when x increases through e.
Thus x = e is a point of local minima and local min. value of f(x) = f(e) = \(\frac{e}{log e}\) = e.

Question 45.
Maximum slope of the curve y = – x³ + 3x² + 9x – 27 is:
(a) 0
(b) 12
(c) 16
(d) 32
Solution:
(b) 12

Given equation of curve be
y = – x³ + 3x² + 9x – 27
∴ slope of curve m = – 3x² + 6x + 9
we want to maximise m.
∴ \(\frac{d m}{d x}\) = – 6x + 6
for maximalminíma, we have \(\frac{d m}{d x}\) = 0
⇒ – 6x + 6 = 0
⇒ x = 1
and \(\frac{d^2 m}{d x^2}\) = – 6
Thus at x = 1;
\(\frac{d^2 m}{d x^2}\) = – 6 < 0
Hence m is maximise when x = 1.
∴ Max value of m = – 3 (1)² + 6 + 9 = 12

Question 46.
If f(x) = Zx³ – 21x² + 36x – 30, then
(a) f(x) has minimum at x = 1
(b) f(x) has maximum at x = 6
(c) f(x) has maximum at x = 1
(d) f(x) has no maxima or minima
Solution:
(c) f(x) has maximum at x = 1

Given f(x) = 2x³ – 21x² + 36x – 30
f’(x) = 6x² – 42x + 36
= 6 (x² – 7x +6)
= 6 (x – 1) (x – 6)
For maxima / minima, f’ (x) = 0
⇒ 6 (x -1) (x – 6)O = 0
⇒ x = 1, 6
∴ f”(x) = 6 (2x – 7)
Now f”(1) = 6 (2 – 7) = – 30 < 0
∴ f(x) is maximise when x = 1
and f”(6) = 6 (12 – 7) = 30 > 0
Thus f(x) is minimise at x = 6.

Question 47.
The least value of the function f(x) = ax + \(\frac{b}{x}\) (x > 0, a > 0, b > 0) is
(a) \(\sqrt{ab}\)
(b) 2 \(\sqrt{ab}\)
(c) ab
(d) 2ab
Solution:
(b) 2 \(\sqrt{ab}\)

Given f(x) = ax + \(\frac{b}{x}\)
∴ f'(x) = a – \(\frac{b}{x^{2}}\)
For critical points ; f'(x) = 0
⇒ a – \(\frac{b}{x^{2}}\) = 0
⇒ x² = \(\frac{b}{a}\)
⇒ x = \(\sqrt{\frac{b}{a}}\) [∵ x > 0]
[Since a > 0, b > 0
∴ \(\sqrt{\frac{b}{a}}\) > 0
∴ x ≠ \(\sqrt{\frac{b}{a}}\)]
∴ least value of f(x) = f (\(\sqrt{\frac{b}{a}}\))
= a \(\sqrt{\frac{b}{a}}\) + \(\frac{b}{\sqrt{\frac{b}{a}}}\)
= \(\sqrt{ab}\) + \(\sqrt{ab}\)
= 2 \(\sqrt{ab}\)

ISC Mathematics Class 12 Solutions – ML Aggarwal Class 12 ISC Solutions

Section A

ML Aggarwal Class 12 Solutions ISC Pdf Chapter 1 Relations and Functions

• Chapter 1 Relations and Functions Ex 1.1
• Chapter 1 Relations and Functions Ex 1.2
• Chapter 1 Relations and Functions Ex 1.3
• Chapter 1 Relations and Functions Ex 1.4
• Chapter 1 Relations and Functions Ex 1.5
• Chapter 1 Relations and Functions Ex 1.6
• Chapter 1 Relations and Functions Chapter Test
• Chapter 1 Relations and Functions MCQs

ML Aggarwal Class 12 Solutions Chapter 2 Inverse Trigonometric Functions

• Chapter 2 Inverse Trigonometric Functions Ex 2.1
• Chapter 2 Inverse Trigonometric Functions Ex 2.2
• Chapter 2 Inverse Trigonometric Functions Ex 2.3
• Chapter 2 Inverse Trigonometric Functions Chapter Test
• Chapter 2 Inverse Trigonometric Functions MCQs

ML Aggarwal Maths for Class 12 Solutions Pdf Chapter 3 Matrices

• Chapter 3 Matrices Ex 3.1
• Chapter 3 Matrices Ex 3.2
• Chapter 3 Matrices Ex 3.3
• Chapter 3 Matrices Ex 3.4
• Chapter 3 Matrices Ex 3.5
• Chapter 3 Matrices Chapter Test
• Chapter 3 Matrices MCQs

Understanding ISC Mathematics Class 12 Solutions Chapter 4 Determinants

• Chapter 4 Determinants Ex 4.1
• Chapter 4 Determinants Ex 4.2
• Chapter 4 Determinants Ex 4.3
• Chapter 4 Determinants Ex 4.4
• Chapter 4 Determinants Ex 4.5
• Chapter 4 Determinants Ex 4.6
• Chapter 4 Determinants Ex 4.7
• Chapter 4 Determinants Chapter Test
• Chapter 4 Determinants MCQs

ML Aggarwal Class 12 ISC Solutions Chapter 5 Continuity and Differentiability

• Chapter 5 Continuity and Differentiability Ex 5.1
• Chapter 5 Continuity and Differentiability Ex 5.2
• Chapter 5 Continuity and Differentiability Ex 5.3
• Chapter 5 Continuity and Differentiability Ex 5.4
• Chapter 5 Continuity and Differentiability Ex 5.5
• Chapter 5 Continuity and Differentiability Ex 5.6
• Chapter 5 Continuity and Differentiability Ex 5.7
• Chapter 5 Continuity and Differentiability Ex 5.8
• Chapter 5 Continuity and Differentiability Ex 5.9
• Chapter 5 Continuity and Differentiability Ex 5.10
• Chapter 5 Continuity and Differentiability Ex 5.11
• Chapter 5 Continuity and Differentiability Ex 5.12
• Chapter 5 Continuity and Differentiability Ex 5.13
• Chapter 5 Continuity and Differentiability Ex 5.14
• Chapter 5 Continuity and Differentiability Chapter Test
• Chapter 5 Continuity and Differentiability MCQs

ML Aggarwal Class 12 ISC Solutions Chapter 6 Indeterminate Forms

• Chapter 6 Indeterminate Forms Ex 6.1
• Chapter 6 Indeterminate Forms Ex 6.2
• Chapter 6 Indeterminate Forms Ex 6.3
• Chapter 6 Indeterminate Forms Ex 6.4
• Chapter 6 Indeterminate Forms Chapter Test
• Chapter 6 Indeterminate Forms MCQs

Class 12 ML Aggarwal Solutions Chapter 7 Applications of Derivatives

• Chapter 7 Applications of Derivatives Ex 7.1
• Chapter 7 Applications of Derivatives Ex 7.2
• Chapter 7 Applications of Derivatives Ex 7.3
• Chapter 7 Applications of Derivatives Ex 7.4
• Chapter 7 Applications of Derivatives Ex 7.5
• Chapter 7 Applications of Derivatives Ex 7.6
• Chapter 7 Applications of Derivatives Ex 7.7
• Chapter 7 Applications of Derivatives Ex 7.8
• Chapter 7 Applications of Derivatives Ex 7.9
• Chapter 7 Applications of Derivatives Chapter Test
• Chapter 7 Applications of Derivatives MCQs

ISC Mathematics Class 12 Solutions Chapter 8 Integrals

• Chapter 8 Integrals Ex 8.1
• Chapter 8 Integrals Ex 8.2
• Chapter 8 Integrals Ex 8.3
• Chapter 8 Integrals Ex 8.4
• Chapter 8 Integrals Ex 8.5
• Chapter 8 Integrals Ex 8.6
• Chapter 8 Integrals Ex 8.7
• Chapter 8 Integrals Ex 8.8
• Chapter 8 Integrals Ex 8.9
• Chapter 8 Integrals Ex 8.10
• Chapter 8 Integrals Ex 8.11
• Chapter 8 Integrals Ex 8.12
• Chapter 8 Integrals Ex 8.13
• Chapter 8 Integrals Ex 8.14
• Chapter 8 Integrals Ex 8.15
• Chapter 8 Integrals Ex 8.16
• Chapter 8 Integrals Ex 8.17
• Chapter 8 Integrals Ex 8.18
• Chapter 8 Integrals Ex 8.19
• Chapter 8 Integrals Chapter Test
• Chapter 8 Integrals MCQs

ISC Maths Class 12 Solutions Chapter 9 Differential Equations

• Chapter 9 Differential Equations Ex 9.1
• Chapter 9 Differential Equations Ex 9.2
• Chapter 9 Differential Equations Ex 9.3
• Chapter 9 Differential Equations Ex 9.4
• Chapter 9 Differential Equations Ex 9.5
• Chapter 9 Differential Equations Ex 9.6
• Chapter 9 Differential Equations Ex 9.7
• Chapter 9 Differential Equations Ex 9.8
• Chapter 9 Differential Equations Chapter Test
• Chapter 9 Differential Equations MCQs

Class 12 ISC Maths Solutions Chapter 10 Probability

• Chapter 10 Probability Ex 10.1
• Chapter 10 Probability Ex 10.2
• Chapter 10 Probability Ex 10.3
• Chapter 10 Probability Ex 10.4
• Chapter 10 Probability Ex 10.5
• Chapter 10 Probability Ex 10.6
• Chapter 10 Probability Ex 10.7
• Chapter 10 Probability Ex 10.8
• Chapter 10 Probability Ex 10.9
• Chapter 10 Probability Ex 10.10
• Chapter 10 Probability Chapter Test
• Chapter 10 Probability MCQs

Section B

ISC Class 12 Maths Solutions Chapter 1 Vectors

• Chapter 1 Vectors Ex 1.1
• Chapter 1 Vectors Ex 1.2
• Chapter 1 Vectors Ex 1.3
• Chapter 1 Vectors Ex 1.4
• Chapter 1 Vectors Ex 1.5
• Chapter 1 Vectors Chapter Test
• Chapter 1 Vectors MCQs

ISC Class 12th Maths Solutions Chapter 2 Three Dimensional Geometry

• Chapter 2 Three Dimensional Geometry Ex 2.1
• Chapter 2 Three Dimensional Geometry Ex 2.2
• Chapter 2 Three Dimensional Geometry Ex 2.3
• Chapter 2 Three Dimensional Geometry Ex 2.4
• Chapter 2 Three Dimensional Geometry Ex 2.5
• Chapter 2 Three Dimensional Geometry Ex 2.6
• Chapter 2 Three Dimensional Geometry Ex 2.7
• Chapter 2 Three Dimensional Geometry Ex 2.8
• Chapter 2 Three Dimensional Geometry Chapter Test
• Chapter 2 Three Dimensional Geometry MCQs

Maths Class 12 ISC Solutions Chapter 3 Applications of Integrals

• Chapter 3 Applications of Integrals Ex 3.1
• Chapter 3 Applications of Integrals Ex 3.2
• Chapter 3 Applications of Integrals Chapter Test
• Chapter 3 Applications of Integrals MCQs

Section C

ISC 12th Maths Solutions Chapter 1 Application of Calculus in Commerce and Economics

• Chapter 1 Application of Calculus in Commerce and Economics Ex 1.1
• Chapter 1 Application of Calculus in Commerce and Economics Ex 1.2
• Chapter 1 Application of Calculus in Commerce and Economics Ex 1.3
• Chapter 1 Application of Calculus in Commerce and Economics Ex 1.4
• Chapter 1 Application of Calculus in Commerce and Economics Chapter Test
• Chapter 1 Application of Calculus in Commerce and Economics MCQs

Maths ISC Class 12 Solutions Chapter 2 Linear Regression

• Chapter 2 Linear Regression Ex 2
• Chapter 2 Linear Regression Chapter Test
• Chapter 2 Linear Regression MCQs

ISC Class 12 Mathematics Solutions Chapter 3 Linear Programming

• Chapter 3 Linear Programming Ex 3.1
• Chapter 3 Linear Programming Ex 3.2
• Chapter 3 Linear Programming Chapter Test
• Chapter 3 Linear Programming MCQs

Case Study Based Questions

• Case Study Based Questions Section A Chapter 1-7
• Case Study Based Questions Section A Chapter 8-10
• Case Study Based Questions Section B
• Case Study Based Questions Section C

Access to comprehensive ML Aggarwal Maths for Class 12 Solutions Chapter 7 Applications of Derivatives Chapter Test encourages independent learning.

ML Aggarwal Class 12 Maths Solutions Section A Chapter 7 Applications of Derivatives Chapter Test

Question 1.
A circular disc is being heated. Due to heat its radius increases at the rate of 005 cm / sec. Find the rate at which its area is increasing when the radius is 3.2 cm. (NCERT)
Solution:
Let r be the radius of circle
Then A = area of circle = πr²
Differentiating w.r.t. t, we have
\(\frac{d A}{d t}\) = 2πr \(\frac{d r}{d t}\) ……….(1)
given, \(\frac{d r}{d t}\) = 0.05 cm/sec ;
r = 3.2 cm
∴ from (1); we have
\(\frac{d A}{d t}\) = (2π × 3.2 × 0.05) cm²/sec
= 0.32 π cm²/sec
Hence, the required rate at which area is increasing be 0.32 π cm²/sec.

Question 2.
The radius of a circular blot of ink is increasing at 0.5 cm per minute. Find the rate of increase of the area when the radius is equal to 5 cm. Find also the approximate change in the area when the radius increases from 5 cm to 5.001 cm.
Solution:
LetA be the area of circular blot of radius r
Then A = πr² ……………(1)
On differentiating eqn. (1) w.r.t. t, we have
\(\frac{d A}{d t}\) = 2πr \(\frac{d r}{d t}\) ……….(2)
given \(\frac{d r}{d t}\) = 0.5 cm / mm ; r = 5 cm
∴ from eqn. (2) ; we have
\(\frac{d A}{d t}\) = (2π × 5 × 0.5) cm²/min
= 5π cm²/min.
Hence the required rate at which area is increasing be 5π cm²/min.
Diff. (1) w.r.t. r, we have
\(\frac{d A}{d r}\) = 2πr
Take r = 5 ;
r + δr = 5.001
∴ δr = dr = 5.001 – 5 = 0.001
Now, dA = \(\frac{d A}{d r}\) dr
= 2πr × δr
= 2π × 5 × 0.001 = 0.01π
∴ δA = 0.01π [∵ δA = dA]
Thus approximate change in area = δA = 0.01 π cm².

Question 3.
A balloon in the form of a right circular cone surmounted by a hemi-sphere, having a diameter equal to the height of the cone, is being inflated by pumping air into it. How fast is its volume changing with respect to its total height h, when h = 9 cm?
Solution:
Let r be the radius of hemisphere and base radius of cone.
It is given that diameter of hemisphere is equal to height of cone.

Let x be the height of cone.
∴ x = 2r
given total height h = x + r
= 2r + r = 3r
∴ r = \(\frac{h}{3}\)
∴ V = Volume of solid
= \(\frac{\pi}{3}\) r²x + \(\frac{2}{3}\) πr³
V = \(\frac{\pi}{3}\left(\frac{h}{3}\right)^2 \times 2 \times \frac{h}{3}+\frac{2}{3} \pi\left(\frac{h}{3}\right)^3\)
= \(\frac{4 \pi}{3}\left(\frac{h}{3}\right)^3\)
= \(\frac{4 \pi}{81} h^3\)
∴ \(\frac{4 \pi}{81}\) 3h²
= \(\frac{4 \pi}{27}\) h²
at h = 9 cm ;
\(\frac{4 \pi}{27}\) × 9²
= 12π cm³ / cm.

Question 4.
Find the equation of the normal to the curve 2y + x² = 3 at the point (1, 1).
Solution:
Given equation of curve be, 2y + x² = 3 …………(1)
Diff. eqn. (1) w.r.t. x; we have
2 \(\frac{d y}{d x}\) + 2x = 0
\(\frac{d y}{d x}\) = – x
∴ slope of Normal to curve at (1, 1)
= \(-\frac{1}{\left(\frac{d y}{d x}\right)_{(1,1)}}=\frac{-1}{-1}\) = – 1
Thus, the required eqn. of normal to given curve at point (1, 1) is given by
y – 1 = 1 (x – 1)
⇒ x = y
⇒ x – y = 0.

Question 5.
Find the equation of the tangent to the curve y = x³ – x² at the point (2, 4). Also find the equation of tangents to the curve which are parallel to x-axis.
Solution:
Given equation of curve be,
y = x³ – x²
Diff. eqn. (1) both sides w.r.t. x, we have
\(\frac{d y}{d x}\) = 3x² – 2x
∴ slope of tangent to curve at point (2, 4) = (\(\frac{d y}{d x}\))_{(2, 4)}
Thus, the eqn. of tangent to curve at point (2, 4) is given by
y – 4 = 8 (x – 2)
⇒ 8x – y – 12 = 0
Let (x₁, y₁) be any point on curve (1).
∴ y₁ = x₁³ – x₁² ………….(2)
Since the tangent to curve is parallel to x – axis.
∴ (\(\frac{d y}{d x}\))_{(x1, y1)} = 0
⇒ 3x₁² – 2x₁ = 0
x₁ = 0, 2/3
∴ from (2);
y₁ = 0, – \(\frac{4}{27}\)
Thus, the points on curve at which tangent is parallel to x-axis are (0, 0) and \(\left(\frac{2}{3},-\frac{4}{27}\right)\).
Now, eqn. of langent to given curve at (0, 0) is given by
y – 0 = 0 (x – 0)
⇒ y = 0
and eqn. of tangent to given curve at point \(\left(\frac{2}{3},-\frac{4}{27}\right)\) is given by
y + \(\frac{4}{27}\) = 0
⇒ 27y + 4 = 0.

Question 6.
Find the equations of the tangents to the curve y = (x³ – 1) (x – 2) at the points where the curve cuts the x-axis.
Solution:
Given curve be,
y = (x³ – 1) (x – 2) …………….(1)
It cuts x-axis
∴ y = 0
⇒ (x³ – 1) (x – 2) = 0
⇒ x = 1, 2
eqn. (1) cuts x-axis at (1, 0) and (2, 0).
Diff. (1) w.r.t. x, we get
\(\frac{d y}{d x}\) = 4x³ – 6x² – 1
∴ slope of tangent to given curve at (1, 0) = \(\left.\frac{d y}{d x}\right]_{(1,0)}\)
= 4 – 6 – 1 = – 3
and slope of to given curve tangent at (2, 0) = \(\left.\frac{d y}{d x}\right]_{(2,0)}\)
= 32 – 24 – 1 = 7
eqn. of tangent at (1, 0) is given by
y – 0 = – 3 (x – 1)
⇒ 3x + y – 3 = 0
and eqn. of tangent at (2, 0) is given by y – 0 = 7 (x – 2)
⇒ 7x – y – 14 = 0.

Question 7.
Find the equation of the tangent to the curve y = 3x³ + 4x² + 2x at the point whose abscissa is – 1. Also find the coordinates of another point on the curve at which the tangent is parallel to that already obtained.
Solution:
Given eqn. of curve be,
y = 3x³ + 4x² + 2x ………..(1)
Duff. eqn. (1) w.r.t. x, we have
\(\frac{d y}{d x}\) = 9x² + 8x + 2
slope of tangent to given curve (1) at x = – 1
= (\(\frac{d y}{d x}\))_{x = – 1}
= 9 – 8 + 2 = 3
When x = – 1
∴ from (1);
y – 3 + 4 – 2 = – 1
Thus the point on curve (1) be (- 1, – 1).
Hence the eqn. of tangent to given curve (1) at p (- 1, – 1) is given by
y + 1 = 3 (x + 1)
⇒ 3x – y + 2 = 0 ………….(2)
Let (x1, y) be any point on curve (1).
∴ y₁ = 3x₁³ + 4x₁² + 2x₁ ………..(3)
Thus slope of tangent to curve (1) at (x₁, y₁) = (\(\frac{d y}{d x}\))_{(x1, y1)}
= 9x₁² + 8x₁ + 2
Since the tangent is parallel to eqn. (2) whose slope be \(\frac{-3}{-1}\) = 3
Thus, 9x₁² + 8x₁ + 2 = 3
⇒ 9x₁² + 8x₁ – 1 = 0
⇒ (x₁ + 1) (9x₁ – 1) = 0
⇒ x₁ = – 1, \(\frac{1}{9}\)
when x₁ = \(\frac{1}{9}\)
∴ from (2) ; we have
y₁ = \(\frac{1}{243}+\frac{4}{81}+\frac{2}{9}\)
= \(\frac{1+12+54}{243}\)
⇒ y₁ = \(\frac{67}{243}\)
Hence the required point on given curve be (\(\frac{1}{9}\), \(\frac{67}{243}\)).

Question 8.
Find the slope of the tangent to the curve x = t² + 31 – 8, y = 2t² – 2t – 5 at the point (2, – 1). (NCERT)
Solution:
Given eqn. of curve be;
x = t² + 3t – 8 …………(1)
and y = 2t² – 2t – 5 ………….(2)
Duff. eqn. (1) and eqn. (2) ; w.r.t. r, we have
∴ \(\frac{d x}{d t}\) = 2t + 3;
\(\frac{d y}{d t}\) = 4t – 2
Thus, \(\frac{d y}{d x}\) = \(\frac{\frac{d y}{d t}}{\frac{d x}{d t}}\)
= \(\frac{4 t-2}{2 t+3}\) …………..(3)
at given point (2, – 1);
⇒ t² + 3t – 8 = 2
⇒ t² + 3t – 10 = 0
⇒ (t – 2) (t + 5) = 0
⇒ t = 2, – 5 and y = 1
⇒ – 1 = 2t² – 2t – 5
⇒ 2t² – 2t – 4 = 0
⇒ t² – t – 2 = 0
⇒ (t + 1) (t – 2) = 0
⇒ t = – 1 , 2
Hence the common value of be 2.
∴ slope of tangent to given curve at t = 2 = (\(\frac{d y}{d x}\))_{t = 2}
= \(\frac{8-2}{4+3}=\frac{6}{7}\)

Question 9.
Find the equation of the tangent to the curve x = a sin³ t, y = b cos³ t at any point ‘t’.
Solution:
Curve eqn. of curve be
x = a sin³ t
and y = b cos³ t
Diff. eqn. (1) and eqn. (2) w.r.t. t, we have
\(\frac{d x}{d t}\) = 3a sin² t cos t;
\(\frac{d y}{d t}\) = 3b cos² t (- sin t)
∴ \(\frac{d y}{d x}=\frac{\frac{d y}{d t}}{\frac{d x}{d t}}\)
= \(\frac{-3 b \cos ^2 t \sin t}{3 a \sin ^2 t \cos t}\)
= – \(\frac{b}{a}\) cot t
Now at any point t, we mean the point (a sin³ t, b cos³ t) on given curve.
Thus, required equation of tangent to given curve at (a sin³ t, b cos³ t) is given by
y – b cos³ t = – \(\frac{b}{a}\) cot t (x – a sin³ t)
⇒ ay – ab cos³ t = – bx cot t + ba sin³ t cot t
⇒ ay + bx cot t = ab cos t (sin² t + cos² t)
⇒ ay + bx cot t = ab cos t
⇒ \(\frac{a y}{a b \cos t}+\frac{b x \cot t}{a b \cos t}\) = 1
⇒ \(\frac{y}{b \cos t}+\frac{x}{a \sin t}\) = 1

Question 10.
Find the condition for the curves \(\frac{x^2}{a^2}-\frac{y^2}{b^2}\) = 1 and xy = c² to intersect orthogonally. (NCERT Exampler)
Solution:
Given eqns. of curves be
\(\frac{x^2}{a^2}-\frac{y^2}{b^2}\) = 1 ………(2)
and xy = c²
Let (x₁, y₁) be the point of intersection of two given curves (1) and (2).
Diff. (1) both sides w.r.t. x; we get
\(\frac{2 x}{a^2}-\frac{2 y}{b^2} \frac{d y}{d x}\) = 0
∴ \(\frac{d y}{d x}=\frac{b^2 x}{a^2 y}\)
Thus, m₁ = \(\left(\frac{d y}{d x}\right)_{\left(x_1, y_1\right)}=\frac{b^2 x_1}{a^2 y_1}\)
Diff. (2) both sides w.r.t. x; we get
x \(\frac{d y}{d x}\) + y = 0
⇒ \(\frac{d y}{d x}\) = – \(\frac{y}{x}\)
Now two curves intersect orthogonally iff m₁m₂ = – 1
⇒ \(\left(\frac{b^2 x_1}{a^2 y_1}\right)\left(-\frac{y_1}{x_1}\right)\) = – 1
⇒ b² = a²

Question 11.
Prove that the curves y² = 4ax and xy = c² cut at right angles if c⁴ = 32a⁴.
Solution:
Given curves are y² = 4ax ……….(1)
and xy = c² …………(2)
From (1) and (2) ; we have
\(\frac{c^4}{x^2}\) = 4
⇒ c⁴ – ax³ = 0
⇒ x = \(\left(\frac{c^4}{4 a}\right)^{1 / 3}\)
∴ From (2) ; we have
y = \(\frac{c^2}{x}=\frac{c^2}{c^{4 / 3}}(4 a)^{1 / 3}\)
= \(\frac{(4 a)^{1 / 3}}{c^{-2 / 3}}\)
Thus both curves intersect at \(\mathrm{P}\left[\left(\frac{c^4}{4 a}\right)^{1 / 3}, \frac{(4 a)^{1 / 3}}{c^{-2 / 3}}\right]\)
Diff. (1) w.r.t. x, we have
\(\frac{d y}{d x}=\frac{2 a}{y}\)
∴ m₁ = \(\left.\frac{d y}{d x}\right]_{\mathrm{at} P}=\frac{(2 a) \cdot c^{-2 / 3}}{(4 a)^{1 / 3}}\)
Diff. (2) w.r.t. x, we have
∴ m₂ = \(\frac{d y}{d x}\)]_{at P}
= – \(\frac{(4 a)^{1 / 3}}{c^{-2 / 3}} \times \frac{(4 a)^{1 / 3}}{c^{4 / 3}}\)
= – \(\frac{(4 a)^{2 / 3}}{c^{2 / 3}}\)
Since the given curves cut right angles at P
∴ m₁m₂ = – 1
⇒ \(\frac{(2 a) c^{-2 / 3}}{(4 a)^{1 / 3}} \cdot\left(-\frac{(4 a)^{2 / 3}}{c^{2 / 3}}\right)\) = – 1
⇒ c^4/3 = 2 . 4^1/3 a^4/3
On cubing ; we have
c⁴ = 32 a⁴.

Question 12.
Using differentials, find the approximate values of:
(i) (81.5)^1/4 (NCERT)
(ii) \(\left(\frac{17}{81}\right)^{1 / 4}\) (NCERT)
(iii) (33)^-1/5 (NCERT)
Solution:
(i) Let y = f(x) = x^1/4
Take x = 81,
x + ∆x = 81.5
⇒ ∆x = 0.5
when x = 81, then y (81)^1/4 = 3,
Let dx = ∆x = 0.5
Now y = x^1/4
∴ \(\frac{d y}{d x}=\frac{1}{4} x^{\frac{-3}{4}}\)
∴ \(\left.\frac{d y}{d x}\right]_{x=81}=\frac{1}{4}(81)^{-3 / 4}=\frac{1}{108}\)
∴ dy = \(\frac{d y}{d x}\) dx
= \(\frac{1}{108}\) × 0.5
= \(\frac{0.5}{108}\)
∴ (81.5)^1/4 = y + ∆y
= 3 + \(\frac{0.5}{108}\) = 3.004

(ii) Let y = f(x) = x^1/4
Take x = \(\frac{16}{81}\)
⇒ x + ∆x = \(\frac{17}{81}\)
∴ ∆x = \(\frac{1}{81}\)
when x = \(\frac{16}{81}\)
then y = \(\left(\frac{16}{81}\right)^{1 / 4}=\frac{2}{3}\)
Let ∆x = dx = \(\frac{1}{81}\)
Now y = x^1/4
⇒ \(\frac{d y}{d x}\) = \(\frac{1}{4}\) x^{– 3/4}
⇒ \(\left.\frac{d y}{d x}\right]_{x=\frac{16}{81}}=\frac{1}{4}\left(\frac{16}{81}\right)^{-3 / 4}\)
= \(\frac{1}{4}\left(\frac{2}{3}\right)^{-3}\)
= \(\frac{1}{4}\left(\frac{3}{2}\right)^3=\frac{27}{32}\)
∴ dy = \(\frac{d y}{d x}\) ∆x
= \(\frac{27}{32} \times \frac{1}{81}=\frac{1}{96}\)
= ∆y [∵ dy ≅ ∆y]
∴ \(\left(\frac{17}{81}\right)^{1 / 4}\) = y + ∆y
= \(\frac{2}{3}+\frac{1}{96}\) = 0.677.

(iii) Let y = f(x) = x^{– 1/5}
Take x = 32,
x + ∆x = 33
∆x = 1
When x = 32 then y = (32)^{– 1/5} = \(\frac{1}{2}\)
Let dx = ∆x = 1
Now y = x^{– 1/5}
⇒ \(\frac{d y}{d x}\) = – \(\frac{1}{5}\) x^{– 6/5}
∴ \(\frac{d y}{d x}\)]_{x = 32} = – \(\frac{1}{5}\) x^{– 6/5}
= – \(\frac{1}{320}\)
∴ dy = \(\frac{d y}{d x}\) dx
= – \(\frac{1}{320}\) × 1
= – \(\frac{1}{320}\) = ∆y
∴ (33)^{– 1/5} = y + ∆y
= \(\frac{1}{2}-\frac{1}{320}\) = 0.497.

Question 13.
Find the approximate change in the volume of a cube ofsidex metres caused by increasing the side by 3%.(NCERT)
Solution:
Given the length of each side of cube be x m.
Then volume of cube = V = x³
∴ \(\frac{d V}{d x}\) = 3x²
Let ∆x be the change in x and the corresponding change in V be ∆V.
∴ ∆V = \(\frac{d V}{d x}\) ∆x [∵ ∆V = dV ; ∆x = dx]
= 3x² ∆x
also it is given that \(\frac{\Delta x}{x}\) × 100 = 3
∆x = \(\frac{3 x}{100}\) = 0.03x
∴ ∆V = 3x² × 003x = 0.09 x³ m³.

Question 14.
If f(x) = 3x² + 15x + 5, then find the approximate value of f (3.02). (NCERT)
Solution:
Given f(x) = y = 3x² + 15x + 5
Take x = 3, x + ∆x = 3.02
⇒ ∆x = 0.02
When x = 3 then y = 3.3² + 153 + 5
⇒ y = 27 + 45 + 5 = 77,
Let ∆x = dx = 0.02
Now y = 3x² + 15x + 3
∴ \(\frac{d y}{d x}\) = 6x + 15
⇒ \(\frac{d y}{d x}\)]_{x = 3} = 6 × 3 + 15 = 33
∴ dy = \(\frac{d y}{d x}\) ∆x
= 33 × 0.02
= 0.66 = ∆y
[∵ ∆y ≅ dy]
∴ f (3.02) = y + ∆y
= 77 + 0.66 = 77.66.

Question 15.
Show that the relative error in computing the volume of a sphere, due to an error in measuring its radius, is approximately equal to three times the relative error in the radius.
Solution:
Let r be the radius of sphere
Then V = volume of sphere = \(\frac{4 \Pi}{3}\) r³
Taking logarithm on both sides, we get
log V = log \(\frac{4 \Pi}{3}\) + log r³
log V = log \(\frac{4 \Pi}{3}\) + 3 log r
On differentiating ; we get
\(\frac{1}{V}\) dV = 0 + \(\frac{3}{r}\) dr
[∵ dV = ∆V and dr = ∆r]
\(\frac{\Delta \mathrm{V}}{\mathrm{V}}=3 \times\left(\frac{\Delta r}{r}\right)\)
∴relative error in V = 3 × (relative error in r)
Hence, the relative error in computing the volume of sphere is equal to 3 times the relative error in radius of sphere.

Question 16.
Determine for which values of x, the function f(x) = \(\frac{x}{x^2+1}\) is increasing and for which values of x, it is decreasing. Find also the points on the graph of the function at which the tangent is parallel to x-axis.
Solution:
Given f(x) = \(\frac{x}{x^2+1}\)
Diff. both sides w.r.t. x, we get
f'(x) = \(=\frac{\left(x^2+1\right) \cdot 1-x \cdot 2 x}{\left(x^2+1\right)^2}\)
= \(\frac{1-x^2}{\left(1+x^2\right)^2}\)
Now f'(x) ≥ o iff \(\frac{1-x^2}{\left(1+x^2\right)^2}\) ≥ 0
⇒ 1 – x² ≥ 0
[∵ (1 + x²)² > 0 ∀ x ∈ R]
⇒ x² ≤ 1
⇒ |x| ≤ 1
⇒ – 1 ≤ x ≤ 1
Thus, f(x) is increasing in [- 1, 1]
Now f'(x) ≤ 0 iff \(\frac{1-x^2}{\left(1+x^2\right)^2}\) > 0
⇒ 1 – x² ≤ 0
[∵ (1 + x²)² > 0 ∀ x ∈ R]
⇒ x² ≥ 1
⇒ |x| ≥ 1
⇒ x ≥ 1 or x ≤ – 1
Thus, the function f(x) is decreasing in (- ∞, – 1] ∪ [1, ∞).
Let (x₁, y₁) be any point on given curve
∴ y₁ = \(\frac{x_1}{x_1^2+1}\) …………(2)
Now slope of tangent to given curve at (x₁, y₁) = [f'(x)]_{(x1, y1)}
= \(\left(\frac{d y}{d x}\right)_{\left(x_1, y_1\right)}\)
= \(\frac{1-x_1^2}{\left(1+x_1^2\right)^2}\)
Also tangent to given curve is || to x-axis.
∴ (\(\frac{d y}{d x}\))_{(x1, y1)} = 0
⇒ \(\frac{1-x_1^2}{\left(1+x_1^2\right)^2}\) = 0
⇒ 1 – x₁² = 0
⇒ x₁ = ± 1
When x₁ = 1
∴ from (2) ;
y₁ = \(\frac{1}{1+1}=\frac{1}{2}\)
When x₁ = – 1
∴ from (2) ;
y₁ = \(\frac{-1}{1+1}=-\frac{1}{2}\)
Hence the required points on graph of function at which tangent is || to x-axis be (1, \(\frac{1}{2}\)) and (- 1, – \(\frac{1}{2}\)).

Question 17.
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. (NCERT)
Solution:
Given, f(x) = (x – 2)⁴ (x + 1)³
Diff. both sides w.r.t. x ; we have
f'(x) = (x – 2)⁴ 3 (x + 1)² + (x + 1)³ 4 (x – 2)³
= (x + 1)² (x – 2)³ [3 (x – 2) + 4 (x + 1)]
= (x + 1)² (x – 2)³ (7x – 2)
For critical points, f’ (x) = 0
⇒ (x + 1)² (x – 2)³ (7x – 2) = 0
⇒ x = – 1, 2, \(\frac{2}{7}\)

Case – I:
at x = – 1
When x slightly < – 1
⇒ x + 1 < 0 also x < – 1 < 2
⇒ x – 2 < 0
∴ f’ (x) = (+ve) (- ve) (- ve) = + ve
When x slightly > – 1
⇒ x + 1 > 0, x – 2 < 0
f’ (x) = (+ ve) (- ve) (- ve) = + ve
So f’ (x) does not changes its sign as we move from slightly < – 1 to slightly > – 1.
∴ x = – 1 be a point of neither maxima nor minima.
Hence x = – 1 be a point of inflexion.

Case-II :
at x = 2
When x slightly < 2
⇒ x – 2 < 0 but 7x – 2 > 0
∴f’ (x) = (+ve) (- ve) (+ve) = – ve
When x slightly > 2
⇒ x – 2 > 0 and 7x – 2 > 0
∴ f’ (x) = (+ ve) (+ ve) (+ ve) = + ve
Thus, f ‘(x) changes its sign from -ve to + ve as we move from slightly < 2 to slightly > 2
∴ x = 2 is a point of minima.

Case – III :
at x = \(\frac{2}{7}\)
When x slightly < \(\frac{2}{7}\)
⇒ 7x – 2 < 0 and x – 2 < 0 ∴ f'(x) = (+ ve) (- ve) (- ve) = + ve When x slightly > \(\frac{2}{7}\)
⇒ 7x – 2 > 0 and x – 2 < 0
∴ f’ (x) = (+ ve) (- ve) (+ ve) = – ve
Thus, f’ (x) changes its sign from + ve to – ve as we move from slightly < \(\frac{2}{7}\) to 2 slightly > \(\frac{2}{7}\).
∴ x = \(\frac{2}{7}\) be a point of local maxima.

Question 18.
Find the points of local maximum and minima (if any) of the function :
f(x) = 2 cos x + x in [0, π].
Find also the absolute maximum and minimum values.
Solution:
Given f(x) = 2 cos x + x
Diff. both sides w.r.t. x, we have
f’ (x) = – 2 sin x + 1
f” (x) = – 2 cos x
For maxima/minima, f’ (x) = 0
⇒ – 2 sin x + 1 = 0
⇒ sin x = \(\frac{1}{2}\) = sin \(\frac{\pi}{6}\)
⇒ x = nπ + (- 1)ⁿ \(\frac{\pi}{6}\) ∀ n ∈ I
but x ∈ [0, π]
∴ x = \(\frac{\pi}{6}\), \(\frac{5 \pi}{6}\) ∈ [0, π]
Thus, the critical points are 7t 571
x = 0, \(\frac{\pi}{6}\), \(\frac{5 \pi}{6}\), π
Now at x = \(\frac{\pi}{6}\)
f” (x) = – 2cos \(\frac{\pi}{6}\)
= – 2 × \(\frac{\sqrt{3}}{2}\)
= – √3 < 0
∴ x = \(\frac{\pi}{6}\) be a point of local maxima.
and f” (\(\frac{5 \pi}{6}\)) = – 2 cos (π – \(\frac{\pi}{6}\))
∴ x = \(\frac{5 \pi}{6}\) be a point of local minima.
Also f” (0) = – 2 cos 0 = – 2 < 0 ∴ x = 0 be a point of local maxima, and f” (π) = – 2 cos π = 2 > 0
∴ x = π be a point of local minima.
∴ f(0) = 2 cos 0 + 0 = 2
f(π) = 2 cos π + π = – 2 + π
f(\(\frac{\pi}{6}\)) = 2 cos \(\frac{\pi}{6}\) + \(\frac{\pi}{6}\)
= √3 + \(\frac{\pi}{6}\)
f(\(\frac{5 \pi}{6}\)) = 2 cos \(\frac{5 \pi}{6}\) + \(\frac{5 \pi}{6}\)
= – √3 + \(\frac{5 \pi}{6}\)
∴ absolute minimum value = – √3 + \(\frac{5 \pi}{6}\)
and absolute maximum value = √3 + \(\frac{\pi}{6}\).

Question 19.
Find the maximum value of the function 2x³ – 24x + 107 in the interval [1, 3]. Also find the maximum value of the same function in the interval [- 3, – 1]. (NCERT)
Solution:
Given f(x) = 2x³ – 24x + 107
∴ f (x) = 6x² – 24
= 6 (x – 2) (x + 2)
For critical points we have f’ (x) = 0
⇒ 6 (x – 2) (x + 2) = 0
⇒ x = 2, – 2
if f(x) be defined on [1, 3]
∴ f’ (x) = 0 at x = 2
Now let us compute the values of f(x) at critical point x = 2 and also at the end point of given interval [1, 3].
∴ f(2) = 16 – 48 + 107 = 123 – 48 = 75
f(1) = 2 – 24 + 107 = 109 – 24 = 85
f(3) = 54 – 72 + 107 = 161 – 72 = 89
out of these values, maximum value of f(x) be f(3) = 89
∴ absolute max. value = 89 at x = 3.
if f(x) be defined in [- 3, – 1]
Then f’ (x) = 0 at x = – 2.
Thus we compute the values of f (x) at critical point x = – 2 and also at the end points of given interval [- 3, – 1]
∴ f(- 2) = – 16 + 48 + 107 = 139
f (- 1) = -2 + 24 + 107 = 129
f(- 3) = – 54 + 72 + 107 = 125
Out of these values, maximum value of f (x) be f(- 2) = 139.
Hence, absolute maximum value = 139 at x = – 2.

Question 20.
OABC is a square of side 1 m. A point X in AB and a point Y in BC are such that AX = x m and BY = k x m and BY is longer than AX. For a given value of k, show that the minimum area of AOXY is ((4k – 1)/8k) m².
Solution:
Given OABC be a square s.t OA = AB = BC = OC = 1 m.

Given AX = x m and BY = kx
and BY > AX
From figure, it is clear that
Let S = area of ∆OXY
= area of square OABC – area of ∆OAX – area of ∆XBY – area of ∆OCY
⇒ S = 1² – \(\frac{1}{2}\) × OA × AX – \(\frac{1}{2}\) × BY × BX – \(\frac{1}{2}\) × OC × CY
⇒ S = 1 – \(\frac{1}{2}\) × 1 × x – \(\frac{1}{2}\) × kx (1 – x) – \(\frac{1}{2}\) × 1 × (1 – kx)
⇒ S = 1 – \(\frac{x}{2}-\frac{k x}{2}+\frac{k x^2}{2}-\frac{1}{2}+\frac{k x}{2}\)
S = \(\frac{1}{2}-\frac{x}{2}+\frac{k x^2}{2}\)
On differentiating w.r.t. x, we have
\(\frac{d \mathrm{~S}}{d x}\) = – \(\frac{1}{2}\) + kx ;
\(\frac{d^2 S}{d x^2}\) = k
For maxima/minima, \(\frac{d \mathrm{~S}}{d x}\) = 0
⇒ – \(\frac{1}{2}\) + kx = 0
⇒ x = \(\frac{1}{2 k}\)
∴ \(\left(\frac{d^2 \mathrm{~S}}{d x^2}\right)_{x=\frac{1}{2 k}}\) = k > 0
Hence S is minimise for x = \(\frac{1}{2 k}\)
∴ Minimum area of ∆OXY = S = \(\frac{1}{2}\left[1-\frac{1}{2 k}+k \times \frac{1}{4 k^2}\right]\)
= \(\frac{1}{2}\left[1-\frac{1}{4 k}\right]\)
= \(\left(\frac{4 k-1}{8 k}\right)\) m².

Question 21.
Find the area of the largest isosceles triangle having perimeter 18 metres.
Solution:
Let x metre be the lengths of equal side of isosceles triangle.
Since the perimeter of an isosceles ∆ be 18 cm.
∴ length of base of isosceles A be (18 – 2x) m
Then by Heron’s formula, we have
A = area of isosceles triangle = \(\sqrt{s(s-a)(s-b)(s-c)}\) ………..(1)
given s = \(\frac{a+b+c}{2}=\frac{18}{2}\) = 9
from (1) ; we have
A = \(\sqrt{9(9-x)(9-x)(9-18+2 x)}\)
= \(\sqrt{9(9-x)^2(2 x-9)}\)
Since the sum of two sides of ∆ is greater than third side.
x + x > 18 – 2x
⇒ 4x > 18
⇒ x > \(\frac{9}{2}\)
and x – x< 18 – 2x
⇒ 0 < 18 – 2x
⇒ 2x < 18
⇒ x < 9
Thus \(\frac{9}{2}\) < x < 9.
∴ A = 3 (9 – x) \(\sqrt{2 x-9}\)
Now A is maximum when A² is maximum.
Let P = A² = 9 (9 – x)² (2x – 9)
Diff. both sides w.r.t. x, we have
\(\frac{d P}{d x}\) = 9 [(9 – x)² . 2 + (2x – 9) 2 (9 – x) (- 1)]
= 18 (9 – x) [9 – x – 2x + 9]
= 18 (9 – x) (18 – 3x)
= 54 (9 – x) (6 – x)
= 54 (x² – 15x + 54)
and \(\frac{d^2 P}{d x^2}\) = 54 (2x – 15)
For maxima/minima, \(\frac{d P}{d x}\) = 0
⇒ 54 (x² – 15x + 54) = 0
⇒ (9 – x) (6 – x) = 0
⇒ x = 6, 9
But \(\frac{9}{2}\) < x < 9
∴ x = 6
∴ \(\left(\frac{d^2 \mathrm{P}}{d x^2}\right)_{x=6}\) = 54 (12 – 15) = – 162 < 0
Thus x = 6 is a point of maxima.
∴ P is maximum for x = 6
Thus A is maximum for x = 6.
and Maximum area of isosceles triangle = 3 (9 – 6) \(\sqrt{12-9}\) = 9√3 m².

Question 22.
Divide the number 4 into two positive numbers such that the sum of square of one and the cube of other is minimum.
Solution:
Let the required two positive numbers bex and 4 – x
since the sum of two positive numbers be 4.
Let P = (4 – x)² + x³ ;
Diff. both sides w.r.t. x, we get
\(\frac{d P}{d x}\) = 2 (4 – x) (- 1) + 3x²
∴ \(\frac{d^2 P}{d x^2}\) = 2 + 6x
For maxima/minima, \(\frac{d P}{d x}\) = 0
⇒ – 8 + 2x + 3x² = 0
⇒ (x + 2) (3x – 4) = 0
⇒ x = – 2, \(\frac{4}{3}\)
but x > 0
∴ x = \(\frac{4}{3}\)
Thus, \(\left(\frac{d^2 \mathrm{P}}{d x^2}\right)_{x=\frac{4}{3}}\) = 2 + 6 × \(\frac{4}{3}\)
= 2 + 8
= 10 > 0
∴ x = \(\frac{4}{3}\) be a point of minima.
∴ P is minimise at x = \(\frac{4}{3}\).
Hence, the required numbers are \(\frac{4}{3}\) and 4 – \(\frac{4}{3}\).
i.e. \(\frac{4}{3}\) and \(\frac{8}{3}\).

Question 23.
Find the point on the curve y² = 2x which is nearest to the point A (1, – 4).
Solution:
Let P (x, y) be any point on the curve
y² = 2x ………….(1)
and let the given point be A (1, – 4).
∴ AP² = (x – 1)² + (y + 4)²
and let S = AP².
Then S is max/min. according as AP is maximum/minimum.
Now S = \(\left(\frac{y^2}{2}-1\right)^2\) + (y + 4)²
∴ \(\frac{d S}{d y}\) = 2 \(\left(\frac{y^2}{2}-1\right)\) y + 2 (y + 4)
= y³ + 8
For Max/Min. \(\frac{d S}{d y}\) = 0
∴ y³ = – 8
= (- 2)³
⇒ y = – 2
∴from (1) ; x = 2
Now \(\frac{d^2 S}{d y^2}\) = 3y²
∴ \(\left(\frac{d^2 \mathrm{~S}}{d y^2}\right)_{y=-2}\) = 12 > 0
∴ S is minimise for y = – 2 and x = 2 and hence the required point be (2, – 2).

Question 24.
A wire of length 25 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 lengths of the two pieces so that the combined area of the square and the circle Is minimum ?
Solution:
Let the length of one piece of wire be x m
∴ Length of other piece must be (25 – x) m.
Let ‘x’ metres be made into circle and
(25 – x) metres be made into square.
So perimeter of circle = x = 2πr,
where r = radius of circle
∴ r = \(\frac{x}{2 \pi}\)
thus area of circle = πr²
Further perimeter of square = 4 × side
∴ side of square = \(\frac{25-x}{4}\)
Thus area of square = \(\left(\frac{25-x}{4}\right)^2\)
Let A = combined area of circle and square
= π \(\frac{x^2}{4 \pi^2}+\left(\frac{25-x}{4}\right)^2\)

∴ A is manimise for x = \(\frac{25 \pi}{\pi+4}\)
Hence the length of two pieces are \(\), 25 – \(\frac{25 \pi}{\pi+4}\)
i.e. \(\frac{25 \pi}{\pi+4}\) m and \(\frac{100}{\pi+4}\) m.

Question 25.
Find the dimensions of the rectangle of maximum area that can be inscribed in the portion of the parabola y² = 4px intercepted by the line x = a.
Solution:
Let ABDC be he rectangle that is inscribed in the portion of given parabola y² = 4px intercepted by the line x = a.

Let (x, y) be the coordinates of point A.
So AC = 2y and CD = a – x
∴ S = area of rectangle
= (a – x) 2y
= (a – \(\frac{y^2}{4 p}\)) 2y
Diff., both sides w.r.t. y ; we have
\(\) ;
\(\frac{d^2 \mathrm{~S}}{d y^2}=-\frac{12 y}{4 p}=-\frac{3 y}{p}\)
For maxima / minima, \(\frac{d S}{d y}\) = 0
⇒ 2a – \(\frac{6 y^2}{4 p}\) = 0
⇒ 6y² = 8ap
⇒ y = ± \(\sqrt{\frac{4 a p}{3}}\)
= ± \(\frac{2}{\sqrt{3}} \sqrt{a p}\)
∴ \(\left(\frac{d^2 \mathrm{~S}}{d y^2}\right)_{y=\frac{2}{\sqrt{3}} \sqrt{a p}}=-\frac{3}{p} \times \frac{2}{\sqrt{3}} \sqrt{a p}\) < 0
Thus S is maximise for
y = \(\frac{2}{\sqrt{3}} \sqrt{a p}\)
= \(\frac{2}{3} \sqrt{3 a p}\)
∴ x = \(\frac{y^2}{4 p}\)
= \(\frac{1}{4 p} \times \frac{4}{9}\) × 3ap = \(\frac{a}{3}\)
Thus, the dimensions of the rectangle are a – x and 2y
i.e. a – \(\frac{a}{3}\) and 2 × \(\frac{2}{3} \sqrt{3 a p}\)
i.e. \(\frac{2 a}{3}\) and \(\frac{4}{3} \sqrt{3 a p}\).

Question 26.
An open topped box is to be constructed by removing equal squares from each corner of a 3 metres by 8 metres rectangular sheet of aluminimum and folding up the sides. Find the volume of the largest such box.(NCERT)
Solution:
Given sides of rectangular sheet be 3 m and 8 m.
Let x units be the length of each side of squares of same size removed from each corner of the sheet.
Thus the dimensions of the open box formed by folding up the flaps are;

Length = 8 – 2x;
breadth = 3 – 2x
and height = x
Let V = volume of cuboid = (8 – 2x)(3 – 2x) x
∴ V = (8 – 2x) (3x – 2x²)
= 4x³ – 22x² + 24x
Diff. both sides w.r.t. x, we have
\(\frac{d V}{d x}\) = 12x² – 44x + 24 ;
\(\frac{d^2 V}{d x^2}\) = 24x – 44
For maxima / minima, \(\frac{d V}{d x}\) = 0
⇒ 4 (3x² – 11x + 6) = 0
⇒ (x – 3) (3x – 2) = 0
⇒ x = 3, \(\frac{2}{3}\)
Further x ≠ 3 if x = 3,
breadth = 3 – 6 = – 3, which is not possible
Thus, \(\left(\frac{d^2 \mathrm{~V}}{d x^2}\right)_{x=\frac{2}{3}}\) = 24 × \(\frac{2}{3}\) – 44
= 16 – 44 = – 28 > 0
∴ x = \(\frac{2}{3}\) be a point of maxima.
Thus V is maximum when x = \(\frac{2}{3}\).
∴ Maximum volume of the box = \(\left(8-\frac{4}{3}\right)\left(3-\frac{4}{3}\right) \frac{2}{3}\)
= \(\frac{20}{3} \times \frac{5}{3} \times \frac{2}{3}\)
= \(\frac{200}{27}\) m³.

Question 27.
Show that the maximum volume of a cylinder which can be inscribed in a cone of height h, and semi-vertical angle 30° is \(\frac{4}{81}\) πh³.
Solution:
Let x be the radius of cylinder andy be the height of cylinder
Then, AF = h – y
and FG = x
∴ tan 30° = \(\frac{x}{h-y}\)
⇒ x = (h – y) \(\frac{1}{\sqrt{3}}\) ……….(1)

Let V = volume of cylinder = \(\frac{\pi}{3}\) (h – y)² . y
∴ \(\frac{d V}{d y}\) = \(\frac{\pi}{3}\) [(h – y)² + 2y (h – y) (- 1)]
∴ \(\frac{d^2 V}{d y^2}\) = \(\frac{\pi}{3}\) [- 2 (h – y) – 2 (h – 2y)]
= \(\frac{\pi}{3}\) [- 4h + 6y]
For max/min, \(\frac{d V}{d y}\) = 0
⇒ (h – y) [(h – y) – 2y] = 0
⇒ (h – y) (h – 3y) = 0
⇒ y = h, \(\frac{h}{3}\) [but y ≠ h as cylinder is inscribed in cone]
and \(\left(\frac{d^2 \mathrm{~V}}{d y^2}\right)_{y=\frac{h}{3}}\) = \(\frac{\pi}{3}\) [- 4h + 2h]
= \(\frac{-2 \pi h}{3}\) < 0
∴ V is maximise for y = \(\frac{h}{3}\)
and Required Max. value = \(\frac{\pi}{3}\left(\frac{2 h}{3}\right)^2 \times \frac{h}{3}=\frac{4 \pi}{81}\) h³.

Question 28.
A cylinder is such that the sum of its height and circumference of its base is 10 metres. Find the maximum volume of the cylinder.
Solution:
Let h metres be the height and r metres be the radius of cylinder.
Then h + 2πr = 10 …………(1)
Let V = volume of cylinder = πr²h
V = πr² (10 – 2πr)
Diff, both sides w.r.t. r we have
\(\frac{d V}{d r}\) = π (20r – 6πr²)
∴ \(\frac{d^2 V}{d r^2}\) = π (20 – 12πr)
For maxima / minima, \(\frac{d V}{d r}\) = 0
⇒ π (20r – 6πr²) = 0
⇒ (20 – 6πr) r = 0
⇒ r = 0, \(\frac{10}{3 \pi}\)
Since r > 0
∴ r = \(\frac{10}{3 \pi}\) m
∴ \(\left(20-12 \pi \times \frac{10}{3 \pi}\right)\) = – 20π < 0
Thus, V is maximise at r = \(\frac{10}{3 \pi}\) metre
and h = 10 – \(\frac{20 \pi}{3 \pi}\) = \(\frac{10}{3}\) m
and Maximum volume of cylinder = πr²h
= \(\pi \times\left(\frac{10}{3 \pi}\right)^2 \times \frac{10}{3}\) m³
= \(\frac{1000}{27 \pi}\) m³

Question 29.
Find the area of the largest rectangle in the first quadrant with two sides on x-axis and y- axis and one vertex on the curve)’ = 12 – x².
Solution:

Let P (x, y) be any point on the given curve y = 12 – x² in the first quadrant.
Let M and N be the foot of perpendiculars drawn from P on y-axis and x-axis respectively.
Then, A area of rectangle = xy = x (12 – x²)
Diff. both sides w.r.t. x, we have
\(\frac{d A}{d x}\) = 12 – 3x²
For maxima / minima, \(\frac{d A}{d x}\) = 0
⇒ 12 – 3x² = 0
⇒ x² = 4
⇒ x = ± 2
at x = 2,
\(\frac{d^2 A}{d x^2}\) = – 12 < 0
∴ x = 2 be a point of maxima.
Hence A will be maximum at x = 2.
∴ maximum area of rectangle = 2 (12 – 2²) = 2 (12 – 4) = 16 units.

Question 30.
A tank with rectangular base and rectangular sides, open at the top Is to be constructed so that its depth is 2 m and volume is 8 m³. If building of tank costs ₹ 70 per sq. metre for the base and ₹ 45 per m² for sides, what is the cost of least expensive tank?
Solution:
Letx, x, y be the length, width and height of the tank
∴ area of square base = x2
and area of four walls = 4xy
Given V = volume of open tank = 8x²y ………….(1)
∴ E = 70x² + 45 (4 xy)
= 70x² + 180x . \(\frac{8}{x^2}\) [using (1)]
⇒ E = 70x² + \(\frac{180 \times 8}{x}\)
Given y = depth of tank = 2m
∴ from (1);
8 = x² × 2
⇒ x = 2[∵ x > 0]
Now we check whether x = 2 is point of maximum or minima.
\(\frac{d E}{d x}\) = 140x – \(\frac{180 \times 8}{x^2}\)
\(\frac{d^2 E}{d x^2}\) = 140 + \(\frac{180 \times 16}{x^2}\)
at x = 2,
\(\frac{d^2 E}{d x^2}\) > 0
∴ E is minimum for x = 2.
∴ from (2) ;
least value of E = 70 (2)² + \(\frac{180 \times 8}{2}\) = ₹ 1000.

Access to comprehensive ML Aggarwal Class 12 Solutions ISC Chapter 7 Applications of Derivatives Ex 7.8 encourages independent learning.

ML Aggarwal Class 12 Maths Solutions Section A Chapter 7 Applications of Derivatives Ex 7.8

Question 1.
Find two numbers whose sum is 24 and whose product is as large as possible. (NCERT)
Solution:
Let the one number be x, (0 < x < 24)
∴ other number be 24 – x because sum of two numbers be 24.
Let P = x (24 – x) ……..(1)
Duff. (1) w.r.t. x; we have
\(\frac{d P}{d x}\) = 24 – 2x
For maximum/minima, \(\frac{d P}{d x}\) = 0
⇒ 24 – 2x = 0
⇒ x = 12
∴ \(\frac{d^2 P}{d x^2}\) = – 2
⇒ (\(\frac{d^2 P}{d x^2}\))_{x = 12} = – 2 < 0
Thus, x = 12 be a point of local maxima and x = 12 be the only point of local maxima.
So the local maximum value be the absolute maximum value.
Thus, required numbers are 12 and (24 – 12) i.e. 12.

Question 2.
The sum of two positive numbers is 20. Find the numbers
(i) if the sum of their squares is minimum.
(ii) if the product of the square of one and the cube of the other is maximum.
Solution:
(i) Let the required numbers be x, 20 – x,
since the sum of two positive integers is 20.
Let S = x² + (20 – x)²
Diff. both sides w.r.t. x, we have
\(\frac{d S}{d x}\) = 2x + 2 (20 – x) (- 1)
= 2x – 40 + 2x
= 4x – 40
For maxima/minima, \(\frac{d S}{d x}\) = 0
⇒ 4x – 40 = 0
⇒ x = 10
∴ (\(\frac{d^2 P}{d x^2}\))_{x = 10} = 4 > 0
Thus x = 10 be a point of local minima.
∴ required numbers are 10 and (20 – 10) i.e. 10 and 10.

(ii) Let the required numbers be x and 20 – x
Since the sum of two numbers be 20.
Let P = x² (20 – x)³ ;
Diff. both sides w.r.t. x, we have
\(\frac{d P}{d x}\) = 3x² (20 – x)² (- 1) + (20 – x)³ 2x
= – x (20 – x)² [3x – 2 (20 – x)]
= – x (20 – x)² (5x – 40)
For maxima / minima, \(\frac{d P}{d x}\) = 0
⇒ x (20 – x)² (5x – 40) = 0
⇒ x = 0, 20, 8
But 0 < x < 20 ∴ x = 8
When x slightly < 8,
\(\frac{d P}{d x}\) = (- ve) (+ ve) (+ ve) = – ve
∴ \(\frac{d P}{d x}\) changes its sign from +ve to – ve.
Hence x = 8 be a point of local maxima.
Thus the required numbers are 8 and 20 – 8 i.e. 8 and 12.

Question 3.
The product of two positive numbers is 16. Find the numbers
(i) if their sum is least.
(ii) if the sum of one and the square of the other is least.
Solution:
(i) Let the two numbers be x and \(\frac{16}{x}\),
since the product of two numbers be 16.
Let S = x + \(\frac{16}{x}\)
∴ \(\frac{d S}{d x}\) = 1 – \(\frac{16}{x^2}\)
For maxima/minima, \(\frac{d P}{d x}\) = 0
⇒ 1 – \(\frac{16}{x^2}\) = 0
⇒ x² = 16
⇒ x = ± 4 but x, y ∈ N
∴ x = 4
and \(\frac{d^2 S}{d x^2}\) = \(\frac{32}{x^3}\)
∴ \(\left(\frac{d^2 S}{d x^2}\right)_{x=4}=\frac{32}{64}=\frac{1}{2}\) > 0
∴ x = 4 be a point local minima.
Thus S is least for x = 4
∴ required numbers are 4 and \(\frac{16}{4}\) i.e. 4 and 4.

(ii) Let the required two numbers be x and \(\frac{16}{x}\),
Since the product of two numbers be 16.
Let S = x² + 16/x
\(\frac{d S}{d x}\) = \(\frac{-16}{x^2}\) + 2x
∴ \(\frac{d^2 S}{d x^2}\) = \(\frac{32}{x^3}\) + 2
For maxima/minimum, \(\frac{d S}{d x}\) = 0
⇒ \(\frac{-16}{x^2}\) + 2x = 0
⇒ x³ = 8
⇒ x = 2
at x = 2;
\(\frac{d^2 S}{d x^2}\) = \(\frac{32}{8}\) + 2
= 4 + 2 = 6 > 0
∴ x = 2 be a point of local minima.
Thus S is minimum for x = 2.
Hence the required numbers are 2 and \(\frac{16}{2}\) i.e. 2 and 8.

Question 4.
Find the positive numbers x and y such that x + y = 60 and xy is maximum. (NCERT)
Solution:
Given x + y = 60 …………(1)
where x, y > 0
Let P = xy³
= y³ (60 – y)
∴ \(\frac{d P}{d y}\) = 180y² – 4y³
= 4y² [45 – y]
For maxima/minina, \(\frac{d P}{d y}\) = 0
⇒ 4y² (45 – y) = 0
⇒ y = 0, 45
but y > 0
∴ y = 45
and \(\frac{d^2 P}{d y^2}\) = 12 × 45 × (- 15) = – 8100 < 0
∴ P is maximise for y = 45
∴ from (1) ; x = 15.
Hence the required two numbers be 15, 45.

Question 5.
Find two positive numbers x andy such that their sum is 35 and the product x²y⁵ is maximum. (NCERT)
Solution:
Let the two positive numbers be x and y.
Given x + y = 35 ………..(1)
Let P product = x²y⁵
⇒ P = (35 – y²) y⁵
∴ \(\frac{d P}{d y}\) = (35 – y)² 5y⁴ ± y⁵ 2(35 – y) (- 1)
= y⁴ (35 – y) [5 (35 – y) – 2y]
= y⁴ (35 – y) (175 – 7y)
For maxima/minima, \(\frac{d P}{d y}\) = 0
⇒ y = 0, 35, 25
but y ≠ 0, 35
[∵ y > 0 and further if y = 35 then x = 0]
∴ possible value of y = 25.
When y slightly < 25 ⇒ \(\frac{d P}{d y}\) = (+ve) (+ve) (+ve) = +ve When y slightly > 25
⇒ \(\frac{d P}{d y}\) = (+ve) (+ve) (-ve) = -ve
∴ \(\frac{d P}{d x}\) varies from +ve to -ve
∴ P is maximise for y = 25
∴ from (1); x = 10
∴ Required two numbers are 10, 25.

Question 6.
(i) Find two positive numbers whose sum is 16 and the sum of whose cube is minimum.
Solution:
(i) Let the two numbers be x andy s.t. x, y > 0
also given, x + y = 16
Let S = x<sup3 + y³
= x³ + (16 – x)³
∴ \(\frac{d S}{d x}\) = 3x² + 3 (16 – x)² (- 1)
∴ \(\frac{d^2 S}{d x^2}\) = 3 [2x + 2 (16 – x)] = 96 > 0
For maximaJminima, \(\frac{d S}{d x}\) = 0
⇒ 3 [x² – (16 – x)²] = 0
⇒ x² – (256 + x² – 32x) = 0
⇒ x = \(\frac{256}{32}\) = 8
∴ (\(\frac{d^2 S}{d x^2}\))_{x = 8} = 3 (16 + 16) = 96 > 0
∴ S is minimum for x = 8 and
from (1) ; y = 8
Thus the required two numbers are 8 and 8.

Question 6 (old).
(i) Find two positive numbers whose sum is 15 and the sum of whose squares is minimum. (NCERT)
Solution:
Let the two numbers be x and y s.t. x, y > 0
Given x + y = 15
Let S = x² + y²
= x² + (15 – x)² …………….(2)
∴ \(\frac{d S}{d x}\) = 2x + 2 (15 – x) (- 1)
= 2 [x – 15 + x]
= 2 (2x – 15);
\(\frac{d^2 S}{d x^2}\) = 4
For maxima/minima
∴ \(\frac{d S}{d x}\) = 0
⇒ x = \(\frac{15}{2}\)
∴ \(\frac{d^2 S}{d x^2}\) = 4 > 0
Thus S is minimise for x = \(\frac{15}{2}\)
∴ from (1) ;
y = 15 – \(\frac{15}{2}\) = \(\frac{15}{2}\)
Thus required two positive numbers are \(\frac{15}{2}\) and \(\frac{15}{2}\).

Question 7.
Three numbers are given whose sum is 180 and the ratio of first two of them is 1 : 2. If the product of the numbers is greatest, find the numbers. (ISC 2003)
Solution:
Let the first two numbers be x and 2x, since the ratio of first two of them be 1 : 2.
Also the s um of first three numbers be 180. required numbers are x, 2x, 180 – 3x.
Let P = x × 2x × (180 – 3x)
= 2x² (180 – 3x)
= 6 (60x² – x3)
Diff. both sides w.r.t. x, we have
\(\frac{d P}{d x}\) = 6 [120x – 3x²] ;
\(\frac{d^2 P}{d x^2}\) = 6 (120 – 6x)
For maxima/minima, \(\frac{d P}{d x}\) = 0
⇒ 120x – 3x² = 0
⇒ – 3x [x – 40] = 0
⇒ x = 0, 40
But x > 0
∴ x = 40
∴ (\(\frac{d^2 P}{d x^2}\))_{x = 40} = 6 (120 – 240)
= – 720 < 0
∴ x = 40 be a point of maxima.
Thus product P is maximum for x = 40.
Hence the required numbers are, 40, 2 × 40 and 180 – 120 i.e. 40, 80 and 60.

Question 8.
Find the maximum profit that a company can make, if the profit function is given by p (x) = 41 + 72x – 18x². (NCERT)
Solution:
Given profit function p(x) = 41 + 72x – 18x²
Diff. both sides w.r.t. x, we have
\(\frac{d P}{d x}\) = 72 – 36x
For maxima/minima, \(\frac{d P}{d x}\) = 0
⇒ 72 – 36x = 0
⇒ x = 2
Also, \(\frac{d^2 y}{d x^2}\) = – 36
∴ (\(\frac{d^2 P}{d x^2}\))_{x = 2} = – 36 < 0
Thus x = 2 be a point of maxima.
∴ profit P(x) will be maximum when x = 2.
and Maximum profit = p (2) = 41 + 144 – 72 = 113 units.

Question 9.
The cost (in Rs.) per article C, of manufacturing a certain article is given by the formula C = 5 + \(\frac{48}{x}\) + 3x², where x is the number of articles manufactured per hour. Find the minimum value of C.
Solution:
Given C = 5 + \(\frac{48}{x}\) + 3x²
Diff. both sides w.r.t. x, we have
\(\frac{d C}{d x}\) = – \(\frac{48}{x^2}\) + 6x
∴ \(\frac{d^2 C}{d x^2}\) = \(\frac{96}{x^3}\) + 6
For maxima/minima, we have
\(\frac{d C}{d x}\) = 0
⇒ – \(\frac{48}{x^2}\) + 6x = 0
⇒ 6x³ – 48 = 0
⇒ x = 2
∴ (\(\frac{d^2 P}{d x^2}\))_{x = 2} = \(\frac{96}{8}\) + 6
= 12 + 6 = 18 > 0
Thus x = 2 be a point of minima.
and Minimum value of C = 5 + \(\frac{48}{2}\) + 3 × 2²
= 5 + 24 + 12 = Rs. 41

Question 10.
Prove that of all rectangles with given area, the square has the smallest perimeter.
Solution:
Let x and y be the length and breadth of rectangle.
∴ A = area of rectangle = xy [given] ………..(1)
Let P = perimeter = 2 (x + y)
= 2 [x + \(\frac{A}{x}\)] [using (1)]
∴ \(\frac{d P}{dx}\) = 2 (1 – \(\frac{\mathrm{A}}{x^2}\))
⇒ \(\frac{d^2 \mathrm{P}}{d x^2}=\frac{4 \mathrm{~A}}{x^3}\)
For maxima or minima, \(\frac{d P}{dx}\) = 0
⇒ 1 – \(\frac{\mathrm{A}}{x^2}\) = 0
⇒ x = √A [∵ x > 0]
∴ \(\frac{d^2 \mathrm{P}}{d x^2}=\frac{4 \mathrm{~A}}{\mathrm{~A}^{3 / 2}}=\frac{4}{\sqrt{\mathrm{A}}}\) > 0
∴ P is minimise for x = √A
∴ From (1) ;
y = \(\frac{\mathrm{A}}{\sqrt{\mathrm{A}}}\) = √A
∴ x = y
Hence the rectangle becomes square.
Thus, the rectangle with given area, the square has the smallest perimeter.

Question 11.
Of all rectangles each of which has perimeter 24 cm, find the one having maximum area. Also find that area.
Solution:
Let x cm be the length and y cm be the breadth of rectangle
s.t 2(x + y) = 24
[∵ perimeter of rectangle = 24]
⇒ x + y = 12
Let A = area of rectangle = xy = x (12 – x) [using (1)]
Differentiating w.r.t. x, we have
\(\frac{d A}{d x}\) = 12 – 2x;
\(\frac{d^2 A}{d x^2}\) = – 2
For maxima/minima, \(\frac{d A}{d x}\) = 0
⇒ 12 – 2x = 0
⇒ x = 6
∴ (\(\frac{d^2 A}{d x^2}\))_{x = 6} = – 2 < 0
∴ x = 6 be a point of maxima.
Thus Area A is maximum for x = 6
∴ from (1);
y = 12 – 6 = 6
Therefore x = y = 6 cm.
Thus the square has the maximum area.
and Maximum area = xy = (6 × 6) cm² = 36 cm².

Question 12.
Prove that of all rectangles with given perimeter, the square has the largest area.
Solution:
Let x cm and y cm be the length and breadth of the rectangle.
Let P be the perimeter of rectangle
∴ P = 2x + 2y …………(1)
Let A = area of rectangle = xy
= x (\(\frac{P}{2}\) – x)
⇒ A = \(\frac{P x}{2}\) – x²
Diff, both sides w.r.t. x; we have
\(\frac{d A}{d x}\) = \(\frac{P}{2}\) – 2x ;
\(\frac{d^2 A}{d x^2}\) = – 2
For maxima/minima, \(\frac{d A}{d x}\) = 0
⇒ \(\frac{P}{2}\) – 2x = 0
⇒ x = \(\frac{P}{4}\)
∴ (\(\frac{d^2 A}{d x^2}\))_{x = \(\frac{P}{4}\)} = – 2 < 0
Thus, x = \(\frac{P}{4}\) be a point of maxima
∴ from (1) ;
2y = P – \(\frac{P}{2}\) = \(\frac{P}{2}\)
⇒ y = \(\frac{P}{4}\)
Thus area of the rectangle is maximum when x = y = \(\frac{P}{4}\)
∴ rectangle becomes square.
Thus, Maximum area = \(\frac{P}{4} \times \frac{P}{4}=\frac{P^2}{16}\)
Hence, the rectangle with given perimeter, the square has the largest area.

Question 13.
Prove that the perimeter of a right- angled triangle of given hypotenuse is maximum when the triangle is isosceles.
Solution:
Let ΔABC be right angled Δ at angle B with hypotenuse h.
Let AB = x
∴ BC = \(\sqrt{h^2-x^2}\)
[using pythagoras theorem]

Thus P = perimetcr of ∆ABC = AB + BC + AC
⇒ P = x + \(\sqrt{h^2-x^2}\) + h
Diff. (1) both sides w.r.t. x; we have
\(\frac{d \mathrm{P}}{d x}=1+\frac{1}{2} \frac{1}{\sqrt{h^2-x^2}}(-2 x)\)
= 1 – \(\frac{x}{\sqrt{h^2-x^2}}\)
For maxima / minima, \(\frac{d P}{d x}\) = 0
⇒ 1 – \(\frac{x}{\sqrt{h^2-x^2}}\) = 0
⇒ \(\sqrt{h^2-x^2}\) = x
⇒ h² – x² = x²
⇒ 2x² = h²
⇒ x = ± \(\frac{h}{\sqrt{2}}\)
Since x, h > 0
∴ x = \(\frac{h}{\sqrt{2}}\)

∴ AB = BC
Hence the perimeter of right angle triangle is maximum when the triangle is isosceles.

Question 14.
Two sides of a triangle have lengths a and b, and the angle between them is 6. What value of 9 will maximize the area of the triangle ? Also find the maximum area of the triangle.
Solution:
Let ABC be the given triangle with AC = b
and BC = a be given two sides.
Let θ be the angle between them
Let A = area of ∆ABC
= \(\frac{1}{2}\) ab sin θ

Clearly A is maximum when sin θ is maximum i.e. sin θ = 1
⇒ θ = \(\frac{\pi}{2}\)
and required maximum area = \(\frac{1}{2}\) ab × 1 = \(\frac{1}{2}\) ab

Question 15.
A right-angled triangle with constant area is given. Prove that the hypotenuse of the triangle is least when the triangle is-isosceles.
Solution:
Let ∆ABC be right angled triangle with constant area P.
Let l be the length of the hypotenuse of ∆ABC
∴ AB = l sin θ
and BC = l cos θ
Also ∠ACB = θ.
∴ area of triangle ∆ABC = \(\frac{1}{2}\) × BC × AB
⇒ P = \(\frac{1}{2}\) (l cos θ) (l sin θ) = \(\frac{l^2}{4}\) sin 2θ
⇒ l² = 4P cosec 2θ
Now l is minimise when l² is minimise, i.e. we want to find the value of 0 for which l² is least.
Let L = l² = 4P cosec 2θ
Diff. both sides w.r.t. θ, we have
\(\frac{d \mathrm{~L}}{d \theta}\) = – 8P cosec 2θ cot 2θ
and \(\frac{d^2 \mathrm{~L}}{d \theta^2}\) = – 8P [- 2 cosec³ 2θ + cot 2θ (- 2 cosec 2θ cot 2θ)]
= 16P [cosec³ 2θ + cosec 2θ cot² 2θ]
For maxima/minima \(\frac{d \mathrm{~L}}{d \theta}\) = 0
⇒ cosec 2θ cot 2θ = 0
⇒ \(\frac{\cos 2 \theta}{\sin ^2 2 \theta}\) = 0
⇒ cos 2θ = 0
⇒ 2θ = \(\frac{\pi}{2}\)
⇒ θ = \(\frac{\pi}{4}\)
∵ θ ∈ (0, \(\frac{\pi}{2}\))
At θ = \(\frac{\pi}{4}\),
\(\frac{d^2 \mathrm{~L}}{d \theta^2}\) = 16P [1³ + 1 × 0] = 16P > 0
∴ Lis minimum for θ = \(\frac{\pi}{4}\)
Thus l is least for θ = \(\frac{\pi}{4}\)
∴ AB = l sin \(\frac{\pi}{4}\)
= \(\frac{l}{\sqrt{2}}\)
and BC = l cos θ
= l cos \(\frac{\pi}{4}\)
= \(\frac{l}{\sqrt{2}}\)
∴ AB = BC.
Hence ∆ABC be an isosceles triangle.

Question 16.
The lengths of the sides of a triangle are 9 + x², 9 + x² and 18 – 2x². Calculate:
(i) the area of the triangle in terms of x.
(ii) the value of x for which this area is maximum.
Solution:
(i) Here given lengths of sides of triangle are 9 + x², 9 + x² and 18 – 2x².
Let a = 9 + x² ;
b = 9 + x²;
c = 18 – 2x²
∴ s = \(\frac{a+b+c}{2}\)
= \(\frac{9+x^2+9+x^2+18-2 x^2}{2}\) = 18
Let A = area of ∆ABC
= \(\sqrt{s(s-a)(s-b)(s-c)}\) [By Heron’s formula]
= \(\sqrt{18\left(18-9-x^2\right)\left(18-9-x^2\right)\left(18-18+2 x^2\right)}\)
= \(\sqrt{18\left(9-x^2\right)^2 \times 2 x^2}\)
= 6 (9 – x²) x
= 54x – 6x³

(ii) Diff, both sides w.r.t. x, we have
\(\frac{d A}{d x}\) = 6 [9 – 3x²] ;
\(\frac{d^2 A}{d x^2}\) = 6 (0 – 6x) = – 36x
For maxima / minima, \(\frac{d A}{d x}\) = 0
⇒ \(\left(\frac{d^2 \mathrm{~A}}{d x^2}\right)_{x=\sqrt{3}}\) = – 36√3 < 0
Thus, x = √3 be a point of maxima
and maximum value of area A = 54 × √3 – (6√3)³
= 54√3 – 18√3 = 36√3 sq. units.

Question 17.
The perimeter of a triangle is 8 cm. If one of the sides is 3 cm, what are the lengths of the other sides for maximum area of the triangle ?
Solution:
Let a, b, c be the lengths of sides of triangle then a + b + c = 8
∴ s = \(\frac{a+b+c}{2}=\frac{8}{2}\) = 4
given one of its sides be of length 3 cm,
let a = 3
b + c = 5
⇒ b = 5 – c
Let A = area of the ∆ABC
= \(\sqrt{s(s-a)(s-b)(s-c)}\)
= \(\sqrt{4(4-3)(4-5+c)(4-c)}\)
= \(\sqrt{4(c-1)(4-c)}\)
⇒ A² = 4 (c – 1) (4 – c) = 4 (- c² + 5c – 4)
Now A is maximise when A2 is maximise.
Let P = A² = 4 (- c² + 5c – 4)
On differentiating w.r.t. c, we get
\(\frac{d P}{d c}\) = 4 (- 2c + 5) ;
∴ (\(\frac{d^2 P}{d c^2}\))_{c = \(\frac{5}{2}\)} = – 8 < 0
Thus P is maximum for c = \(\frac{5}{2}\)
Hence Area A is maximum for c = \(\frac{5}{2}\)
Thus the required lengths of sides of triangle are 3, 5 – \(\frac{5}{2}\) and \(\frac{5}{2}\) i.e. 3 cm, \(\frac{5}{2}\) cm and \(\frac{5}{2}\) cm.

Question 18.
A sheet of paper is to contain 18 cm² of printed matter. The margins at the top and bottom are 2 cm each, and at the sides 1 cm each. Find the dimensions of the sheet which require the least amount of paper.
Solution:
Let x cm (x > 0) be the one dimension of the page then the other dimension be \(\frac{18}{x}\) cm, since the area of given page be 18 sq. cm.

Let A cm² be the area of printed page.
Then, A = (x – 2) (\(\frac{18}{x}\) – 4)
= 18 – 4x – \(\frac{36}{x}\) + 8
= 26 – 4x – \(\frac{36}{x}\)
On differentiating w.r.t. x, we have
\(\frac{d \mathrm{~A}}{d x}=-4+\frac{36}{x^2}\)
\(\frac{d^2 \mathrm{~A}}{d x^2}=-\frac{72}{x^3}\)
For maxima / minima, \(\frac{d A}{d x}\) = 0
⇒ – 4 + \(\frac{36}{x^2}\) = 0
⇒ x² = 9
⇒ x = 3
∴ (\(\frac{d^2 A}{d x^2}\))_{x = 3} = \(-\frac{72}{27}=-\frac{8}{3}\) < 0
Thus A is maximum for x = 3
∴ the dimensions of the printed page be x and \(\frac{18}{x}\) i.e. 3 cm and 6 cm
Hence, the required dimensions of the sheet that requires the least amount of paper be (x + 2) and (\(\frac{18}{x}\) + 4) cm i.e. 5 cm and 10 cm.

Question 19.
Show that of all rectnagles inscribed in a given fixed circle the square has the maximum area.
Solution:
Let x and y be the length and breadth of the rectangle respectively.
Which is inscribed in a circle of radius r
∴ x² + y² = (2r)²
x² + y² = 4r² …………(1)
Thus A = area of rectangle = xy
= x \(\sqrt{4 r^2-x^2}\) [using (1)]

For max/ minima, \(\frac{d A}{d x}\) = 0
⇒ 4r² – 2x² = 0
⇒ x = √2r
[∵ x > 0]
∴ \(\left(\frac{d^2 \mathrm{~A}}{d x^2}\right)_{x=\sqrt{2} r}=\frac{-\sqrt{2} r\left(8 r^2\right)}{\left(2 r^2\right)^{3 / 2}}\) < 0
∴ A is maximum for x = √2r
∴ From (1) ;
2r² + y² = 4r²
⇒ y = √2r
Hence A is maximise for x = y = √2r
i.e. rectangle becomes square.

Question 20.
(i) The sum of perimeters of a circle and a square is k, where k is some constant. Prove that the sum of their areas is least when the side of the square is double the radius of the circle. (NCERT)
(ii) Given the sum of perimeters of a circle and a square, show that the sum of areas is least when the diameter of the circle is equal to the side of the square.
Solution:
(i) Let x be the side of square and y be the radius of circle.
Let P be the sum of their perimeters.
Then P = 4x + 2πy
Let A combined area of square and circle
⇒ A = x² + πy² [using (1)]
= x² + π \(\left(\frac{\mathrm{P}-4 x}{2 \pi}\right)^2\) [using (1)]
⇒ A = x² + \(\frac{1}{4 \pi}\) (P – 4x)²
∴ \(\frac{d A}{d x}\) = 2x + \(\frac{1}{2 \pi}\) (P – 4x) (- 4)
For maxima / minima, we have \(\frac{d A}{d x}\) = 0
⇒ 4r² – 2x² = 0
⇒ x = √2r [∵ x > 0]
∴ \(\left(\frac{d^2 \mathrm{~A}}{d x^2}\right)_{x=\sqrt{2} r}=\frac{-\sqrt{2} r\left(8 r^2\right)}{\left(2 r^2\right)^{3 / 2}}\) < 0
∴ A is minimum for x = √2r
∴ From (1) ;
2r² + y = 4r²
⇒ y = √2r
Hence A is maximise for x = y = √2r
i.e. rectangle becomes square.

Question 20 (old).
(i) The sum of perimeters of a circle and a square is k, where A is some constant. Prove that the sum of their arcas is least when the side of the square is double the radius of the circle. (NCERT)
(ii) Given the sum of perimeters of a circle and a square, show that the sum of areas is least when the diameter of the circle is equal to the side of the square.
Solution:
(i) Let x be the side of square and y be the radius of circle.
Let P be the sum of their perimeters.
Then P = 4x + 2πy …………..(1)
Let A = combined area of square and circle
⇒ A = x² + πy²
= x² + π \(\left(\frac{P-4 x}{2 \pi}\right)^2\) [using (1)]
⇒ A = x² + \(\frac{1}{4 \pi}\) (P – 4x)²
∴ \(\frac{d A}{d x}\) = 2x + \(\frac{1}{2 \pi}\) (P – 4x) (- 4)
For maxima/minima, we have \(\frac{d A}{d x}\) = 0
2x – \(\frac{2}{\pi}\) (P – 4x) = 0
⇒ 2πx – 2P + 8x = 0
⇒ x = \(\frac{2 P}{2 \pi+8}=\frac{P}{\pi+4}\)
Also \(\frac{d^2 P}{d x^2}\) = 2 + \(\frac{16}{2 \pi}\) > 0
∴ P is minimum for x = \(\frac{9}{\pi+4}\)
∴ from (1) ;
2πy = P – \(\frac{4 \mathrm{P}}{\pi+4}\)
⇒ 2πy = \(\frac{\pi \mathrm{P}}{\pi+4}\)
⇒ y = \(\frac{\mathrm{P}}{2(\pi+4)}=\frac{x}{2}\)
⇒ x = 2y
i.e. side of square = diameter of circle.
Hence A is minimum when side of the square is equal to the diameter of the circle.

(ii) Let x be the length of each side of square and r be the radius of circle.
Then P = sum of perimeters of a circle and as square = 2πr + 4x …………(1)
Let A = combined area of square and circle
⇒ A = x² + r²
= x² + π \(\left(\frac{P-4 x}{2 \pi}\right)^2\) [using (1)]
⇒ A = x² + \(\frac{1}{4 \pi}\) (P – 4x)² ;
Diff. both sides w.r.t. x, we have
\(\frac{d A}{d x}\) = 2x + \(\frac{1}{4 \pi}\) × 2 (P – 4x) (- 4)
and \(\frac{d^2 A}{d x^2}\) = 2 + (- \(\frac{2}{\pi}\)) (- 4)
= 2 + \(\frac{8}{\pi}\)
For maxima and minima, \(\frac{d A}{d x}\) = 0
⇒ 2x – \(\frac{2}{\pi}\) (P – 4x) = 0
⇒ 2πx – 2P + 8x = 0
⇒ (π + 4) x = P
⇒ x = \(\frac{P}{\pi+4}\)
Now, \(\left(\frac{d^2 \mathrm{~A}}{d x^2}\right)_{x=\frac{\mathrm{P}}{\pi+4}}\) = 2 + \(\frac{8}{\pi}\)
Hence A is least when x = \(\frac{P}{\pi+4}\)
⇒ (π + 4) x = 2πr + 4x [using (1)]
⇒ πx = 2πr
⇒ x = 2r
Thus, diameter of circle is equal to side of square.

Question 21.
A wire 10 metres long is cut into two parts. One part is bent into the shape of a circle and the other into the shape of an equilateral triangle. How should the wire be cut so that the combined area of the two figures is as small as possible?
Solution:
Let r metre be the radius of circle and a metres be the side of equilateral triangle.
Then 2πr + 3a = 10 ………….( 1)
Let A combined area of circle and equilateral triangle
⇒ A = πr² + \(\frac{\sqrt{3}}{4}\) a²
= πr² + \(\frac{\sqrt{3}}{4}\left(\frac{10-2 \pi r}{3}\right)^2\) [using (1)]
On differentiating both sides w.r.t. r, we have
\(\frac{d A}{d r}\) = 2πr + \(\frac{\sqrt{3}}{36}\) 2 (10 – 2πr) (- 2π)
= 2πr – \(\frac{\sqrt{3}}{9}\) π (10 – 2πr)
∴ \(\frac{d^2 A}{d x^2}\) = 2π – \(\frac{\sqrt{3} \pi}{9}\) (0 – 2π)
= 2π + \(\frac{2 \sqrt{3}}{9}\) π²
For maxima / minima, \(\frac{d A}{d r}\) = 0
⇒ 2πr – \(\frac{\sqrt{3} \pi}{9}\) (10 – 2πr) = 0
⇒ 18πr – 10√3π + 2√3π²r = 0
⇒ (18 + 2√3π) r = 10√3
⇒ r = \(\frac{10 \sqrt{3}}{18+2 \sqrt{3} \pi}\)
⇒ r = \(\frac{10 \sqrt{3}}{2 \sqrt{3}(\pi+3 \sqrt{3})}\)
= \(\frac{5}{\pi+2 \sqrt{3}}\) m
Also, at r = \(\frac{5}{\pi+3 \sqrt{3}}\) ;
\(\frac{d^2 A}{d x^2}\) = 2π + \(\frac{2 \sqrt{3}}{9}\) π² > 0
∴ A is least for r = \(\frac{5}{\pi+3 \sqrt{3}}\) m
Thus, length of piece of wire bent into the form of circle = 2πr = \(\frac{10 \pi}{\pi+3 \sqrt{3}}\) m
and length of piece of wire bent into form of equilateral triangle = \(\left(10-\frac{10 \pi}{\pi+3 \sqrt{3}}\right)\) m
= \(\frac{30 \sqrt{3}}{\pi+3 \sqrt{3}}\) m

Question 22.
A wire of length 36 cm is cut into two pieces. One of the piece is turned into the form of a square and the other in the form of an equilateral triangle. Find the lengh of each piece so that the sum of the areas of the two figures be minimum.
Solution:
Let the length of one piece of wire be x m
∴ other piece must be (36 – x) m.
Let x metres be made into square and (36 – x) m be made into equilateral ∆.
So perimeter of square = 4 × side = x
∴ side of square = \(\frac{x}{4}\)
∴ area of square = \(\left(\frac{x}{4}\right)^2\)
Further perimeter of equilateral ∆ = 3 × side
= 36 – x
∴ side of equilateral ∆ = 12 – \(\frac{x}{3}\)
∴ area of equilateral ∆ = \(\frac{\sqrt{3}}{4}\) (side)²
= \(\frac{\sqrt{3}}{4}\left(12-\frac{x}{3}\right)^2\)
∴ A = combined area of square and equilateral ∆ = \(\frac{x^2}{16}+\frac{\sqrt{3}}{4}\left(12-\frac{x}{3}\right)^2\)
∴ \(\frac{d A}{d x}\) = \(\frac{x}{8}+\frac{\sqrt{3}}{2}\left(12-\frac{x}{3}\right)\left(-\frac{1}{3}\right)\)
\(\frac{d A}{d x}\) = \(\frac{x}{8}+\frac{x}{6 \sqrt{3}}\) – 2
For maxima / minima, \(\frac{d A}{d x}\) = 0

Hence the length of two pieces are \(\frac{144}{4+3 \sqrt{3}}\) m and \(\left(36-\frac{144}{x+3 \sqrt{3}}\right)\) m i.e. \(\frac{108 \sqrt{3}}{4+3 \sqrt{3}}\) m.

Question 22.
A window is in the form of a rectangle surrounded by a semicircular opening. The total perimeter of the window is 10 m. Find the dimensions of the window to adniit maximum light through the whole opening.
Solution:
Let the width and height of the window be 2x metre and y metre respectively.
Given total perimeter of window = 10 m
⇒ 2x + 2y + πx = 10 ………..(1)
Let A = total area of window
⇒ A = 2xy + \(\frac{\pi x^2}{2}\)
⇒ A = x (10 – πx – 2x) + \(\frac{\pi x^2}{2}\)
∴ \(\frac{d A}{d x}\) = 10 – (π + 2) 2x + πx

For maxima/minima, \(\frac{d A}{d x}\) = 0
⇒ 10 – 2πx – 4x + πx = 0
⇒ 10 – 4x – πx = 0
⇒ x = \(\frac{10}{\pi+4}\)
∴ \(\frac{d^2 A}{d x^2}\) = – 2 (π + 2)
= – π – 4
at x = \(\frac{10}{\pi+4}\);
\(\frac{d^2 A}{d x^2}\) = – π – 4 < 0
Hence A is maximum when x = \(\frac{10}{\pi+4}\)
∴ from (1) ;
2y = 10 – \(\frac{10}{\pi+4}\)
= \(\frac{10 \pi+40-10 \pi-20}{\pi+4}\)
⇒ y = \(\frac{10}{\pi+4}\)
Thus, dimensions of the window are 2x and y
i.e. \(\frac{2 \times 10}{\pi+4}\) and \(\frac{10}{\pi+4}\)
i.e. \(\frac{10}{\pi+4}\) and \(\frac{10}{\pi+4}\).

Question 23.
A window is in the form of a rectangle above which there is a semicircle. If the perimeter of the window is p cm, show that the window will admit maximum possible light only when the radius of semicircle is \(\frac{p}{\pi+4}\) cm.
Solution:
Let r cm be the radius of semi-circle and x cm be the side BC of rectangle as shown in figure shown alongside.
Then p perimeter of combined figure
⇒ p = πr + 2x + 2r ……….(1)

Let A = area of the window.
⇒ A = \(\frac{1}{2}\) πr² + 2rx
= \(\frac{\pi r^2}{2}+2 r \frac{(p-\pi r-2 r)}{2}\) ………(2) [using (1)]
Now maximum light wil be admitted through the window if the area of the window is maximum.
For this we have to maximise A.
Diff. eqn. (2) both sides w.r.t. r, we have
\(\frac{d A}{d r}\) = πr
For maxima/minima, \(\frac{d A}{d r}\) = 0
⇒ πr + p – 2πr – 4r = 0
⇒ p – πr – 4r = 0
⇒ r = \(\frac{p}{\pi+4}\)
∴ at r = \(\frac{p}{\pi+4}\),
\(\frac{d^2 A}{d x^2}\) = – (π + 4) < 0
Thus A is least for r = \(\frac{p}{\pi+4}\)
Thus, from eqn. (1); we have
2x = p – \(\frac{(\pi+2) p}{\pi+4}\)
= \(\frac{p \pi+4 p-\pi p-2 p}{\pi+4}\)
⇒ x = \(\frac{p}{\pi+4}\)
Hence, maximum light is admitted when radius of semi-circle be r = \(\frac{p}{\pi+4}\) cm.

Question 24.
A rectangular area of 9000 m² is to be surrounded by a fence with two opposite sides being made of brick and the other two of wood. One metre of wooden fencing costs 25 while one metre of brick walling costs 10. What is the least amount of money that must be alloted for the construction of such a fence?
Solution:
Let x be the length of rectangular area and
y be the breadth of rectangula area
∴ xy = 9000 m²
Let C = amount of money that must be allotted for the construction of such a fence.
∴ C = 25 × 2x + 2y × 10
= 50x + 20 × \(\frac{9000}{x}\) [using (1)]
Diff. both sides w.r.t. x; we have
\(\frac{d C}{d x}\) = 50 – \(\frac{180000}{x^2}\) ;
\(\frac{d^2 \mathrm{C}}{d x^2}=\frac{360000}{x^3}\)
For maxima / minima, \(\frac{d C}{d x}\) = 0
⇒ 50 – \(\frac{180000}{x^2}\) = 0
⇒ x² = \(\frac{18000}{50}\) = 3600
⇒ x = 60 (∵ x > 0)
∴ from (1) ;
⇒ y = \(\frac{9000}{60}\) = 150
Now \(\frac{d^2 \mathrm{C}}{d x^2}=\frac{360000}{x^3}\)
∴ \(\left(\frac{d^2 \mathrm{C}}{d x^2}\right)_{x=60}=\frac{360000}{60 \times 60 \times 60}\)
= \(\frac{360}{216}=\frac{5}{3}\) > 0
Thus C is minimise for x = 60 and y = 150
∴ required amount of money = C
= ₹ (50 × 60 + 20 × 150)
= ₹ 6000.

Question 25.
(i) Find the point on the parabola y² = 4x which is nearest to the point (2, – 8).
(ii) Find the point on the curve x² = 8y which is nearest to the point (2, 4).
Sol.
(I) Let P(x, y) be any point on the curve
y² = 4x …………(1)
Let A (2, – 8) be the given point.
∴ |AP| = \(\sqrt{(x-2)^2+(y+8)^2}\)
⇒ AP² = (x – 2)² + (y + 8)²
= (\(\frac{y^2}{4}\) – 4)² + (y + 8)² [using (1)]
Let S = AP²,
Then S is maximum or minimum according as AP is maximum or minimum.
Now, S = (\(\frac{y^2}{4}\) – 4)² + + (y + 8)²
∴ \(\frac{d S}{d y}\) = 2 (\(\frac{y^2}{4}\) – 2) (\(\frac{y}{2}\)) + 2 (y + 8)
∴ \(\frac{d S}{d y}\) = \(\frac{y^3}{4}\) – 2y + 2y + 16
= \(\frac{y^3}{4}\) + 16
∴ \(\frac{d^2 \mathrm{~S}}{d y^2}=\frac{3 y^2}{4}\)
For max./minima, \(\frac{d S}{d y}\) = 0
⇒ y³ = – 64
⇒ y³ = – 64
⇒ y = – 4
∴ (\(\frac{d^2 S}{d y^2}\))_{y = – 4} = \(\frac{3}{4}\) (16) = 12 > 0
∴ S is minimise for y = – 4
and from (1); x = 4
Hence the point (4, – 4) on curve y³² = 4x is nearest to the given point (2, – 8).

(ii) Let P(x, y) be any point on the curve
x² = 8y …………………..(1)
and let the given point be A (2, 4).
∴ AP² = (x – 2)² + (y – 4)²
= (x – 2)² + \(\left(\frac{x^2}{8}-4\right)^2\) [using (1)]
Let S = AP²,
Now S is max./min.
according as AP is max./min.
∴ S = (x – 2)² + \(\left(\frac{x^2}{8}-4\right)^2\)
∴ \(\frac{d S}{d x}\) = 2 (x – 2) + 2 \(\left(\frac{x^2}{8}-4\right) \frac{2 x}{8}\)
= 2x – 4 + \(\frac{x^3}{16}\) – 2x
∴ \(\frac{d \mathrm{~S}}{d x}=\frac{x^3}{16}\) – 4
∴ \(\frac{d^2 \mathrm{~S}}{d x^2}=\frac{3 x^2}{16}\)
For max./min., \(\frac{d S}{d x}\) = 0
⇒ x³ = 8
⇒ x = 4
when x = 4,
\(\left(\frac{d^2 \mathrm{~S}}{d x^2}\right)=\frac{3 \times 16}{16}\) = 3 > 0
∴ S is minimize for x = 4
∴ from (1) ; y = 2
∴ Required point on given curve be (4, 2).

Question 26.
Find the maximum area of an isosceles triangle inscribed in the ellipse \(\frac{x^2}{16}+\frac{y^2}{9}\) = 1 with its vertex at one end of the major axis.
Solution:
Let any point P on the ellipse

\(\frac{x^2}{16}+\frac{y^2}{9}\) = 1 is (4 cos θ, 3 sin θ)
∴ A = Area of ∆ \(\frac{1}{2}\) (PP’) (AB)
= \(\frac{1}{2}\) (6 sin θ) (4 – 4 cos θ)
= 12 sin θ (1 – cos θ)
= 12 [sin θ – \(\frac{1}{2}\) sin 2θ]
∴ \(\frac{d \mathrm{~A}}{d \theta}\) = 12 [cos θ – cos 2θ]
For Maxima / Minima, \(\frac{d \mathrm{~A}}{d \theta}\) = 0
⇒ cos 2θ – cos θ = 0
⇒ 2 cos θ – cos θ – 1 = 0
⇒ (cos θ – 1) (2 cos θ + 1) = 0
⇒ cos θ = 1 or
cos θ = – \(\frac{1}{2}\) = cos (π – \(\frac{\pi}{3}\))
⇒ θ = 0 or θ = \(\frac{2 \pi}{3}\)
When θ = 0, the point P coincide with A (4, 0)
When θ = \(\frac{2 \pi}{3}\),
Now \(\frac{d^2 \mathrm{~A}}{d \theta^2}\) = 12 (- sin θ + 2 sin 2θ)
⇒ \(\left(\frac{d^2 \mathrm{~A}}{d \theta^2}\right)_{\theta=\frac{2 \pi}{3}}=12\left(-\sin \frac{2 \pi}{3}+2 \sin \frac{4 \pi}{3}\right)\)
= \(\left(-\frac{\sqrt{3}}{2}+2\left(\frac{-\sqrt{3}}{2}\right)\right)\)
= – 18√3 < 0
∴ A is maximise at θ = \(\frac{\pi}{3}\)
Hence maximum area = A
= 12 \(\left[\sin \frac{2 \pi}{3}-\frac{1}{2} \sin \frac{4 \pi}{3}\right]\)
= 12 \(\left[\frac{\sqrt{3}}{2}+\frac{1}{2} \times \frac{\sqrt{3}}{2}\right]\)
= 9√3 sq. units

Question 27.
Show that the surface area of a closed cuboid with square base and given volume is minimum when it is a cube.
Solution:
Let x, x, y are the length, breadth and height of closed cuboid.
∴ area of base square = x² ;
area of four walls = 4xy
Let S = area of cuboid = 2x² + 4xy ……….(1)
Further, V = volume of cuboid = x²y ………..(2)
∴ S = 2x² + 4x \(\left(\frac{\mathrm{V}}{x^2}\right)\) [using (1)]
∴ \(\frac{d \mathrm{~S}}{d x}=4 x-\frac{4 \mathrm{~V}}{x^2}\)
∴ \(\frac{d^2 \mathrm{~S}}{d x^2}=4+\frac{8 \mathrm{~V}}{x^3}\)
For Max/Min, \(\frac{d S}{d x}\) = 0
⇒ 4x = \(\frac{4 V}{x^2}\)
⇒ x³ = V
⇒ x = (V)^1/3
at x = (V)^1/3
⇒ \(\frac{d^2 S}{d x^2}\) = 4 + \(\frac{8 V}{V}\) = 12 > 0
∴ S is minimise for x = \(\sqrt[3]{v}\)
∴ y = \(\frac{\mathrm{V}}{x^2}=\frac{\mathrm{V}}{\mathrm{V}^{2 / 3}}\)
= V^1/3
Hence all the dimensions of cuboid are equal.
∴ Cuboid because cube.
Hence the surface area of a closed cuboid with square base and given volume is minimum when it is a cube.

Question 28.
An open box with a square base is to be made of given iron sheet of area 27 sq.m. Show that the maximum volume of the box is 13.5 cu. m .
Solution:
Let x, x and y are the length, breadth and height of square box.
It is given that
x² + 4xy = 27 sq. m
and V = volume of box = x²y
⇒ V = x² \(\left(\frac{27-x^2}{4 x}\right)\)
= \(\frac{27}{4} x-\frac{x^3}{4}\)
∴ \(\frac{d V}{d x}=\frac{27}{4}-\frac{3 x^2}{4}\)
For maximaJminima,
⇒ \(\frac{d V}{d x}\) = 0
⇒ \(\frac{27}{4}-\frac{3 x^2}{4}\) = 0
⇒ 27 = 3x²
⇒ at x = 3; (∵ x > 0)
Thus V is maximise volume x = 3
∴ y = \(\frac{27-9}{4 \times 3}=\frac{3}{2}\)
∴ Maximum volume of box = x²y
=3² × \(\frac{3}{2}\)
= \(\frac{27}{2}\) cu.m = 13.5 cu.m.

Question 28 (old).
An open box with a square base is to be made of given ¡ron sheet of area 27 sq. m. Show that the maximum volume of the box is 13.5 cu. m.
Solution:
Let x be the side of the square base and y be the height of the box.
Thus area of square base = x²
and area of four walls = 4 xy
given area of square box + area of four val1s = 27 m²
⇒ x² + 4xy = 27
Let V = volume of box = x²y = x² \(\left(\frac{27-x^2}{4 x}\right)\) [using (1)]
V = \(\frac{1}{4}\) (27x – x³)
Diff, both sides w.r,t. x, we have
\(\frac{d V}{d x}\) = \(\frac{1}{4}\) (27 – 3x²) ;
\(\frac{d^2 V}{d x^2}\) = \(\frac{1}{4}\) (- 6x)
For maxima / minima, \(\frac{d V}{d x}\) = 0
⇒ 27 – 3x² = 0
⇒ x² = 9
⇒ x = 3 (∵ x > 0)
∴ \(\left(\frac{d^2 \mathrm{~V}}{d x^2}\right)_{x=3}=-\frac{3}{2} \times 3=-\frac{9}{2}\) < 0
Thus V is minimise for x = 3
∴ from (1) ; we have
y = \(\frac{27-x^2}{4 x}\)
= \(\frac{27-9}{12}\)
= \(\frac{18}{12}\)
= \(\frac{3}{2}\)
Hence, the required maximum volume of box = x²y
= (3² × \(\frac{3}{2}\)) cu.m = 13.5 cu.m.

Question 29.
The volume of a closed metal box with a square base is 4096 cm³. The cost of polishing the outer surface of the box is Rs. 4 per cm². Find the dimensions of the box for the minimum cost of polishing it. (ISC 2019)
Solution:
Let the dimensions of closed metal box are x cm, x cm and y cm respectively.
Then volume of closed metal box = x²y
and given volume of box = 4096 cm³
∴ x²y = 4096
Let S = surface area of closed metal box = 2x² + 4xy.
Given, the cost of polishing the outer surface of box be Rs. 4 per cm².
Let C be the cost of polishing the whole metal box
Then C = 4(2x² + 4xy) = 8(x² + 2xy)
C = 8(x² + 2x × \(\frac{4096}{x^2}\)) [using eqn. (1)]
∴ \(\frac{d \mathrm{C}}{d x}=8\left(2 x-\frac{2 \times 4096}{x^2}\right)\)
= 16 \(\left(x-\frac{4096}{x^2}\right)\)
For maximum / minima, \(\frac{d C}{d x}\) = 0
⇒ x³ = 4096
⇒ x = 16 (∵ x > 0)
Now \(\frac{d^2 \mathrm{C}}{d x^2}=16\left(1+\frac{2 \times 4096}{x^3}\right)\)
∴ \(\left(\frac{d^2 c}{d x^2}\right)_{x=16}=16\left(1+\frac{2 \times 4096}{16^3}\right)\)
= 48 > 0
Thus x = 16 be a point of minima
4096
∴ from (1) ;
y = \(\frac{4096}{16^2}\) = 16
Thus, cost is minimum when x = 16 cm and y = 16 cm.

Question 30.
An open tank with a square base and vertical sides is to be constructed from a metal sheet so as to hold a given quantity of water. Show that the cost of material will be least when the depth of the tank is half of its width.
Solution:
Let x, x and y are the length, width and height of an open tank
∴ Volume of tank V = x²y (given) …………..(1)
Now it is given that, the tank to hold a given quantity of water
∴ V = constant.
To minimise the cost of material. we have to minimise the surface area of the tank.
Let S = x² + 4xy
= x² + 4x (\(\frac{\mathrm{~V}}{x^2}\)) (using (1))
⇒ \(\frac{d \mathrm{~S}}{d x}=2 x-\frac{4 \mathrm{~V}}{x^2}\)
∴ \(\frac{d^2 \mathrm{~S}}{d x^2}=2+\frac{8 \mathrm{~V}}{x^3}\)
For max./minima, \(\frac{d S}{d x}\) = 0
⇒ x = (2V)^1/3
and \(\frac{d^2 S}{d x^2}\) = 2 + \(\frac{8 V}{2 V}\) = 6 > 0
∴ S is minimise for x³ = 2V.
i.e. x³ = 2x²y
⇒ x = 2y
⇒ y = \(\frac{x}{2}\)
i.e. S is minimum when depth of the tank is half of its width.

Question 31.
A tank with rectangular sides, open at the top, is to be constructed so that its depth is 2 m and volume is 8 m3. 1f building of tank costs Rs. 70 per square metre for the base and Rs. 45 per square metre for the sides, what is the cost of least expensive tank ?
Solution:
Let x, x, y be the length, width and height of the tank
∴ area of square base = x² and
area of four walls = 4xy
Given V = volume of open tank = 8 x²y …………(1)
∴ E = 70x² + 45 (4xy)
= 70x² + 180x . \(\frac{8}{x^2}\) [using (1)]
⇒ E = 70x² + \(\frac{180 \times 8}{x}\) ……….(2)
Given y = depth of tank = 2m
∴ from (1) ;
8 = x² × 2
⇒ x = 2[∵ x > 0]
Now we check whether x = 2 is point of maximum or minima.
\(\frac{d E}{d x}\) = 140x – \(\frac{180 \times 8}{x^2}\)
\(\frac{d^2 \mathrm{E}}{d x^2}=140+\frac{180 \times 16}{x^2}\)
at x = 2,
\(\frac{d^2 E}{d x^2}\) > 0
∴ E is minimum for x = 2.
∴ from (2) ; we have
least value of E = 70 (2)² + \(\frac{180 \times 8}{2}\) = ₹ 1000.

Question 32.
A rectangular sheet of tin 45 cm by 24 cm is to be made into a box without top by cutting off squares from the corners and folding up the flaps. What should be the side of the in square in order that the volume of the box is maximum ? (NCERT)
Solution:
Let x be the side of the square that is cut off from each comer of the plate.
Then sides of the base are 45 – 2x, 24 – 2x, x cm.
∴ Volume of the box = V = (45 – 2x) (24 – 2x) x
∴ V = (45 – 2x) (24x – 2x²)
= 2x (12 – x) (45 – 2x)
∴ V = 2x [540 – 24x – 45x + 2x²]
= 2x [2x² – 69x + 540]
∴ \(\frac{d V}{d x}\) = 2 [6x² – 138x + 540]
= 4 [3x² – 69x + 270]

For max/minima, \(\frac{d V}{d x}\) = 0
3x² – 69x + 270 = 0
∴ x = \(\frac{69 \pm 39}{6}\) = 5, 18
But x ≠ 18.
Thus, x = 5,
\(\frac{d^2 V}{d x^2}\) = 4 (6x – 69)
∴ (\(\frac{d^2 V}{d x^2}\))_{x = 5} = 4 (- 39)
= – 156 < 0
∴ V is maximise for x = 5.
Hence the volume of the box is maximum when the side of square is 5 cm and max. volume = 35 × 14 × 5 = 2450 cm³.

Question 33.
Show that the semi-vertical angle of a cone of the maximum volume and of given slant height is cos^-1 \(\frac{1}{\sqrt{3}}\).
Solution:
Let θ be the semi-vertical angle of cone and / be the given slant height of cone.
∴ r = radius of cone = l sin θ
and h = height of cone = l cos θ
Let V = volume of cone = \(\frac{1}{3}\) πr²h
= \(\frac{1}{3}\) π (l sin θ)² (l cos θ)

⇒ \(\frac{d V}{d \theta}\) = \(\frac{\pi l^3}{3}\) [sin² θ (- sin θ) + 2 cos² θ sin θ]
= \(\frac{\pi l^3}{3}\) sin θ [2 cos² θ – sin² θ]
∴ \(\frac{d^2 \mathrm{~V}}{d \theta^2}\) = \(\frac{\pi l^3}{3}\) [- 3 sin² θ cos θ + 2 cos³ θ – 4 cos θ sin² θ]
For Max/minima, \(\frac{d V}{d \theta}\) = 0
⇒ sin θ (2 cos² θ – sin² θ) = 0
∴ sin θ = 0
⇒ θ = 0, π which is not possible, (∵ in this case, r = 0)
∴ tan² θ = 2
⇒ tan θ = √2 [∵ θ lies between 0 and π/2]
∴ θ = tan^-1 (√2)
at tan θ = √2,

sin θ = \(\frac{\sqrt{2}}{\sqrt{3}}\)
and cos θ = \(\frac{1}{\sqrt{3}}\)
\(\frac{d^2 \mathrm{~V}}{d \theta^2}=\frac{\pi l^3}{3}\left[2 \times \frac{1}{3 \sqrt{3}}-7 \times \frac{1}{\sqrt{3}} \times \frac{2}{3}\right]\)
= \(\frac{\pi l^3}{3} \times\left(-\frac{12}{3 \sqrt{3}}\right)\) < 0
∴ V is maximise for θ = tan^-1 (√2) i.e. cos^-1 \(\frac{1}{\sqrt{3}}\)

Question 34.
Show that a right circular cylinder of given volume, which is open at the top, has minimum total surface area if its height is equal to radius of its base.
Solution:
Let h be the height and r be the radius of cylinder
V = πr²h (given) ………….(1)
Let S = Total surface area
= πr² + 2πrh
= πr² + 2πr \(\frac{\mathrm{V}}{\pi r^2}\) [using (1)]
∴ S = πr² + \(\frac{2 V}{r}\)
∴ \(\frac{d S}{d r}\) = 2πr – \(\frac{2 V}{r^{2}}\)
For Maxima / Minima, \(\frac{d S}{d r}\) = 0
⇒ 2πr = \(\frac{2 V}{r^{2}}\)
⇒ r = \(\left(\frac{\mathrm{V}}{\pi}\right)^{1 / 3}\)
Now \(\frac{d^2 S}{d r^2}\) = 2π + \(\frac{4 V}{r^{3}}\)
∴ at r = \(\left(\frac{\mathrm{V}}{\pi}\right)^{1 / 3}\) ;
\(\frac{d^2 S}{d r^2}\) = 2π + \(\frac{4 \mathrm{~V} \times \pi}{\mathrm{V}}\) = 6π > 0
∴ S is minimise for r = \(\left(\frac{\mathrm{V}}{\pi}\right)^{1 / 3}\)
∴ From (1) ; we have
h = \(\frac{\mathrm{V}}{\pi\left(\frac{\mathrm{V}}{\pi}\right)^{2 / 3}}\)
= \(\frac{\mathrm{V}^{1 / 3}}{\pi^{1 / 3}}=\left(\frac{\mathrm{V}}{\pi}\right)^{1 / 3}\)
Hence S is minimise when radius of cylinder is equal to height of cylinder.

Question 34 (old).
Show that the height of a closed right circular cylinder of given surface and maximum volume is equal to the diameter of base.
Solution:
Let h be the height and r be the radius of closed cylinder.
Let S = surface area of cylinder
⇒ S = 2πr + 2πrh ……………(1) (given)
Let V = volume of cylinder = πr²h
= πr² \(\left[\frac{\mathrm{S}-2 \pi r^2}{2 \pi r}\right]\)
⇒ V = \(\frac{r}{2}\) [S – 2πr²]
⇒ \(\frac{d V}{d r}\) = \(\frac{1}{2}\) [S – 6πr²]
For MaximaJMinima, \(\frac{d V}{d r}\) = 0
⇒ S = 6πr²
∴ \(\frac{d^2 V}{d r^2}\) = \(\frac{1}{2}\) (- 12πr)
= – 6πr
= – 6π\(\sqrt{\frac{\mathrm{S}}{6 \pi}}\) < 0 [∵ r > 0]
Thus V is maximise when S = 6πr²
2πr² + 2πrh = 6πr² [using (1)]
⇒ 2πrh = 4πr²
⇒ h = 2r
Thus height of cylinder = diameter of closed cylinder.

Question 35.
(i) Show that a closed cylindrical vessel of given volume has the least (total) surface area when its height is twice its radius.
(ii) A closed right circular cylinder has a volume of 2156 cu. cm. What will be the radius of its base so that its total surface area is minimum ? Find the height of the cylinder when its total surface area is minimum. Take π = \(\frac{22}{7}\). (ISC 2004)
Solution:
(1) Let h be the height and r be the radius of closed cylinder
V = πr²h (given) …………(1)
and Let S = surface area of cylinder = 2πr² + 2πrh
⇒ S = 2πr² + 2πr \(\left(\frac{\mathrm{V}}{\pi r^2}\right)\)
= 2πr² + \(\frac{2 V}{r}\)

∴ S is minimise for r³ = \(\frac{V}{2 \pi}\)
i.e. V = 2πr³
⇒ 2πr³ = πr²h
⇒ h = 2r [using (1)]
i.e. height of cylinder diameter of cylinder.

(ii) Let r be the radius of the base of cylinder
and h be the height of cylinder.
Then volume of cylinder = 2156 cm³
⇒ πr²h = 2156
⇒ h = \(\frac{2156}{\pi r^2}\) ……….(1)
Let S = Total surface area of cylinder
∴ S = 2πr² + 2πrh
S = 2πr (r + \(\frac{2156}{\pi r^2}\)) [using (1)]
∴ \(\frac{d \mathrm{~S}}{d r}=2 \pi\left(2 r-\frac{2156}{\pi r^2}\right)\)
For maxima/minima, \(\frac{d S}{d r}\) = 0
⇒ 2r – \(\frac{2156}{\pi r^2}\) = 0
⇒ r³ = \(\frac{2156}{2 \pi}=\frac{1078}{22}\) × 7
= 49 × t = 7³
⇒ r = 7 cm
∴ \(\left(\frac{d^2 \mathrm{~S}}{d r^2}\right)_{r=7}=2 \pi\left(2+\frac{4312}{\pi \times 7^3}\right)\)
= 2π \(\left(2+\frac{4312}{22 \times 49}\right)\)
= 12π > 0
Thus, S is minimise for r = 7
Hence the required radius of base of cylinder = 7 cm.

Question 36.
Show that the radius of a closed right circular cylinder of given surface area and maximum volume is equal to half of its height. (ISC 2020)
Solution:
Let r be the radius and h be the height of right circular cylinder.
Let S be given surface area of cylinder.
∴ S = 2πr² + 2πrh
Let V = volume of cylinder = πr²h
⇒ V = πr² \(\left(\frac{\mathrm{S}-2 \pi r^2}{2 \pi r}\right)\)
[∵ from (1) ;
h = \(\frac{S-2 \pi r^2}{2 \pi r}\)]
⇒ V = \(\frac{r}{2}\) (S – 2πr²)
= \(\frac{ S r}{2}\) – πr³
∴ \(\frac{d V}{d r}\) = \(\frac{S}{2}\) – 3πr²
For maxima/minima, \(\frac{d V}{d r}\) = 0
⇒ \(\frac{S}{2}\) = 3πr²
⇒ S = 6πr²
⇒ 2πr² + 2πrh = 6πr²
⇒ 2πrh = πr²
⇒ h = 2r
⇒ r = \(\frac{h}{2}\)
∴ \(\frac{d^2 V}{d x^2}\) = – 6πr
⇒ \(\frac{d^2 V}{d x^2}\) < 0
Thus V is maximise when r = \(\frac{h}{2}\)
i.e. V is maximum when radius of right circular cylinder is half of its height.

Question 36 (old).
Show that the height of the cone of maximum volume that can be inscribed in a sphere of radius 12 cm is 16 cm.
Solution:
Let ABC be the cone of largest volume that can be inscribed in the sphere, it is understandable that for maximum value, the axis of the cone must be along the diameter of sphere. Let DE = x,
∴ radius of cone = \(\sqrt{r^2-x^2}\) =BE
and height of cane = r + x
Then volume of cone
⇒ V = \(\frac{1}{3}\) π (r² – x²) (r + x)
⇒ \(\frac{d \mathrm{~V}}{d x}=\frac{\pi}{3}\) [r² – 2rx – 3x²]
⇒ \(\frac{d^2 V}{d x^2}\) = \(\frac{\pi}{3}\) (- 6x – 2r)
For Max./minima, \(\frac{d V}{d x}\) = 0
⇒ 3x² + 2rx – r² = 0
⇒ (x + r) (3x – r) = 0
⇒ x = \(\frac{r}{3}\) [∵ r ≠ x]
at x = \(\frac{r}{3}\),
⇒ \(\frac{d^2 V}{d x^2}\) = \(\frac{\pi}{3}\) (- 4r) < 0
∴ V is maximum for x = \(\frac{r}{3}\)
∴ Max. volume of cone = \(\frac{\pi}{3}\left(r^2-\frac{r^2}{9}\right)\left(r+\frac{r}{3}\right)\)
= \(\frac{32 \pi r^3}{81}\)
Thus Max. value of cone = \(\frac{8}{27}\left(\frac{4}{3} \pi r^3\right)\)
= \(\frac{8}{27}\) (volume of sphere)
∴ height of cone = x + r
= \(\frac{r}{3}\) + r = \(\frac{4r}{3}\)
= \(\frac{4 \times 12}{3}\) = 16 cm then.

[Since given radius of sphere = r = 12 cm]

Question 37.
Find the maximum volume of the cylinder which can be inscribed in a sphere of radius k 3√3 cm. (Leave the answer in terms of it). (ISC 2013)
Solution:
Let h be the height and R be the radius of the cylinder that is inscribed in a sphere of radius r.
In ∆EFA,
R² = r² – (\(\frac{h}{2}\))² ………..(1)

Let V = volume of under
= πr²h
= πh [r² – \(\left(\frac{h}{2}\right)^2\)]
∴ \(\frac{d V}{d h}\) = π \(\left[r^2-\frac{3}{4} h^2\right]\)
∴ \(\frac{d^2 V}{d h^2}\) = – \(\frac{3}{2}\) πh
For maxima / minima, \(\frac{d V}{d h}\) = 0
⇒ r² – \(\frac{3}{4}\) h² = 0
⇒ h = \(\frac{2 r}{\sqrt{3}}\)
∴ \(\left(\frac{d^2 \mathrm{~V}}{d h^2}\right)_{h=\frac{2 r}{\sqrt{3}}}=-\frac{3}{2} \pi \times \frac{2 r}{\sqrt{3}}\)
= – \(\frac{3 \pi r}{\sqrt{3}}\) < 0
∴ V is maximise h = \(\frac{2 r}{\sqrt{3}}\)
∴ Max. volume = π × \(\frac{2 r}{\sqrt{3}}\left[r^2-\frac{1}{4} \times \frac{4 r^2}{3}\right]\)
= \(\frac{2 r \pi}{\sqrt{3}} \times \frac{2 r^2}{3}=\frac{4 \pi r^3}{3 \sqrt{3}}\)
Given radius of sphere r = 3√3 cm
∴ Max. value = \(\frac{4 \pi}{3 \sqrt{3}} \times(3 \sqrt{3})^3\)
= 108π cm³.

Question 38.
A cone is inscribed in a sphere of radius 12 cm. If the volume of the cone Is maximum, find its height.
Solution:
Let ABC be the cone of largest volume that can be inscribed in the sphere, it is understandable that for maximum value, the axis of the cone must be along the diameter of sphere.
Let DE = x,
radius of cone = \(\sqrt{r^2-x^2}\) = BE
and height of cane = r + x
Then volume of cone
⇒ V = \(\frac{1}{3}\) π (r² – x²)
⇒ \(\frac{d V}{d x}\) = \(\frac{\pi}{3}\) [r² – 2rx – 3x²]
⇒ \(\frac{d^2 V}{d x^2}\) = \(\frac{\pi}{3}\) [- 6x – 2r]
For Max./minima, \(\frac{d V}{d x}\) = 0
⇒ 3x² + 2rx – r² = 0
⇒ (x + r) (3x – r) = 0
⇒ x = \(\frac{r}{3}\) [∵ r ≠ x]
at x = \(\frac{r}{3}\),
\(\frac{d^2 V}{d x^2}\) = \(\frac{\pi}{3}\) (- 4r) < 0
∴ V is maximise for x = \(\frac{r}{3}\)
∴ Max. volume of cone = \(\frac{\pi}{3}\left(r^2-\frac{r^2}{9}\right)\left(r+\frac{r}{3}\right)=\frac{32 \pi r^3}{81}\)
Thus Max. volume of cone = \(\frac{8}{27}\left(\frac{4}{3} \pi r^3\right)\)
= \(\frac{8}{27}\) (volume of the sphere)
∴ height of cone = x + r
= \(\frac{r}{3}\) + r
= \(\frac{4 r}{3}=\frac{4 \times 12}{3}\)
= 16 cm then.

[Since given radius of sphere = r = 12 cm].

Question 39.
Show that the altitude of a right circular cone of maximum curved surface area which can be inscribed in a sphere of radius r is \(\frac{4 r}{3}\).
Solution:
Let PAB be a cone of maximum curved area that can be inscribed in a sphere of radius r.
Now it is obvious that. For maximum volume, axis of cone must lie along diameter of sphere.

∴ PM height of cone = OP + OM = r + x
In right angled ∆OAM,
AM² = OA² – OM²
= r² – x²
Let S = curved surface area of cone
= π (AM) l
= π \(\sqrt{r^2-x^2}\left[\sqrt{(r+x)^2+r^2-x^2}\right]\)
⇒ V = S² = π² (r² – x²) (2r² + 2rx)
To maximise S it is convenient to maximise S² = V
∴ \(\frac{d V}{d x}\) = π² [2r³ – 4r²x – 6rx²]
and \(\frac{d^2 V}{d x^2}\) = π² [- 4r² – 12rx]
For maxima / minima, \(\frac{d V}{d x}\) = 0
⇒ 2r³ – 4r²x – 6rx² = 0
⇒ – 2r (3x² + 2rx – r²) = 0
⇒ (x + r) (3x – r) = 0
⇒ x = – r, \(\frac{r}{3}\) [but x > 0]
∴ x = \(\frac{r}{3}\)
∴ \(\left(\frac{d^2 \mathrm{~V}}{d x^2}\right)_{x=\frac{r}{3}}\)
= – 8π²r² < 0
Thus curved surface area of cone is maximise when x = \(\frac{r}{3}\)
:. altitude of cone = r + x = r + \(\frac{r}{3}\) = \(\frac{4 r}{3}\)

Question 40.
Show that the height of a right circular cylinder of greatest volume which can be inscribed in a right circular cone of height h and radius r is one-third of the height of the cone and the greatest volume of the cylinder is \(\frac{4}{9}\) times the volume of the cone.
Solution:
Let PAB be the cone with height OP = h
and OA = OB = r
Let a cylinder of base radius OM’ = ON’ = x
and height = OO’ be inscribed in cone.
Now ∆POB ~ ∆NN’B

Thus, \(\frac{\mathrm{OP}}{\mathrm{NN}^{\prime}}=\frac{\mathrm{OB}}{\mathrm{N}^{\prime} \mathrm{B}}\)
⇒ \(\frac{h}{\mathrm{NN}^{\prime}}=\frac{r}{r-x}\)
NN’ = \(\frac{h(r-x)}{r}\) = height of cylinder
V = volume of right circular cylinder
= πx² (NN’)
⇒ V = πx² \(\frac{h}{r}\) (r – x)
= \(\frac{\pi h}{r}\) (rx² – x³)
∴ \(\frac{d \mathrm{~V}}{d x}=\frac{\pi h}{r}\) (2rx – 3x²)
and \(\frac{d^2 V}{d x^2}\) = \(\frac{\pi h}{r}\) (2r – 6x)
For maxima / minima, \(\frac{d V}{d x}\) = 0
⇒ \(\frac{\pi h}{r}\) (2rx – 3x²) = 0
⇒ 2rx = 3x²
⇒ x = \(\frac{2 r}{3}\)
and \(\frac{d^2 V}{d x^2}\) = \(\frac{\pi h}{r}\) (2r – 4r)
= – 2πh < 0 [∵ x > 0]
Thus V is maximise when x = \(\frac{2 r}{3}\)
∴ NN’ = height of cylinder
= \(\frac{h}{r}\left[r-\frac{2 r}{3}\right]=\frac{h}{3}\)
Hence volume of right circular cylinder which is inscribed in right circular cone be greatest when height of cylinder = \(\frac{1}{3}\) height
of cone and greatest volume of cylinder
= πx²\(\frac{h}{3}\)
= \(\pi\left(\frac{2 r}{3}\right)^2 \frac{h}{3}\)
= \(\frac{4}{27}\) πr²h
= \(\frac{4}{9}\) (\(\frac{1}{3}\) πr²h)
= \(\frac{4}{9}\) (Volume of cone).

Question 41.
The sum of the surface areas of a rectangular parallelopiped with sides x, 2x and \(\frac{x}{3}\) a sphere is given to be constant, prove that the sum of their volumes is minimum if x is equal to three times the radius of the sphere. Also find the minimum value of the sum of their volumes.
Solution:
S = sum of areas of || piped and sphere
⇒ S = 2 (x × 2x + 2x . \(\frac{x}{3}\) + x . \(\frac{x}{3}\)) + 4πy²
where y = radius of sphere
⇒ S = 2 (2x² + \(\frac{2}{3}\) x² + \(\frac{x^{2}}{3}\)) + 4πy²
⇒ S = 6x² + 4πy² …………..(1)
Let V = volume of ||piped + Volume of sphere
= x (2x) \(\frac{x}{3}\) + \(\frac{4}{3}\) πy³

Thus V is minimise when x = 3y
∴ V = \(\frac{4}{3} \pi y^3+\frac{2}{3} x^3\)
= \(\frac{4 \pi}{3}\left(\frac{x}{3}\right)^3+\frac{2}{3} x^3\)
V = \(\frac{4 \pi}{81} x^3+\frac{2}{3} x^3\)
= \(\frac{2}{3} x^3\left(1+\frac{2 \pi}{27}\right)\).

Question 42.
A manufacturer plans to construct a cylindrical can to hold one cubic metre of oil. If the cost of constructing top and bottom of the can is twice the cost of constructing the side, what are the dimensions ofthe most economical can?
Solution:
Let r be the radius of cylindrical can and h be the height of cylindrical can.
Then 1 = πr²h volume of cylindrical can ………..(1)
Let Rs. p be the cost of constructing the side
then cost of constructing top and bottom of can be Rs. 2p each.
Let C = total cost of constructing the cylindrical can
⇒ C = 2πr² × 2p + 2πrh × p
⇒ C = 4pπr² + 2πrp × \(\frac{1}{\pi r^2}\) [using eqn. (1)]
⇒ C = 4pπr² + \(\frac{2 p}{r}\)
On differentiating w.r.t. r, we have
\(\frac{d C}{d r}\) = 8pπr – \(\frac{2 p}{r^{2}}\) ;
\(\frac{d^2 \mathrm{C}}{d r^2}=8 p \pi+\frac{4 p}{r^3}\)
For maxima / minima,
\(\frac{d C}{d r}\) = 0
⇒ 8pπr = \(\frac{2 p}{r^{2}}\)
⇒ r³ = \(\frac{1}{4 \pi}\)
⇒ r = \(\left(\frac{1}{4 \pi}\right)^{1 / 3}\)
at r = \(\left(\frac{1}{4 \pi}\right)^{1 / 3}\) ;
\(\frac{d^2 C}{d r^2}\) = 8πp + 4p × 4π
= 24pπ > 0
Thus C is maximise for r = \(\left(\frac{1}{4 \pi}\right)^{1 / 3}\)
∴ from (1) ; we have
h = \(\frac{1}{\pi r^2}\)
= \(\frac{1}{\pi\left(\frac{1}{4 \pi}\right)^{2 / 3}}=\frac{(4 \pi)^{2 / 3}}{\pi}\)
⇒ h = \(\left(\frac{16}{\pi}\right)^{1 / 3}\)
Hence, the required dimensions of the most economical can be r = \(\left(\frac{1}{4 \pi}\right)^{1 / 3}\) metre and
h = \(\left(\frac{16}{\pi}\right)^{1 / 3}\) metre.

[Since given radius of sphere = r = 12 cm]