Ncert Solutions For Class 12 Maths Ex 7.1

Ncert Solutions For Class 12 Maths Chapter 7 Ex 7.1

Question 1:

By the method of inspection obtain an integral (or anti – derivative) of the sin 3x.

Answer:

As the derivative is sin 3x and x is the function of the anti – derivative of sin 3x.

\(\frac{d}{dx} (cos\; 3x) = – 3 sin\; 3x \\ sin\; 3x = – \frac{1}{3} \frac{d}{dx} (cos\; 3x) \\ sin\; 3x = \frac{d}{dx} (- \frac{1}{3} cos\; 3x) \\ Hence,\; the\; anti – derivative\; of\; sin\; 3x \;is\; (- \frac{1}{3} cos\; 3x)\)

Question 2:

By the method of inspection obtain an integral (or anti – derivative) of the cos 2x.

Answer:

As the derivative is cos 2x and x is the function of the anti – derivative of cos 2x

\(\frac{d}{dx} (sin\; 2x) = – 2 cos\; 2x \\ cos\; 2x = \frac{1}{2} \frac{d}{dx} (sin\; 2x) \\ cos\; 2x = \frac{d}{dx} (\frac{1}{2} (sin\; 2x)) \\ Hence,\; the\; anti – derivative\; of\; sin\; 2x \;is\; (- \frac{1}{2} cos\; 2x)\)

 

 

Question 3:

By the method of inspection obtain an integral (or anti – derivative) of the e5x.

Answer:

As the derivative is e5x and x is the function of the anti – derivative of e5x

\(\frac{d}{dx} (e ^{5x}) = 5 e ^{5x} \\ e ^{5x} = \frac{1}{5} \frac{d}{dx} (e^{5x}) \\ e ^{5x} = \frac{d}{dx} (\frac{1}{5} e^{5x}) \\ Hence,\; the\; anti – derivative\; of\; e ^{5x} is \frac{1}{5} e^{5x}\)

Question 4:

By the method of inspection obtain an integral (or anti – derivative) of the (mx + n) 2.

Answer:

As the derivative is (mx + n) 2 and x is the function of the anti – derivative of (mx + n) 2

\(\frac{d}{dx} (mx + n)^{3} = 3m (mx + n) ^{2}\\ (mx + n) ^{2} = \frac{1}{3m} \frac{d}{dx} (mx + n) ^{3} \\ (mx + n) ^{2} = \frac{d}{dx} (\frac{1}{3m} (mx + n) ^{3}) \\ Hence,\; the\; anti – derivative\; of\; (mx + n) ^{2} is \frac{1}{3m} (mx + n) ^{3}\)

Question 5:

By the method of inspection obtain an integral (or anti – derivative) of the sin 3x – 5 e 2x

Answer:

As the derivative is (sin 3x – 5 e 2x) and x is the function of the anti – derivative of (sin 3x – 5 e 2x)

\(\frac{d}{dx} (- \frac{1}{3} cos\; 3x – \frac{5}{2} e^{2x}) = sin 3x – 5 e^{2x} \\ Hence,\; the\; anti – derivative\; of\; sin\; 3x – 5 e^{2x} \;is\; (- \frac{1}{3} cos\; 3x – \frac{5}{2} e^{2x})\)

Question 6:

By the method of inspection obtain an integral of the \(\int (4 e^{2u} + 1) du\)

Answer:

Integral of \((4 e^{2u} + 1)\) and u is the function of the integral \((4 e^{2u} + 1)\).

\(\int (4 e^{2u} + 1) du \\ 4 \int e^{2u} du + \int 1 du \\ 4 (\frac{e^{2u}}{2}) + u + c \\ 2 e^{2u} + u + c \\ Where\; c\; is\; the\; constant.\)

Question 7:

By the method of inspection obtain an integral of the \(\int u^{2} (1 – \frac{1}{u^{2}}) du\)

Answer:

Integral of \(u^{2} (1 – \frac{1}{u^{2}})\) and u is the function of the integral \(u^{2} (1 – \frac{1}{u^{2}})\)

\(\int u^{2} (1 – \frac{1}{u^{2}}) du \\ \int (u^{2} – 1) du \\ \frac{u^{3}}{3} – u + c \\ Where\; c\; is\; the\; constant\)

Question 8:

By the method of inspection obtain an integral of the \(\int (a u^{2} + b u + c) du\)

Answer:

Integral of \(a u^{2} + b u + c\) and u is the function of the integral \(a u^{2} + b u + c\)

\(\int (a u^{2} + b u + c) du \\ a \int (u^{2}) du + b \int u du + c \int 1 du \\ a (\frac{u^{3}}{3}) + b (\frac{u^{2}}{2}) + cu + C \\ \\ Where\; C\; is\; the\; constant\)

Question 9:

By the method of inspection obtain an integral of the \(\int (a u^{2} + e^{u}) du\)

Answer:

Integral of \(a u^{2} + e^{u}\) and u is the function of the integral \(a u^{2} + e^{u}\)

\(\int (a u^{2} + e^{u}) du \\ a \int (u^{2}) du + \int e^{u} du \\ a (\frac{u^{3}}{3}) + e^{u} + C \\ \\ Where\; C\; is\; the\; constant\)

Question 10:

By the method of inspection obtain an integral of the \(\int (\sqrt{u} + \frac{1}{\sqrt{u}}) ^{2} du\)

Answer:

Integral of \((\sqrt{u} + \frac{1}{\sqrt{u}}) ^{2}\) and u is the function of the integral \((\sqrt{u} + \frac{1}{\sqrt{u}}) ^{2}\)

\((\sqrt{u} + \frac{1}{\sqrt{u}}) ^{2} \\ \int (u + \frac{1}{u} – 2) du \\ \int u du + \int \frac{1}{u} du – 2 \int 1 du \\ \frac{u^{2}}{2} + log \left | u \right | – 2 u + C \\ Where\; C\; is\; the\; constant\)

Question 11:

By the method of inspection obtain an integral of the \(\int \frac{u^{3} + 4 u^{2} + 4}{u ^{2}} du\)

Answer:

Integral of and u is the function of the integral \(\frac{u^{3} + 4 u^{2} + 4}{u ^{2}}\)

\(\int \frac{u^{3} + 4 u^{2} + 4}{u ^{2}} du \\ \int u du + 4 \int 1 du + \int \frac{4}{u^{2}} du \\ \frac{u ^{2}}{2} + 4 u + \frac{4}{x} + C \\ Where\; C\; is\; the\; constant\)

Question 12:

By the method of inspection obtain an integral of the \(\frac{u^{3} + 4 u + 4}{\sqrt{u}}\)

Answer:

Integral of \(\frac{u^{3} + 4 u + 4}{\sqrt{u}}\) and u is the function of the integral \(\frac{u^{3} + 4 u + 4}{\sqrt{u}}\)

\(\int \frac{u^{3} + 4 u + 4}{\sqrt{u}} du \\ \int (u ^{\frac{5}{2}} + 4 u ^{\frac{1}{2}} + 4 u^{- \frac{1}{2}}) \\ = \frac{u ^{\frac{7}{2}}}{\frac{7}{2}} + \frac{4 (u ^{\frac{3}{2}})}{\frac{3}{2}} + \frac{4 (u ^{\frac{1}{2}})}{\frac{1}{2}} + C \\ = \frac{2}{7} (u ^{\frac{7}{2}}) + \frac{8}{3} (u ^{\frac{3}{2}}) + 8 u ^{\frac{1}{2}} + C \\ = \frac{2}{7} (u ^{\frac{7}{2}}) + \frac{8}{3} (u ^{\frac{3}{2}}) + 8 \sqrt{u} + C \\ Where\; C\; is\; the\; constant\)

Question 13:

By the method of inspection obtain an integral of the \(\frac{u^{3} – u^{2} + u + 1}{u – 1}\)

Answer:

Integral of \(\frac{u^{3} – u^{2} + u + 1}{u – 1}\) and u is the function of the integral \(\frac{u^{3} – u^{2} + u + 1}{u – 1}\)

\(\int \frac{u^{3} – u^{2} + u + 1}{u – 1} du \\ On\; divinding,\; we\; get\; \\ \int (u^{2} + 1) du \\ \int u^{2} du + \int 1 du \\ \frac{u^{3}}{3} + u + C Where\; C\; is\; the\; constant\)

Question 14:

By the method of inspection obtain an integral of the \((1 – u) \sqrt{u}\)

Answer:

Integral of \((1 – u) \sqrt{u}\) and u is the function of the integral \((1 – u) \sqrt{u}\)

\(\int (1 + u) \sqrt{u}\; du \\ \int (\sqrt{u} + u^{\frac{3}{2}}) du \\ \int u^{\frac{1}{2}} du + \int u^{\frac{3}{2}} du \\ \frac{u^{\frac{3}{2}}}{\frac{3}{2}} + \frac{u^{\frac{5}{2}}}{\frac{5}{2}} + C \\ \frac{2}{3} u^{\frac{3}{2}} + \frac{2}{5} u^{\frac{5}{2}} + C \\ Where\; C\; is\; the\; constant\)

Question 15:

By the method of inspection obtain an integral of the \(\sqrt{u} (3u^{2} + 2u + 5)\)

Answer:

Integral of \(\sqrt{u} (3u^{2} + 2u + 5)\) and u is the function of the integral \(\sqrt{u} (3u^{2} + 2u + 5)\)

\(\int \sqrt{u} (3u^{2} + 2u + 5) du \\ \int (3u ^{\frac{5}{2}} + 2u ^{\frac{3}{2}} + 5u ^{\frac{1}{2}}) du \\ 3 \int u ^{\frac{5}{2}} du + 2 \int u ^{\frac{3}{2}} du + 5 \int u ^{\frac{1}{2}} du \\ 3 (\frac{u ^{\frac{7}{2}}}{\frac{7}{2}}) + 2 (\frac{u ^{\frac{5}{2}}}{\frac{5}{2}}) + 5 (\frac{u ^{\frac{3}{2}}}{\frac{3}{2}}) + C \\ \frac{6}{7} u ^{\frac{7}{2}} + \frac{4}{5} u ^{\frac{5}{2}} + \frac{10}{3} u ^{\frac{3}{2}} + C \\ Where\; C\; is\; the\; constant\)

Question 16:

By the method of inspection obtain an integral of the \(2 u – 2 cos\; u + e ^{u}\)

Answer:

Integral of \(2 u – 2 cos\; u + e ^{u}\) and u is the function of the integral \(2 u – 2 cos\; u + e ^{u}\)

\(\int (2 u – 2 cos\; u + e ^{u}) du \\ 2 \int u du – 2 \int cos\; u du + \int e ^{u} du \\ 2 \frac{u ^{2}}{2} – 2 (sin u) + e ^{u} + C \\ u ^{2} – 2 sin\; u + e ^{u} + C Where\; C\; is\; the\; constant\)

Question 17:

By the method of inspection obtain an integral of the \((4 v^{2} + 2 sin v + 6 \sqrt{v})\)

Answer:

Integral of \((4 v^{2} + 2 sin v + 6 \sqrt{v})\) and v is the function of the integral \((4 v^{2} + 2 sin v + 6 \sqrt{v})\)

\(\int (4 v^{2} + 2 sin v + 6 \sqrt{v})\; dv \\ 4 \int v^{2} dv + 2 \int sin v dv + 6 \int v^{\frac{1}{2}} \\ \frac{4 v^{3}}{3} + 2 (- cos\; v) + 6 (\frac{v ^{\frac{3}{2}}}{\frac{3}{2}}) + C \\ \frac{4}{3} v^{3} – 2 cos\; v + 4 v ^{\frac{3}{2}} + C \\ Where\; C\; is\; the\; constant\)

Question 18:

By the method of inspection obtain an integral of the \(sec\; \Theta (tan\; \Theta + sec\; \Theta)\)

Answer:

Integral of \(sec\; \Theta (tan\; \Theta + sec\; \Theta)\) and \(\Theta\) is the function of the integral \(sec\; \Theta (tan\; \Theta + sec\; \Theta)\)

\(\int sec\; \Theta (tan\; \Theta + sec\; \Theta) d\Theta \\ \int (sec\; \Theta\; tan\; \Theta + sec ^{2}\; \Theta) d\Theta \\ sec\; \Theta\; + tan\; \Theta + K \\ Where\; K\; is\; the\; constant\)

Question 19:

By the method of inspection obtain an integral of the \(\frac{sec ^{2}\; \Theta}{cosec ^{2}\; \Theta}\)

Answer:

Integral of \(\frac{sec ^{2}\; \Theta}{cosec ^{2}\; \Theta}\) and \(\frac{3 – 2 sin\; \Theta}{cos ^{2}\; \Theta}\)c is the function of the integral \(\frac{sec ^{2}\; \Theta}{cosec ^{2}\; \Theta}\)

\(\int \frac{sec ^{2}\; \Theta}{cosec ^{2}\; \Theta} d\Theta \\ \int \frac{\frac{1}{cos ^{2}\; \Theta}}{\frac{1}{sin ^{2}\; \Theta}} d\Theta \\ \int \frac{sin ^{2}\; \Theta}{cos ^{2}\; \Theta} d\Theta \\ \int (tan ^{2}\; \Theta) d\Theta \\ \int (sec ^{2}\; \Theta – 1) d\Theta \\ \int sec ^{2}\; \Theta d\Theta – \int 1 d\Theta \\ tan \Theta – \Theta + K \\ Where\; K\; is\; the\; constant\)

Question 20:

By the method of inspection obtain an integral of the \(\frac{3 – 2 sin\; \Theta}{cos ^{2}\; \Theta}\)

Answer:

Integral of \(\frac{3 – 2 sin\; \Theta}{cos ^{2}\; \Theta}\) and \(\frac{3 – 2 sin\; \Theta}{cos ^{2}\; \Theta}\) is the function of the integral \(\frac{3 – 2 sin\; \Theta}{cos ^{2}\; \Theta}\)

\(\int \frac{3 – 2 sin\; \Theta}{cos ^{2}\; \Theta} d\Theta \\ \int (\frac{3}{cos ^{2}\; \Theta} – \frac{2 sin\; \Theta}{cos ^{2}\; \Theta}) d\Theta \\ 3 \int sec ^{2}\; \Theta\; d\Theta – 2 \int tan\; \Theta \; sec\; \Theta d\Theta \\ 3 tan\; \Theta – 2 sec\; \Theta + K \\ Where\; K\; is\; the\; constant\)

Question 21:

Which of the following below is an integral of \(\sqrt{u} + \frac{1}{\sqrt{u}}\):

\((a) \frac{1}{3} u^{\frac{1}{3}} + 2 u^{\frac{1}{2}} + C \\ (b) \frac{2}{3} u^{\frac{2}{3}} + \frac{1}{2} u^{2} + C \\ (c) \frac{2}{3} u^{\frac{3}{2}} + 2 u^{\frac{1}{2}} + C \\ (d) \frac{3}{2} u^{\frac{3}{2}} + \frac{1}{2} u^{\frac{1}{2}} + C\)

Answer:

Integral of \(\sqrt{u} + \frac{1}{\sqrt{u}}\) and u is the function of the integral \(\sqrt{u} + \frac{1}{\sqrt{u}}\)

\(\int \sqrt{u} + \frac{1}{\sqrt{u}} du \\ \int u ^{\frac{1}{2}} du + \int u^{- \frac{1}{2}} du \\ \frac{u ^{\frac{3}{2}}}{\frac{3}{2}} + \frac{u ^{\frac{1}{2}}}{\frac{1}{2}} + C \\ \frac{3}{2} u ^{\frac{3}{2}} + 2 u ^{\frac{1}{2}} \\ Option\; c\; is\; correct\)

Question 22:

\(Suppose\; \frac{d}{dr} f (r) = 4 r^{3} – \frac{3}{r^{4}},\; in\; such\; a\; way\; that\; f (2) = 0,\; then\; f (r)\; is\\ (a) r^{4} + \frac{1}{r ^{3}} – \frac{129}{8} (b) r^{3} + \frac{1}{r ^{4}} + \frac{129}{8} (c) r^{4} + \frac{1}{r ^{3}} + \frac{129}{8} (d) r^{3} + \frac{1}{r ^{4}} – \frac{129}{8}\)

Answer:

Given,

\(\frac{d}{dr} f (r) = 4 r^{3} – \frac{3}{r^{4}} \\ Integral\; of\; 4 r^{3} – \frac{3}{r^{4}} = f (r) \\ f (r) = \int 4 r^{3} – \frac{3}{r^{4}}\; dr \\ f (r) = 4 \int r^{3} dr – 3 \int (r ^{- 4}) dr \\ f (r) = 4 \frac{r ^{4}}{4} – 3 \frac{r ^{- 3}}{- 3} + K \\ f (r) = r ^{4} + \frac{1}{r ^{3}} + K \\ And, \\ f (2) = 0 \\ f (2) = 2 ^{4} + \frac{1}{2 ^{3}} + K = 0 \\\) \(16 + \frac{1}{8} + K = 0 \\ K = – \frac{129}{8} \\ f (r) = r ^{4} + \frac{1}{r ^{3}} – \frac{129}{8} \\ Option\; (a)\; is\; correct\)