NCERT Solutions For Class 12 Maths Chapter 7 Exercise 7.1

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Chapter 7 Ex.7.1 Question 1

Find an anti-derivative (or integral) of the following functions by the method of inspection, \(\sin 2x\).

Solution

\[\begin{align}&\Rightarrow\; \frac{d}{{dx}}\left( {\cos 2x} \right) = - 2\sin 2x\\&\Rightarrow\; \sin 2x = - \frac{1}{2}\frac{d}{{dx}}\left( {\cos 2x} \right)\\&\Rightarrow\; \sin 2x = \frac{d}{{dx}}\left( { - \frac{1}{2}\cos 2x} \right)\end{align}\]

Thus, the anti-derivative of \(\rm{sin}\) \(2x \)is \( - \frac{1}{2}\cos 2x\)

Chapter 7 Ex.7.1 Question 2

Find an anti-derivative (or integral) of the following functions by the method of inspection, \(\cos 3x\).

Solution

\[\begin{align}& \Rightarrow\; \frac{d}{{dx}}\left( {\sin 3x} \right) = 3\cos 3x\\& \Rightarrow\; \cos 3x = \frac{1}{3}\frac{d}{{dx}}\left( {\sin 3x} \right)\\ &\Rightarrow\; \cos 3x = \frac{d}{{dx}}\left( {\frac{1}{3}\sin 3x} \right)\end{align}\]

Thus, the anti-derivative of \(\cos 3x\) is \(\frac{1}{3}\sin 3x\)

Chapter 7 Ex.7.1 Question 3

Find an anti-derivative (or integral) of the following functions by the method of inspection, \({e^{2x}}\).

Solution

\[\begin{align} &\Rightarrow\; \frac{d}{{dx}}\left( {{e^{2x}}} \right) = 2{e^{2x}}\\& \Rightarrow\; {e^{2x}} = \frac{1}{2}\frac{d}{{dx}}\left( {{e^{2x}}} \right)\\& \Rightarrow\; {e^{2x}} = \frac{d}{{dx}}\left( {\frac{1}{2}{e^{2x}}} \right)\end{align}\]

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

Chapter 7 Ex.7.1 Question 4

Find an anti-derivative (or integral) of the following functions by the method of inspection, \({\left( {ax + b} \right)^2}\).

Solution

\[\begin{align}&\frac{d}{{dx}}{\left( {ax + b} \right)^3} = 3a{\left( {ax + b} \right)^2}\\ &\Rightarrow\; {\left( {ax + b} \right)^2} = \frac{1}{{3a}}\frac{d}{{dx}}{\left( {ax + b} \right)^3}\\ &\Rightarrow\; {\left( {ax + b} \right)^2} = \frac{d}{{dx}}\left( {\frac{1}{{3a}}{{\left( {ax + b} \right)}^3}} \right)\end{align}\]

Thus, the anti-derivative \({\left( {ax + b} \right)^2}\)of is \(\frac{1}{{3a}}{\left( {ax + b} \right)^3}\)

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Chapter 7 Ex.7.1 Question 5

Find an anti-derivative (or integral) of the following functions by the method of inspection, \(\sin 2x - 4{e^{3x}}\)

Solution

\(\frac{d}{{dx}}\left( { - \frac{1}{2}\cos 2x - \frac{4}{3}{e^{3x}}} \right) = \sin 2x - 4{e^{3x}}\)

Thus, the anti-derivative of \(\sin 2x - 4{e^{3x}}\) is \(\left( { - \frac{1}{2}\cos 2x - \frac{4}{3}{e^{3x}}} \right)\)

Chapter 7 Ex.7.1 Question 6

Find:

\(\int {\left( {4{e^{3x}} + 1} \right)dx} \)

Solution

\[\begin{align}\int {\left( {4{e^{3x}} + 1} \right)dx}& = 4\int {{e^{3x}}dx + \int {1\;dx} } \\& = 4\left( {\frac{{{e^{3x}}}}{3}} \right) + x + C\\ &= \frac{4}{3}{e^{3x}} + x + C\end{align}\]

Chapter 7 Ex.7.1 Question 7

Find:

\(\int {{x^2}\left( {1 - \frac{1}{{{x^2}}}} \right)dx} \)

Solution

\[\begin{align}\int {{x^2}\left( {1 - \frac{1}{{{x^2}}}} \right)dx}& = \int {\left( {{x^2} - 1} \right)} dx\\ &= \int {{x^2}dx - \int {1\;dx} } \\ &= \frac{{{x^3}}}{3} - x + C\end{align}\]

Chapter 7 Ex.7.1 Question 8

Find:

\(\int {\left( {a{x^2} + bx + c} \right)dx} \)

Solution

\[\begin{align}\int {\left( {a{x^2} + bx + c} \right)dx} &= a\int {{x^2}dx + b\int {x\;dx + c\int {1dx} } } \\ &= a\left( {\frac{{{x^3}}}{3}} \right) + b\left( {\frac{{{x^2}}}{2}} \right) + cx + C\\& = \frac{{a{x^3}}}{3} + \frac{{b{x^2}}}{2} + cx + C\end{align}\]

Chapter 7 Ex.7.1 Question 9

Find:

\(\int {\left( {2{x^2} + {e^x}} \right)} \;dx\)

Solution

\[\begin{align}\int {\left( {2{x^2} + {e^x}} \right)} \;dx& = 2\int {{x^2}dx + \int {{e^x}dx} } \\ &= 2\left( {{{\frac{x}{3}}^3}} \right) + {e^x} + C\\& = \frac{2}{3}{x^3} + {e^x} + C\end{align}\]

Chapter 7 Ex.7.1 Question 10

Find:

\(\int {{{\left( {\sqrt x - \frac{1}{{\sqrt x }}} \right)}^2}dx} \)

Solution

\[\begin{align}\int {{{\left( {\sqrt x - \frac{1}{{\sqrt x }}} \right)}^2}dx}& = \int {\left( {x + \frac{1}{x} - 2} \right)} \\& = \int {x\;dx + \int {\frac{1}{x}dx - 2\int {1dx} } } \\ &= \frac{{{x^2}}}{2} + \log \left| x \right| - 2x + C\end{align}\]

Chapter 7 Ex.7.1 Question 11

Find:

\(\int {\frac{{{x^3} + 5{x^2} - 4}}{{{x^2}}}} dx\)

Solution

\[\begin{align}\int {\frac{{{x^3} + 5{x^2} - 4}}{{{x^2}}}} dx &= \int {\left( {x + 5 - 4{x^{ - 2}}} \right)} dx\\ &= \int {x\;dx + 5\int {1\;dx - 4\int {{x^{ - 2}}dx} } } \\ &= \frac{{{x^2}}}{2} + 5x - 4\left( {\frac{{{x^{ - 1}}}}{{ - 1}}} \right) + C\\ &= \frac{{{x^2}}}{2} + 5x + \frac{4}{x} + C\end{align}\]

Chapter 7 Ex.7.1 Question 12

Find:

\(\begin{align} \int {\frac{{{x^3} + 3x + 4}}{{\sqrt x }}} \end{align} dx\)

Solution

\[\begin{align}\int {\frac{{{x^3} + 3x + 4}}{{\sqrt x }}} dx &= \int {\left( {{x^{\frac{5}{2}}} + 3{x^{\frac{1}{2}}} + 4{x^{ - \frac{1}{2}}}} \right)} dx\\& = \frac{{{x^{\frac{7}{2}}}}}{{\frac{7}{2}}} + \frac{{3\left( {{x^{\frac{3}{2}}}} \right)}}{{\frac{3}{2}}} + \frac{{4\left( {{x^{\frac{1}{2}}}} \right)}}{{\frac{1}{2}}} + C\\& = \frac{2}{7}{x^{\frac{7}{2}}} + 2{x^{\frac{3}{2}}} + 8{x^{\frac{1}{2}}} + C\\& = \frac{2}{7}{x^{\frac{7}{2}}} + 2{x^{\frac{3}{2}}} + 8\sqrt x + C\end{align}\]

Chapter 7 Ex.7.1 Question 13

Find:

\(\begin{align} \int {\frac{{{x^3} - {x^2} + x - 1}}{{x - 1}}} dx \end{align}\)

Solution

\[\begin{align}\int {\frac{{{x^3} - {x^2} + x - 1}}{{x - 1}}} dx &= \int {\left[ {\frac{{\left( {{x^2} + 1} \right)\left( {x - 1} \right)}}{{x - 1}}} \right]} dx\\ &= \int {\left( {{x^2} + 1} \right)dx} \\& = \int {{x^2}dx + \int {1\;dx} } \\& = \frac{{{x^3}}}{3} + x + C\end{align}\]

Chapter 7 Ex.7.1 Question 14

Find:

\(\int {\left( {1 - x} \right)\sqrt x\;dx} \)

Solution

\[\begin{align}\int {\left( {1 - x} \right)\sqrt x\;dx} &= \int {\left( {\sqrt x - {x^{\frac{3}{2}}}} \right)} dx\\ &= \int {{x^{\frac{1}{2}}}dx - } \int {{x^{\frac{3}{2}}}dx} \\& = \frac{{{x^{\frac{3}{2}}}}}{{\frac{3}{2}}} - \frac{{{x^{\frac{5}{2}}}}}{{\frac{5}{2}}} + C\\ &= \frac{2}{3}{x^{\frac{3}{2}}} - \frac{2}{5}{x^{\frac{5}{2}}} + C\end{align}\]

Chapter 7 Ex.7.1 Question 15

Find:

\(\int {\sqrt x \left( {3{x^2} + 2x + 3} \right)} dx\)

Solution

\[\begin{align}\int {\sqrt x \left( {3{x^2} + 2x + 3} \right)} dx &= \int {\left( {3{x^{\frac{5}{2}}} + 2{x^{\frac{3}{2}}} + 3{x^{\frac{1}{2}}}} \right)dx} \\& = 3\int {{x^{\frac{5}{2}}}} dx + 2\int {{x^{\frac{3}{2}}}} dx + 3\int {{x^{\frac{1}{2}}}} dx\\& = 3\left( {\frac{{{x^{\frac{7}{2}}}}}{{\frac{7}{2}}}} \right) + 2\left( {\frac{{{x^{\frac{5}{2}}}}}{{\frac{5}{2}}}} \right) + 3\left( {\frac{{{x^{\frac{3}{2}}}}}{{\frac{3}{2}}}} \right) + C\\& = \frac{6}{7}{x^{\frac{7}{2}}} + \frac{4}{5}{x^{\frac{5}{2}}} + 2{x^{\frac{3}{2}}} + C\end{align}\]

Chapter 7 Ex.7.1 Question 16

Find:

\(\int {\left( {2x - 3\cos x + {e^x}} \right)}\; dx\)

Solution

\[\begin{align}\int {\left( {2x - 3\cos x + {e^x}} \right)}\;dx &= 2\int {x\;dx - 3\int {\cos x\;dx + \int {{e^x}\;dx} } } \\ &= \frac{{2{x^2}}}{2} - 3\left( {\sin x} \right) + {e^x} + C\\ &= {x^2} - 3\sin x + {e^x} + C\end{align}\]

Chapter 7 Ex.7.1 Question 17

Find:

\(\int {\left( {2{x^2} - 3\sin x + 5\sqrt x } \right)} dx\)

Solution

\[\begin{align}\int {\left( {2{x^2} - 3\sin x + 5\sqrt x } \right)} dx &= 2\int {{x^2}dx - 3\int {\sin x\;dx} + 5\int {{x^{\frac{1}{2}}}dx} } \\ &= \frac{{2{x^3}}}{3} - 3\left( { - \cos x} \right) + 5\left( {\frac{{{x^{\frac{3}{2}}}}}{{\frac{3}{2}}}} \right) + C\\ &= \frac{2}{3}{x^3} + 3\cos x + \frac{{10}}{3}{x^{\frac{3}{2}}} + C\end{align}\]

Chapter 7 Ex.7.1 Question 18

Find:

\(\int {\sec x\left( {\sec x + \tan x} \right)} dx\)

Solution

\[\begin{align}\int {\sec x\left( {\sec x + \tan x} \right)} dx &= \int {\left( {{{\sec }^2}x + \sec x\tan x} \right)} dx\\ &= \int {{{\sec }^2}xdx} + \int {\sec x\tan xdx} \\ &= \tan x + \sec x + C\end{align}\]

Chapter 7 Ex.7.1 Question 19

Find:

\(\begin{align} \int {\frac{{{{\sec }^2}x}}{{\cos e{c^2}x}}} dx \end{align}\)

Solution

\[\begin{align}\int {\frac{{{{\sec }^2}x}}{{\cos e{c^2}x}}} dx &= \int {\frac{{\frac{1}{{{{\cos }^2}x}}}}{{\frac{1}{{{{\sin }^2}x}}}}} dx\\&= \int {\frac{{{{\sin }^2}x}}{{{{\cos }^2}x}}dx} \\&= \int {{{\tan }^2}x\;dx} \\&= \int {\left( {{{\sec }^2}x - 1} \right)dx} \\&= \int {{{\sec }^2}x\;dx - \int {1\;dx} } \\&= \tan x - x + C\end{align}\]

Chapter 7 Ex.7.1 Question 20

Find:

\(\begin{align} \int {\frac{{2 - 3\sin x}}{{{{\cos }^2}x}}} dx \end{align}\)

Solution

\[\begin{align}\int {\frac{{2 - 3\sin x}}{{{{\cos }^2}x}}} dx &= \int {\left( {\frac{2}{{{{\cos }^2}x}} - \frac{{3\sin x}}{{{{\cos }^2}x}}} \right)} dx\\ &= \int {2{{\;\sec }^2}x\;dx - 3\int {\tan x\sec x\;dx} } \\ &= 2\tan x - 3\sec x + C\end{align}\]

Chapter 7 Ex.7.1 Question 21

The anti-derivative of \(\left( {\sqrt x + \frac{1}{{\sqrt x }}} \right)\) equals

\(\begin{align}&\left( A \right)\,\,\,\,\frac{1}{3}{x^{\frac{1}{3}}} + 2{x^{\frac{1}{2}}} + C\\&\left( B \right)\,\,\,\,\frac{2}{3}{x^{\frac{2}{3}}} + \frac{1}{2}{x^2} + C\\&\left( C \right)\,\,\,\,\frac{2}{3}{x^{\frac{3}{2}}} + 2{x^{\frac{1}{2}}} + C\\&\left( D \right)\,\,\,\,\frac{3}{3}{x^{\frac{3}{2}}} + \frac{1}{2}{x^{\frac{1}{2}}} + C\end{align}\)

Solution

\[\begin{align}\left( {\sqrt x + \frac{1}{{\sqrt x }}} \right)& = \int {{x^{\frac{1}{2}}}} dx + \int {{x^{ - \frac{1}{2}}}dx} \\ &= \frac{{{x^{\frac{3}{2}}}}}{{\frac{3}{2}}} + \frac{{{x^{\frac{1}{2}}}}}{{\frac{1}{2}}} + C\\& = \frac{2}{3}{x^{\frac{3}{2}}} + 2{x^{\frac{1}{2}}} + C\end{align}\]

Thus, the correct option is C.

Chapter 7 Ex.7.1 Question 22

If \(\frac{d}{{dx}}f\left( x \right) = 4{x^3} - \frac{3}{{{x^4}}}\) such that\(f\left( 2 \right) = 0\), then\(f\left( x \right)\) is

\(\begin{align}&\left( A \right)\,\,\,\,{x^4} + \frac{1}{{{x^3}}} - \frac{{129}}{8}\\&\left( B \right)\,\,\,\,{x^3} + \frac{1}{{{x^4}}} + \frac{{129}}{8}\\&\left( C \right)\,\,\,\,{x^4} + \frac{1}{{{x^3}}} + \frac{{129}}{8}\\&\left( D \right)\,\,\,\,{x^3} + \frac{1}{{{x^4}}} - \frac{{129}}{8}\end{align}\)

Solution

Given, \(\frac{d}{{dx}}f\left( x \right) = 4{x^3} - \frac{3}{{{x^4}}}\)

Anti-derivative of \(4{x^3} - \frac{3}{{{x^4}}} = f\left( x \right)\)

Therefore,

\[\begin{align}f\left( x \right) &= \int {4{x^3} - \frac{3}{{{x^4}}}} dx\\f\left( x \right)& = 4\int {{x^3}dx - 3\int {\left( {{x^{ - 4}}} \right)dx} } \\f\left( x \right)& = 4\left( {\frac{{{x^4}}}{4}} \right) - 3\left( {\frac{{{x^{ - 3}}}}{{ - 3}}} \right) + C\\f\left( x \right)& = {x^4} + \frac{1}{{{x^3}}} + C\end{align}\]

Also,

\[\begin{align}& \Rightarrow\; f\left( 2 \right) = 0\\& \Rightarrow\; f\left( 2 \right) = {\left( 2 \right)^4} + \frac{1}{{{{\left( 2 \right)}^3}}} + C = 0\\& \Rightarrow\; 16 + \frac{1}{8} + C = 0\\& \Rightarrow\; C = - \left( {16 + \frac{1}{8}} \right)\\& \Rightarrow\; C = - \frac{{129}}{8}\\& \Rightarrow\; f\left( x \right) = {x^4} + \frac{1}{{{x^3}}} - \frac{{129}}{8}\end{align}\]

Thus, the correct option is A.

  
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