NCERT Solutions For Class 12 Maths Chapter 5 Exercise 5.2


Chapter 5 Ex.5.2 Question 1

Differentiate the function with respect to x.

\(\sin \left( {{x^2} + 5} \right)\)

 

Solution

Video Solution

 

Let \(f\left( x \right) = \sin \left( {{x^2} + 5} \right),{\text{ }}u\left( x \right) = {x^2} + 5\) and \(v\left( t \right) = \sin t\)

Then, \(\left( {vou} \right)\left( x \right) = v\left( {u\left( x \right)} \right) = v\left( {{x^2} + 5} \right) = \tan \left( {{x^2} + 5} \right) = f\left( x \right)\)

Thus, \(f\) is a composite of two functions.

Put \(t = u\left( x \right) = {x^2} + 5\)

Then, we get

\[\begin{align}\frac{{dv}}{{dt}} &= \frac{d}{{dt}}\left( {\sin t} \right) = \cos t = \cos \left( {{x^2} + 5} \right)\\\frac{{dt}}{{dx}} &= \frac{d}{{dx}}\left( {{x^2} + 5} \right) = \frac{d}{{dx}}\left( {{x^2}} \right) + \frac{d}{{dx}}\left( 5 \right) = 2x + 0 = 2x\end{align}\]

By chain rule of derivative,

\(\frac{{df}}{{dx}} = \frac{{dv}}{{dt}}.\frac{{dt}}{{dx}} = \cos \left( {{x^2} + 5} \right) \times 2x = 2x\cos \left( {{x^2} + 5} \right)\)

Alternate method:

\[\begin{align}\frac{d}{{dx}}\left[ {\sin \left( {{x^2} + 5} \right)} \right]& = \cos \left( {{x^2} + 5} \right).\frac{d}{{dx}}\left( {{x^2} + 5} \right)\\ &= \cos \left( {{x^2} + 5} \right).\left[ {\frac{d}{{dx}}\left( {{x^2}} \right) + \frac{d}{{dx}}\left( 5 \right)} \right]\\& = \cos \left( {{x^2} + 5} \right).\left[ {2x + 0} \right]\\ &= 2x\cos \left( {{x^2} + 5} \right)\end{align}\]

Chapter 5 Ex.5.2 Question 2

Differentiate the function with respect to x

\(\cos \left( {\sin x} \right)\)

 

Solution

Video Solution

 

Let \(f\left( x \right) = \cos \left( {\sin x} \right),\;u\left( x \right) = \sin x\) and \(v\left( t \right) = \cos t\)

Then, \(\left( {vou} \right)\left( x \right) = v\left( {u\left( x \right)} \right) = v\left( {\sin x} \right) = \cos \left( {\sin x} \right) = f\left( x \right)\)

Here, \(f\) is a composite function of two functions.

Put \(t = u\left( x \right) = \sin x\)

\[\begin{align}\therefore \frac{{dv}}{{dt}}& = \frac{d}{{dt}}\left[ {\cos t} \right] = - \sin t = - \sin \left( {\sin x} \right)\\\frac{{dt}}{{dx}} &= \frac{d}{{dx}}\left( {\sin x} \right) = \cos x\end{align}\]

By chain rule,

\(\frac{{df}}{{dx}} = \frac{{dv}}{{dt}}.\frac{{dt}}{{dx}} = - \sin \left( {\sin x} \right).\cos x = - \cos x\sin \left( {\sin x} \right)\)

Alternate method:

\[\begin{align}\frac{d}{{dx}}\left[ {\cos \left( {\sin x} \right)} \right] &= - \sin \left( {\sin x} \right).\frac{d}{{dx}}\left( {\sin x} \right)\\ &= - \sin \left( {\sin x} \right) \times \cos x\\& = - \cos x\sin \left( {\sin x} \right)\end{align}\].

Chapter 5 Ex.5.2 Question 3

Differentiate the function with respect to x

\(\sin \left( {ax + b} \right)\)

 

Solution

Video Solution

 

Let \(f\left( x \right) = \sin \left( {ax + b} \right),\;u\left( x \right) = ax + b\) and \(v\left( t \right) = \sin t\)

Then, \(\left( {vou} \right)\left( x \right) = v\left( {u\left( x \right)} \right) = v\left( {ax + b} \right) = \sin \left( {ax + b} \right) = f\left( x \right)\)

Here, \(f\) is a composite function of two functions \(u\) and \(v\).

Put, \(t = u\left( x \right) = ax + b\)

Thus,

\[\begin{align}\frac{{dv}}{{dt}} &= \frac{d}{{dt}}\left( {\sin t} \right) = \cos t = \cos \left( {ax + b} \right)\\\frac{{dt}}{{dx}} &= \frac{d}{{dx}}\left( {ax + b} \right) = \frac{d}{{dx}}\left( {ax} \right) + \frac{d}{{dx}}\left( b \right) = a + 0 = a\end{align}\]

Hence, by chain rule, we get

\(\frac{{df}}{{dx}} = \frac{{dv}}{{dt}}.\frac{{dt}}{{dx}} = \cos \left( {ax + b} \right).a = a\cos \left( {ax + b} \right)\)

Alternate method:

\[\begin{align}\frac{d}{{dx}}\left[ {\sin \left( {ax + b} \right)} \right] &= \cos \left( {ax + b} \right).\frac{d}{{dx}}\left( {ax + b} \right)\\& = \cos \left( {ax + b} \right).\left[ {\frac{d}{{dx}}\left( {ax} \right) + \frac{d}{{dx}}\left( b \right)} \right]\\ &= \cos \left( {ax + b} \right).\left( {a + 0} \right)\\ &= a\cos \left( {ax + b} \right)\end{align}\]

Chapter 5 Ex.5.2 Question 4

Differentiate the function with respect to x

\(\sec \left( {\tan \left( {\sqrt x } \right)} \right)\)

 

Solution

Video Solution

 

Let \(f\left( x \right) = \sec \left( {\tan \left( {\sqrt x } \right)} \right),\;u\left( x \right) = \sqrt x ,v\left( t \right) = \tan t\) and \(w\left( s \right) = \sec s\)

Then, \(\left( {wovou} \right)\left( x \right) = w\left[ {v\left( {u\left( x \right)} \right)} \right] = w\left[ {v\left( {\sqrt x } \right)} \right] = w\left( {\tan \sqrt x } \right) = \sec \left( {\tan \sqrt x } \right) = f\left( x \right)\)

Here, \(f\) is a composite function of three functions u, v and w.

Put, \(s = v\left( t \right) = \tan t\) and \(t = u\left( x \right) = \sqrt x \)

Then,

\[\begin{align}\frac{{dw}}{{ds}}& = \frac{d}{{ds}}\left( {\sec s} \right)\\ &= \sec s\tan s\\ &= \sec \left( {\tan t} \right).\tan \left( {\tan t} \right) \quad \left[ {s = \tan t} \right]\\ &= \sec \left( {\tan \sqrt x } \right).\tan \left( {\tan \sqrt x } \right) \quad \left[ {t = \sqrt x } \right]\end{align}\]

Now,

\[\begin{align}\frac{{ds}}{{dt}} &= \frac{d}{{dt}}\left( {\tan t} \right) = {\sec ^2}t = {\sec ^2}\sqrt x \\\frac{{dt}}{{dx}} &= \frac{d}{{dx}}\left( {\sqrt x } \right) = \frac{d}{{dx}}\left( {{x^{\frac{1}{2}}}} \right) = \frac{1}{2}.{x^{\frac{1}{2} - 1}} = \frac{1}{{2\sqrt x }}\end{align}\]

Hence, by chain rule, we get

\[\begin{align}\frac{d}{{dx}}\left[ {\sec \left( {\tan \sqrt x } \right)} \right] &= \frac{{dw}}{{ds}}.\frac{{ds}}{{dt}}.\frac{{dt}}{{dx}}\\ &= \sec \left( {\tan \sqrt x } \right).\tan \left( {\tan \sqrt x } \right).{\sec ^2}\sqrt x .\frac{1}{{2\sqrt x }}\\ &= \frac{1}{{2\sqrt x }}{\sec ^2}\sqrt x \sec \left( {\tan \sqrt x } \right)\tan \left( {\tan \sqrt x } \right)\\ &= \frac{{{{\sec }^2}\sqrt x \sec \left( {\tan \sqrt x } \right)\tan \left( {\tan \sqrt x } \right)}}{{2\sqrt x }}\end{align}\]

Alternate method:

\[\begin{align}\frac{d}{{dx}}\left[ {\sec \left( {\tan \sqrt x } \right)} \right] &= \sec \left( {\tan \sqrt x } \right).\tan \left( {\tan \sqrt x } \right).\frac{d}{{dx}}\left( {\tan \sqrt x } \right)\\ &= \sec \left( {\tan \sqrt x } \right).\tan \left( {\tan \sqrt x } \right).{\sec ^2}\left( {\sqrt x } \right).\frac{d}{{dx}}\left( {\sqrt x } \right)\\& = \sec \left( {\tan \sqrt x } \right).\tan \left( {\tan \sqrt x } \right).{\sec ^2}\left( {\sqrt x } \right).\frac{1}{{2\sqrt x }}\\& = \frac{{\sec \left( {\tan \sqrt x } \right).\tan \left( {\tan \sqrt x } \right).{{\sec }^2}\left( {\sqrt x } \right)}}{{2\sqrt x }}\end{align}\]

Chapter 5 Ex.5.2 Question 5

Differentiate the function with respect to x

\(\frac{{\sin \left( {ax + b} \right)}}{{\cos \left( {cx + d} \right)}}\)

 

Solution

Video Solution

 

Given, \(f\left( x \right) = \frac{{\sin \left( {ax + b} \right)}}{{\cos \left( {cx + d} \right)}}\), where \(g\left( x \right) = \sin \left( {ax + b} \right)\) and \(h\left( x \right) = \cos \left( {cx + d} \right)\)

\(\therefore f = \frac{{g'h - gh'}}{{{h^2}}}\)

Consider \(g\left( x \right) = \sin \left( {ax + b} \right)\)

Let \(u\left( x \right) = ax + b,\;v\left( t \right) = \sin t\)

Then \(\left( {vou} \right)\left( x \right) = v\left( {u\left( x \right)} \right) = v\left( {ax + b} \right) = \sin \left( {ax + b} \right) = g\left( x \right)\)

\(\therefore g\) is a composite function of two functions, and v.

Put, \(t = u\left( x \right) = ax + b\)

\[\begin{align}\frac{{dv}}{{dt}} &= \frac{d}{{dt}}\left( {\sin t} \right) = \cos t = \cos \left( {ax + b} \right)\\\frac{{dt}}{{dx}} = \frac{d}{{dx}}\left( {ax + b} \right) &= \frac{d}{{dx}}\left( {ax} \right) + \frac{d}{{dx}}\left( b \right) = a + 0 = a\end{align}\]

Thus, by chain rule, we get

\(g' = \frac{{dg}}{{dx}} = \frac{{dv}}{{dt}}.\frac{{dt}}{{dx}} = \cos \left( {ax + b} \right).a = a\cos \left( {ax + b} \right)\)

Consider \(h\left( x \right) = \cos \left( {cx + d} \right)\)

Let \(p\left( x \right) = cx + d,\;q\left( y \right) = \cos y\)

Then, \(\left( {qop} \right)\left( x \right) = q\left( {p\left( x \right)} \right) = q\left( {cx + d} \right) = \cos \left( {cx + d} \right) = h\left( x \right)\)

\(\therefore h\) is a composite function of two functions, and q.

Put, \(y = p\left( x \right) = cx + d\)

\[\begin{align}\frac{{dq}}{{dy}} &= \frac{d}{{dy}}\left( {\cos y} \right) = - \sin y = - \sin \left( {cx + d} \right)\\\frac{{dy}}{{dx}} &= \frac{d}{{dx}}\left( {cx + d} \right) = \frac{d}{{dx}}\left( {cx} \right) + \frac{d}{{dx}}\left( d \right) = c\end{align}\]

Using chain rule, we get

\[\begin{align}h' = \frac{{dh}}{{dx}} = \frac{{dq}}{{dy}}.\frac{{dy}}{{dx}}\\ = - \sin \left( {cx + d} \right) \times c\\ = - c\sin \left( {cx + d} \right)\end{align}\]

Therefore,

\[\begin{align}f' &= \frac{{a\cos \left( {ax + b} \right).\cos \left( {cx + d} \right) - \sin \left( {ax + b} \right)\left\{ { - c\sin \left( {cx + d} \right)} \right\}}}{{{{\left[ {\cos \left( {cx + d} \right)} \right]}^2}}}\\ &= \frac{{a\cos \left( {ax + b} \right)}}{{\cos \left( {cx + d} \right)}} + c\sin \left( {ax + b} \right).\frac{{\sin \left( {cx + d} \right)}}{{\cos \left( {cx + d} \right)}} \times \frac{1}{{\cos \left( {cx + d} \right)}}\\ &= a\cos \left( {ax + b} \right)\sec \left( {cx + d} \right) + c\sin \left( {ax + b} \right)\tan \left( {cx + d} \right)\sec \left( {cx + d} \right)\end{align}\]

Chapter 5 Ex.5.2 Question 6

Differentiate the function with respect to x

\(\cos {x^3}.{\sin ^2}\left( {{x^5}} \right)\)

 

Solution

Video Solution

 

Given, \(\cos {x^3}.{\sin ^2}\left( {{x^5}} \right)\)

\[\begin{align}\frac{d}{{dx}}\left[ {\cos {x^3}.{{\sin }^2}\left( {{x^5}} \right)} \right] &= {\sin ^2}\left( {{x^5}} \right) \times \frac{d}{{dx}}\left( {\cos {x^3}} \right) + \cos {x^3} \times \frac{d}{{dx}}\left[ {{{\sin }^2}\left( {{x^5}} \right)} \right]\\ &= {\sin ^2}\left( {{x^5}} \right) \times \left( { - \sin {x^3}} \right) \times \frac{d}{{dx}}\left( {{x^3}} \right) + \cos {x^3} \times 2\sin \left( {{x^5}} \right).\frac{d}{{dx}}\left[ {\sin {x^5}} \right]\\ &= - \sin {x^3}{\sin ^2}\left( {{x^5}} \right) \times 3{x^2} + 2\sin {x^5}\cos {x^3}.\cos {x^5} \times \frac{d}{{dx}}\left( {{x^5}} \right)\\ &= - 3{x^2}\sin {x^3}.{\sin ^2}\left( {{x^5}} \right) + 2\sin {x^5}\cos {x^5}\cos {x^3} \times 5{x^4}\\ &= 10{x^4}\sin {x^5}\cos {x^5}\cos {x^3} - 3{x^2}\sin {x^3}{\sin ^2}\left( {{x^5}} \right)\end{align}\]

Chapter 5 Ex.5.2 Question 7

Differentiate the function with respect to \(x \)

\(2\sqrt {\cot \left( {{x^2}} \right)} \)

 

Solution

Video Solution

 

\[\begin{align}\frac{d}{{dx}}\left[ {2\sqrt {\cot \left( {{x^2}} \right)} } \right] &= 2.\frac{1}{{2\sqrt {\cot \left( {{x^2}} \right)} }} \times \frac{d}{{dx}}\left[ {\cot \left( {{x^2}} \right)} \right]\\ &= \sqrt {\frac{{\sin \left( {{x^2}} \right)}}{{\cos \left( {{x^2}} \right)}}} \times - \cos e{c^2}\left( {{x^2}} \right) \times \frac{d}{{dx}}\left( {{x^2}} \right)\\ &= \sqrt {\frac{{\sin \left( {{x^2}} \right)}}{{\cos \left( {{x^2}} \right)}}} \times \frac{{ - 1}}{{{{\sin }^2}\left( {{x^2}} \right)}} \times \left( {2x} \right)\\& = \frac{{ - 2x}}{{\sin {x^2}\sqrt {\cos {x^2}\sin {x^2}} }}\\ &= \frac{{ - 2\sqrt 2 x}}{{\sin {x^2}\sqrt {2\sin {x^2}\cos {x^2}} }}\\ &= \frac{{ - 2\sqrt 2 x}}{{\sin {x^2}\sqrt {\sin 2{x^2}} }}\end{align}\]

Chapter 5 Ex.5.2 Question 8

Differentiate the function with respect to x

\(\cos \left( {\sqrt x } \right)\)

 

Solution

Video Solution

 

Let \(f\left( x \right) = \cos \left( {\sqrt x } \right)\)

Also, let \(u\left( x \right) = \sqrt x \) and, \(v\left( t \right) = \cos t\)

Then,

\[\begin{align}\left( {vou} \right)\left( x \right) &= v\left( {u\left( x \right)} \right)\\ & = v\left( {\sqrt x } \right)\\ &= \cos \sqrt x \\ &= f\left( x \right)\end{align}\]

Since, \(f\) is a composite function of u and v.

\(t = u\left( x \right) = \sqrt x \)

Then,

\[\begin{align}\frac{{dt}}{{dx}}& = \frac{d}{{dx}}\left( {\sqrt x } \right) = \frac{d}{{dx}}\left( {{x^{\frac{1}{2}}}} \right) = \frac{1}{2}{x^{\frac{{ - 1}}{2}}}\\ &= \frac{1}{{2\sqrt x }}\end{align}\]

And,

\[\begin{align}\frac{{dv}}{{dt}}& = \frac{d}{{dt}}\left( {\cos t} \right) = - \sin t\\& = - \sin \left( {\sqrt x } \right)\end{align}\]

Using chain rule, we get

\[\begin{align}\frac{{dt}}{{dx}} &= \frac{{dv}}{{dt}}.\frac{{dt}}{{dx}}\\ &= - \sin \left( {\sqrt x } \right).\frac{1}{{2\sqrt x }}\\& = - \frac{1}{{2\sqrt x }}\sin \left( {\sqrt x } \right)\\& = - \frac{{\sin \left( {\sqrt x } \right)}}{{2\sqrt x }}\end{align}\]

Alternate method:

\[\begin{align}\frac{d}{{dx}}\left[ {\cos \left( {\sqrt x } \right)} \right] &= - \sin \left( {\sqrt x } \right).\frac{d}{{dx}}\left( {\sqrt x } \right)\\& = - \sin \left( {\sqrt x } \right) \times \frac{d}{{dx}}\left( {{x^{\frac{1}{2}}}} \right)\\ &= - \sin \sqrt x \times \frac{1}{2}{x^{\frac{{ - 1}}{2}}}\\ &= \frac{{ - \sin \sqrt x }}{{2\sqrt x }}\end{align}\]

Chapter 5 Ex.5.2 Question 9

Prove that the function \(f\) given by

\(f\left( x \right) = \left| {x - 1} \right|, x \in {\bf{R}}\) is not differentiable at \(x = 1\).

 

Solution

Video Solution

 

Given, \(f\left( x \right) = \left| {x - 1} \right|,\;x \in {\bf{R}}\)

It is known that a function \(f\) is differentiable at a point \(x = c\) in its domain if both

\(\mathop {\lim }\limits_{h \to {0^ - }} \frac{{f\left( c \right) - f\left( {c - h} \right)}}{h}\) and \(\mathop {\lim }\limits_{h \to {0^ + }} \frac{{f\left( {c + h} \right) - f\left( c \right)}}{h}\) are finite and equal.

To check the differentiability of the given function at \(x = 1\),

Consider LHD at \(x = 1\)

\[\begin{align}\mathop {\lim }\limits_{h \to {0^ - }} \frac{{f\left( 1 \right) - f\left( {1 - h} \right)}}{h} &= \mathop {\lim }\limits_{h \to {0^ - }} \frac{{f\left| {1 - 1} \right| - \left| {1 - h - 1} \right|}}{h}\\ &= \mathop {\lim }\limits_{h \to {0^ - }} \frac{{0 - \left| h \right|}}{h}\\ &= \mathop {\lim }\limits_{h \to {0^ - }} \frac{{ - h}}{h}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {h < 0 \Rightarrow \; \left| h \right| = - h} \right)\\ &= - 1\end{align}\]

Consider RHD at \(x = 1\)

\[\begin{align}\mathop {\lim }\limits_{h \to {0^ + }} \frac{{f\left( {1 + h} \right) - f\left( 1 \right)}}{h}& = \mathop {\lim }\limits_{h \to {0^ + }} \frac{{f\left| {1 + h - 1} \right| - \left| {1 - 1} \right|}}{h}\\& = \mathop {\lim }\limits_{h \to {0^ + }} \frac{{\left| h \right| - 0}}{h}\\& = \mathop {\lim }\limits_{h \to {0^ + }} \frac{h}{h}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {h > 0 \Rightarrow \; \left| h \right| = h} \right)\\ &= 1\end{align}\]

Since LHD and RHD at \(x = 1\) are not equal,

Therefore, \(f\) is not differentiable at \(x = 1\).

Chapter 5 Ex.5.2 Question 10

Prove that the greatest integer function defined by \(f\left( x \right) = \left[ x \right],0 < x < 3\) is not differentiable at \(x = 1\) and \(x = 2\).

 

Solution

Video Solution

 

Given, \(f\left( x \right) = \left[ x \right],0 < x < 3\)

It is known that a function \(f\) is differentiable at a point x = c in its domain if both

\(\mathop {\lim }\limits_{h \to {0^ - }} \frac{{f\left( c \right) - f\left( {c - h} \right)}}{h}\) and \(\mathop {\lim }\limits_{h \to {0^ + }} \frac{{f\left( {c + h} \right) - f\left( c \right)}}{h}\) are finite and equal.

At \(x = 1\),

Consider the LHD at \(x = 1\)

\[\begin{align}\mathop {\lim }\limits_{h \to {0^ - }} \frac{{f\left( 1 \right) - f\left( {1 - h} \right)}}{h}& = \mathop {\lim }\limits_{h \to {0^ - }} \frac{{\left[ 1 \right] - \left[ {1 - h} \right]}}{h}\\ &= \mathop {\lim }\limits_{h \to {0^ - }} \frac{{1 - 0}}{h}\\ &= \mathop {\lim }\limits_{h \to {0^ - }} \frac{1}{h}\\& = \infty \end{align}\]

Consider RHD at \(x = 1\)

\[\begin{align}\mathop {\lim }\limits_{h \to {0^ + }} \frac{{f\left( {1 + h} \right) - f\left( 1 \right)}}{h} &= \mathop {\lim }\limits_{h \to {0^ + }} \frac{{\left[ {1 + h} \right] - \left[ 1 \right]}}{h}\\ &= \mathop {\lim }\limits_{h \to {0^ + }} \frac{{1 - 1}}{h}\\& = \mathop {\lim }\limits_{h \to {0^ + }} 0\\ &= 0\end{align}\]

Since LHD and RHD at \(x = 1\) are not equal,

Hence, \(f\) is not differentiable at \(x = 1\).

To check the differentiability of the given function at \(x = 2\),

Consider LHD at \(x = 2\)

\[\begin{align}\mathop {\lim }\limits_{h \to {0^ - }} \frac{{f\left( 2 \right) - f\left( {2 - h} \right)}}{h}& = \mathop {\lim }\limits_{h \to {0^ - }} \frac{{\left[ 2 \right] - \left[ {2 - h} \right]}}{h}\\& = \mathop {\lim }\limits_{h \to {0^ - }} \frac{{2 - 1}}{h}\\ &= \mathop {\lim }\limits_{h \to {0^ - }} \frac{1}{h}\\& = \infty \end{align}\]

Now, consider RHD at \(x = 2\)

\[\begin{align}\mathop {\lim }\limits_{h \to {0^ + }} \frac{{f\left( {2 + h} \right) - f\left( 2 \right)}}{h}& = \mathop {\lim }\limits_{h \to {0^ + }} \frac{{\left[ {2 + h} \right] - \left[ 2 \right]}}{h}\\& = \mathop {\lim }\limits_{h \to {0^ + }} \frac{{2 - 2}}{h}\\ &= \mathop {\lim }\limits_{h \to {0^ + }} 0\\& = 0\end{align}\]

Since, LHD and RHD at \(x = 2\) are not equal.

Hence, \(f\) is not differentiable at \(x = 2\).

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