Exercise 5.2 (Revised) - Chapter 5 - Continuity & Differentiability - Ncert Solutions class 12 - Maths

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Chapter 5: Continuity & Differentiability NCERT Solutions Class 12 Maths

Ex 5.2 Question 1
Differentiate the functions with respect to $x \sin \left(x^2+5\right)$
$
y=\sin \left(x^2+5\right)
$

We need to find derivative of $y$, w.r.t.x
$
\begin{aligned}
\frac{d y}{d x} & =\frac{d\left(\sin \left(x^2+5\right)\right)}{d x} \\
& =\cos \left(x^2+5\right) \times \frac{d\left(x^2+5\right)}{d x} \\
& =\cos \left(x^2+5\right) \times\left(\frac{d\left(x^2\right)}{d x}+\frac{d(5)}{d x}\right) \\
& =\cos \left(x^2+5\right) \times\left(2 x^{2-1}+0\right) \\
& =\cos \left(x^2+5\right) \times 2 x \\
& =2 x \cos \left(x^2+5\right)
\end{aligned}
$

Ex 5.2 Question 2:

Differentiate the functions with respect to $x$.
$\cos (\sin x)$

Answer :
Let $f(x)=\cos (\sin x), u(x)=\sin x$, and $v(t)=\cos t$
Then. $($ vou $)(x)=v(u(x))=v(\sin x)=\cos (\sin x)=f(x)$
Thus, $f$ is a composite function of two functions.
Put $t=u(x)=\sin x$
$
\begin{aligned}
& \therefore \frac{d v}{d t}=\frac{d}{d t}[\cos t]=-\sin t=-\sin (\sin x) \\
& \frac{d t}{d x}=\frac{d}{d x}(\sin x)=\cos x
\end{aligned}
$

By chain rule, $\frac{d f}{d x}=\frac{d v}{d t} \cdot \frac{d t}{d x}=-\sin (\sin x) \cdot \cos x=-\cos x \sin (\sin x)$
Alternate method
$
\frac{d}{d x}[\cos (\sin x)]=-\sin (\sin x) \cdot \frac{d}{d x}(\sin x)=-\sin (\sin x) \cdot \cos x=-\cos x \sin (\sin x)
$

Ex 5.2 Question 3:

Differentiate the functions with respect to $x$.
$
\sin (a x+b)
$

Answer :
Let $f(x)=\sin (a x+b), u(x)=a x+b$, and $v(t)=\sin t$
Then, $($ vou $)(x)=v(u(x))=v(a x+b)=\sin (a x+b)=f(x)$
Thus, $f$ is a composite function of two functions, $u$ and $v$.
Put $t=u(x)=a x+b$
Therefore,
$
\begin{aligned}
& \frac{d v}{d t}=\frac{d}{d t}(\sin t)=\cos t=\cos (a x+b) \\
& \frac{d t}{d x}=\frac{d}{d x}(a x+b)=\frac{d}{d x}(a x)+\frac{d}{d x}(b)=a+0=a
\end{aligned}
$

Hence, by chain rule, we obtain
$
\frac{d f}{d x}=\frac{d v}{d t} \cdot \frac{d t}{d x}=\cos (a x+b) \cdot a=a \cos (a x+b)
$

Alternate method
$
\begin{aligned}
\frac{d}{d x}[\sin (a x+b)] & =\cos (a x+b) \cdot \frac{d}{d x}(a x+b) \\
& =\cos (a x+b) \cdot\left[\frac{d}{d x}(a x)+\frac{d}{d x}(b)\right] \\
& =\cos (a x+b) \cdot(a+0) \\
& =a \cos (a x+b)
\end{aligned}
$

Ex 5.2 Question 4:

Differentiate the functions with respect to $x$.
$
\sec (\tan (\sqrt{x}))
$

Answer :
Let $f(x)=\operatorname{scc}(\tan \sqrt{x}), u(x)=\sqrt{x}, v(t)=\tan t$, and $w(s)=\operatorname{scc} s$
Then, $($ wovou $)(x)=w[v(u(x))]=w[v(\sqrt{x})]=w(\tan \sqrt{x})=\sec (\tan \sqrt{x})=f(x)$
Thus, $f$ is a composite function of three functions, $u, v$, and $w$.
Put $s=v(t)=\tan t$ and $t=u(x)=\sqrt{x}$
$
\begin{aligned}
& \text { Then, } \frac{d v}{d s}=\frac{d}{d s}(\sec s)=\sec s \tan s=\sec (\tan t) \cdot \tan (\tan t) \quad[s=\tan t] \\
& =\sec (\tan \sqrt{x}) \cdot \tan (\tan \sqrt{x}) \quad[t=\sqrt{x}] \\
& \frac{d s}{d t}=\frac{d}{d t}(\tan t)=\sec ^2 t=\sec ^2 \sqrt{x} \\
& \frac{d t}{d x}=\frac{d}{d x}(\sqrt{x})=\frac{d}{d x}\left(x^{\frac{1}{2}}\right)=\frac{1}{2} \cdot x^{\frac{1}{2}-1}=\frac{1}{2 \sqrt{x}} \\
&
\end{aligned}
$

Hence, by chain rule, we obtain
$
\frac{d t}{d x}=\frac{d w}{d s} \cdot \frac{d s}{d t} \cdot \frac{d t}{d x}
$

$
\begin{aligned}
& =\sec (\tan \sqrt{x}) \cdot \tan (\tan \sqrt{x}) \times \sec ^2 \sqrt{x} \times \frac{1}{2 \sqrt{x}} \\
& =\frac{1}{2 \sqrt{x}} \sec ^2 \sqrt{x} \sec (\tan \sqrt{x}) \tan (\tan \sqrt{x}) \\
& =\frac{\sec ^2 \sqrt{x} \sec (\tan \sqrt{x}) \tan (\tan \sqrt{x})}{2 \sqrt{x}}
\end{aligned}
$

Alternate method
$
\begin{aligned}
\frac{d}{d x}[\sec (\tan \sqrt{x})] & =\sec (\tan \sqrt{x}) \cdot \tan (\tan \sqrt{x}) \cdot \frac{d}{d x}(\tan \sqrt{x}) \\
& -\sec (\tan \sqrt{x}) \cdot \tan (\tan \sqrt{x}) \cdot \sec ^2(\sqrt{x}) \cdot \frac{d}{d x}(\sqrt{x}) \\
& -\sec (\tan \sqrt{x}) \cdot \tan (\tan \sqrt{x}) \cdot \sec ^2(\sqrt{x}) \cdot \frac{1}{2 \sqrt{x}} \\
& =\frac{\sec (\tan \sqrt{x}) \cdot \tan (\tan \sqrt{x}) \sec ^2(\sqrt{x})}{2 \sqrt{x}}
\end{aligned}
$

Ex 5.2 Question 5:

Differentiate the functions with respect to $x$.
$
\frac{\sin (a x+b)}{\cos (c x+d)}
$

Answer :
The given function is $f(x)=\frac{\sin (a x+b)}{\cos (c x+d)}=\frac{g(x)}{h(x)}$, where $g(x)=\sin (a x+b)$ and
$
\begin{aligned}
& h(x)=\cos (c x+d) \\
& \therefore f^{\prime}=\frac{g^{\prime} h-g h^{\prime}}{h^2}
\end{aligned}
$

Consider $g(x)=\sin (a x+b)$
Let $u(x)=a x+b, v(t)=\sin t$
Then, $($ vou $)(x)=v(u(x))=v(a x+b)=\sin (a x+b)=g(x)$
$\therefore g$ is a composite function of two functions, $u$ and $v$.
Put $t=u(x)=a x+b$
$
\begin{aligned}
& \frac{d v}{d t}=\frac{d}{d t}(\sin t)=\cos t=\cos (a x+b) \\
& \frac{d t}{d x}=\frac{d}{d x}(a x+b)=\frac{d}{d x}(a x)+\frac{d}{d x}(b)=a+0=a
\end{aligned}
$

Therefore, by chain rule, we obtain
$
g^{\prime}=\frac{d g}{d x}=\frac{d v}{d t} \cdot \frac{d t}{d x}=\cos (a x+b) \cdot a=a \cos (a x+b)
$

Consider $h(x)=\cos (c x+d)$
Let $p(x)=c x+d, q(y)=\cos y$
Then, $(g \circ p)(x)=q(p(x))=q(c x+d)=\cos (c x+d)=h(x)$
$\therefore h$ is a composite function of two functions, $p$ and $q$.
$
\begin{aligned}
& \text { Put } y=p(x)=c x+d \\
& \frac{d q}{d y}=\frac{d}{d y}(\cos y)=-\sin y=-\sin (c x+d) \\
& \frac{d y}{d x}=\frac{d}{d x}(c x+d)=\frac{d}{d x}(c x)+\frac{d}{d x}(d)=c
\end{aligned}
$

Therefore, by chain rule, we obtain
$
\begin{aligned}
& h^{\prime}=\frac{d h}{d x}=\frac{d q}{d y} \cdot \frac{d y}{d x}=-\sin (c x+d) \times c=-c \sin (c x+d) \\
& \therefore f^{\prime}=\frac{a \cos (a x+b) \cdot \cos (c x+d)-\sin (a x+b)\{-c \sin (c x+d)\}}{[\cos (c x \mid d)]^2} \\
& =\frac{a \cos (a x+b)}{\cos (c x+d)}+c \sin (a x+b) \cdot \frac{\sin (c x+d)}{\cos (c x+d)} \times \frac{1}{\cos (c x+d)} \\
& =a \cos (a x+b) \sec (c x+d)+c \sin (a x+b) \tan (c x+d) \sec (c x+d) \\
&
\end{aligned}
$

Ex 5.2 Question 6:

Differentiate the functions with respect to $x$.
$
\cos x^3 \cdot \sin ^2\left(x^5\right)
$

Answer:
The given function is $\cos x^3 \cdot \sin ^2\left(x^5\right)$.
$
\begin{aligned}
& \frac{d}{d x}\left[\cos x^3 \cdot \sin ^2\left(x^5\right)\right]=\sin ^2\left(x^5\right) \times \frac{d}{d x}\left(\cos x^3\right)+\cos x^3 \times \frac{d}{d x}\left[\sin ^2\left(x^5\right)\right] \\
& =\sin ^2\left(x^5\right) \times\left(-\sin x^3\right) \times \frac{d}{d x}\left(x^3\right)+\cos x^3 \times 2 \sin \left(x^5\right) \cdot \frac{d}{d x}\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 \cdot \cos x^5 \times \frac{d}{d x}\left(x^5\right) \\
& =-3 x^2 \sin x^3 \cdot \sin ^2\left(x^5\right)+2 \sin x^5 \cos x^5 \cos x^3 \cdot \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{aligned}
$

Ex 5.2 Question 7:

Differentiate the functions with respect to $x$.
$
2 \sqrt{\cot \left(x^2\right)}
$

Answer:
$
\begin{aligned}
& \frac{d}{d x}\left[2 \sqrt{\cot \left(x^2\right)}\right] \\
& =2 \cdot \frac{1}{2 \sqrt{\cot \left(x^2\right)}} \times \frac{d}{d x}\left[\cot \left(x^2\right)\right] \\
& =\sqrt{\frac{\sin \left(x^2\right)}{\cos \left(x^2\right)}} \times-\operatorname{cosec}^2\left(x^2\right) \times \frac{d}{d x}\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(2 x) \\
& =\frac{-2 x}{\sqrt{\cos x^2} \sqrt{\sin x^2} \sin x^2} \\
& =\frac{-2 \sqrt{2} x}{\sqrt{2 \sin x^2 \cos x^2} \sin x^2} \\
& =\frac{-2 \sqrt{2} x}{\sin x^2 \sqrt{\sin 2 x^2}}
\end{aligned}
$

Ex 5.2 Question 8:

Differentiate the functions with respect to $x$.
$
\cos (\sqrt{x})
$

Answer:
Let $f(x)=\cos (\sqrt{x})$
Also, let $u(x)=\sqrt{x}$
And, $v(t)=\cos t$
Then, $($ vou $)(x)=v(u(x))$
$
\begin{aligned}
& =v(\sqrt{x}) \\
& =\cos \sqrt{x} \\
& =f(x)
\end{aligned}
$

Clearly, $f$ is a composite function of two functions, $u$ and $v$, such that $t=u(x)=\sqrt{x}$

Then, $\frac{d t}{d x}=\frac{d}{d x}(\sqrt{x})=\frac{d}{d x}\left(x^{\frac{1}{2}}\right)=\frac{1}{2} x^{-\frac{1}{2}}$
$
=\frac{1}{2 \sqrt{x}}
$

And, $\frac{d v}{d t}=\frac{d}{d t}(\cos t)=-\sin t$
$
=-\sin (\sqrt{x})
$

By using chain rule, we obtain
$
\begin{aligned}
& \frac{d t}{d x}=\frac{d v}{d t} \cdot \frac{d t}{d x} \\
& =-\sin (\sqrt{x}) \cdot \frac{1}{2 \sqrt{x}} \\
& =-\frac{1}{2 \sqrt{x}} \sin (\sqrt{x}) \\
& =-\frac{\sin (\sqrt{x})}{2 \sqrt{x}}
\end{aligned}
$

Alternate method
$
\begin{aligned}
\frac{d}{d x}[\cos (\sqrt{x})] & =-\sin (\sqrt{x}) \cdot \frac{d}{d x}(\sqrt{x}) \\
& =-\sin (\sqrt{x}) \times \frac{d}{d x}\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{aligned}
$

Ex 5.2 Question 9:

Prove that the function $f$ given by $f(x)=x-1 \mid, x \in \mathbf{R}$ is not differentiable at $x=1$.

Answer :
The given function is $f(x)=|x-1|, x \in \mathbf{R}$

$\text { It is known that a function } f \text { is differentiable at a point } x=c \text { in its domain if both }$

$\lim _{h \rightarrow 0^{-}} \frac{f(c+h)-f(c)}{h} \text { and } \lim _{h \rightarrow 0^{+}} \frac{f(c+h)-f(c)}{h} \text { are finite and equal. }$

To check the differentiability of the given function at $x=1$, consider the left hand limit of $f$ at $x=1$
$
\begin{aligned}
& \lim _{h \rightarrow 0^{-}} \frac{f(1+h)-f(1)}{h}=\lim _{h \rightarrow 0^{-}} \frac{|1+h-1|-|1-1|}{h} \\
& =\lim _{h \rightarrow 0^{-}} \frac{|h|-0}{h}=\lim _{h \rightarrow 0^{-}} \frac{-h}{h} \quad(h<0 \Rightarrow|h|=-h) \\
& =-1
\end{aligned}
$

Consider the right hand limit of $f$ at $x=1$
$
\begin{aligned}
& \lim _{h \rightarrow 0^{+}} \frac{f(1+h)-f(1)}{h}-\lim _{h \rightarrow 0^{+}} \frac{|1+h-1|-|1-1|}{h} \\
& =\lim _{h \rightarrow 0^{+}} \frac{|h|-0}{h}=\lim _{h \rightarrow 0^{+}} \frac{h}{h} \quad(h>0 \Rightarrow|k|=h) \\
& =1
\end{aligned}
$

Since the left and right hand limits of $f$ at $x=1$ are not equal, $f$ is not differertiable at $x=1$

Ex 5.2 Question 10:

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

Answer :
The given function $f$ is $f(x)=[x], 0<x<3$
It is known that a function $f$ is differentiable at a point $x=c$ in its domain if both $\lim _{h \rightarrow 0} \frac{f(c+h)-f(c)}{h}$ and $\lim _{h \rightarrow 0^{+}} \frac{f(c+h)-f(c)}{h}$ are finite and equal.

To check the differentiability of the given function at $x=1$, consider the left hand limit of $f$ at $x=1$
$
\begin{aligned}
& \lim _{h \rightarrow 0^{-}} \frac{f(1+h)-f(1)}{h}=\lim _{h \rightarrow 0^{-}} \frac{[1+h]-[1]}{h} \\
& =\lim _{h \rightarrow 0^{-}} \frac{0-1}{h}=\lim _{h \rightarrow 0^{-}} \frac{-1}{h}=\infty
\end{aligned}
$

Consider the right hand limit of $f$ at $x=1$
$
\begin{aligned}
& \lim _{h \rightarrow 0^{+}} \frac{f(1+h)-f(1)}{h}=\lim _{h \rightarrow 0^{+}} \frac{[1+h]-[1]}{h} \\
& =\lim _{h \rightarrow 0^{+}} \frac{1-1}{h}=\lim _{h \rightarrow 0^{+}} 0=0
\end{aligned}
$

Since the left and right hand limits of $f$ at $x=1$ are not equal, $f$ is not differentiable at $x=1$