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  • Explain solution rd sharma class 12 chapter 9 Differentiability exercise Multiple choice question, question 24

Explain solution rd sharma class 12 chapter 9 Differentiability exercise Multiple choice question, question 24

Answers (1)

(a)

HINTS: understand the definition of continuity and differentiability.

GIVEN:  f(x)=\left\{\begin{array}{cl} \frac{1-\cos x}{x \sin x} & x \neq 0 \\ \frac{1}{2} & x=0 \end{array}\right.

SOLUTION:

At x=0

\begin{aligned} \lim _{x \rightarrow 0^{-}} f(x) &=\lim _{h \rightarrow 0} f(0-h)=\lim _{h \rightarrow 0} \frac{1-\cos (-h)}{-n \sin (-h)} \\ \end{aligned}

\begin{aligned} &=\lim _{h \rightarrow 0} \frac{1-\cos (-h)}{-h \sin (-h)} \\ &=\lim _{h \rightarrow 0} \frac{1-\cos h}{h \sin h} \\ \end{aligned}

\begin{aligned} &=\lim _{h \rightarrow 0} \frac{\pm \sinh }{\sin h+h \cosh } \\ &=\lim _{h \rightarrow 0} \frac{\frac{1}{\sinh +h \cosh }}{\sinh } \\ &=\lim _{h \rightarrow 0} \frac{1}{1+h \operatorname{coth}}\\ &=1 \end{aligned}

RHL=

\begin{aligned} \lim _{x \rightarrow 0^{+}} f(x) &=\lim _{h \rightarrow 0} f(0+h)=\lim _{h \rightarrow 0} \frac{1-\cos -h}{-n \sin -h} \\ \end{aligned}

\begin{aligned} &=\lim _{h \rightarrow 0} \frac{0-\sin h}{\sinh +h \cosh } \\ &=\lim _{h \rightarrow 0} \frac{1}{\frac{\sinh }{\sin h}+\frac{\cosh }{\sinh }} \\ &=1\end{aligned}

As LHL=RHL

f(x) is not continuous, so f(x) is not differentiable

Now,

LHD=

\begin{aligned} & \lim _{x \rightarrow 0^{-}} \frac{f(x)-f(0)}{x-0}=\lim _{h \rightarrow 0} \frac{f(0-h)-f(0)}{-h} \\ =& \lim _{h \rightarrow 0} \frac{\frac{1-(\cos (-h)}{-h \sin (-h)}-\frac{1}{2}}{-h} \\ \end{aligned}

\begin{aligned} =& \lim _{h \rightarrow 0} \frac{\frac{1-\cos h}{h \sinh }-\frac{1}{2}}{-h} \\ =& \lim _{h \rightarrow 0} \frac{1-\cos h}{-\sinh }+\lim _{h \rightarrow 0} \frac{1}{2 h}=\infty \end{aligned}

RHD=

\begin{aligned} &\lim _{x \rightarrow 0^{-1}} \frac{f(x)-f(0)}{x-0}=\lim _{h \rightarrow 0} \frac{f(h)-f(0)}{h} \\ \end{aligned}

\begin{aligned} &=\lim _{h \rightarrow 0} \frac{\frac{1-\cos h}{h \sinh }-\frac{1}{2}}{-a} \\ &=\lim _{h \rightarrow 0} \frac{1-\cos h}{-\sinh }-\lim _{h \rightarrow 0} \frac{1}{2 h}=\infty\end{aligned}

F(x) is continuous and differential

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