#### please solve rd sharma class 12 chapter 9 Differentiability exercise Multiple choice question, question 4 maths textbook solution

: (b)

HINTS: Understand the  definition of continuity and differentiability

SOLUTION:

$f(x)= \begin{cases}\frac{1 x+21}{\tan ^{-1}(x+2)} & x \neq-2 \\ 2 & x=-2\end{cases}$

Check the continuity at x=-2

\begin{aligned} \lim _{x \rightarrow-2} f(x) &=\lim _{h \rightarrow 0} f(-2-h) \\ \end{aligned}

\begin{aligned}&=\lim _{h \rightarrow 0} \frac{1}{\frac{\tan^{-1} h}{h}} \\ \end{aligned}

\begin{aligned}&=-1 \quad\left[\operatorname{as}\lim_{h \rightarrow 0} \frac{\tan ^{-1} h}{h}=1\right] \\ \end{aligned}

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

\begin{aligned} &=\lim _{h \rightarrow 0} \frac{-2+h+2}{\tan ^{-1}(-2+h+2)} \quad[|x+2|=x+2=x>2] \\ \end{aligned}

\begin{aligned} &=\lim _{h \rightarrow 0} \frac{h}{\tan ^{-1} h} \\ &=\lim _{h \rightarrow 0} \frac{1}{\frac{\tan ^{-1} h}{h}} \\ &=1 \end{aligned}

As $\lim _{x \rightarrow-2^{-}} f(x) \text { and } \lim _{x \rightarrow-2^{+}} f(x)$

f(x) is not continuous at x=-2