#### Need solution for rd sharma maths class 12 chapter 9 Differentiability exercise Multiple choice question, question 18

(b)

HINTS: composition of function is continuous of function

GIVEN: $f(x)=1+|\cos x|$

SOLUTION:

Let $g(x)=1+|x|, h(x)=\cos x$ as $1+|x|$ and cos x are continuous function

Therefore, composition  of function ,that is  is also continuous.

$f(x)=g o h(x)=g(h(x))$

Now,

cos x is  differentiable

$g(x)=1+|x|= \begin{cases}1+x & x>1 \\ 1-x & x<1\end{cases}$

At, x=1

LHD=  $\lim _{x \rightarrow 1} \frac{g(x)-g(1)}{x-1}$

\begin{aligned} &=\lim _{h \rightarrow 0} \frac{g(1-h)-g(1)}{-h} \\ &=\lim _{h \rightarrow 0} \frac{1-(1-h)-2}{-h} \\ &=\infty \end{aligned}

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

\begin{aligned} &=\lim _{h \rightarrow 0} \frac{1-(1+h)-2}{h} \\ &=\lim _{h \rightarrow 0} \frac{h}{h} \\ &=1\end{aligned}

As LHD $\neq$ RHD

g(x) is not differentiable at x=1

therefore f(x) is also with differentiation at x=1

at x=1

\begin{aligned} &1+\cos x=1 \\ &\cos x=0 \\ &x=n \pi \end{aligned}

Hence f(x) is continuous everywhere.