#### Please solve RD Sharma class 12 chapter 9 Differentiability exercise Fill in the blanks question  5 maths textbook solution

Answer: $\pm \pi$

Hint: If $f(x)$ is differentiable at all $x\in R$, we must show$f'(x)$exists at all $x\in R$

Given: $f(x)=\cos ^{-1}(\cos x)$

Solution:

$f(x)=\cos ^{-1}(\cos x)=\left\{\begin{array}{l} 2 \pi+x, x \in(2 \pi,-\pi) \\ -x, x \in(-\pi, 0) \\ x, x \in(0, \pi) \\ 2 \pi-x, x \in(\pi, 2 \pi) \\ 2 \pi+x, x \in(2 \pi, 3 \pi) \end{array}\right\}$

\begin{aligned} &\text { At } x=\pi, \\ &\begin{array}{l} \lim _{x \rightarrow \pi^{-}} f(x)=\lim _{x \rightarrow \pi^{-}} x=\pi \\\\ \lim _{x \rightarrow \pi^{+}} f(x)=\lim _{x \rightarrow \pi^{+}}(2 \pi-x)=2 \pi-\pi=\pi \end{array} \end{aligned}

\begin{aligned} &\mathrm{LHS}=\mathrm{RHS} \Rightarrow f(x) \text { is continuous. } \\\\ &\text { At } x=\pi \\\\ &\qquad f^{\prime}(x)=\left\{\begin{array}{l} 1, x \in(0, \pi) \\\\ 0-1, x \in(\pi, 2 \pi) \end{array}\right\} \end{aligned}

\begin{aligned} &\lim _{x \rightarrow \pi^{-}} f^{\prime}(x)=1 \\\\ &\lim _{x \rightarrow \pi^{+}} f^{\prime}(x)=-1 \end{aligned}

$\mathrm{LHS} \neq \mathrm{RHS} \Rightarrow f(x)$  is not differentiable.

$\therefore \quad At , x=\pm \pi f(x)$ is continuous but not differentiable.

At $x=-\pi$

\begin{aligned} &\lim _{x \rightarrow \pi^{-}} f(x)=\lim _{x \rightarrow \pi-2 \pi+x}=\lim _{h \rightarrow 0} 2 \pi+(\pi-h) \\\\ &=\lim _{h \rightarrow 0} 2 \pi+\pi-h=3 \pi-0=3 \pi \end{aligned}

$\lim _{x \rightarrow \pi^{+}} f(x)=\lim _{x \rightarrow \pi^{+}}-x=\lim _{h \rightarrow 0}-(\pi+h)$

\begin{aligned} &=\lim _{h \rightarrow 0}-\pi-h \\\\ &=-\pi-0 \end{aligned}

\begin{aligned} &=-\pi \\ &\text { L.H.S } \neq \text { R. H.S } \end{aligned}

$f(x)$ is not continuous at $x=-\pi$

Since we know that not continuous  ⇒not differentiable at a point

So $f(x)$ is also not differentiable at $x=-\pi$ as

$f(x)$ is not continuous at $x=-\pi$