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

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Answer: \pm \pi

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

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


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

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