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The function  \mathrm{f(x)=\sin ^{-1}(\cos x)} is

Option: 1

discontinuous at x=0
 


Option: 2

continuous at x=0
 


Option: 3

differentiable at x=0
 


Option: 4

none of these


Answers (1)

best_answer

We have,

\mathrm{(\text { LHL at } x=0)= \lim _{x \rightarrow 0^{-}} f(x)=\lim _{h \rightarrow 0} f(0-h) \\ }

\mathrm{= \lim _{h \rightarrow 0} \sin ^{-1}(\cos (-h))=\lim _{h \rightarrow 0} \sin ^{-1}(\cos h) }
\mathrm{=\sin ^{-1} 1=\pi / 2 }

\mathrm{\text { RHL at } x=0)= \lim _{x \rightarrow 0^{+}} f(x)=\lim _{h \rightarrow 0} f(0+h) \\ }

\mathrm{ =\lim _{h \rightarrow 0} \sin ^{-1}(\cos h)=\sin ^{-1}(1)=\pi / 2 }
\mathrm{and, ~f(0)=\sin ^{-1}(\cos 0)=\sin ^{-1}(1)=\pi / 2. }
\mathrm{\therefore(~ LHL~ at ~x=0)=(~ RHL at ~\mathrm{x}=0)=f(0) }
\mathrm{So, ~f(x) ~is ~continuous~ at ~x=0 } 
\mathrm{Now, ~f^{\prime}(x)=\frac{-\sin x}{\sqrt{1-\cos ^2 x}}=\frac{\sin x}{|\sin x|}:=\left\{\begin{array}{l}\frac{-\sin x}{-\sin x}=1, x<0 \\ \frac{-\sin x}{\sin x}=-1, x>0\end{array}\right. }
\mathrm{\therefore \quad( LHD~ at~ x=0)=1~ and (RHD~ at~ x=0)=-1 }
\mathrm{Hence, ~f(x)~ is ~not~ differentiable~ at~ x=0 }

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manish painkra

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