A) continuous at x = 0
B) discontinuous at x = 0
C) differentiable at x = 0
D) None of the above
Correct Answer: A
Solution :
Since, \[f(x)={{\sin }^{-1}}(\cos x)\] \[\therefore \]\[\underset{x\to 0}{\mathop{\lim }}\,{{\sin }^{-1}}(\cos x)={{\sin }^{-1}}(\cos 0)\] \[={{\sin }^{-1}}(1)=\frac{\pi }{2}\]and\[f(0)=\frac{\pi }{2}\] \[\therefore \] Function is continuous at x = 0. But \[f'(x)=\frac{-\sin x}{\sqrt{1-{{\cos }^{2}}x}}=-\frac{\sin x}{|\sin x|}\] \[=\left\{ \begin{matrix} \frac{-\sin x}{-\sin x}=1, & x0 \\ \end{matrix} \right.\] \[\therefore \]f(x) is not differentiable at x = 0.
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