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

$\begin{array}{1 1}(a)\;\text{continuous at x=0}&(b)\;\text{discontinuous at x=0}\\(c)\;\text{differentiable at x=0}&(d)\;\text{None of these}\end{array}$

1 Answer

Here $f(0^-)=f(0^+)=f(0)=\large\frac{\pi}{2}$
$\therefore f$ is continuous at $x=0$
But $f'(x)=\large\frac{-\sin x}{\sqrt{1-\cos^2x}}$
$\qquad\qquad=\large\frac{-\sin x}{|\sin x|}$
$\qquad\qquad=\left\{\begin{array}{1 1}\large\frac{-\sin x}{-\sin x}=1&x<0\\\large\frac{-\sin x}{\sin x}=-1&x>0\end{array}\right.$
$f(x)$ is not differentiable at $x=0$
Hence (a) is the correct answer.
answered Dec 23, 2013 by sreemathi.v

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