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If \mathrm{[x]} denotes the integral part of \mathrm{x} and in (0, \pi) ,we define
\mathrm{f(x)=\left\{\begin{array}{cr} {\left[\frac{2\left(\sin x-\sin ^n x\right)+\left|\sin x-\sin ^n x\right|}{2\left(\sin x-\sin ^n x\right)-\left|\sin x-\sin ^n x\right|}\right],} & x \neq \frac{\pi}{2} \\ 3, & x=\frac{\pi}{2} \end{array}\right.}
Then for \mathrm{n>1},

Option: 1

\mathrm{f(x)} is continuous but not differentiable at \mathrm{x=\frac{\pi}{2}}


Option: 2

both continuous and differentiable at \mathrm{x=\frac{\pi}{2}}


Option: 3

 neither continuous nor differentiable at \mathrm{x=\frac{\pi}{2}}


Option: 4

\mathrm{\lim _{x \rightarrow \frac{\pi}{2}} f(x) \, \text{exist but}\quad \lim _{x \rightarrow \frac{\pi}{2}} f(x) \neq f\left(\frac{\pi}{2}\right)}


Answers (1)

best_answer

For \mathrm{0<x<\frac{\pi}{2}\: or \: \frac{\pi}{2}<x<\pi, 0<\sin x<1}

\mathrm{\therefore \quad \text { for } n>1, \sin x>\sin ^n x}

\mathrm{\therefore \quad f(x)=\left[\frac{3\left(\sin x-\sin ^n x\right)}{\sin x-\sin ^n x}\right]=3, x \neq \frac{\pi}{2} \text { and } f(x)=3, \quad x=\frac{\pi}{2}}

Thus in \mathrm{(0, \pi), f(x)=3}.

Hence \mathrm{f(x)} is continuous and differentiable at \mathrm{x=\frac{\pi}{2}}.

Posted by

vinayak

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