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If \mathrm{[x]} denotes the integral part of \mathrm{x} and \mathrm{f(x)=[n+p \sin x], 0<x<\pi, n \in \boldsymbol{I}} and \mathrm{p} is a prime number, then the number of points where \mathrm{f(x)}  is not differentiable is

Option: 1

\mathrm{p-1}


Option: 2

\mathrm{p}


Option: 3

\mathrm{2 p-1}


Option: 4

\mathrm{2 p+1}


Answers (1)

best_answer

\mathrm{[x]} is not differentiable at integral points.

Also \mathrm{[n+p \sin x]=n+[p \sin x]}
\mathrm{\therefore \quad[p \sin x]}  is not differentiable, where \mathrm{p \sin x}  is an integer.

But \mathrm{p } is prime and \mathrm{0<\sin x \leq 1 \quad[\mathrm{Q} 0<x<\pi] }
\mathrm{\therefore \quad p \sin x } is an integer only when \mathrm{\sin x=\frac{r}{p}}, where \mathrm{0< r\leq p} and \mathrm{r \in N}

For \mathrm{r=p, \sin x=1 \Rightarrow x=\frac{\pi}{2}\: in \: (0, \pi)}
For \mathrm{0<r<p, \sin x=\frac{r}{p}}

\mathrm{\therefore \quad x=\sin ^{-1} \frac{r}{p} \text { or } \pi-\sin ^{-1} \frac{r}{p}}
Number of such values of \mathrm{x=p-1+p-1=2 p-2}.

\mathrm{\therefore } Total number of points where \mathrm{f\left ( x \right ) } is not differentiable \mathrm{=1+2 p-2=2 p-1}.

Posted by

Riya

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