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  • Let <img alt="\mathrm{f(x)=[n+p \sin x], x \in(0, \pi), n \in Z, p}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7Bf%28x%29%3D%5Bn&plus;p%20%5Csin%20x%5D%2C%20x%20%5Cin%

Let \mathrm{f(x)=[n+p \sin x], x \in(0, \pi), n \in Z, p}  is a prime number and \mathrm{[x]}= the greatest integer less than or equal to \mathrm{x}. The number of points at which \mathrm{f(x)}  is not differentiable is

Option: 1

\mathrm{p}


Option: 2

\mathrm{p-1}


Option: 3

\mathrm{2p+1}


Option: 4

\mathrm{2p-1}


Answers (1)

best_answer

\mathrm{[n+p \sin x]} is not differentiable at those points where \mathrm{ n+p \sin x} is an integer. As \mathrm{p}  is a prime number, \mathrm{n+p \sin x}  is an integer if \mathrm{\sin x=1,-1, \frac{r}{p}}, that is, \mathrm{x=\frac{\pi}{2},-\frac{\pi}{2}, \sin ^{-1} \frac{r}{p}, \pi-\sin ^{-1} \frac{r}{p}}, where \mathrm{0 \leq r \leq p-1}. But \mathrm{x \neq-\frac{\pi}{2}, 0}.

\therefore the function is not differentiable at \mathrm{x=\frac{\pi}{2}, \sin ^{-1} \frac{r}{p}, \pi-\sin ^{-1} \frac{r}{p}}, where \mathrm{0<r \leq p-1}. So, the required number of points =\mathrm{1+2(p-1)=2 p-1}.

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manish

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