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  • Let <img alt="f(x)=[n+p \sin x], x \in(0, \pi), n \in Z" src="https://entrancecorner.

Let f(x)=[n+p \sin x], x \in(0, \pi), n \in Z 

and P is a prime number, where[ ] denotes the greatest integer function. Then number of points wheref(x) is not differentiable is 

 

Option: 1

2p+1


Option: 2

2p-1


Option: 3

2p


Option: 4

2p-2


Answers (1)

best_answer

f(x)  is not differentiable at those points where \mathrm{n}+\mathrm{p} \sin \mathrm{x}  is integer. As p is a prime number.

n+p \sin x is an integer if \sin x=1,-1, r / p

x=\frac{\pi}{2},-\frac{\pi}{2}, \sin ^{-1} \frac{r}{p}, \pi-\sin ^{-1} \frac{r}{p}

Where 0 \leq r \leq p-1

But x \neq-\frac{\pi}{2}, 0 

 Function is not differentiable at


x=\frac{\pi}{2}, \sin ^{-1} \frac{r}{p}, \pi-\sin ^{-1} \frac{r}{p} \text { where } 0<r \leq p-1

So the number of points are =1+2(p-1)=2 p-1

 

Posted by

Devendra Khairwa

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