If -x- denotes the integral part of x and fx-n-p sin x- 0 - x - - n - I and p is a prime number- then the number of points where fx is not differentiable is

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If [x] deno...

Question

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

p - 1

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2 p - 1

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p

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2 p + 1

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Solution

The correct option is B $2 p - 1$
For X belongs to $(0, π) sin x$ lies in $(0, 1) [sin x]$ is discontinuous at $\frac{π}{2}$ sum of differentiable at it and now considering $[n + sin x]$ where $sin x$ is just shifted by $n$ units along $y - a x i s$ but still the values lie in $(n, n + 1)$ and here too the curve will be discontinuous/not differentiable at 1 point at $\frac{π}{2}$
now taking $p$ ( a prime no ) into consideration
for $p = 1,$ we have $[sin x]$ with 1 discontinuous/not differentiable points.
for $p = 2,$ we have $[2 sin x]$ with 3 discontinuous/not differentiable points.
for $p = 3,$ we have $[3 sin x]$ with 5 discontinuous/not differentiable points.
for $p = p,$ it follow to $(2 p - 1)$ discontinuous/not differentiable points.
Hence, the answer is $2 p - 1.$

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If [x] denotes the integral part of x and f(x)=[n+psinx],0<x<π,n∈I and p is a prime number, then the number of points where f(x) is not differentiable is

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