Question

The number of points where the function $f\left(x\right)=\{\begin{array}{cc}1+\left[\cos \frac{\pi x}{2}\right],& 1<x\le 2\\ 1-\left\{x\right\},& 0\le x<1\\ |\sin \pi x|,& -1\le x<0\end{array}$ and $f\left(1\right)=0$ is continuous but non-differentiable is/are (where [.] and {.} represent greatest integer and fractional part functions, respectively)

• A. 0
• B. 1
• C. 2
• D. None of these

Answer 1

Answer: (B)

1

For 1<x<2 $⟹\frac{\pi }{2}$ < $\frac{\pi x}{2}$ < $\pi$

So, [ $cos\left(\frac{\pi x}{2}\right)$ ] = - 1

f(x) = 1 + $\left[\cos \right(\frac{\pi x}{2}\left)\right]=1-1=0$

For 0 < x < 1:

{x} =x

f(x)=1-{x} = 1-x

For -1 < x < 0:

f(x)=sin$|\pi x|=-sin\pi x$

Checking continuity and differentiability at x=1:

At $x=1^{-}$

f(x) =0, f'(x)= -1 and

At $x=1^{+}$

f(x)=0, f'(x)=0

So, f(x) is continuous at x= 1 but not differentiable.

When we proceed similarly for x=0, we will find that the function is not continuous at x=0.

Hence, answer is B