Let `f(x)=min{1,cos x,1-sinx}, -pi le x le pi`, Then, f(x) is

Options

A. not continuous at `x=pi//2`
B. continuous but not differentiable at x=0
C. neither continuous nor differentiable at `x=pi//2`
D. none of these

Answer

B

Explanation

We have
`f(x)=min {1,cosx,1-sin x}`
`Rightarrow f(x)={{:(,cos x,-pi//2 le x le 0),(,1-sin x,0 lt xle pi//2),(,cos x, pi//2 lt x le pi):}`
We find that
`underset(x to 0^(-))lim f(x)=underset(x to 0^(-))lim cos x=1`
`underset(x to 0^(+))lim f(x)=underset(x to 0)lim 1-sinx=1`
`and f(0)=cos 0=1`
`therefore underset(x to 0^(-))lim f(x)=underset(x to 0^(+))lim f(x)=f(0)`
So, f(x) is continuous at x=0
Now,
`("LHD at x=0")=((d)/(dx)(cos x))_(x=0)=0`
`("RHD at x=0")=((d)/(dx)(1-sin x))_(x=0)=-1`
`therefore ("LHD at x=0") ne ("RHD at x=0")`
Hence, f(x) is not differentiable at x=0. However it is continuous there at.
At `x=pi//2` also, f(x) is continuous but not differentiable