`if f(x) ={{:(-x=(pi)/(2),xle -(pi)/(2)),(- cos x, -(pi)/(2)lt x ,le 0),( x-1, 0 lt x le 1),("in"x, x gt1):}`

Options

A. f(x) is continuous at `x=-(pi)/(2)`
B. f(x) is not differentiable at x=0
C. f(x) is differentiable at `x=1, -(3)/(2)`
D. f(x) is discontinuous at x=0

Answer

D

Explanation

Clearly, `underset(x to -(pi)/(2))lim f(x)=underset(x to -(pi)/(2))lim (-x-(pi)/(2))=0` `underset(x to -(pi)/(2^(+)))lim f(x)=underset(x to -(pi)/(2))lim cos(-(pi)/(2))=0`
and `f(-(pi)/(2))=-(-(pi)/(2))-(pi)/(2)=0`
`"Thus", underset(x to -(pi)/(2))li f(x)=f(-(pi)/(2))=underset(x to -(pi)/(2^(+)))lim f(x)`
So, f(x) is continuous at `x=-(pi)/(2)`
`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 (x-1)=-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 is continuous at x=0
`f'(x)={{:(,-1,x lt -(pi)/(2)),(,sin x,-(pi)/(2)lt x lt 0),(,1,0 lt x lt 1),(,(1)/(x),x gt 1):}`
Clearly, f(x) is not differentiable at x=0. However, it is differentiable at x=1, `-(3)/(2)`