Let `f(x)={{:([x],x cancelinI),(x-1, x in I):}`( where, [.] denotes the greatest integer function) and `g(x)={{:(sinx+cosx",", xlt0),(1",",xge 0):}` Then for f(g(x)) at x = 0

Options

A. `underset(xrarr0)(lim)g(g(x))` exists but not continuous
B. continuous but not differentiable at x = 0
C. differentiable at x = 0
D. `underset(xrarr0)(lim)f(g(x))` does not exist

Answer

C

Explanation

`f(x)={{:([x], x cancelinI),(x-1, x inI):}` (where, [.] denotes the greatest integer function) and `g(x)={{:(sinx+cosx"," ,xlt0),(1",",xge0):}`. Then for f(g(x)) at x = 0
For continuity at x = 0, `f(g(0))=f(1)=1-1=0`
`f(g(0^(+)))=f(sin(0^(-))+cos(0^(-)))=f((0^(-))+(1^(-)))=f(1^(-))=[1^(-)]=0`
Thus f(x) is continuous at x = 0
For differentiability at x = 0
For differentiability at x = 0,
`f'(g(0^(+)))=underset(hrarr0)(lim)(f(g(h))-f(g(0)))/(h)`
`=underset(hrarr0)(lim)(0-0)/(h)=0`
`f'(g(0^(-)))=underset(hrarr0)(lim)([1^(-)]-0)/(-h)=0`
`thereforef(x)` is differentiable at x = 0.