Let f(x) and g(x) be two functions given by `f(x)=-1|x-1|,-1 le x le 3` and `g(x)=2-|x+1|,-2 le x le 2`
Then,

Options

A. fog is differentiable at x=-1 and gof is differentiable at x=1
B. for is differentiable at x=-1 and gof is not differentiable at x=1
C. fog is differentiable at x=1 and gof is differentiable at x=-1
D. none of these

Answer

D

Explanation

We have
`fog(x)={{:(,1+x,-2 le x lt -1),(,-x-1,-1le x lt 0),(,x-1,0 le x le 2):}`
`and gof(x)={{:(,2+x,-1 le x lt 0),(,2-x,0 le x lt 1),(,x,1 le x lt 2),(,4-x,2 le x le 3):}`
We observe that
`("LHD of fog(x) at x=-1")=((d)/(dx)(1+x))_(atx=-1)=1`
`("RHD of fog(x) at x=-1")=((d)/(dx)(-x-1))_(atx=-1)=-1`
Clearly, `("LHD of fog(x) at x=-1")ne ("RHD of fog(x) at x=-1")`
So, fog is not differentiable at x=-1
Clearly, fog is a polynomial function on [0,2].
So, it is differentiable at `x =1 in [0,2]`
Also, `("LHD of gof(x) at x=1")=((d)/(dx)(2-x))_("at x=1")=-1`
`("RHD of gof(x) at x=1")=((d)/(dx)(x))_("at x=1")=1`
Clearly, these two are not same, So, gof is not differentiable at x=1.