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

none of these

We have,

$fog(x)=\left\{\begin{array}{lc} 1+x, & -2 \leq x<-1 \\ -x-1, & -1 \leq x<0 \\ x-1, & 0 \leq x \leq 2
\end{array}\right.$

and,

$gof(x)=\left\{\begin{array}{lr} 2+x, & -1 \leq x<0 \\ 2-x, & 0 \leq x<1 \\ x, & 1 \leq x<2 \\ 4-x, & 2 \leq x \leq 3 \end{array}\right.$

We observe that

(LHD of fog(x) at x = -1) = $\left(\frac{d}{d x}(1+x)\right)_{\text {at } x=-1}=1$

(RHD of fog(x) at x = -1) = $\left(\frac{d}{d x}(-x-1)\right)_{\text {at } x=1}=-1$

Clearly,

(LHD of fog(x) at x = -1) ≠ (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) = $\left(\frac{d}{d x}(2-x)\right)_{\text {at } x=1}=-1$

and,

(RHD of gof(x) at x = 1) = $\left(\frac{d}{d x}(x)\right)_{\text {at } x=1}=1$.

Clearly, these two are not same. So, gof is not differentiable at x = 1.