Let $$f\left( x \right) = x\left| x \right|$$ and $$g\left( x \right) = \sin x.$$
Statement-1: gof is differentiable at $$x=0$$ and its derivative is continuous at that point.
Statement-2: gof is twice differentiable at $$x=0$$.
A
Statement-1 is true, Statement-2 is true; Statement-2 is not a correct explanation for Statement-1.
B
Statement-1 is true, Statement-2 is false
C
Statement-1 is false, Statement-2 is true
D
Statement-1 is true, Statement-2 is true Statement-2 is a correct explanation for Statement-1
Explanation
Given that $$f\left( x \right) = x\left| x \right|\,\,$$ and $$\,\,g\left( x \right) = \sin x$$
So that go
$$f\left( x \right) = g\left( {f\left( x \right)} \right)$$
$$ = g\left( {x\left| x \right|} \right) = \sin x\left| x \right|$$