JEE Main · 2009 · 12 Apr 2009 · Offline · Continuity and Differentiability

Here $f(x)=x|x|$ equals $-x^2$ for $x<0$ and $x^2$ for $x\ge 0$, so $g\circ f(x)=\sin (x|x|)$ is $-\sin (x^2)$ on the left and $\sin (x^2)$ on the right. Its first derivative is $-2x\cos (x^2)$ and $2x\cos (x^2)$ respectively, and both sides give $0$ at the origin with a matching value nearby, so the composite is differentiable there with a continuous derivative, confirming the first statement. The second derivative, however, tends to $-2$ from the left and $+2$ from the right at $x=0$, so the function is not twice differentiable there and the second statement is false.