Let f(x)=begincasesx^2sin 1/x,&x≠0\\[2mm]0,&x=0endcases, g(x)=f(x)+|x| and F(x)=∫0^xf(t) dt. Then

(A) True: $\frac{f(h)-f(0)}{h}=h\sin\frac1h\to0$, so $f'(0)=0$. (B) False: for $x\ne0$, $f'(x)=2x\sin\frac1x-\cos\frac1x$, and $\cos\frac1x$ has no limit as $x\to0$, so $f'$ is discontinuous at $0$. (C) True: $g'(0^+)=f'(0)+1=1$ while $g'(0^-)=f'(0)-1=-1$; since $|x|$ has a corner and $f$ is differentiable, the sum has the corner. (D) True: $f$ is continuous, so $F'(x)=f(x)$ by the fundamental theorem, and $F''(0)=f'(0)=0$ exists. (F would fail to be thrice differentiable at 0, since $f'$ is discontinuous there, but that is not asked.) Hence the correct options are (A), (C) and (D).