a) $0 \le F(x)$ for all $x$.   

b) $\lim\_{x \to 0} \frac {F(x)}{x} = 1$  

c) $\lim\_{x \to \infty} F(x)= \infty $   

d) $F$ has an inverse and $(F^{-1})'(0)=\frac{1}{\sin 1}$   

What I did: for (a) I went for the derivative $F'(x)=\sin\frac{1}{1+x^2}$, I saw that it's bigger than zero and it's decreasing, could just infer that the function $F(x)$ is increasing, and not $F(x)\ge0$ for all x, so I left it aside.   

(b) I took the limit and it was $0$, so it's wrong.   

(c) I think I thought a little too broad here and just said that $\sin\frac{1}{1+x^2}$, is like $\frac{1}{x^2}$, which converges if the integral was from $1 \to \infty$, not from zero.   

(d) I tried to do this $F((F^{-1})'(0))$, which is supposed to give me $0$, but I didn't get zero. so I assumed its wrong.   

 In the end the answer was (d), I would really appreciate an explanation of why it is true, hence $F((F^{-1})'(0))\ne0$, also I would appreciate any tips about my thinking and the way I tried to solve the question.   
 Thanks in advance.