If $f(x)≤x$ , then $f′(x)≤1$? I'm studying Calculus and having a trouble solving this question.

1) If $f(x)\leq x$, then $f′(x)\leq 1$ for all $x$?

2) What if $f(0)=0$, $f′(x)$ exists for all $x$?

I could easily find the counter example for 1) (Therefore it is false)

But I'm not sure about 2)

If $f(0)=0$ and $f′(x)$ exists for all $x$  &amp;
$f(x)\leq x$ , then $f′(x)\leq 1$ for all $x$?

Please leave a comment if you don't mind :)
Hint: Consider $$f(x)=x-A \sin^2 x $$
for large $A$.
Let for example $f(x)=x\sin x$, then $f(0)=0$ and for any real $x$ $f(x)\leq x$. Then the derivative will look like this: $f'(x)=\sin x+x\cos x$ and so will exist for all $x$. Taking $x=2\pi n$, where $n$ is positive integer we get $f'(x)=2\pi n$, which evidently has no upper limitation.