Given conditions on differentiability, what can we say about second derivative? Let $f:[0,10] \rightarrow [10,30]$ be a continuous and twice differentiable function such that $f(0)=10$ and $f(10) =20$. Suppose $\mid f'(x) \mid \leqslant 1 \; \forall x \in [0,10]$. Then the value of $f"(5)$ is
$(a) 0$
$(b) 1/2$
$(c) 1$
$(d) Can't be determined$
My trial:
We need the function to be twice differentiable. So let $f"(x)=c$ where $c$ is a constant.
Then $f(x) =\frac{cx^{2}}{2} + dx + e$ where $c,d,e \in \mathbb{R}$
Also, $f(0) =10$ implies $e=10$
and $f(10) = 20 \;$ implies $ \; 50c + 10d =10$ or $5c + d=1$
Now, $f'(5) = 5c +d = 1$ (suprise, shawty!)
therefore $c = \frac{1-d}{5} , d \in \mathbb{R}$
How to proceed? what values of $d$ to chooses such that $\mid f'(x) \mid \leqslant 1$
Any other easier approach is also welcome. Thanks. Can we somehow use MVT or Brower's Fixed Point here?

Hint
One can prove that $f'(x)=1$ for all $x\in [0,10]$.
