Checking increasing or decreasing function of twice differentiable function

Question

If $f:R \to R$ is a twice differentiable function such that $f''(x)>0$ for all $x \in R$ and $f(\frac{1}{2})=\frac{1}{2}$ and $f(1)=1$, then

(A) $f'(1) \le 0$

(B) $\frac{1}{2}\le f'(1) \le 1$

(C) $0\le f'(1) \le \frac{1}{2}$

(D) $f'(1) \ge 1$

My approach is as follow

$f'(x)=\frac{f(1)-f(\frac{1}{2})}{1-\frac{1}{2}}$=1, it is a single choice so I am not sure between the option (B) and (D) how do we check as $f'(x)$ is an increasing function given $f''(x)>0$

Answer 1

by langarange $$f'(c)=1$$ as you have found for some c in $(1/2,1)$

again $$f''(k)=\frac{f'(1)-f'(c)}{1-c}>0$$

whch means $$f'(1)>f'(c)=1$$

Answer 2

As $x$ went from (1/2) to 1, $f(x)$ also went from (1/2) to 1.

Therefore, the averate rate of change was 1 in this interval.

But $f''(x)$ given as always positive. Therefore, $f'(x)$ is always increasing. Since the average rate of change in the interval [1/2, 1] is 1, $f'(x)$ must have had that value somewhere in the interval [1/2, 1].

Therefore, (again since $f'(x)$ is always increasing), $f'(x)$ must be $> 1$ at $x = 1$.