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

$f'(1)>1$

Let h(x) = f(x) - x

It is given that f is twice differentiable so it is everywhere continuous and differentiable.

Consider the function h(x) defined on [0, 1].

Clearly, h(x) is continuous on [0, 1] and differentiable on (0, 1).

Now, $h\left(\frac{1}{2}\right)=f\left(\frac{1}{2}\right)-\frac{1}{2}=\frac{1}{2}-\frac{1}{2}=0$ and $h(1)=f(1)-1=1-1=0$

∴  $h\left(\frac{1}{2}\right)=h(1)$

Thus, h(x) satisfies the conditions of Rolle's theorem. Consequently there exists $\alpha \in(0,1)$ such that

$h'(\alpha)=0 \Rightarrow f'(\alpha)-1=0 \Rightarrow f'(\alpha)=1$

Now, f''(x) > 0 for all $x \in R$

⇒ f'(x) is increasing on R

$\Rightarrow f'(1)>f'(\alpha)$        [∵ α ∈ (0, 1)]

$\Rightarrow f'(1)>1$