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

$f\text{'}\left(1\right)>1$

$\text{ As per LMVT }\frac{\text{f}\left(1\right)\text{-f}\left({\displaystyle \frac{1}{2}}\right)}{1-{\displaystyle \frac{1}{2}}}\text{=}{f}^{\text{'}}\left(c\right)\Rightarrow f^{\text{'}}\left(c\right)=1\text{ for some }c\in \left(\frac{1}{2},1\right)$

$\text{ And as }{f}^{\text{'}\text{'}}\left(x\right)>0\Rightarrow f^{\text{'}}\left(x\right)\text{ is increases }$

$\text{ So }{f}^{\text{'}}\left(x\right)>1\text{ for }x>C$

$\text{ As }1>c,f^{\text{'}}\left(1\right)>f^{\text{'}}\left(c\right)=1$