Let f be a twice differentiable function on $(1,6)$  if $f\left(2\right)=8$  , $f^{\text{'}}\left(2\right)=5,f^{\text{'}}\left(x\right)\ge 1$  and $f^{\text{'}\text{'}}\left(x\right)\ge 4\forall x\in (1,6)$  then

f be twice differentiable function
 $\because f^{\text{'}}\left(x\right)\ge 1$$\frac{f\left(5\right)-f\left(2\right)}{3}\ge 1$$[\because LMVT]$  
 $f\left(5\right)\ge 3+f\left(2\right)$  $\Rightarrow f\left(5\right)\ge 3+8\Rightarrow f\left(5\right)\ge 11$
 And also  $f^{\text{'}\text{'}}\left(x\right)\ge 4$  $\frac{f^{\text{'}}\left(5\right)-f^{\text{'}}\left(2\right)}{5-2}\ge 4$
$\Rightarrow f^{\text{'}}\left(5\right)-f^{\text{'}}\left(2\right)\ge 12$$\Rightarrow f^{\text{'}}\left(5\right)\ge 12+f^{\text{'}}\left(2\right)\ge 12+5\ge 17$
$\therefore f\left(5\right)+f^{\text{'}}\left(5\right)\ge 28$