Let f(x) be a polynomial function such that $f\left(x\right)+f^{\text{'}}\left(x\right)+f^{"}\left(x\right)=x^5+64$ . Then, the value of  $\underset{x\to 1}{\lim }\frac{f\left(x\right)}{x-1}$

 $L=\underset{x\to 1}{\lim }\frac{f\left(x\right)}{x-1}\left(finite\text{ value}\right)$
$\therefore f\left(1\right)=0\text{ and L=}\underset{x\to 1}{\text{lim}}{f}^{\text{'}}\left(1\right)$
Now    $f\left(x\right)+f^{\text{'}}\left(x\right)+f^{"}\left(x\right)=x^5+64$ ….(1)
So, clearly f(x) is polynomial of degree 5.
Differentiating $f^{\text{'}}\left(x\right)+f^{"}\left(x\right)+f^{"\text{'}}\left(x\right)=5x^4$
$\begin{array}{l}\Rightarrow f^{"}\left(x\right)+f^{"\text{'}}\left(x\right)+f^{iv}\left(x\right)=20x^3\text{ }\mathrm{....}\text{(2)}\\ \Rightarrow f^{"\text{'}}\left(x\right)+f^{iv}\left(x\right)+f^v\left(x\right)=60x^2\mathrm{...}\left(3\right)\end{array}$
From (2)-(3), we get
$f^v\left(x\right)-f^{"}\left(x\right)=60x^2\text{-20}{x}^3\mathrm{...}\left(3\right)$
Also, from (3), differentiating two more times we get
 $f^v\left(x\right)=120$
Now, putting x=1 in (4), we get
$\Rightarrow f^{\text{'}}\left(1\right)=-15$