Let $f\left(x\right)$ be a polynomial function such that $f\left(x\right)+f^1\left(x\right)+f^{11}\left(x\right)=x^5+64$ Then the value of  $\left|\underset{x\to 1}{lim}\frac{f\left(x\right)}{x-1}\right|=k$  then the sum of the  digits of k    

$$\begin{gathered} L=\underset{x\to 1}{\mathrm{l}t}\frac{f\left|x\right|}{x-1}=finite, \therefore f\left(1\right)=0 \\ =L=\underset{x\to 1}{\mathrm{l}t}{f}^1\left(x\right)=f^1\left(1\right) \\ Now, f\left(x\right)+f^1\left(x\right)+f^{11}\left(x\right)=x^5+6x\to 1 \end{gathered}$$

Clearly $f\left(x\right)$ is a polynomial of degree 5

$$\begin{gathered} f^1\left(x\right)+f^{11}\left(x\right)+f^{111}\left(x\right)=5x^4\to 2 \\ {f}^{11}\left(x\right)+f^{111}\left(x\right)+f^1\left(x\right)=20x^3\to 3 \\ {f}^{111}\left(x\right)+f^{1V}\left(x\right)+f^v\left(x\right)=60x^2\to 4 \\ From 3-4 \Rightarrow f^{11}\left(x\right)-f^V\left(x\right)=20x^3-60x^2 \\ {f}^{11}\left(x\right)-120=20x^3-60x^2 (f^2\left(x\right)=120) \\ {f}^{11}\left(x\right)=20x^3-60x^2+120 \\ x=1\Rightarrow f^{11}\left(1\right)=-40+120=80 \\ From 1 f\left(1\right)+f^1\left(1\right)+f^{11}\left(1\right)=1+64 \\ 0+f^1\left(1\right)+80=65 \\ {f}^1\left(1\right)=-15 \end{gathered}$$