JEE Main 2022 (Online) 25th June Morning ShiftQuestion 1
Let $f(x)$ be a polynomial function such that $f(x)+f'(x)+f''(x)=x^{5}+64$. Then, the value of $\lim_{x\to1}\frac{f(x)}{x-1}$ is equal to :
Limits & Continuity
Solution
$f$ has degree 5 and $f=(1+D+D^2)^{-1}(x^5+64)=(1-D)(1+D^3+\dots)P$ with $P=x^5+64$. $f=P-P'+P'''-P''''=x^5-5x^4+60x^2-120x+64$. $f(1)=1-5+60-120+64=0$, so $\lim\frac{f(x)}{x-1}=f'(1)=5-20+120-120=-15$. Option (A).