JEE Main · 2022 · 25 Jun 2022 · Shift 1 · Limits

Since ${\lim }_{x\to 1}\frac{f(x)}{x-1}$ exists, we must have $f(1)=0$. By L'Hôpital's rule, the limit equals $f^{\prime }(1)$. From $f(x)+f^{\prime }(x)+f^{\prime \prime }(x)=x^5+64$, setting $x=1$: $f(1)+f^{\prime }(1)+f^{\prime \prime }(1)=65$. Differentiating the original equation repeatedly and evaluating at $x=1$: $f^{\prime }(1)+f^{\prime \prime }(1)+f^{\prime \prime \prime }(1)=5$, $f^{\prime \prime }(1)+f^{\prime \prime \prime }(1)+f^{(4)}(1)=20$, $f^{\prime \prime \prime }(1)+f^{(4)}(1)+f^{(5)}(1)=60$. Since $f$ is a polynomial of degree 5, $f^{(5)}(x)=120$. From the last equation: $f^{\prime \prime \prime }(1)+f^{(4)}(1)=-60$. Subtracting from the third: $f^{\prime \prime }(1)=80$. From $f(1)+f^{\prime }(1)+f^{\prime \prime }(1)=65$ with $f(1)=0$: $f^{\prime }(1)=65-80=-15$.