Question

Let $f ( x )$ be a polynomial function such that $f(x)+f^{\prime}(x)+f^{\prime \prime}(x)=x^{5}+64$. Then, the value of $\displaystyle\lim _{x \rightarrow 1} \frac{f(x)}{x-1}$

• A. -15
• B. -60
• C. 60
• D. 15

Answer 1

Answer: (A)

$\underset{x \rightarrow 1}{\text{Lt}} \frac{f(x)}{x-1}=f^{\prime}(1)($ and $f(1)=0)$
$f(x)+f^{\prime}(x)+t^{\prime \prime}(x)=x^{5}+64$
$f^{\prime}(x)+f^{\prime \prime}(x)+f^{\prime \prime \prime}(x)=5 x^{4}$
$f^{\prime \prime}(x)+f^{\prime \prime \prime}(x)+f^{i v}(x)=20 x^{3}$
$f^{\prime \prime \prime}(x)+f^{\text {iv }}(x)+f^{v}(x)=60 x^{2}$
$\therefore f^{v}(x)-f^{\prime \prime}(x)=60 x^{2}-20 x^{3}$
$\Rightarrow 120-f^{\prime \prime}(1)=40 \Rightarrow f^{\prime \prime}(1)=80$
Also $f(1)+f^{\prime}(1)+f^{\prime \prime}(1)=65 \Rightarrow f^{\prime}(1)=-15 .$