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 $\underset{x\to 1}{\lim }\frac{f(x)}{x-1}$

• A. -15
• B. -60
• C. 60
• D. 15

Answer 1

Answer: (A)

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