Question

Let $f\left(x\right)=a{x}^2+\frac{3b}{x}\forall x\ne 0$ then $\frac{f^{{}^{\prime \prime }}\left(x\right)}{f\left(x\right)}=$

• A. $\frac{-2}{x^2}$
• B. $\frac{2}{x^2}$
• C. $\frac{1}{x^2}$
• D. $\frac{-1}{x^2}$

Answer 1

Answer: (B)

$\frac{2}{x^2}$

$f\left(x\right)=a{x}^2+\frac{3b}{x}$
$\Rightarrow f^{\prime }\left(x\right)=2ax-\frac{3b}{x^2}$
$\Rightarrow f^{\prime \prime }\left(x\right)=2a+\frac{6b}{x^3}=\frac{2}{x^2}(a{x}^2+\frac{3b}{x})=\frac{2}{x^2}f\left(x\right)$
$\therefore \frac{f^{\prime \prime }\left(x\right)}{f(x}=\frac{2}{x^2}$