Question

Let $g(x)=x^3 \ln \left(x^2 f(x)\right)$ where $f(x)$ is a differentiable positive function on $(0, \infty)$ satisfying $f(2)=\frac{1}{4}$ and $f^{\prime}(2)=-3$, then $g^{\prime}(2)$ equals

• A. 77
• B. -77
• C. 88
• D. -88

Answer 1

Answer: (D)

imodsC We have $g(x)=x^3 \ln \left(x^2 f(x)\right)$
$\therefore$ On differentiating b.t.s. w.r.t. $x$, we get
$g^{\prime}(x)=\frac{x^3}{x^2 f(x)}\left(x^2 f^{\prime}(x)+2 x f(x)\right)+3 x^2 \ln \left(x^2 f(x)\right) $
$\text { Hence } g^{\prime}(2)=\frac{2}{f(2)}\left(4 f^{\prime}(2)+4 f(2)\right)+12 \ln (4 f(2))$
$=8\left(4 \times-3+4 \times \frac{1}{4}\right)+12 \ln \left(4 \times \frac{1}{4}\right)=8(-12+1)=-88$