Question

Let $f:R\to R$ be a continuously differentiable function such that $f\left(2\right)=6$ and $f^{\prime }\left(2\right)=\frac{1}{48}.$
If $\underset{6}{\overset{f\left(x\right)}{\int }}4t^{3}dt=(x-2)g\left(x\right),$ then $\underset{x\to 2}{lim}g\left(x\right)$ is equal to

• A. $12$
• B. $18$
• C. $24$
• D. $36$

Answer 1

The correct option is B $18$

$\underset{6}{\overset{f\left(x\right)}{\int }}4t^{3}dt=(x-2)g\left(x\right)$
$\Rightarrow g\left(x\right)=\frac{\underset{6}{\overset{f\left(x\right)}{\int }}4t^{3}dt}{x-2};x\ne 2$
$\Rightarrow \underset{x\to 2}{lim}g\left(x\right)=\underset{x\to 2}{lim}\frac{\underset{6}{\overset{f\left(x\right)}{\int }}4t^{3}dt}{x-2}$
Apply L'Hospital Rule,
$\Rightarrow \underset{x\to 2}{lim}g\left(x\right)=\underset{x\to 2}{lim}\frac{4{\left(f\right(x\left)\right)}^3\cdot f^{\prime }\left(x\right)}{1}$
$\Rightarrow \underset{x\to 2}{lim}g\left(x\right)$$=4{\left(f\right(2\left)\right)}^3\cdot f^{\prime }\left(2\right)=4\cdot 6^3\cdot \frac{1}{48}=18$