Question:
Let $f: R \rightarrow R$ be a continuously differentiable function wuch that $f(2)=6$ and $f(2)=\frac{1}{48}$.
If $\int_{6}^{\mathrm{f}(x)} 4 \mathrm{t}^{3} \mathrm{dt}=(\mathrm{x}-2) \mathrm{g}(\mathrm{x})$, then $\lim _{x \rightarrow 2} \mathrm{~g}(\mathrm{x})$ is equal to :
Correct Option: , 4
Solution:
$\lim _{x \rightarrow 2} g(x)=\lim _{x \rightarrow 2} \frac{\int_{6}^{f(x)} 4 t^{3} d t}{x-2}$
$=\lim _{x \rightarrow 2} \frac{4 . f^{3}(x) \cdot f^{\prime}(x)}{1}$
$=4 \mathrm{f}^{3}(2) \mathrm{f}^{\prime}(2)=18$