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