Let $f: R \rightarrow R$ 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:

JEE Main · 2019 · 12 Apr 2019 · Shift 1 · Limits

Let

f : R \to R

be a continuously differentiable function such that

f (2) = 6

and

f^{'} (2) = \frac{1}{48}

. If

\int_{6}^{f (x)} 4 t^{3} d t = (x - 2) g (x)

, then

lim_{x \to 2} g (x)

is equal to:

Answer: (B)

Writing

g (x) = \frac{1}{x - 2} \int_{6}^{f (x)} 4 t^{3} d t

, the limit as

x \to 2

is of the form

\frac{0}{0}

since

f (2) = 6

makes the integral vanish. Differentiating top and bottom (L'Hopital), and using the fundamental theorem of calculus with the chain rule,

x \to 2 lim g (x) = x \to 2 lim 4 [f (x)]^{3} f^{'} (x) = 4 f (2)^{3} f^{'} (2) = 4 \times 6^{3} \times \frac{1}{48} = 4 \times 216 \times \frac{1}{48} = 18

JEE Main · 2019 · 12 Apr 2019 · Shift 1 · Single correct option

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