Let f:R→R be a continuous and differentiable function such that f2=6 and f^(')2=1/48. If ∫6f(x)4t^3dt=x-2gx, then limx→2gx is equal to

Correct answer: (B) 18. limx→2gx=limx→2(∫6fx4t^3dt)/(x-2) 0/0 form. By using L'Hospital rule =limx→2(4.f^3(x)⋅f^(')x)/1 =4f^32f^(')2=4×6×6×6×1/48=18.

Q45 / 90 · JEE Main 2019 (12 Apr Shift 1) · Mathematics

Let $f : R \to R$ be a continuous and 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 $\underset{x \to 2}{l i m} g (x)$ is equal to

A
$24$
Correct answer
B
$18$
C
$12$
D
$36$

Solution

$\underset{x \to 2}{l i m} g (x) = \underset{x \to 2}{l i m} \frac{\int_{6}^{f (x)} 4 t^{3} d t}{x - 2} (\frac{0}{0} form)$ .

By using L'Hospital rule
$= \underset{x \to 2}{l i m} \frac{4 . f^{3} (x) \cdot f^{'} (x)}{1}$
$= 4 f^{3} (2) f^{'} (2) = 4 \times 6 \times 6 \times 6 \times \frac{1}{48} = 18$ .

Hence, the correct answer is option: 'B'.

Let f:R→R be a continuous and differentiable function such that f2=6 and f'2=148. If ∫6f(x)4t3dt=x-2gx, then limx→2gx is equal to

Solution

Relevant Questions

Let $f : R \to R$ be a continuous and 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 $\underset{x \to 2}{l i m} g (x)$ is equal to