QJEEM19Mac7c3: JEE Main 2019 Mathematics - Class 12 JEEM19Mac7c3 Solution: Let f : R R be a continuously differentiable function such that …

Question

Verified Solution • 2025–26 Session moderate

Let f : R

\to

R be a continuously differentiable function such that f(2) = 6 and f'(2) =

{1 \over {48}}

. If

\int\limits_6^{f\left( x \right)} {4{t^3}} dt

= (x - 2)g(x), then

\mathop {\lim }\limits_{x \to 2} g\left( x \right)

is equal to :

Accepted Answer

Given

\int\limits_6^{f(x)} {4{x^3}dx} = g(x).(x - 2)

\Rightarrow g(x) =

{{\int\limits_0^{f\left( x \right)} {4{x^3}dx} } \over {x - 2}}

\therefore

\mathop {\lim }\limits_{x \to 2} g(x) =

\mathop {\lim }\limits_{x \to 2} {{\int\limits_0^{f\left( x \right)} {4{x^3}dx} } \over {x - 2}}

At x = 2 this limit is in

{0 \over 0}

form.

So we can use L'Hopital's rule. Use leibniz intgral rule to differentiate the integration.

\Rightarrow \mathop {\lim }\limits_{x \to 2} {{4{f^3}(x).f'(x)} \over 1} = 4 \times {6^3} \times {1 \over {48}} = 18

Step-by-Step Verified Solution