JEE MAIN - Mathematics (2019 - 12th April Morning Slot - No. 10)
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 :
18
36
12
24
Explanation
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.
$$ \Rightarrow \mathop {\lim }\limits_{x \to 2} {{4{f^3}(x).f'(x)} \over 1} = 4 \times {6^3} \times {1 \over {48}} = 18$$
$$ \Rightarrow g(x) = $$ $${{\int\limits_0^{f\left( x \right)} {4{x^3}dx} } \over {x - 2}}$$
$$ \therefore $$ $$\mathop {\lim }\limits_{x \to 2} g(x) = $$$$\mathop {\lim }\limits_{x \to 2} {{\int\limits_0^{f\left( x \right)} {4{x^3}dx} } \over {x - 2}}$$
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.
$$ \Rightarrow \mathop {\lim }\limits_{x \to 2} {{4{f^3}(x).f'(x)} \over 1} = 4 \times {6^3} \times {1 \over {48}} = 18$$
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