JEE MAIN - Mathematics (2019 - 9th April Evening Slot - No. 20)
If f : R $$ \to $$ R is a differentiable function and
f(2) = 6,
then $$\mathop {\lim }\limits_{x \to 2} {{\int\limits_6^{f\left( x \right)} {2tdt} } \over {\left( {x - 2} \right)}}$$ is :-
then $$\mathop {\lim }\limits_{x \to 2} {{\int\limits_6^{f\left( x \right)} {2tdt} } \over {\left( {x - 2} \right)}}$$ is :-
2f'(2)
24f'(2)
0
12f'(2)
Explanation
$$\mathop {\lim }\limits_{x \to 2} {{\int\limits_6^{f\left( x \right)} {2tdt} } \over {\left( {x - 2} \right)}}$$
This is $${0 \over 0}$$ form so we use L – Hospital Rule.
= $$\mathop {\lim }\limits_{x \to 2} {{2f\left( x \right).f'\left( x \right) - 0} \over 1}$$
= $${2f\left( 2 \right).f'\left( 2 \right)}$$
= 12f'(2)
Note : Newton - Leibnitz rule of differentiation under integration
$${d \over {dx}}\int\limits_{\alpha \left( x \right)}^{\beta \left( x \right)} {f\left( u \right)} du$$
= $$f\left( {\beta \left( x \right)} \right)\beta '\left( x \right) - f\left( {\alpha \left( x \right)} \right)\alpha '\left( x \right)$$
This is $${0 \over 0}$$ form so we use L – Hospital Rule.
= $$\mathop {\lim }\limits_{x \to 2} {{2f\left( x \right).f'\left( x \right) - 0} \over 1}$$
= $${2f\left( 2 \right).f'\left( 2 \right)}$$
= 12f'(2)
Note : Newton - Leibnitz rule of differentiation under integration
$${d \over {dx}}\int\limits_{\alpha \left( x \right)}^{\beta \left( x \right)} {f\left( u \right)} du$$
= $$f\left( {\beta \left( x \right)} \right)\beta '\left( x \right) - f\left( {\alpha \left( x \right)} \right)\alpha '\left( x \right)$$
Comments (0)
