JEE MAIN - Mathematics (2005 - No. 52)
Let $$f:R \to R$$ be a differentiable function having $$f\left( 2 \right) = 6$$,
$$f'\left( 2 \right) = \left( {{1 \over {48}}} \right)$$. Then $$\mathop {\lim }\limits_{x \to 2} \int\limits_6^{f\left( x \right)} {{{4{t^3}} \over {x - 2}}dt} $$ equals :
$$f'\left( 2 \right) = \left( {{1 \over {48}}} \right)$$. Then $$\mathop {\lim }\limits_{x \to 2} \int\limits_6^{f\left( x \right)} {{{4{t^3}} \over {x - 2}}dt} $$ equals :
$$24$$
$$36$$
$$12$$
$$18$$
Explanation
$$\mathop {\lim }\limits_{x \to 2} \int\limits_6^{f\left( x \right)} {{{4{t^3}} \over {x - 2}}} dt$$
$$ = \mathop {\lim }\limits_{x \to 0} {{\int\limits_6^{f\left( x \right)} {4{t^3}dt} } \over {x - 2}}$$
This limit resembles a derivative because the fraction has the form $0/0$ as $x \to 2$ since both the numerator (integral from $6$ to $f(x)$) and the denominator ($x - 2$) are zero when $x = 2$.
Applying $$L'$$ Hospital rule
$$\mathop {\lim }\limits_{x \to 2} {{\left[ {4f{{\left( x \right)}^3}f'\left( x \right)} \right]} \over 1}$$
$$ = 4{\left( {f\left( 2 \right)} \right)^3}f'\left( 2 \right)$$
$$ = 4 \times {6^3} \times {1 \over {48}} = 18$$
$$ = \mathop {\lim }\limits_{x \to 0} {{\int\limits_6^{f\left( x \right)} {4{t^3}dt} } \over {x - 2}}$$
This limit resembles a derivative because the fraction has the form $0/0$ as $x \to 2$ since both the numerator (integral from $6$ to $f(x)$) and the denominator ($x - 2$) are zero when $x = 2$.
Applying $$L'$$ Hospital rule
$$\mathop {\lim }\limits_{x \to 2} {{\left[ {4f{{\left( x \right)}^3}f'\left( x \right)} \right]} \over 1}$$
$$ = 4{\left( {f\left( 2 \right)} \right)^3}f'\left( 2 \right)$$
$$ = 4 \times {6^3} \times {1 \over {48}} = 18$$
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