QJEEM05Maddac: JEE Main 2005 Mathematics - Class 12 JEEM05Maddac Solution: Let be a differentiable function having , . Then equals :

Question

Verified Solution • 2025–26 Session moderate

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 :

Accepted Answer

\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

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