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Question

If f:RR be a differentiable function having f(2)=6,f'(2)=148. Then limx26f(x)4t3dtx-2=


A

18

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B

12

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C

36

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D

24

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Solution

The correct option is A

18


Explanation for the correct answer:

Step 1: Using Leibnitz's Theorem,

According to the theorem: ddxv(x)u(x)f(t)dt=fu(x)ddxu(x)-fv(x)ddxv(x)

Given, f(2)=6,f'(2)=148

To find, limx26f(x)4t3dtx-2

Now it is clear from Leibnitz's Theorem,

The form of the limit is 00

Step 2: Using L-hospital's Rule,

L-hospital's Rule: limxcf(x)g(x)=limxcf'(x)g'(x)

limx26f(x)4t3dtx-2

Applying, L-hospital's rule,

=limx24f(x)3ddxf(x)-4×63ddx(6)1

=limx24(f(x))3f'(x)-01=4×63×1481=4×21648=18

Hence, the correct answer is an option (A).


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