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Definite Integration question

2005 · Shift 0 · Q110

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Mathematics
Definite Integration
2005 · Shift 0 · Q110

Definite Integration question

2005 · Shift 0 · Q110

JEE MainMathematicsDefinite IntegrationMCQ+4 / −1

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 :

A
$24$
B
$36$
C
$12$
D
$18$

View written solutionFree

Correct answer: D

We need to evaluate

L=\lim_{x\to 2}\int_{6}^{f(x)} \frac{4t^3}{x-2}\,dt.

Since $x$ is the variable of the limit and $t$ is the integration variable, for fixed $x\neq 2$ we can take $\frac{1}{x-2}$ outside the integral:

L=\lim_{x\to 2} \frac{1}{x-2}\int_6^{f(x)}4t^3\,dt.

Evaluate the inner integral:

\int 4t^3\,dt=t^4.

So,

\int_6^{f(x)}4t^3\,dt=f(x)^4-6^4=f(x)^4-1296.

Hence,

L=\lim_{x\to 2}\frac{f(x)^4-1296}{x-2}.

Since $f(2)=6$ , we have

1296=6^4=f(2)^4.

Thus,

L=\lim_{x\to 2}\frac{f(x)^4-f(2)^4}{x-2}.

This is exactly the derivative of the function $g(x)=f(x)^4$ at $x=2$ :

L=g'(2).

Differentiate using the chain rule:

g'(x)=4f(x)^3f'(x).

Therefore,

L=g'(2)=4\,f(2)^3\,f'(2).

Substitute the given values:

L=4\cdot 6^3\cdot \frac{1}{48}.

Now $6^3=216$ , so

L=4\cdot 216\cdot \frac{1}{48}=\frac{864}{48}=18.

Therefore, the value of the limit is

\boxed{18}.

Option check:

A: $24$ ✗
B: $36$ ✗
C: $12$ ✗
D: $18$ ✓

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