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Definite Integration question

2019 · 12 Apr · Shift 1 · Q33
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Definite Integration question

2019 · 12 Apr · Shift 1 · Q33

JEE MainMathematicsDefinite IntegrationMCQ+4 / −1
Let f : R →\to→ R be a continuously differentiable function such that f(2) = 6 and f'(2) = 148{1 \over {48}}481​. If ∫6f(x)4t3dt\int\limits_6^{f\left( x \right)} {4{t^3}} dt6∫f(x)​4t3dt= (x - 2)g(x), then lim⁡x→2g(x)\mathop {\lim }\limits_{x \to 2} g\left( x \right)x→2lim​g(x) is equal to :
  1. A
    18
  2. B
    36
  3. C
    12
  4. D
    24
View written solutionFree

Correct answer: A

  1. We are given
∫6f(x)4t3 dt=(x−2)g(x)\int_6^{f(x)} 4t^3\,dt=(x-2)g(x)∫6f(x)​4t3dt=(x−2)g(x)

and we need to find

lim⁡x→2g(x).\lim_{x\to 2} g(x).x→2lim​g(x).
  1. First evaluate the integral explicitly:
∫4t3 dt=t4.\int 4t^3\,dt=t^4.∫4t3dt=t4.

So,

∫6f(x)4t3 dt=f(x)4−64=f(x)4−1296.\int_6^{f(x)}4t^3\,dt=f(x)^4-6^4=f(x)^4-1296.∫6f(x)​4t3dt=f(x)4−64=f(x)4−1296.

Hence,

(x−2)g(x)=f(x)4−1296.(x-2)g(x)=f(x)^4-1296.(x−2)g(x)=f(x)4−1296.

Therefore,

g(x)=f(x)4−1296x−2.g(x)=\frac{f(x)^4-1296}{x-2}.g(x)=x−2f(x)4−1296​.
  1. Now take the limit as x→2x\to 2x→2:
lim⁡x→2g(x)=lim⁡x→2f(x)4−1296x−2.\lim_{x\to 2} g(x)=\lim_{x\to 2}\frac{f(x)^4-1296}{x-2}.x→2lim​g(x)=x→2lim​x−2f(x)4−1296​.

Since f(2)=6f(2)=6f(2)=6, this becomes

lim⁡x→2f(x)4−64x−2.\lim_{x\to 2}\frac{f(x)^4-6^4}{x-2}.x→2lim​x−2f(x)4−64​.

This is the derivative of f(x)4f(x)^4f(x)4 at x=2x=2x=2.

  1. Let
F(x)=f(x)4.F(x)=f(x)^4.F(x)=f(x)4.

Then by chain rule,

F′(x)=4f(x)3f′(x).F'(x)=4f(x)^3 f'(x).F′(x)=4f(x)3f′(x).

So,

lim⁡x→2g(x)=F′(2)=4[f(2)]3f′(2).\lim_{x\to 2} g(x)=F'(2)=4[f(2)]^3 f'(2).x→2lim​g(x)=F′(2)=4[f(2)]3f′(2).

Substitute the given values:

=4⋅63⋅148.=4\cdot 6^3\cdot \frac{1}{48}.=4⋅63⋅481​.

Now,

63=216,6^3=216,63=216,

so

4⋅216⋅148=86448=18.4\cdot 216\cdot \frac{1}{48}=\frac{864}{48}=18.4⋅216⋅481​=48864​=18.
  1. Therefore,
lim⁡x→2g(x)=18.\boxed{\lim_{x\to 2} g(x)=18}.x→2lim​g(x)=18​.
  1. Option check:
  • A: 181818 ✅
  • B: 363636 ❌
  • C: 121212 ❌
  • D: 242424 ❌

Thus the correct option is A.

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