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Definite Integration question

2019 · 9 Apr · Shift 2 · Q41
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Definite Integration question

2019 · 9 Apr · Shift 2 · Q41

JEE MainMathematicsDefinite IntegrationMCQ+4 / −1
If f : R →\to→ R is a differentiable function and f(2) = 6, then lim⁡x→2∫6f(x)2tdt(x−2)\mathop {\lim }\limits_{x \to 2} {{\int\limits_6^{f\left( x \right)} {2tdt} } \over {\left( {x - 2} \right)}}x→2lim​(x−2)6∫f(x)​2tdt​ is :-
  1. A
    2f'(2)
  2. B
    24f'(2)
  3. C
    0
  4. D
    12f'(2)
View written solutionFree

Correct answer: D

  1. We need to evaluate
L=lim⁡x→2∫6f(x)2t dtx−2L=\lim_{x\to 2}\frac{\int_6^{f(x)}2t\,dt}{x-2}L=x→2lim​x−2∫6f(x)​2tdt​

given that fff is differentiable and f(2)=6f(2)=6f(2)=6.

  1. First compute the integral in terms of f(x)f(x)f(x):
∫2t dt=t2\int 2t\,dt=t^2∫2tdt=t2

So,

∫6f(x)2t dt=[t2]6f(x)=f(x)2−36\int_6^{f(x)}2t\,dt=\left[t^2\right]_6^{f(x)}=f(x)^2-36∫6f(x)​2tdt=[t2]6f(x)​=f(x)2−36

Therefore,

L=lim⁡x→2f(x)2−36x−2L=\lim_{x\to 2}\frac{f(x)^2-36}{x-2}L=x→2lim​x−2f(x)2−36​
  1. Since f(2)=6f(2)=6f(2)=6, we have
36=f(2)236=f(2)^236=f(2)2

Hence,

L=lim⁡x→2f(x)2−f(2)2x−2L=\lim_{x\to 2}\frac{f(x)^2-f(2)^2}{x-2}L=x→2lim​x−2f(x)2−f(2)2​
  1. Use factorization:
f(x)2−f(2)2=(f(x)−f(2))(f(x)+f(2))f(x)^2-f(2)^2=(f(x)-f(2))(f(x)+f(2))f(x)2−f(2)2=(f(x)−f(2))(f(x)+f(2))

Thus,

L=lim⁡x→2(f(x)−f(2)x−2)(f(x)+f(2))L=\lim_{x\to 2}\left(\frac{f(x)-f(2)}{x-2}\right)(f(x)+f(2))L=x→2lim​(x−2f(x)−f(2)​)(f(x)+f(2))
  1. Now use differentiability of fff at x=2x=2x=2:
lim⁡x→2f(x)−f(2)x−2=f′(2)\lim_{x\to 2}\frac{f(x)-f(2)}{x-2}=f'(2)x→2lim​x−2f(x)−f(2)​=f′(2)

and by continuity of differentiable functions,

lim⁡x→2(f(x)+f(2))=f(2)+f(2)=12\lim_{x\to 2}(f(x)+f(2))=f(2)+f(2)=12x→2lim​(f(x)+f(2))=f(2)+f(2)=12
  1. Therefore,
L=f′(2)⋅12=12f′(2)L=f'(2)\cdot 12=12f'(2)L=f′(2)⋅12=12f′(2)
  1. Checking options:
  • A: 2f′(2)2f'(2)2f′(2) ❌
  • B: 24f′(2)24f'(2)24f′(2) ❌
  • C: 000 ❌
  • D: 12f′(2)12f'(2)12f′(2) ✅

So the correct answer is Option D.

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