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Definite Integration question

2024 · 30 Jan · Shift 1 · Q40
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Definite Integration question

2024 · 30 Jan · Shift 1 · Q40

JEE MainMathematicsDefinite IntegrationMCQ+4 / −1
Let f:[−π2,π2]→Rf:\left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \rightarrow \mathbf{R}f:[−2π​,2π​]→R be a differentiable function such that f(0)=12f(0)=\frac{1}{2}f(0)=21​. If the lim⁡x→0x∫0xf(t)dtex2−1=α\lim_{x \rightarrow 0} \frac{x \int_0^x f(\mathrm{t}) \mathrm{dt}}{\mathrm{e}^{x^2}-1}=\alphax→0lim​ex2−1x∫0x​f(t)dt​=α, then 8α28 \alpha^28α2 is equal to :
  1. A
    4
  2. B
    2
  3. C
    1
  4. D
    16
View written solutionFree

Correct answer: B

  1. We need to evaluate
α=lim⁡x→0x∫0xf(t) dtex2−1\alpha=\lim_{x\to 0}\frac{x\int_0^x f(t)\,dt}{e^{x^2}-1}α=x→0lim​ex2−1x∫0x​f(t)dt​

given that fff is differentiable and f(0)=12f(0)=\frac12f(0)=21​.

  1. Since fff is differentiable, it is continuous at 000. Hence for small xxx,
∫0xf(t) dt∼f(0) x=12x.\int_0^x f(t)\,dt \sim f(0)\,x = \frac12 x.∫0x​f(t)dt∼f(0)x=21​x.

So the numerator behaves like

x∫0xf(t) dt∼x⋅12x=12x2.x\int_0^x f(t)\,dt \sim x\cdot \frac12 x = \frac12 x^2.x∫0x​f(t)dt∼x⋅21​x=21​x2.

Also,

ex2−1∼x2e^{x^2}-1 \sim x^2ex2−1∼x2

as x→0x\to 0x→0.

Therefore,

α=lim⁡x→0x∫0xf(t) dtex2−1=12.\alpha=\lim_{x\to 0}\frac{x\int_0^x f(t)\,dt}{e^{x^2}-1} =\frac{1}{2}.α=x→0lim​ex2−1x∫0x​f(t)dt​=21​.
  1. Now compute:
8α2=8(12)2=8⋅14=2.8\alpha^2 = 8\left(\frac12\right)^2 = 8\cdot \frac14 = 2.8α2=8(21​)2=8⋅41​=2.
  1. Checking options:
  • A: 444 ❌
  • B: 222 ✅
  • C: 111 ❌
  • D: 161616 ❌

Hence the correct option is B.

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