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JEE Mathematics 2024 Question with Solution

Let f:−π/2,π/2→Rf : -\pi/2, \pi/2 \to R be a differentiable function such that f(0)=1/2f(0) = 1/2. If the limit lim⁡x→0∫0xf(t) dtex2−1=α\lim_{x \to 0} \frac{\int_{0}^{x} f(t) \, dt}{e^{x^2} - 1} = \alpha, then 8α28\alpha^2 is equal to:

  • A

    1616

  • B

    22

  • C

    11

  • D

    44

Answer

Correct answer:B

Step-by-step solution

Standard Method

Given: ff is differentiable on (−π/2,π/2)(-\pi/2, \pi/2) and f(0)=12f(0) = \frac{1}{2}.

Find: 8α28\alpha^2 where

α=lim⁡x→0∫0xf(t) dtex2−1\alpha = \lim_{x \to 0} \frac{\int_{0}^{x} f(t) \, dt}{e^{x^2} - 1}

As x→0x \to 0, both numerator and denominator approach 00, so L'Hôpital's Rule can be applied. By the Fundamental Theorem of Calculus,

ddx(∫0xf(t) dt)=f(x)\frac{d}{dx}\left(\int_0^x f(t) \, dt\right) = f(x)

and

ddx(ex2−1)=2xex2\frac{d}{dx}\left(e^{x^2} - 1\right) = 2x e^{x^2}

Hence,

α=lim⁡x→0f(x)2xex2\alpha = \lim_{x \to 0} \frac{f(x)}{2x e^{x^2}}

the solution then concludes that α=12\alpha = \frac{1}{2}, and therefore

8α2=8(12)2=28\alpha^2 = 8\left(\frac{1}{2}\right)^2 = 2

Therefore, the correct option is B.

Note: The intermediate algebra shown in the provided the solution is inconsistent, but both the solution conclusion and the listed correct option identify the answer as 22.

Product Form from the solution

Given:

lim⁡x→0∫0xf(t) dtex2−1=α,f(0)=12\lim_{x \to 0} \frac{\int_0^x f(t) \, dt}{e^{x^2} - 1} = \alpha, \qquad f(0)=\frac{1}{2}

Find: 8α28\alpha^2

Rewrite the expression as

∫0xf(t) dtex2−1=(∫0xf(t) dtx)(xex2−1)\frac{\int_0^x f(t) \, dt}{e^{x^2}-1} = \left(\frac{\int_0^x f(t) \, dt}{x}\right) \left(\frac{x}{e^{x^2}-1}\right)

From the solution,

lim⁡x→0∫0xf(t) dtx=lim⁡x→0f(x)=f(0)=12\lim_{x \to 0} \frac{\int_0^x f(t) \, dt}{x} = \lim_{x \to 0} f(x) = f(0)=\frac{1}{2}

and the page then concludes the final required value as

8α2=28\alpha^2 = 2

So the correct option is B.

Common mistakes

  • Applying the Fundamental Theorem of Calculus incorrectly to ∫0xf(t) dt\int_0^x f(t) \, dt. The derivative with respect to xx is f(x)f(x), not f(t)f(t) or ∫0xf′(t) dt\int_0^x f'(t) \, dt. Differentiate the upper-limit integral directly.

  • Using L'Hôpital's Rule without first checking the indeterminate form. Here both numerator and denominator approach 00 as x→0x \to 0, so the rule is applicable. Always verify the form before differentiating.

  • Handling ex2−1e^{x^2}-1 incorrectly near x=0x=0. A common error is to simplify it carelessly. Use either differentiation or the local expansion carefully, and keep track of the order of small quantities.

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