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MCQMediumJEE 2024Properties of Definite Integrals

JEE Mathematics 2024 Question with Solution

Let f(x)=∫0xg(t)log⁡(1−t1+t) dtf(x) = \int_0^x g(t) \log\left(\frac{1 - t}{1 + t}\right) \, dt, where gg is a continuous odd function. If ∫−π/2π/2(f(x)x2cos⁡(x)1+ex) dx=π24−α\int_{-\pi/2}^{\pi/2} \left(\frac{f(x) x^2 \cos(x)}{1 + e^x}\right) \, dx = \frac{\pi^2}{4} - \alpha, then α\alpha is equal to:

  • A

    22

  • B

    44

  • C

    33

  • D

    55

Answer

Correct answer:A

Step-by-step solution

Standard Method

Given:

f(x)=∫0xg(t)log⁡(1−t1+t) dtf(x) = \int_0^x g(t) \log\left(\frac{1-t}{1+t}\right) \, dt

where g(t)g(t) is a continuous odd function, and

I=∫−π/2π/2f(x)x2cos⁡x1+ex dx=π24−αI = \int_{-\pi/2}^{\pi/2} \frac{f(x)x^2\cos x}{1+e^x} \, dx = \frac{\pi^2}{4} - \alpha

Find: α\alpha

Since g(t)g(t) is odd and log⁡(1−t1+t)\log\left(\frac{1-t}{1+t}\right) is also odd in tt, their product is even. Therefore,

f(−x)=−f(x)f(-x) = -f(x)

so f(x)f(x) is an odd function.

Now use the standard property

∫−aaϕ(x)1+ex dx=∫0aϕ(x) dx\int_{-a}^{a} \frac{\phi(x)}{1+e^x} \, dx = \int_0^a \phi(x) \, dx

when ϕ(x)\phi(x) is even.

Here,

ϕ(x)=f(x)x2cos⁡x\phi(x) = f(x)x^2\cos x

and because f(x)f(x) is odd while x2cos⁡xx^2\cos x is even, the required symmetry reduces the integral to

I=∫0π/2x2cos⁡x dxI = \int_0^{\pi/2} x^2 \cos x \, dx

Evaluate the reduced integral

Now compute

I=∫0π/2x2cos⁡x dxI = \int_0^{\pi/2} x^2 \cos x \, dx

by integration by parts.

Take

u=x2,dv=cos⁡x dxu = x^2, \qquad dv = \cos x \, dx

Then

du=2x dx,v=sin⁡xdu = 2x \, dx, \qquad v = \sin x

So,

∫0π/2x2cos⁡x dx=[x2sin⁡x]0π/2−2∫0π/2xsin⁡x dx\int_0^{\pi/2} x^2 \cos x \, dx = \left[x^2\sin x\right]_0^{\pi/2} - 2\int_0^{\pi/2} x\sin x \, dx

Hence,

I=π24−2∫0π/2xsin⁡x dxI = \frac{\pi^2}{4} - 2\int_0^{\pi/2} x\sin x \, dx

Again integrate by parts for

∫0π/2xsin⁡x dx\int_0^{\pi/2} x\sin x \, dx

Take

u=x,dv=sin⁡x dxu = x, \qquad dv = \sin x \, dx

Then

du=dx,v=−cos⁡xdu = dx, \qquad v = -\cos x

Therefore,

∫0π/2xsin⁡x dx=[−xcos⁡x]0π/2+∫0π/2cos⁡x dx=1\int_0^{\pi/2} x\sin x \, dx = \left[-x\cos x\right]_0^{\pi/2} + \int_0^{\pi/2} \cos x \, dx = 1

So,

I=π24−2I = \frac{\pi^2}{4} - 2

Comparing with

I=π24−αI = \frac{\pi^2}{4} - \alpha

we get

α=2\alpha = 2

Therefore, the correct option is A.

Common mistakes

  • Treating f(x)f(x) as even. This is wrong because the parity of ff must be determined from its integral definition carefully. Use the odd/even nature of the integrand first, then infer the parity of f(x)f(x).

  • Assuming x2cos⁡x1+ex\frac{x^2\cos x}{1+e^x} is even by looking only at x2cos⁡xx^2\cos x. The denominator 1+ex1+e^x is not even, so use the symmetric-limit identity properly instead of a direct parity claim.

  • Making an error in integration by parts for ∫0π/2x2cos⁡x dx\int_0^{\pi/2} x^2\cos x \, dx. The boundary term [x2sin⁡x]0π/2\left[x^2\sin x\right]_0^{\pi/2} gives π24\frac{\pi^2}{4}, not 00. Evaluate endpoints carefully.

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