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JEE Mathematics 2024 Question with Solution

For 0<a<10 < a < 1, find the value of the integral ∫0πdx1−2acos⁡(x)+a2\int_{0}^{\pi} \frac{dx}{1 - 2a \cos(x) + a^2}:

  • A

    π1−a2\frac{\pi}{1 - a^2}

  • B

    π1+a2\frac{\pi}{1 + a^2}

  • C

    π2π+a2\frac{\pi^2}{\pi + a^2}

  • D

    π2π−a2\frac{\pi^2}{\pi - a^2}

Answer

Correct answer:A

Step-by-step solution

Standard Method

Given:

I=∫0πdx1−2acos⁡x+a2,0<a<1I = \int_{0}^{\pi} \frac{dx}{1 - 2a\cos x + a^2}, \quad 0 < a < 1

Find: The value of II.

From the solution, the intended evaluation uses the standard trigonometric integral form. The denominator is rewritten as

1−2acos⁡x+a2=(1+a2)−2acos⁡x1 - 2a\cos x + a^2 = (1+a^2) - 2a\cos x

A known result is

∫0πdxα−βcos⁡x=πα2−β2(α>∣β∣)\int_{0}^{\pi} \frac{dx}{\alpha - \beta \cos x} = \frac{\pi}{\sqrt{\alpha^2 - \beta^2}} \quad (\alpha > |\beta|)

Here,

α=1+a2,β=2a\alpha = 1+a^2, \quad \beta = 2a

So,

α2−β2=(1+a2)2−(2a)2=(1−a2)2=1−a2\sqrt{\alpha^2 - \beta^2} = \sqrt{(1+a^2)^2 - (2a)^2} = \sqrt{(1-a^2)^2} = 1-a^2

since 0<a<10 < a < 1.

Therefore,

I=π1−a2I = \frac{\pi}{1-a^2}

Hence, the value of the integral is π1−a2\frac{\pi}{1-a^2}. This matches option A.

Discrepancy note: the solution labels the correct option as B, but it also states the final value as π1−a2\frac{\pi}{1-a^2}, which corresponds to option A in the given options. The value concluded in the working is taken as authoritative.

Using the rewritten denominator

Given:

I=∫0πdx1−2acos⁡x+a2,0<a<1I = \int_{0}^{\pi} \frac{dx}{1 - 2a\cos x + a^2}, \quad 0 < a < 1

Find: The exact value of the integral.

The denominator can be rewritten as

1−2acos⁡x+a2=(1+a2)−2acos⁡x1 - 2a\cos x + a^2 = (1+a^2) - 2a\cos x

Now compare with the standard form

∫0πdxp−qcos⁡x=πp2−q2,p>∣q∣\int_{0}^{\pi} \frac{dx}{p - q\cos x} = \frac{\pi}{\sqrt{p^2-q^2}}, \quad p>|q|

Taking

p=1+a2,q=2ap = 1+a^2, \qquad q = 2a

we get

p2−q2=(1+a2)2−4a2p^2-q^2 = (1+a^2)^2 - 4a^2 =1+2a2+a4−4a2= 1 + 2a^2 + a^4 - 4a^2 =1−2a2+a4= 1 - 2a^2 + a^4 =(1−a2)2= (1-a^2)^2

Therefore,

p2−q2=(1−a2)2=1−a2\sqrt{p^2-q^2} = \sqrt{(1-a^2)^2} = 1-a^2

because $$0

Common mistakes

  • Using the incorrect interval from the solution, namely [0,π/2][0,\pi/2], is wrong because the question clearly asks for integration over [0,π][0,\pi]. Always use the bounds from the question text, not a mismatched intermediate statement.

  • Matching the final value to the wrong option label is a common error here. The solution says option B, but the value π1−a2\frac{\pi}{1-a^2} is actually option A in the given list. Always verify the option text itself.

  • Applying the standard formula without checking p>∣q∣p>|q| can lead to sign mistakes. Here p=1+a2p=1+a^2 and q=2aq=2a, and for 0<a<10<a<1 we have 1+a2>2a1+a^2>2a, so the formula is valid.

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