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JEE Advanced 2019 Paper 1 · Mathematics · Indefinite and Definite Integrals · hard

Consider the following four definite integrals and their claimed values. Which of the claims is/are correct?

One or more options correct

Consider the following four definite integrals and their claimed values. Which of the claims is/are correct?
  1. A. ∫02π∣sin⁡x−cos⁡x∣ dx=42\displaystyle\int_0^{2\pi}|\sin x-\cos x|\,dx=4\sqrt2
  2. B. ∫−11x2ln⁡ ⁣(1+x1−x)dx=0\displaystyle\int_{-1}^{1}x^2\ln\!\left(\frac{1+x}{1-x}\right)dx=0
  3. C. ∫01ln⁡(1+x)1+x2 dx=π4ln⁡2\displaystyle\int_0^{1}\frac{\ln(1+x)}{1+x^2}\,dx=\frac{\pi}{4}\ln2
  4. D. ∫0∞dx(1+x2)(1+x3)=π4\displaystyle\int_0^{\infty}\frac{dx}{(1+x^2)(1+x^3)}=\frac{\pi}{4}
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Answer: A, B, D

(A) Write sin⁡x−cos⁡x=2sin⁡(x−π4)\sin x-\cos x=\sqrt2\sin\left(x-\frac\pi4\right). Over a full period of length 2π2\pi, the integral of ∣sin⁡∣|\sin| is 44, so the value is 424\sqrt2, true. (B) The factor x2x^2 is even and ln⁡1+x1−x\ln\frac{1+x}{1-x} is odd, so the integrand is odd on a symmetric interval and the integral is 00, true. (C) Put x=tan⁡θx=\tan\theta: the integral becomes ∫0π/4ln⁡(1+tan⁡θ) dθ\int_0^{\pi/4}\ln(1+\tan\theta)\,d\theta; using θ→π4−θ\theta\to\frac\pi4-\theta and (1+tan⁡θ)(1+tan⁡(π4−θ))=2(1+\tan\theta)(1+\tan(\frac\pi4-\theta))=2 gives π8ln⁡2\frac\pi8\ln2, not π4ln⁡2\frac\pi4\ln2, so false. (D) Substituting x→1xx\to\frac1x turns the integrand into x3(1+x2)(1+x3)\frac{x^3}{(1+x^2)(1+x^3)}; adding it to the original gives 1+x3(1+x2)(1+x3)=11+x2\frac{1+x^3}{(1+x^2)(1+x^3)}=\frac1{1+x^2}, so 2I=∫0∞dx1+x2=π22I=\int_0^\infty\frac{dx}{1+x^2}=\frac\pi2 and I=π4I=\frac\pi4, true.
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Original question written in the JEE Advanced 2019 Paper 1 pattern and checked twice for the correct answer. Not an official question paper. How we write and check questions · Corrections policy · Report a mistake