JEE Advanced 2007 Paper 1Question 1
Match the integrals in Column I with the values in Column II and indicate your answer by darkening the appropriate bubbles in the $4\times4$ matrix given in the ORS.
Column I: (A) $\int_{-1}^{1}\frac{dx}{1+x^2}$; (B) $\int_{0}^{1}\frac{dx}{\sqrt{1-x^2}}$; (C) $\int_{2}^{3}\frac{dx}{1-x^2}$; (D) $\int_{1}^{2}\frac{dx}{x\sqrt{x^2-1}}$.
Column II: (p) $\frac{1}{2}\log\left(\frac{2}{3}\right)$; (q) $2\log\left(\frac{2}{3}\right)$; (r) $\frac{\pi}{3}$; (s) $\frac{\pi}{2}$.
Definite Integration
AnswerA-s,B-s,C-p,D-r
Solution
Concept. Each integral is a standard form; evaluate it at the limits. Formulas used. (1) $\int\frac{dx}{1+x^2}=\tan^{-1}x$, $\int\frac{dx}{\sqrt{1-x^2}}=\sin^{-1}x$(2) $\int\frac{dx}{1-x^2}=\frac12\ln\Big|\frac{1+x}{1-x}\Big|$(3) $\int\frac{dx}{x\sqrt{x^2-1}}=\sec^{-1}x$ ($x>1$)Step 1 - (A). By (1), $\tan^{-1}1-\tan^{-1}(-1)=\frac\pi4+\frac\pi4=\frac\pi2$: A $\to$ s. Step 2 - (B). By (1), $\sin^{-1}1-\sin^{-1}0=\frac\pi2$: B $\to$ s. Step 3 - (C). By (2), $$\frac12\Big(\ln\Big|\frac{4}{-2}\Big|-\ln\Big|\frac{3}{-1}\Big|\Big)=\frac12(\ln2-\ln3)=\frac12\log\frac23.$$ C $\to$ p. Step 4 - (D). By (3), $\sec^{-1}2-\sec^{-1}1=\frac\pi3-0=\frac\pi3$: D $\to$ r. Answer: A-s,B-s,C-p,D-r