IIT-JEE 2007 · Paper 1 · Mathematics Q66

MathematicsMatrix match+6Question 66 of 66

Match the integrals in Column I with the values in Column II and indicate your answer by darkening the appropriate bubbles in the 4 × 4 matrix given in the ORS.

Column I

A
$\displaystyle\int_{-1}^1\frac{dx}{1+x^2}$
B
$\displaystyle\int_0^1\frac{dx}{\sqrt{1-x^2}}$
C
$\displaystyle\int_2^3\frac{dx}{1-x^2}$
D
$\displaystyle\int_1^2\frac{dx}{x\sqrt{x^2-1}}$

Column II

P
$\frac12\log\left(\frac23\right)$
Q
$2\log\left(\frac23\right)$
R
$\frac\pi3$
S
$\frac\pi2$

Answer and worked solution

Official answer: A → S; B → S; C → P; D → R

A: $\left[\tan^{-1}x\right]_{-1}^1=\frac\pi4+\frac\pi4=\frac\pi2$ . → S.
B: $\left[\sin^{-1}x\right]_0^1=\frac\pi2$ . → S.
C: $\displaystyle\int\frac{dx}{1-x^2}=\frac12\log\left|\frac{1+x}{1-x}\right|$ , so the value is $\frac12\left(\log\frac42-\log\frac31\right)=\frac12\log\frac23$ . → P.
D: $\left[\sec^{-1}x\right]_1^2=\frac\pi3-0=\frac\pi3$ . → R.

Answer: A → S; B → S; C → P; D → R

Topics:Fundamental theorem of calculus Standard integrals

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IIT-JEE 2007 · Paper 1, Mathematics question 66