Question

Observe the following Lists

List-I	List-II
A: ${\int }_{-2}^2\frac{1}{4+x^2}dx$	1) $\frac{\pi }{3}$
B: ${\int }_1^2\frac{1}{x\sqrt{x^2-1}}dx$	2) 0
C: ${\int }_0^{\pi }\cos 3x.\cos 2xdx$	3) $\frac{\pi }{4}$
	4) $\frac{\pi }{2}$

• A. A-3, B-1, C-4
• B. A-3, B-1, C-2
• C. A-1, B-3, C-2
• D. A-4, B-1, C-2

Answer 1

The correct option is B A-3, B-1, C-2

A) ${\int }_{-2}^2\frac{1}{4+x^2}dx={\left[\frac{1}{2}{\tan }^{-1}\frac{x}{2}\right]}_{-2}^2$
$=[\frac{1}{2}{\tan }^{-1}1-\frac{1}{2}{\tan }^{-1}(-1)]=\frac{\pi }{8}+\frac{\pi }{8}=\frac{\pi }{4}$
B) ${\int }_1^2\frac{1}{x\sqrt{x^2-1}}dx={\left[{\sec }^{-1}x\right]}_1^2$
$=[{\sec }^{-1}2-{\sec }^{-1}1]=[\frac{\pi }{3}-0]=\frac{\pi }{3}$
C) ${\int }_0^{\pi }\cos 3x\cos 2xdx=\frac{1}{2}{\int }_0^{\pi }(\cos x+\cos 5x)dx$
$=\frac{1}{10}{[5\sin x+\sin 5x]}_0^{\pi }=0$