Question

Match the statements of Column $I$ with values of Column $II$

	Column I		Column II
A.	$|4\sin x-1|<\sqrt{5},x\in [0,\pi ]$ then domain is	1.	$[0,\frac{\pi }{4}]\cup [\frac{3\pi }{4},\pi ]$
B.	$4{\sin }^{2}x-8\sin x+3\le 0,[0,2\pi ],$ then domain is	2.	$[\frac{3\pi }{2},2\pi ]\cup \left\{0\right\}$
C.	$\left|\tan x\right|\le 1$ and $x\in [0,\pi ],$ then domain is	3.	$[0,\frac{3\pi }{10})$
D.	$\cos x-\sin x\ge 1$ and $[0,2\pi ],$ then domain is	4.	$[\frac{\pi }{6},\frac{5\pi }{6}]$

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

Answer 1

The correct option is A $A-3,B-4,C-1,D-2$

A) $|4\sin x-1|<\sqrt{5}\Rightarrow -\sqrt{5}<4\sin x-1<\sqrt{5}\Rightarrow \frac{1-\sqrt{5}}{4}<\sin x<\frac{1+\sqrt{5}}{4}\Rightarrow -{18}^{0}Forx\in [0,\pi ]$

Domain is $x\in [0,\frac{3\pi }{10})$

B) $4{\sin }^{2}x-8\sin x+3\le 0\Rightarrow (2\sin x-1)(2\sin x-3)\le 0$

$\Rightarrow \frac{1}{2}\le \sin x\le \frac{3}{2}$

$\sin x\le \frac{3}{2}$ gives no solution as $\sin x\in [-1,1]$

And $\sin x\le \frac{1}{2}\Rightarrow x\in [2n\pi +\frac{\pi }{6},2n\pi +\frac{5\pi }{6}]$

As $x\in [0,2\pi ]$

Therefore, domain is $x\in [\frac{\pi }{6},\frac{5\pi }{6}]$

C) $\left|\tan x\right|\le 1\Rightarrow -1\le \tan x\le 1$

$\Rightarrow {\tan }^{-1}-1\le x\le {\tan }^{-1}1\Rightarrow n\pi -\frac{\pi }{4}\le x\le n\pi +\frac{\pi }{4}$

For $x\in [0,\pi ]$

Domain is $x\in [0,\frac{\pi }{4}]\cup [\frac{3\pi }{4},\pi ]$

D) $\cos x-\sin x\ge 1\Rightarrow \frac{1}{\sqrt{2}}\cos x-\frac{1}{\sqrt{2}}\sin x\ge \frac{1}{\sqrt{2}}$

$\Rightarrow \cos (x+\frac{\pi }{4})\ge \frac{1}{\sqrt{2}}\Rightarrow x+\frac{\pi }{4}\ge 2n\pi \pm \frac{\pi }{4}$

For $x\in [0,2\pi ]$

Domain is $x\in [\frac{3\pi }{2},2\pi ]\cup \left\{0\right\}$