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Question

Match the statements of Column $$ I $$ with values of Column $$II$$
Column IColumn II
A.$$\left|4\sin x-1\right|<\sqrt{5},x\in [0,\pi]$$ then domain is1.$$\displaystyle \left[0, \frac{\pi}{4}\right]\cup \left[ \frac{3\pi}{4},\pi\right]$$
B.$$ 4\sin ^{2}x -8\sin x+3\leq 0,[0,2\pi],$$ then domain is2.$$\displaystyle \left[ \frac{3\pi}{2},2\pi\right] \cup \left\{ 0\right\}$$
C.$$ \left|\tan x\right| \leq 1$$ and $$x\in [0,\pi],$$ then domain is3.$$\displaystyle \left.\left[ 0,\frac{3\pi}{10}\right.\right)$$
D.$$ \cos x -\sin x \geq 1$$ and $$[0,2\pi],$$ then domain is4.$$\displaystyle \left[ \frac{\pi}{6},\frac{5\pi}{6}\right]$$


A
A3,B4,C1,D2
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B
A4,B3,C1,D2
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C
A3,B4,C2,D1
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D
A3,B2,C1,D4
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Solution

The correct option is A $$A- 3, B- 4, C- 1, D- 2$$
A) $$\left| 4\sin { x } -1 \right| <\sqrt { 5 } \quad \Rightarrow -\sqrt { 5 } <4\sin { x } -1<\sqrt { 5 } \\ \Rightarrow \cfrac { 1-\sqrt { 5 }}{ 4 } <\sin { x } <\cfrac { 1+\sqrt { 5 }}{ 4 } \Rightarrow -18^{ 0 }For \ x\in\left[ 0,\pi \right] $$
Domain is $$x\in\left[ 0,\cfrac { 3\pi}{ 10 }\right) $$

B) $$4\sin ^{ 2 }{ x } -8\sin { x } +3\le 0\Rightarrow \left( 2\sin { x } -1 \right) \left( 2\sin { x } -3 \right) \le 0$$
$$\Rightarrow \cfrac { 1 }{ 2 } \le \sin { x } \le \cfrac { 3 }{ 2 } $$
$$\sin { x } \le \cfrac { 3 }{ 2 } $$ gives no solution as $$\sin { x } \in \left[ -1,1 \right] $$
And $$\sin { x } \le \cfrac { 1 }{ 2 } \Rightarrow x\in \left[ 2n\pi +\cfrac { \pi}{ 6 } ,2n\pi +\cfrac { 5\pi}{ 6 }\right] $$
As $$x\in \left[ 0,2\pi\right] $$
Therefore, domain is $$x\in \left[ \cfrac { \pi}{ 6 } ,\cfrac { 5\pi}{ 6 }\right] $$

C) $$\left| \tan { x }\right| \le 1\Rightarrow -1\le \tan { x } \le 1$$
$$\Rightarrow \tan ^{ -1 }{ -1 } \le x\le \tan ^{ -1 }{ 1 } \Rightarrow n\pi -\cfrac { \pi}{ 4 } \le x\le n\pi +\cfrac { \pi}{ 4 } $$
For $$x\in \left[ 0, \pi\right] $$
Domain is $$x\in \left[ 0,\cfrac { \pi}{ 4 }\right] \cup \left[ \cfrac { 3\pi}{ 4 } ,\pi\right] $$

D) $$\cos { x } -\sin { x } \ge 1\Rightarrow \cfrac { 1 }{ \sqrt { 2 }} \cos { x } -\cfrac { 1 }{ \sqrt { 2 }} \sin { x } \ge \cfrac { 1 }{ \sqrt { 2 }} $$
$$\Rightarrow \cos { \left( x+\cfrac { \pi}{ 4 }\right)} \ge \cfrac { 1 }{ \sqrt { 2 }} \Rightarrow x+\cfrac { \pi}{ 4 } \ge 2n\pi \pm \cfrac { \pi}{ 4 } $$
For $$x\in \left[ 0,2\pi\right] $$
Domain is $$x\in \left[ \cfrac { 3\pi}{ 2 } ,2\pi\right] \cup \left\{ 0 \right\} $$

Maths

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