Consider the following lists: has two elements [2022]

Question

Consider the following lists: [2022]

The correct option is:

Smart Achievers · Accepted Answer

(2)

We have (I)

${x \in [\frac{- 2 π}{3}, \frac{2 π}{3}] : \cos x + \sin x = 1}$

$\cos x + \sin x = 1$

$\Rightarrow \frac{1}{\sqrt{2}} \cos x + \frac{1}{\sqrt{2}} \sin x = \frac{1}{\sqrt{2}}$

$\Rightarrow \sin (\frac{π}{4} + x) = \frac{1}{\sqrt{2}}$

$\Rightarrow \frac{π}{4} + x = n π + {(- 1)}^{n} \frac{π}{4}$

$\Rightarrow x = n π + {(- 1)}^{n} \frac{π}{4} - \frac{π}{4}$ $∴$ $x has 2 elements \to (P)$

We have (II)

${x \in [\frac{- 5 π}{18}, \frac{5 π}{18}] : \sqrt{3} \tan 3 x = 1};$ $\sqrt{3} \tan 3 x = 1$

$\Rightarrow \tan 3 x = \frac{1}{\sqrt{3}}$

$x = (6 n + 1) \frac{π}{18}, n \in ℤ \Rightarrow 3 x = n π + \frac{π}{6}$

$or x = \frac{n π}{3} + \frac{π}{18}$ $∴$ $x has 2 elements \to (P)$

We have (III)

${x \in [\frac{- 6 π}{5}, \frac{6 π}{5}] : 2 \cos 2 x = \sqrt{3}}$

$2 \cos (2 x) = \sqrt{3}$

$\Rightarrow \cos 2 x = \frac{\sqrt{3}}{2}$ $\Rightarrow 2 x = 2 n π \pm \frac{π}{6};$ $n \in Z$

$or x = n π \pm \frac{π}{12};$ $n \in Z$

$\Rightarrow x \in {\pm \frac{π}{12}, - π \pm \frac{π}{12}, π \pm \frac{π}{12}}$

$∴$ $x has 6 elements \to (T)$

We have (IV)

${x \in [\frac{- 7 π}{4}, \frac{7 π}{4}] : \sin x - \cos x = 1}$

$\sin x - \cos x = 1$

$\Rightarrow \frac{1}{\sqrt{2}} \sin (x) - \frac{1}{\sqrt{2}} \cos (x) = \frac{1}{\sqrt{2}}$

$\Rightarrow \sin (x - \frac{π}{4}) = \frac{1}{\sqrt{2}}$

$\Rightarrow x - \frac{π}{4} = n π + {(- 1)}^{n} \frac{π}{4}$

$\Rightarrow x = n π + {(- 1)}^{n} \frac{π}{4} + \frac{π}{4}$

$\Rightarrow x \in {\frac{π}{2}, \frac{- 3 π}{2}, - π, π}$

$∴$ $x has 4 elements \to (R)$

Want Us to Email you About Special Offers & Updates?