Question

Consider the following lists:
 

List I		List II
I	$\left\{x \in\left[-\frac{2 \pi}{3}, \frac{2 \pi}{3}\right]: \cos x+\sin x=1\right\}$	P	has two elements
II	$\left\{x \in\left[-\frac{5 \pi}{18}, \frac{5 \pi}{18}\right]: \sqrt{3} \tan 3 x=1\right\} $	Q	has three elements
III	$ \left\{x \in\left[-\frac{6 \pi}{5}, \frac{6 \pi}{5}\right]: 2 \cos (2 x)=\sqrt{3}\right\} $	R	has four elements
IV	$ \left\{x  \in\left[-\frac{7 \pi}{4}, \frac{7 \pi}{4}\right]: \sin x-\cos x=1\right\}$	S	has five elements
		T	has six elements

The correct option is:

• A. (I) → (P); (II) → (S); (III) → (P); (IV) → (S)
• B. (I) → (P); (II) → (P); (III) → (T); (IV) → (R)
• C. (I) → (Q); (II) → (P); (III) → (T); (IV) → (S)
• D. (I) → (Q); (II) → (S); (III) → (P); (IV) → (R)

Answer 1

The Correct Option is B

\[ \cos x + \sin x = 1 \Rightarrow \sqrt{2} \cos \left(x - \frac{\pi}{4}\right) = 1 \] \[ \Rightarrow \cos \left(x - \frac{\pi}{4}\right) = \frac{1}{\sqrt{2}} \] \[ \Rightarrow x - \frac{\pi}{4} = 2n\pi \pm \frac{\pi}{4} \] \[ \Rightarrow x = 2n\pi \pm \frac{\pi}{4} \] For \(n = 0\), we have \(x = \frac{\pi}{4}\), and for \(n = 1, 2, 3, \dots\), no solution in \(\left[-\frac{2\pi}{3}, \frac{2\pi}{3}\right]\).

Thus, (I) $\rightarrow$ (P). \[ \tan 3x = \frac{1}{\sqrt{3}} \Rightarrow 3x = n\pi + \frac{\pi}{6} \Rightarrow x = \frac{n\pi + \frac{\pi}{6}}{3} \] For \(n = 0\), we get \(x = \frac{\pi}{18}\); for \(n = 1\), we get \(x = \frac{5\pi}{18}\); and for \(n = 2, 3, \dots\), there are no solutions in the given interval. Thus, (II) $\rightarrow$ (P).

\[ \cos 2x = \frac{\sqrt{3}}{2} \Rightarrow 2x = 2n\pi \pm \frac{\pi}{6} \Rightarrow x = n\pi \pm \frac{\pi}{12} \] For \(n = -1\), we get \(x = \frac{-13\pi}{12}, \frac{-11\pi}{12}\). Thus, (III) $\rightarrow$ (T).

\[ \cos \left(\frac{\pi}{4} + x\right) = \frac{-1}{\sqrt{2}} \Rightarrow \cos \left(\frac{\pi}{4} + x\right) = \cos \left(\frac{3\pi}{4}\right) \] \[ \Rightarrow \frac{\pi}{4} + x = 2n\pi \pm \frac{3\pi}{4} \Rightarrow x = 2n\pi \pm \frac{3\pi}{4} - \frac{\pi}{4} \] For \(n = 0\), we get \(x = \frac{\pi}{2}, -\pi\);
for \(n = 1\), we get \(x = \pi\);
and for \(n = -1\),
we get \(x = \frac{3\pi}{2}\).

Thus, (IV) $\rightarrow$ (R).

So, the correct option is (B): (I) → (P); (II) → (P); (III) → (T); (IV) → (R)