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

The correct option is:

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

Answer 1

Answer: (B)

(I) $\left\{x\in \left[\frac{-2\pi }{3},\frac{2\pi }{3}\right]:\cos x+\sin x=1\right\}$
$\cos x+\sin x=1$
$\Rightarrow \frac{1}{\sqrt{2}}\cos x+\frac{1}{\sqrt{2}}\sin x=\frac{1}{\sqrt{2}}$
$\Rightarrow \cos \left(x-\frac{\pi }{4}\right)=\cos \frac{\pi }{4}$
$\Rightarrow x-\frac{\pi }{4}=2n\pi \pm \frac{\pi }{4};n\in Z$
$\Rightarrow x=2n\pi ;x=2n\pi +\frac{\pi }{2};n\in Z$
$\Rightarrow x\in \left\{0,\frac{\pi }{2}\right\}$ in given range has two solutions
(II) $\left\{x\in \left[\frac{-5\pi }{18},\frac{5\pi }{18}\right]:\sqrt{3}\tan 3x=1\right\}$
$\sqrt{3}\tan 3x=1\Rightarrow \tan 3x=\frac{1}{\sqrt{3}}\Rightarrow 3x=n\pi +\frac{\pi }{6}$
$\Rightarrow x=(6n+1)\frac{\pi }{18};n\in Z$
$\Rightarrow x\in \left\{\frac{\pi }{18},\frac{-5\pi }{18}\right\}$ in given range has two solutions
(III) $\left\{x\in \left[-\frac{6\pi }{5},\frac{6\pi }{5}\right]:2\cos (2x)=\sqrt{3}\right\}$
$2\cos 2x=\sqrt{3}$
$\Rightarrow \cos 2x=\frac{\sqrt{3}}{2}=\cos \frac{\pi }{6}$
$\Rightarrow 2x=2n\pi \pm \frac{\pi }{6};n\in Z$
$\Rightarrow x=n\pi \pm \frac{\pi }{12};n\in Z$
$x\in \left\{\pm \frac{\pi }{12},\pi \pm \frac{\pi }{12},-\pi \pm \frac{\pi }{12}\right\}$
Six solutions in given range
(IV) $\left\{x\in \left[-\frac{7\pi }{4},\frac{7\pi }{4}\right]:\sin x-\cos x=1\right\}$
$\cos x-\sin x=-1$
$\Rightarrow \cos \left(x+\frac{\pi }{4}\right)=\frac{-1}{\sqrt{2}}=\cos \frac{3\pi }{4}$
$\Rightarrow x+\frac{\pi }{4}=2n\pi \pm \frac{3\pi }{4};n\in Z$
$\Rightarrow x=2n\pi +\frac{\pi }{2}\text{ or }x=2n\pi -\pi ;n\in Z$
$\Rightarrow x\in \left\{\frac{\pi }{2},\frac{-3\pi }{2},\pi ,-\pi \right\}$ four solutions in given range