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) $$\left\{x \in \left[-\frac{7\pi}{4}, \frac{7\pi}{4}\right] : \sin x - \cos x = 1\right\}$$	(S) has five elements
	(T) has six elements

The correct option is:

Given code for the number of elements
$$P=2,\;Q=3,\;R=4,\;S=5,\;T=6$$

Case I : $$\cos x+\sin x=1,\;x\in\left[-\frac{2\pi}{3},\frac{2\pi}{3}\right]$$

Write $$\cos x+\sin x=\sqrt2\sin\left(x+\frac{\pi}{4}\right)$$.
Hence $$\sqrt2\sin\left(x+\frac{\pi}{4}\right)=1$$ gives
$$\sin\left(x+\frac{\pi}{4}\right)=\frac1{\sqrt2}\,.$$

For $$\sin\theta=\frac1{\sqrt2}$$, $$\theta=\frac{\pi}{4}+2k\pi$$ or $$\theta=\frac{3\pi}{4}+2k\pi$$.
So

$$x+\frac{\pi}{4}=\frac{\pi}{4}+2k\pi\;\Longrightarrow\;x=2k\pi$$
$$x+\frac{\pi}{4}=\frac{3\pi}{4}+2k\pi\;\Longrightarrow\;x=\frac{\pi}{2}+2k\pi$$

Inside $$\left[-\frac{2\pi}{3},\frac{2\pi}{3}\right]\;(=-120^\circ\text{ to }120^\circ)$$ we have
$$x=0,\;\;x=\frac{\pi}{2}$$ only. Thus there are $$2$$ solutions → code $$P$$.

Case II : $$\sqrt3\tan3x=1,\;x\in\left[-\frac{5\pi}{18},\frac{5\pi}{18}\right]$$

The equation is $$\tan3x=\frac1{\sqrt3}=\tan\frac{\pi}{6}$$.
Hence $$3x=\frac{\pi}{6}+k\pi\;\;(k\in\mathbb Z)$$ giving $$x=\frac{\pi}{18}+\frac{k\pi}{3}$$.

Check which integers $$k$$ keep $$x$$ in $$[-\frac{5\pi}{18},\,\frac{5\pi}{18}]$$:

k = -1 : $$x=-\frac{5\pi}{18}$$ (left end, allowed)
k = 0 : $$x=\frac{\pi}{18}$$ (inside interval)
k = 1 : $$x=\frac{7\pi}{18}$$ (>$$\frac{5\pi}{18}$$, rejected)
k ≤ -2 or k ≥ 2 give still larger magnitudes, rejected.

Thus exactly $$2$$ solutions → code $$P$$.

Case III : $$2\cos(2x)=\sqrt3,\;x\in\left[-\frac{6\pi}{5},\frac{6\pi}{5}\right]$$

Rewriting, $$\cos(2x)=\frac{\sqrt3}{2}=\cos\frac{\pi}{6}$$.
Therefore $$2x=2n\pi\pm\frac{\pi}{6}\;\Longrightarrow\;x=n\pi\pm\frac{\pi}{12}\;(n\in\mathbb Z).$$

Scan through the interval $$-\,\frac{6\pi}{5}(-216^\circ)\;\text{to}\;\frac{6\pi}{5}(216^\circ).$$

n = -1 : $$x=-\pi\pm\frac{\pi}{12}\;=\;-\,\frac{13\pi}{12},\;-\frac{11\pi}{12}$$ (both inside)
n = 0 : $$x=\pm\frac{\pi}{12}$$ (inside)
n = +1 : $$x=\pi\pm\frac{\pi}{12}\;=\;\frac{11\pi}{12},\;\frac{13\pi}{12}$$ (inside)
n = ±2 already lie outside ±$$\frac{6\pi}{5}$$.

Total number of solutions = $$2+2+2 = 6$$ → code $$T$$.

Case IV : $$\sin x-\cos x=1,\;x\in\left[-\frac{7\pi}{4},\frac{7\pi}{4}\right]$$

Write $$\sin x-\cos x=\sqrt2\sin\left(x-\frac{\pi}{4}\right)$$.
Thus $$\sqrt2\sin\left(x-\frac{\pi}{4}\right)=1$$ gives
$$\sin\left(x-\frac{\pi}{4}\right)=\frac1{\sqrt2}$$.

Hence

$$x-\frac{\pi}{4}=\frac{\pi}{4}+2k\pi\;\Longrightarrow\;x=\frac{\pi}{2}+2k\pi$$
$$x-\frac{\pi}{4}=\frac{3\pi}{4}+2k\pi\;\Longrightarrow\;x=\pi+2k\pi$$

Check inside $$-\,\frac{7\pi}{4}(-315^\circ)\;\text{to}\;\frac{7\pi}{4}(315^\circ):$$

k = -1 : $$x=-\frac{3\pi}{2},\;x=-\pi$$ (both inside)
k = 0 : $$x=\frac{\pi}{2},\;x=\pi$$ (inside)
k = 1 or k ≤ -2 give magnitudes >$$\frac{7\pi}{4}$$, rejected.

Total solutions = $$4$$ → code $$R$$.

Summary of matches
(I) → P  (II) → P  (III) → T  (IV) → R

Option B states exactly this mapping, so

Correct option: Option B which is: (I) → (P); (II) → (P); (III) → (T); (IV) → (R)