JEE Advanced 2022 — Paper 1 — Question 15

Step 1: For (I): Rewrite $\cos x + \sin x = 1$ as $\sqrt{2}\sin\left(x+\frac{\pi}{4}\right)=1$ → $\sin\left(x+\frac{\pi}{4}\right)=\frac{1}{\sqrt{2}}$. Step 2: General solution: $x+\frac{\pi}{4}=n\pi+(-1)^n\frac{\pi}{4}$ → $x = n\pi+(-1)^n\frac{\pi}{4}-\frac{\pi}{4}$. For even $n$: $x = n\pi$; for odd $n$: $x = n\pi-\frac{\pi}{2}$. Step 3: In interval $\left[-\frac{2\pi}{3},\frac{2\pi}{3}\right]$, solutions: $0$ and $\frac{\pi}{2}$. So 2 elements → (I) → P. Step 4: For (II): $\sqrt{3}\tan 3x = 1$ → $\tan 3x = \frac{1}{\sqrt{3}}$ → $3x = n\pi+\frac{\pi}{6}$ → $x = \frac{n\pi}{3}+\frac{\pi}{18}$. Step 5: In interval $\left[-\frac{5\pi}{18},\frac{5\pi}{18}\right]$, solutions: $-\frac{5\pi}{18}$ ($n=-1$) and $\frac{\pi}{18}$ ($n=0$). So 2 elements → (II) → P. Step 6: For (III): $2\cos 2x = \sqrt{3}$ → $\cos 2x = \frac{\sqrt{3}}{2}$ → $2x = 2n\pi \pm \frac{\pi}{6}$ → $x = n\pi \pm \frac{\pi}{12}$. Step 7: In interval $\left[-\frac{6\pi}{5},\frac{6\pi}{5}\right]$, solutions from $n=0,\pm1$ give six values: $\pm\frac{\pi}{12}$, $\pi\pm\frac{\pi}{12}$, $-\pi\pm\frac{\pi}{12}$. So 6 elements → (III) → T. Step 8: For (IV): $\sin x - \cos x = 1$ → $\sqrt{2}\sin\left(x-\frac{\pi}{4}\right)=1$ → $\sin\left(x-\frac{\pi}{4}\right)=\frac{1}{\sqrt{2}}$ → $x-\frac{\pi}{4}=n\pi+(-1)^n\frac{\pi}{4}$ → $x = n\pi+(-1)^n\frac{\pi}{4}+\frac{\pi}{4}$. For even $n$: $x = n\pi+\frac{\pi}{2}$; for odd $n$: $x = n\pi$. In interval $\left[-\frac{7\pi}{4},\frac{7\pi}{4}\right]$, solutions: $-\frac{3\pi}{2}, -\pi, \frac{\pi}{2}, \pi$ → 4 elements → (IV) → R. Hence option B is correct.