JEE Advanced 2022 · Paper 1, Mathematics question 15

(I) $\sin(x+\frac\pi4)=\frac1{\sqrt2}$ gives $x=2k\pi$ or $\frac\pi2+2k\pi$. In $[-\frac{2\pi}3,\frac{2\pi}3]$: $x=0,\frac\pi2$. Two elements → P.

(II) $\tan3x=\frac1{\sqrt3}$ gives $x=\frac\pi{18}+\frac{k\pi}3$. In $[-\frac{5\pi}{18},\frac{5\pi}{18}]$: $\frac\pi{18},-\frac{5\pi}{18}$. Two elements → P.

(III) $\cos2x=\frac{\sqrt3}2$ gives $x=\pm\frac\pi{12}+k\pi$. In $[-\frac{6\pi}5,\frac{6\pi}5]$: $\pm\frac\pi{12},\pm\frac{11\pi}{12},\pm\frac{13\pi}{12}$. Six elements → T.

(IV) $\sin(x-\frac\pi4)=\frac1{\sqrt2}$ gives $x=\frac\pi2+2k\pi$ or $\pi+2k\pi$. In $[-\frac{7\pi}4,\frac{7\pi}4]$: $\frac\pi2,-\frac{3\pi}2,\pi,-\pi$. Four elements → R.

(I)→P, (II)→P, (III)→T, (IV)→R.