JEE Main 2026 · 2 Apr · Shift 1, Mathematics question 12

Range. $\sin^2x+\sin x\cos x=\frac12+\frac1{\sqrt2}\sin\left(2x-\frac\pi4\right)\in[-0.21,\,1.21]$, so $a=0$ or $1$.

$a=0$: $\sin x=0$ ($x=-\pi,0,\pi$) or $\tan x=-1$ ($x=-\frac\pi4,\frac{3\pi}4$): 5 values.

$a=1$: $\sin x\cos x=\cos^2x$, so $\cos x=0$ ($x=\pm\frac\pi2$) or $\tan x=1$ ($x=\frac\pi4,-\frac{3\pi}4$): 4 values.

$n(S)=9$.

(D)