JEE Advanced 2026 · Paper 1, Mathematics question 14

(P) With $s=\sin^2x$: $s^3+(1-s)^2=1\implies s(s+2)(s-1)=0$, so $\sin^2x=0$ or $1$. In $[-\pi,\pi]$: $-\pi,0,\pi,\pm\frac\pi2$, i.e. 5 → 5.

(Q) With $c=\cos^2x$: $1-c+c^3=1\implies c=0$ or $1$. In $\left[-\frac\pi2,\frac\pi2\right]$: $\pm\frac\pi2,0$, i.e. 3 → 3.

(R) $\frac{1+\cos x}2-(1-\cos^2x)=\frac12\implies2\cos^2x+\cos x-2=0$, so $\cos x=\frac{\sqrt{17}-1}4\approx0.78$. Two solutions in $[-\pi,\pi]$ → 2.

(S) $3(1-\cos x)-\cos3x=3\implies-3\cos x-(4\cos^3x-3\cos x)=0\implies\cos x=0$. In $[-2\pi,2\pi]$: $\pm\frac\pi2,\pm\frac{3\pi}2$, i.e. 4 → 4.

P→5, Q→3, R→2, S→4.