JEE Main · 2026 · 23 Jan 2026 · Shift 1 · Area Under Curves

On $[0,\pi /4]$ the larger of $\sin x,\cos x$ is $\cos x$; the two are equal at $x=\pi /4$ and again at $x=5\pi /4$, so $\sin x$ is the larger on $[\pi /4,5\pi /4]$ and $\cos x$ on $[5\pi /4,3\pi /2]$. Because the area is bounded by the $x$-axis and the curve dips below it on $(\pi ,3\pi /2)$, each piece is taken in absolute value: These four integrals equal $\frac{1}{\sqrt{2}}$, $1+\frac{1}{\sqrt{2}}$, $1-\frac{1}{\sqrt{2}}$ and $1-\frac{1}{\sqrt{2}}$, whose sum is $A=3$. Hence $A+A^2=3+9=12.$