JEE Main 2026 (Online) 23rd January Morning ShiftQuestion 1
Let the area of the region bounded by the curve $y=\max\{\sin x,\cos x\}$, $x=0$, $x=\frac{3\pi}{2}$, and the $x$-axis be $A$. Then $A+A^2$ is equal to $\_\_\_\_\_\_$ .
Application of Integrals
Answer$12$
Solution
Areas: $\int_0^{\pi/4}\cos x=\tfrac{\sqrt2}2$; $\int_{\pi/4}^{\pi}\sin x=1+\tfrac{\sqrt2}2$; on $(\pi,\tfrac{5\pi}4)$ ($\sin$ larger) and $(\tfrac{5\pi}4,\tfrac{3\pi}2)$ ($\cos$ larger) each region has area $1-\tfrac{\sqrt2}2$. $A=3$, $A+A^2=12$.