JEE Main 2026 (Online) 8th April Evening ShiftQuestion 1
Let $f(x)=\begin{cases}\frac{1}{3}, & x\le\frac{\pi}{2}\\ \frac{b(1-\sin x)}{(\pi-2x)^{2}}, & x>\frac{\pi}{2}\end{cases}$. If $f$ is continuous at $x=\pi/2$, then the value of $\int_{0}^{3b-6}\left|x^{2}+2x-3\right|dx$ is :
Limits & Continuity
Solution
Put $x=\frac\pi2+h$: $\frac{b(1-\cos h)}{4h^2}\to\frac b8=\frac13\Rightarrow b=\frac83$, so $3b-6=2$. $x^2+2x-3=(x+3)(x-1)<0$ on $(0,1)$, $>0$ on $(1,2)$. $\int_0^1(3-2x-x^2)dx=3-1-\frac13=\frac53$; $\int_1^2(x^2+2x-3)dx=\frac73$. Sum $=4$: option D.