JEE Main 2026 Apr 8 Shift 2, Mathematics Q19: Continuity of Function

Find $b$: Write $1-\sin x=1-\cos \left(\frac{\pi }{2}-x\right)=2{\sin }^2\left(\frac{\pi }{4}-\frac{x}{2}\right)$.

Let $h=x-\frac{\pi }{2}$, so $\pi -2x=-2h$ and as $x\to \pi /2^{+}$ we get $h\to 0^{+}$. Also $\frac{\pi }{4}-\frac{x}{2}=-\frac{h}{2}$.

$\underset{x\to \pi /2^{+}}{\lim }\frac{b(1-\sin x)}{(\pi -2x{)}^2}=\underset{h\to 0}{\lim }\frac{2b{\sin }^2(h/2)}{4h^2}=\frac{2b}{4}\cdot \frac{1}{4}=\frac{b}{8}$

Continuity: $\frac{b}{8}=\frac{1}{3}⟹b=\frac{8}{3}$.

Upper limit: $3b-6=8-6=2$.

Evaluate the integral: $x^2+2x-3=(x+3)(x-1)$, which is negative on $(0,1)$ and positive on $(1,2)$.

${\int }_0^2|(x+3)(x-1)|dx=-{\int }_0^1(x^2+2x-3)dx+{\int }_1^2(x^2+2x-3)dx$

$=-{\left[\frac{x^3}{3}+x^2-3x\right]}_0^1+{\left[\frac{x^3}{3}+x^2-3x\right]}_1^2$

$=-\left(\frac{1}{3}+1-3\right)+\left(\frac{8}{3}+4-6-\left(\frac{1}{3}+1-3\right)\right)$

$=-\left(-\frac{5}{3}\right)+\left(\frac{2}{3}+\frac{5}{3}\right)=\frac{5}{3}+\frac{7}{3}=4$.