JEE Main 2026 — Continuity and Differentiability Question with Solution

Since $f(x)$ is continuous at $x=\pi /2$, the right-hand limit must equal the value of the function at $x=\pi /2$.

${\lim }_{x\to \pi /2^{+}}f(x)=f(\pi /2)$

${\lim }_{x\to \pi /2^{+}}\frac{b(1-\sin x)}{(\pi -2x{)}^2}=\frac{1}{3}$

Let $x=\pi /2+h$. As $x\to \pi /2^{+}$, $h\to 0^{+}$.

${\lim }_{h\to 0^{+}}\frac{b(1-\sin (\pi /2+h))}{(\pi -2(\pi /2+h){)}^2}=\frac{1}{3}$

${\lim }_{h\to 0^{+}}\frac{b(1-\cos h)}{(-2h{)}^2}=\frac{1}{3}$

${\lim }_{h\to 0^{+}}\frac{b(1-\cos h)}{4h^2}=\frac{1}{3}$

Using the standard limit ${\lim }_{h\to 0}\frac{1-\cos h}{h^2}=\frac{1}{2}$, we get:

$\frac{b}{4}\times \frac{1}{2}=\frac{1}{3}\Rightarrow \frac{b}{8}=\frac{1}{3}\Rightarrow b=\frac{8}{3}$

The upper limit of the integral is $3b-6=3\left(\frac{8}{3}\right)-6=2$.

The integral to evaluate is $I={\int }_0^2|x^2+2x-3|dx$.

Factoring the quadratic expression gives $x^2+2x-3=(x+3)(x-1)$.

For $x\in [0,1]$, $(x+3)(x-1)\le 0$, so $|x^2+2x-3|=3-2x-x^2$.

For $x\in [1,2]$, $(x+3)(x-1)\ge 0$, so $|x^2+2x-3|=x^2+2x-3$.

Splitting the integral at $x=1$:

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

Evaluating the first integral:

${\int }_0^1(3-2x-x^2)dx={\left[3x-x^2-\frac{x^3}{3}\right]}_0^1=3-1-\frac{1}{3}=\frac{5}{3}$

Evaluating the second integral:

${\int }_1^2(x^2+2x-3)dx={\left[\frac{x^3}{3}+x^2-3x\right]}_1^2$

$=\left(\frac{8}{3}+4-6\right)-\left(\frac{1}{3}+1-3\right)=\left(\frac{8}{3}-2\right)-\left(\frac{1}{3}-2\right)=\frac{7}{3}$

Adding the two parts:

$I=\frac{5}{3}+\frac{7}{3}=\frac{12}{3}=4$

Answer: $4$