JEE Main 2026 — Mathematics Calculus

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

$$\lim_{x \to \pi/2^+} f(x) = f(\pi/2)$$x→π/2+lim​f(x)=f(π/2)

$$\lim_{x \to \pi/2^+} \dfrac{b(1-\sin x)}{(\pi-2x)^2} = \dfrac{1}{3}$$x→π/2+lim​(π−2x)2b(1−sinx)​=31​

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

$$\lim_{h \to 0^+} \dfrac{b(1-\sin(\pi/2+h))}{(\pi-2(\pi/2+h))^2} = \dfrac{1}{3}$$h→0+lim​(π−2(π/2+h))2b(1−sin(π/2+h))​=31​

$$\lim_{h \to 0^+} \dfrac{b(1-\cos h)}{(-2h)^2} = \dfrac{1}{3}$$h→0+lim​(−2h)2b(1−cosh)​=31​

$$\lim_{h \to 0^+} \dfrac{b(1-\cos h)}{4h^2} = \dfrac{1}{3}$$h→0+lim​4h2b(1−cosh)​=31​

Using the standard limit $\displaystyle \lim_{h \to 0} \dfrac{1-\cos h}{h^2} = \dfrac{1}{2}$h→0lim​h21−cosh​=21​, we get:

$$\dfrac{b}{4} \times \dfrac{1}{2} = \dfrac{1}{3} \Rightarrow \dfrac{b}{8} = \dfrac{1}{3} \Rightarrow b = \dfrac{8}{3}$$4b​×21​=31​⇒8b​=31​⇒b=38​

The upper limit of the integral is $3b - 6 = 3\left(\dfrac{8}{3}\right) - 6 = 2$3b−6=3(38​)−6=2.

The integral to evaluate is $\displaystyle I = \displaystyle\int_{0}^{2} |x^2+2x-3|\,dx$I=∫02​∣x2+2x−3∣dx.

Factoring the quadratic expression gives $x^2+2x-3 = (x+3)(x-1)$x2+2x−3=(x+3)(x−1).

For $x \in [0, 1]$x∈[0,1], $(x+3)(x-1) \leq 0$(x+3)(x−1)≤0, so $|x^2+2x-3| = 3 - 2x - x^2$∣x2+2x−3∣=3−2x−x2.

For $x \in [1, 2]$x∈[1,2], $(x+3)(x-1) \geq 0$(x+3)(x−1)≥0, so $|x^2+2x-3| = x^2+2x-3$∣x2+2x−3∣=x2+2x−3.

Splitting the integral at $x = 1$x=1:

$$I = \displaystyle\int_{0}^{1} (3 - 2x - x^2)\,dx + \displaystyle\int_{1}^{2} (x^2+2x-3)\,dx$$I=∫01​(3−2x−x2)dx+∫12​(x2+2x−3)dx

Evaluating the first integral:

$$\displaystyle\int_{0}^{1} (3 - 2x - x^2)\,dx = \left[ 3x - x^2 - \dfrac{x^3}{3} \right]_0^1 = 3 - 1 - \dfrac{1}{3} = \dfrac{5}{3}$$∫01​(3−2x−x2)dx=[3x−x2−3x3​]01​=3−1−31​=35​

Evaluating the second integral:

$$\displaystyle\int_{1}^{2} (x^2+2x-3)\,dx = \left[ \dfrac{x^3}{3} + x^2 - 3x \right]_1^2$$∫12​(x2+2x−3)dx=[3x3​+x2−3x]12​

$$= \left( \dfrac{8}{3} + 4 - 6 \right) - \left( \dfrac{1}{3} + 1 - 3 \right) = \left( \dfrac{8}{3} - 2 \right) - \left( \dfrac{1}{3} - 2 \right) = \dfrac{7}{3}$$=(38​+4−6)−(31​+1−3)=(38​−2)−(31​−2)=37​

Adding the two parts:

$$I = \dfrac{5}{3} + \dfrac{7}{3} = \dfrac{12}{3} = 4$$I=35​+37​=312​=4

Answer: $4$4