JEE Main 2026 — Mathematics Calculus

Let $\displaystyle I = \displaystyle\int_{-2}^{2} \left(|\sin x| + [x \sin x]\right) dx = I_1 + I_2$I=∫−22​(∣sinx∣+[xsinx])dx=I1​+I2​, where

$\displaystyle I_1 = \displaystyle\int_{-2}^{2} |\sin x|\, dx$I1​=∫−22​∣sinx∣dx and $\displaystyle I_2 = \displaystyle\int_{-2}^{2} [x \sin x]\, dx$I2​=∫−22​[xsinx]dx.

Evaluating $I_1$I1​:

Since $|\sin x|$∣sinx∣ is even, $\displaystyle I_1 = 2\displaystyle\int_{0}^{2} |\sin x|\, dx$I1​=2∫02​∣sinx∣dx.

For $x \in [0, 2]$x∈[0,2], $2 < \pi$2<π, so $\sin x \geq 0$sinx≥0 and $|\sin x| = \sin x$∣sinx∣=sinx.

$$I_1 = 2\displaystyle\int_{0}^{2} \sin x\, dx = 2\left[-\cos x\right]_0^2 = 2(1 - \cos 2) = 2 - 2\cos 2$$I1​=2∫02​sinxdx=2[−cosx]02​=2(1−cos2)=2−2cos2

Evaluating $I_2$I2​:

Let $g(x) = x\sin x$g(x)=xsinx. Then $g(-x) = (-x)\sin(-x) = x\sin x = g(x)$g(−x)=(−x)sin(−x)=xsinx=g(x), so $g(x)$g(x) is even, and hence $[x\sin x]$[xsinx] is also even.

$$I_2 = 2\displaystyle\int_{0}^{2} [x\sin x]\, dx$$I2​=2∫02​[xsinx]dx

Behaviour of $y = x\sin x$y=xsinx on $[0, 2]$[0,2]:

At $x = 0$x=0, $y = 0$y=0. At $x = 2$x=2, $y = 2\sin 2 \approx 1.818$y=2sin2≈1.818.

$y' = \sin x + x\cos x > 0$y′=sinx+xcosx>0 on $[0, 2]$[0,2], so $y$y is strictly increasing on this interval.

Therefore, $x\sin x$xsinx takes each value in $[0, 1.818]$[0,1.818] exactly once, and crosses $1$1 at some unique $\alpha \in (0, 2)$α∈(0,2) with $\alpha\sin\alpha = 1$αsinα=1.

For $x \in [0, \alpha)$x∈[0,α): $0 \leq x\sin x < 1 \Rightarrow [x\sin x] = 0$0≤xsinx<1⇒[xsinx]=0

For $x \in [\alpha, 2]$x∈[α,2]: $1 \leq x\sin x < 2 \Rightarrow [x\sin x] = 1$1≤xsinx<2⇒[xsinx]=1

$$I_2 = 2\left(\displaystyle\int_{0}^{\alpha} 0\, dx + \displaystyle\int_{\alpha}^{2} 1\, dx\right) = 2(2 - \alpha) = 4 - 2\alpha$$I2​=2(∫0α​0dx+∫α2​1dx)=2(2−α)=4−2α

Combining:

$$I = I_1 + I_2 = (2 - 2\cos 2) + (4 - 2\alpha) = 6 - 2\cos 2 - 2\alpha$$I=I1​+I2​=(2−2cos2)+(4−2α)=6−2cos2−2α

Given $I = 2(3 - \cos 2) + \beta = 6 - 2\cos 2 + \beta$I=2(3−cos2)+β=6−2cos2+β, so:

$$6 - 2\cos 2 - 2\alpha = 6 - 2\cos 2 + \beta \Rightarrow \beta = -2\alpha$$6−2cos2−2α=6−2cos2+β⇒β=−2α

Now:

$$\beta \sin\left(\dfrac{\beta}{2}\right) = -2\alpha \sin\left(\dfrac{-2\alpha}{2}\right) = -2\alpha \sin(-\alpha) = 2\alpha\sin\alpha$$βsin(2β​)=−2αsin(2−2α​)=−2αsin(−α)=2αsinα

Since $\alpha\sin\alpha = 1$αsinα=1:

$$\beta \sin\left(\dfrac{\beta}{2}\right) = 2 \cdot 1 = 2$$βsin(2β​)=2⋅1=2

Hence, the correct option is $(2)\ 2$(2) 2.