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JEE Main 2026 · 4 Apr · Shift 1, Mathematics question 19

Let ∫−22(∣sin⁡x∣+[xsin⁡x])dx=2(3−cos⁡2)+β\displaystyle\int_{-2}^2\left(|\sin x|+[x\sin x]\right)dx=2(3-\cos2)+\beta, where [⋅][\cdot] is the greatest integer function. Then βsin⁡(β2)\beta\sin\left(\dfrac\beta2\right) equals:

Choose one option
Answer and worked solution

Official answer: B

First part. ∫−22∣sin⁡x∣ dx=2(1−cos⁡2)\displaystyle\int_{-2}^2|\sin x|\,dx=2(1-\cos2).

Second part. xsin⁡xx\sin x is even and lies in [0, 2sin⁡2≈1.82][0,\,2\sin2\approx1.82] on [0,2][0,2]. Its integer part is 1 for x>x1x>x_1, where x1sin⁡x1=1x_1\sin x_1=1, so this integral is 2(2−x1)2(2-x_1).

Compare. The total is 6−2cos⁡2−2x16-2\cos2-2x_1, so β=−2x1\beta=-2x_1.

Value. βsin⁡β2=2x1sin⁡x1=2\beta\sin\frac\beta2=2x_1\sin x_1=2.

(B)

Topics:Properties of definite integralsGraphs of functions
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