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← JEE Main 2026 · 8 Apr · Shift 2
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JEE Main 2026 · 8 Apr · Shift 2, Mathematics question 19

Let f(x)={13,x≤π2b(1−sin⁡x)(π−2x)2,x>π2f(x)=\begin{cases}\dfrac13, & x\le\dfrac\pi2\\[2mm] \dfrac{b(1-\sin x)}{(\pi-2x)^2}, & x>\dfrac\pi2\end{cases}. If ff is continuous at x=π/2x=\pi/2, then the value of ∫03b−6∣x2+2x−3∣dx\displaystyle\int_0^{3b-6}\left|x^2+2x-3\right|dx is :

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Answer and worked solution

Official answer: D

Continuity. Put x=π2+hx=\tfrac\pi2+h: b(1−cos⁡h)4h2→b8=13\dfrac{b(1-\cos h)}{4h^2}\to\dfrac b8=\dfrac13, so b=83b=\tfrac83 and 3b−6=23b-6=2.

Integral. x2+2x−3=(x+3)(x−1)x^2+2x-3=(x+3)(x-1) changes sign at 1:
∫01(3−2x−x2) dx+∫12(x2+2x−3) dx=53+73=4.\int_0^1(3-2x-x^2)\,dx+\int_1^2(x^2+2x-3)\,dx=\frac53+\frac73=4.

(D)

Topics:ContinuityProperties of definite integrals
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