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JEE MAIN 2026 · Limit, Continuity and Differentiability

JEE Main 2026 Session 2 Question Paper: 8 Apr, Shift 2 (B.E./B.Tech), question 19

Updated
Single correct+4 / -1Answer matches NTA’s final key

Let f(x)={13,x≤π2b(1−sin⁡x)(π−2x)2,x>π2f(x) = \begin{cases} \dfrac13, & x \le \dfrac{\pi}{2} \\[2mm] \dfrac{b(1 - \sin x)}{(\pi - 2x)^2}, & x > \dfrac{\pi}{2} \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:

Choose one

Solution

Solution

Find bb. Put x=π2+hx = \frac{\pi}{2} + h. Then 1−sin⁡x=1−cos⁡h≈h221 - \sin x = 1 - \cos h \approx \frac{h^2}{2} and (π−2x)2=4h2(\pi - 2x)^2 = 4h^2, so the limit is b8\frac{b}{8}.

Continuity: b8=13⇒b=83\frac{b}{8} = \frac13 \Rightarrow b = \frac83, and 3b−6=23b - 6 = 2.

Integrate. x2+2x−3=(x+3)(x−1)x^2 + 2x - 3 = (x + 3)(x - 1) is negative on [0,1)[0, 1) and positive on (1,2](1, 2]:

∫01(3−2x−x2) dx+∫12(x2+2x−3) dx=(3−1−13)+(73+3−3)=53+73=4.\int_0^1 (3 - 2x - x^2)\,dx + \int_1^2 (x^2 + 2x - 3)\,dx = \left(3 - 1 - \tfrac13\right) + \left(\tfrac73 + 3 - 3\right) = \tfrac53 + \tfrac73 = 4.

Answer: 4.

Answer checked against the official answer key. The solution is our own working.

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