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JEE Main 2026 · 4 Apr · Shift 2, Mathematics question 25

Let ff be a twice differentiable function such that f(x)=∫0xtan⁡(t−x) dt−∫0xf(t)tan⁡t dtf(x)=\displaystyle\int_0^x\tan(t-x)\,dt-\int_0^xf(t)\tan t\,dt, x∈(−π2,π2)x\in\left(-\dfrac\pi2,\dfrac\pi2\right). Then f′′(π6)+12f′(−π6)+f(π6)f''\left(\dfrac\pi6\right)+12f'\left(-\dfrac\pi6\right)+f\left(\dfrac\pi6\right) is equal to ________.

Answer and worked solution

Official answer: 5

First integral. ∫0xtan⁡(t−x) dt=ln⁡cos⁡x\displaystyle\int_0^x\tan(t-x)\,dt=\ln\cos x.

Differential equation. Differentiating, f′=−tan⁡x(1+f)f'=-\tan x(1+f), so 1+f=Ccos⁡x1+f=C\cos x; f(0)=0f(0)=0 gives f=cos⁡x−1f=\cos x-1.

Value.
f′′(π6)+12f′(−π6)+f(π6)=−32+12⋅12+(32−1)=5.f''\left(\tfrac\pi6\right)+12f'\left(-\tfrac\pi6\right)+f\left(\tfrac\pi6\right)=-\tfrac{\sqrt3}2+12\cdot\tfrac12+\left(\tfrac{\sqrt3}2-1\right)=5.

Answer: 5

Topics:Leibniz ruleVariable separable
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