Method of DifferentiationmediumPYQ · JEE Main · 4 Apr 2026 · Shift 2 (Afternoon)Free

Integral Functional Equation: Combined Value = 5 | JEE Main 2026

JEE Maths question with a full step-by-step solution.

Question
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^{x}f(t)\tan t\,dt, x∈(−π2,π2)x\in\left(-\dfrac{\pi}{2},\dfrac{\pi}{2}\right). Then f′′(π6)+12f′(−π6)+f(π6)f''\left(\dfrac{\pi}{6}\right)+12f'\left(-\dfrac{\pi}{6}\right)+f\left(\dfrac{\pi}{6}\right) is equal to
Solution
Answer: 5 (± 0.01)
Step 1: Differentiate f(x)=∫0xtan⁡(t−x) dt−∫0xf(t)tan⁡t dtf(x)=\int_0^x\tan(t-x)\,dt-\int_0^x f(t)\tan t\,dt: boundary term of first integral =tan⁡(x−x)=0=\tan(x-x)=0, second gives f(x)tan⁡xf(x)\tan x. ∴\therefore
f′(x)=−tan⁡x−f(x)tan⁡x=−(1+f(x))tan⁡x.f'(x)=-\tan x-f(x)\tan x=-(1+f(x))\tan x.
 Step 2: y=f(x)y=f(x), separate:
dyy+1=−tan⁡x dx⇒ln⁡∣y+1∣=ln⁡∣cos⁡x∣+const⇒y+1=ccos⁡x.\frac{dy}{y+1}=-\tan x\,dx\Rightarrow \ln|y+1|=\ln|\cos x|+\text{const}\Rightarrow y+1=c\cos x.
 Step 3: f(0)=0f(0)=0: 0+1=c⇒c=10+1=c\Rightarrow c=1. ∴\therefore
f(x)=cos⁡x−1,f′(x)=−sin⁡x,f′′(x)=−cos⁡x.f(x)=\cos x-1,\qquad f'(x)=-\sin x,\qquad f''(x)=-\cos x.
 Step 4:
f ⁣(π6)=32−1,f′ ⁣(−π6)=sin⁡π6=12,f′′ ⁣(π6)=−32.f\!\left(\frac\pi6\right)=\frac{\sqrt3}{2}-1,\quad f'\!\left(-\frac\pi6\right)=\sin\frac\pi6=\frac12,\quad f''\!\left(\frac\pi6\right)=-\frac{\sqrt3}{2}.
 Step 5:
f′′ ⁣(π6)+12f′ ⁣(−π6)+f ⁣(π6)=−32+6+32−1=5.f''\!\left(\frac\pi6\right)+12f'\!\left(-\frac\pi6\right)+f\!\left(\frac\pi6\right)=-\frac{\sqrt3}{2}+6+\frac{\sqrt3}{2}-1=5.
 Correct answer: 5
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