JEE Main 2026 (Online) 4th April Evening ShiftQuestion 1
Let $f$ be a twice differentiable function such that $f(x)=\int_{0}^{x}\tan(t-x)\,dt-\int_{0}^{x}f(t)\tan t\,dt$, $x\in\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$. Then $f''\left(\frac{\pi}{6}\right)+12f'\left(-\frac{\pi}{6}\right)+f\left(\frac{\pi}{6}\right)$ is equal to $\_\_\_\_\_\_$ .
Differential Equation
Answer$5$
Solution
Concept. Evaluate the first integral, differentiate, and solve the linear equation with IF $\sec x$. Formulas used. (1) $\int\tan u\,du=-\ln\cos u$(2) $\frac{d}{dx}\int_0^xg(t)\,dt=g(x)$Step 1 - First integral. Put $u=t-x$: $$\int_0^x\tan(t-x)\,dt=\int_{-x}^{0}\tan u\,du=\ln\cos x.$$ So $f(x)=\ln\cos x-\int_0^xf(t)\tan t\,dt$, and $f(0)=0$. Step 2 - Differentiate. By (2), $f'=-\tan x-f\tan x$, i.e. $f'+f\tan x=-\tan x$. Step 3 - Solve. With IF $\sec x$: $(f\sec x)'=-\sec x\tan x$, so $f\sec x=-\sec x+C$. $f(0)=0$ gives $C=1$: $$f(x)=\cos x-1,\quad f'(x)=-\sin x,\quad f''(x)=-\cos x.$$ Step 4 - Value. $$f''\left(\tfrac\pi6\right)+12f'\left(-\tfrac\pi6\right)+f\left(\tfrac\pi6\right)=-\frac{\sqrt3}{2}+6+\frac{\sqrt3}{2}-1=5.$$ Answer: 5