Question

Let $f$ be a twice differentiable function such that $f(x) = \int_0^x \tan(t-x) dt - \int_0^x f(t) \tan t dt, x \in (-\frac{\pi}{2}, \frac{\pi}{2})$. Then $f''(\frac{\pi}{6}) + 12 f'(-\frac{\pi}{6}) + f(\frac{\pi}{6})$ is equal to ________.

Answer 1

Correct Answer: 5

Given \[ f(x)=\int_0^x \tan(t-x)\,dt-\int_0^x f(t)\tan t\,dt \] Differentiate both sides: \[ f'(x)= -\tan x-f(x)\tan x \] \[ f'(x)=-(1+f(x))\tan x \] This is linear differential equation: \[ f'(x)+f(x)\tan x=-\tan x \] Integrating factor: \[ I.F.=e^{\int \tan x\,dx}=\sec x \] Multiply throughout: \[ \frac{d}{dx}(f(x)\sec x)=-\tan x\sec x \] Integrate: \[ f(x)\sec x=-\sec x+C \] \[ f(x)=-1+C\cos x \] At $x=0$ \[ f(0)=0 \] \[ 0=-1+C \] \[ C=1 \] So \[ f(x)=\cos x-1 \] Now \[ f'(x)=-\sin x \] \[ f''(x)=-\cos x \] Substitute values: \[ f''\left(\frac{\pi}{6}\right)=-\frac{\sqrt{3}}{2} \] \[ f'\left(-\frac{\pi}{6}\right)=\frac{1}{2} \] \[ f\left(\frac{\pi}{6}\right)=\frac{\sqrt{3}}{2}-1 \] Hence \[ f''\left(\frac{\pi}{6}\right)+12f'\left(-\frac{\pi}{6}\right)+f\left(\frac{\pi}{6}\right) \] \[ = -\frac{\sqrt{3}}{2}+12\cdot\frac{1}{2}+\frac{\sqrt{3}}{2}-1 \] \[ = -\frac{\sqrt{3}}{2}+6+\frac{\sqrt{3}}{2}-1 \] \[ =5 \] Final Answer: \[ \boxed{5} \]