Let f be a twice differentiable function such that [f(x) = _0^x (t-x),dt - _0^x f(t) t,dt,…

Question

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

Answer: 5

Accepted Answer

Simplify $\int_0^x \tan(t-x)\,dt$. Substitute $u=t-x$: when $t=0$, $u=-x$; when $t=x$, $u=0$.
$$\int_0^x\tan(t-x)\,dt = \int_{-x}^0 \tan u\,du = [-\ln(\cos u)]_{-x}^0 = -\ln(\cos 0)+\ln(\cos(-x)) = \ln(\cos x).$$

So the equation becomes
$$f(x) + \int_0^x f(t)\tan t\,dt = \ln(\cos x).$$

Differentiate both sides w.r.t. $x$:
$$f'(x) + f(x)\tan x = -\tan x \implies f'(x) = -(f(x)+1)\tan x.$$

At $x=0$: from the original equation $f(0)=0$.

Let $y(x)=f(x)+1$. Then $y'+y\tan x=0 \implies \dfrac{y'}{y}=-\tan x \implies \ln|y|=\ln(\cos x)+C \implies y=C\cos x$.

Using $f(0)=0 \Rightarrow y(0)=1=C\cos 0 \Rightarrow C=1$. So
$$f(x)+1=\cos x \implies f(x)=\cos x - 1.$$

Then $f'(x)=-\sin x$, $f''(x)=-\cos x$.
$$f''\!\left(\frac{\pi}{6}\right) = -\frac{\sqrt3}{2},\qquad f'\!\left(-\frac{\pi}{6}\right)=\frac12 \implies 12f'\!\left(-\frac{\pi}{6}\right)=6,$$
$$f\!\left(\frac{\pi}{6}\right)=\frac{\sqrt3}{2}-1.$$

Sum: $-\frac{\sqrt3}{2}+6+\frac{\sqrt3}{2}-1 = 5$.

Solutions · 1