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

Step 1: Understanding the Concept:
We have an integral equation defining $f(x)$. We will apply the Leibniz Rule to differentiate under the integral sign to convert it into a solvable ordinary differential equation.

Step 2: Key Formula or Approach:
Leibniz Rule for differentiation of integrals:
$\frac{d}{dx} \int_{a(x)}^{b(x)} g(x, t) dt = g(x, b(x)) \cdot b'(x) - g(x, a(x)) \cdot a'(x) + \int_{a(x)}^{b(x)} \frac{\partial g}{\partial x}(x, t) dt$.

Step 3: Detailed Explanation:
Given $f(x) = \int_0^x \tan(t-x) dt - \int_0^x f(t) \tan t dt$.
Differentiate with respect to $x$:
$f'(x) = \left[ \tan(x-x) \cdot 1 - \tan(0-x) \cdot 0 + \int_0^x \frac{\partial}{\partial x}(\tan(t-x)) dt \right] - f(x) \tan x \cdot 1$
$f'(x) = [0 + \int_0^x -\sec^2(t-x) dt] - f(x) \tan x$
Evaluate the remaining integral:
$\int_0^x -\sec^2(t-x) dt = [-\tan(t-x)]_{t=0}^{t=x} = -\tan(x-x) - (-\tan(0-x)) = 0 + \tan(-x) = -\tan x$.
Thus, the equation simplifies to:
$f'(x) = -\tan x - f(x) \tan x = -(1 + f(x)) \tan x$.
This is a separable and linear differential equation. Rearrange it:
$f'(x) + f(x) \tan x = -\tan x$.
The integrating factor is $\text{I.F.} = e^{\int \tan x dx} = e^{\ln|\sec x|} = \sec x$.
Multiply through by $\sec x$:
$\frac{d}{dx}(f(x) \sec x) = -\tan x \sec x$.
Integrate both sides:
$f(x) \sec x = -\sec x + C \implies f(x) = -1 + C \cos x$.
To find $C$, use the original integral equation at $x=0$:
$f(0) = \int_0^0 \dots dt - \int_0^0 \dots dt = 0$.
Substitute into our solution:
$0 = -1 + C \cos(0) \implies C = 1$.
So, $f(x) = \cos x - 1$.
Now compute the required terms:
$f'(x) = -\sin x$
$f''(x) = -\cos x$
Calculate at the requested points:
$f\left(\frac{\pi}{6}\right) = \cos\left(\frac{\pi}{6}\right) - 1 = \frac{\sqrt{3}}{2} - 1$
$f'\left(-\frac{\pi}{6}\right) = -\sin\left(-\frac{\pi}{6}\right) = \frac{1}{2}$
$f''\left(\frac{\pi}{6}\right) = -\cos\left(\frac{\pi}{6}\right) = -\frac{\sqrt{3}}{2}$
Now sum them up:
$f''\left(\frac{\pi}{6}\right) + 12 f'\left(-\frac{\pi}{6}\right) + f\left(\frac{\pi}{6}\right) = -\frac{\sqrt{3}}{2} + 12\left(\frac{1}{2}\right) + \left(\frac{\sqrt{3}}{2} - 1\right)$
$= -\frac{\sqrt{3}}{2} + 6 + \frac{\sqrt{3}}{2} - 1 = 5$.

Step 4: Final Answer:
The value of the expression is 5.