Question

If \[ f(x)=\int_{0}^{x}\tan(t-x)\,dt+\int_{0}^{x}f(t)\tan t\,dt \] then the value of \[ f''\left(\frac{\pi}{6}\right)-12f'\left(-\frac{\pi}{6}\right)+f\left(\frac{\pi}{6}\right) \] is

• A. \(-7-\frac{16}{3\sqrt3}\)
• B. \(7+\frac{5}{3\sqrt3}\)
• C. \(7-\frac{16}{3\sqrt3}\)
• D. \(\frac{1}{3\sqrt3}\)

Hint:

Integral equations often reduce to differential equations after differentiation.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
To find the derivatives of \( f(x) \), we apply the Leibniz Rule for differentiation under the integral sign. Note that in the first integral, the variable \( x \) appears inside the integrand.
Step 2: Key Formula or Approach:
Differentiating \( f(x) \) using Leibniz Rule: \[ f'(x) = \tan(x-x) + \int_0^x \frac{\partial}{\partial x} \tan(t-x) \, dt + f(x) \tan x \] \[ f'(x) = \int_0^x -\sec^2(t-x) \, dt + f(x) \tan x \]
Step 3: Detailed Explanation:
1. Evaluate the integral in \( f'(x) \): \( \int_0^x -\sec^2(t-x) \, dt = [-\tan(t-x)]_0^x = -(0 - \tan(-x)) = -\tan x \). So, \( f'(x) = -\tan x + f(x) \tan x = \tan x (f(x) - 1) \). 2. Differentiate again for \( f''(x) \): \( f''(x) = \sec^2 x (f(x) - 1) + \tan x f'(x) \). 3. At \( x = 0, f(0) = 0 \). Using \( f'(x) \), \( f'(0) = 0 \). 4. Solve the linear differential equation \( \frac{df}{f-1} = \tan x \, dx \): \( \ln|f-1| = \ln|\sec x| + C \). Since \( f(0)=0, C=0 \). \( f(x) - 1 = -\sec x \implies f(x) = 1 - \sec x \). 5. Substitute \( f(x), f'(x), f''(x) \) into the required expression and evaluate at \( \pm \pi/6 \).
Step 4: Final Answer:
The result of the expression is \( 7 - \frac{16}{3\sqrt{3}} \).