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

Concept: Differentiate the functional integral equation and convert it into a differential equation.
Step 1: Simplify the first integral \[ f(x)=\int_{0}^{x}\tan(t-x)\,dt+\int_{0}^{x}f(t)\tan t\,dt \] Let \(u=t-x\). Then \[ \int_{0}^{x}\tan(t-x)dt = \int_{-x}^{0}\tan u\,du \] \[ =\ln|\sec u|\,\Big|_{-x}^{0} \] \[ =-\ln(\sec x) \] Thus \[ f(x)=-\ln(\sec x)+\int_{0}^{x}f(t)\tan t\,dt \]
Step 2: Differentiate \[ f'(x)=-\tan x+f(x)\tan x \] \[ f'(x)=\tan x(f(x)-1) \]
Step 3: Form differential equation \[ \frac{df}{dx}=(f-1)\tan x \] \[ \frac{df}{f-1}=\tan x\,dx \] \[ \ln|f-1|=-\ln(\cos x)+C \] \[ f-1=C\sec x \]
Step 4: Find constant Given \(f(0)=0\) \[ 0-1=C\sec0 \] \[ C=-1 \] Thus \[ f(x)=1-\sec x \]
Step 5: Derivatives \[ f'(x)=-\sec x\tan x \] \[ f''(x)=-(\sec x\tan^2x+\sec^3x) \]
Step 6: Evaluate \[ f\left(\frac{\pi}{6}\right)=1-\frac{2}{\sqrt3} \] \[ f'\left(-\frac{\pi}{6}\right)=\frac{2}{3} \] \[ f''\left(\frac{\pi}{6}\right)=-\frac{10}{3\sqrt3} \]
Step 7: Substitute \[ f''\left(\frac{\pi}{6}\right)-12f'\left(-\frac{\pi}{6}\right)+f\left(\frac{\pi}{6}\right) \] \[ =-\frac{10}{3\sqrt3}-12\left(\frac23\right)+1-\frac{2}{\sqrt3} \] \[ =-7-\frac{16}{3\sqrt3} \]