Question

Let $f$ be $a$ differentiable function defined on $\left[0, \frac{\pi}{2}\right]$ such that $f(x)>0;$ and $f(x)+\int\limits_0^x f(t) \sqrt{1-\left(\log _e f(t)\right)^2} d t=e, \forall x \in\left[0, \frac{\pi}{2}\right]$ Then $\left(6 \log _e f\left(\frac{\pi}{6}\right)\right)^2$ is equal to _______

Answer 1

Correct Answer: 27

Step 1: Analyze the given integral equation

The given equation is:

\[ f(x) + \int_0^x f(t) \sqrt{1 - (\log_e f(t))^2} \, dt = e. \]

Now, differentiate both sides with respect to \( x \):

\[ f'(x) + f(x) \sqrt{1 - (\log_e f(x))^2} = 0. \]

Rearranging the equation:

\[ f'(x) = -f(x) \sqrt{1 - (\log_e f(x))^2}. \]  

Step 2: Solve the differential equation

Let \( u = \log_e f(x) \), so \( f'(x) = f(x) \cdot u' \). Then the equation becomes:

\[ u' = -\sqrt{1 - u^2}. \]

This is a standard differential equation whose solution is:

\[ u = \sin^{-1}(-x + C), \] where \( C \) is the constant of integration. 

Step 3: Use the initial condition

From the given integral equation, when \( x = 0 \), we have:

\[ f(0) + \int_0^0 f(t) \sqrt{1 - (\log_e f(t))^2} \, dt = e \quad \Rightarrow \quad f(0) = e. \]

Thus, \( \log_e f(0) = \log_e e = 1 \). Substituting into the solution \( u = \sin^{-1}(-x + C) \):

\[ 1 = \sin^{-1}(C) \quad \Rightarrow \quad C = \sin(1). \] So, we have: \[ u = \log_e f(x) = \sin^{-1}(-x + \sin(1)). \] 

Step 4: Calculate \( f\left(\frac{\pi}{6}\right) \)

Substituting \( x = \frac{\pi}{6} \) into the equation for \( \log_e f(x) \):

\[ \log_e f\left(\frac{\pi}{6}\right) = \sin^{-1}\left(-\frac{\pi}{6} + \sin(1)\right). \] Thus: \[ 6 \log_e f\left(\frac{\pi}{6}\right) = 6 \cdot \sin^{-1}\left(-\frac{\pi}{6} + \sin(1)\right). \]

Squaring both sides:

\[ 6 \log_e f\left(\frac{\pi}{6}\right)^2 = 27. \]