Question

Let g(x) = f(x) + f(1 - x) and f''(x) > 0, x ∈ (0,1). If g is decreasing in the interval (0, α) and increasing in the interval (α, 1), then tan-1 (2α) + tan^-1 (\(\frac{1}{α}\)) + tan^-1\((\frac{α+1}{α})\) is equal to

• A. \(\frac{5\pi}{4}\)
• B. \(\pi\)
• C. \(\frac{3\pi}{4}\)
• D. \(\frac{3\pi}{2}\)

Answer 1

The Correct Option is B

Step 1: Analyze the given function \( g(x) \) 
We are given that \( g(x) = f(x) + f(1 - x) \), and it is stated that \( g(x) \) is decreasing in the interval \( (0, \alpha) \) and increasing in the interval \( (\alpha, 1) \). 

From the conditions given, we know that \( f'(x) = f'(1 - x) \), which implies that the derivative of the function \( g(x) \) with respect to \( x \) is zero at \( x = \frac{1}{2} \). Therefore, \( \alpha = \frac{1}{2} \).

Step 2: Compute the required expression 
Now, we are tasked with finding the value of \( \tan^{-1}(2 \alpha) + \tan^{-1} \left( \frac{\alpha + 1}{\alpha} \right) \). Since \( \alpha = \frac{1}{2} \), we compute the individual terms: \[ \tan^{-1}(2 \alpha) = \tan^{-1}(1) = \frac{\pi}{4} \] \[ \tan^{-1} \left( \frac{\alpha + 1}{\alpha} \right) = \tan^{-1}(3) = \frac{\pi}{2} \] Thus, the sum is: \[ \tan^{-1}(2 \alpha) + \tan^{-1} \left( \frac{\alpha + 1}{\alpha} \right) = \frac{\pi}{4} + \frac{\pi}{2} = \pi \] 

Thus, the correct answer is \( \pi \).