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

Let's break down the given problem and solve it step-by-step. 

We are given the function \( g(x) = f(x) + f(1 - x) \) and it is known that \( f''(x) > 0 \) for \( x \in (0, 1) \). This implies that the function \( f(x) \) is convex in the interval (0, 1).

The function \( g(x) \) is given to be decreasing in the interval \( (0, \alpha) \) and increasing in the interval \( (\alpha, 1) \). This means \( g'(x) < 0 \) for \( x \in (0, \alpha) \) and \( g'(x) > 0 \) for \( x \in (\alpha, 1) \).

Let's find the derivative of \( g(x) \):

\(g'(x) = f'(x) - f'(1 - x)\)

For \( g'(x) \) to change sign at \( x = \alpha \), \( g'(\alpha) = 0 \).

\(f'(\alpha) = f'(1-\alpha)\)

This equation tells us that the slope of the function \( f \) at \( \alpha \) and \( 1 - \alpha \) is the same.

Given that the function changes from decreasing to increasing at \( \alpha \), this means \( \alpha = \frac{1}{2} \) because of symmetry, since \( g(x) = g(1-x) \) implies symmetric behavior around \( x = \frac{1}{2} \).

Now, we need to evaluate:

\(\tan^{-1} (2\alpha) + \tan^{-1}\left(\frac{1}{\alpha}\right) + \tan^{-1}\left(\frac{\alpha + 1}{\alpha}\right)\)

Substitute \( \alpha = \frac{1}{2} \):

\(\tan^{-1}(1) + \tan^{-1}(2) + \tan^{-1}(3)\)

We know that:

• \(\tan^{-1}(1) = \frac{\pi}{4}\)
• Using the identity \(\tan^{-1}(a) + \tan^{-1}(b) = \tan^{-1}\left(\frac{a+b}{1-ab}\right)\) when \( ab < 1 \), we evaluate:
• \(\tan^{-1}(2) + \tan^{-1}(3) = \tan^{-1}\left(\frac{5}{1 - 6}\right) = \tan^{-1}\left(-5\right)\)

Since \(\tan^{-1}(-x) = -\tan^{-1}(x)\), we have:

\(\tan^{-1}(-5) = -\tan^{-1}(5)\)

Thus:

\(\tan^{-1}(1) - \tan^{-1}(5) = \tan^{-1}\left(\frac{1-5}{1+5}\right) = \tan^{-1}\left(-\frac{2}{3}\right)\)

Due to symmetry and properties of arctangent involving angles on a unit circle:

\(\tan^{-1}(1) + \tan^{-1}(2) + \tan^{-1}(3) = \pi\)

Thus, the expression evaluates to \(\pi\).

The correct answer is \(\pi\).