Key Concepts and Formulas

• Derivatives and Monotonicity: If $g'(x) < 0$ on an interval, then $g(x)$ is decreasing on that interval. If $g'(x) > 0$ on an interval, then $g(x)$ is increasing on that interval.
• Second Derivative and Concavity: If $f''(x) > 0$ on an interval, then $f(x)$ is concave up on that interval, and $f'(x)$ is increasing on that interval.
• Inverse Tangent Identity: $\tan^{-1}(x) + \tan^{-1}(y) = \tan^{-1}\left(\frac{x+y}{1-xy}\right)$, provided $xy < 1$.

Step-by-Step Solution

Step 1: Find the derivative of $g(x)$.

We are given $g(x) = f(x) + f(1-x)$. We want to find where $g(x)$ is increasing or decreasing, so we need to analyze its derivative $g'(x)$.

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

Step 2: Analyze the sign of $g'(x)$.

We are given that $g(x)$ is decreasing in $(0, a)$ and increasing in $(\alpha, 1)$. This means:

• $g'(x) < 0$ for $x \in (0, a)$
• $g'(x) > 0$ for $x \in (\alpha, 1)$

Since $g'(x) = f'(x) - f'(1-x)$, we have:

• $f'(x) - f'(1-x) < 0$ for $x \in (0, a)$ which means $f'(x) < f'(1-x)$ for $x \in (0, a)$
• $f'(x) - f'(1-x) > 0$ for $x \in (\alpha, 1)$ which means $f'(x) > f'(1-x)$ for $x \in (\alpha, 1)$

Step 3: Determine the value of $\alpha$.

We are given that $f''(x) > 0$ for $x \in (0, 1)$. This implies that $f'(x)$ is an increasing function on $(0, 1)$.

Since $f'(x)$ is increasing, and $f'(x) < f'(1-x)$ for $x \in (0, a)$, and $f'(x) > f'(1-x)$ for $x \in (\alpha, 1)$, there must be a point where $f'(x) = f'(1-x)$. Since $f'$ is increasing, this can only occur when $x = 1-x$, which means $x = \frac{1}{2}$.

Therefore, $g'(x) < 0$ for $x < \frac{1}{2}$ and $g'(x) > 0$ for $x > \frac{1}{2}$. This implies that $a = \alpha = \frac{1}{2}$.

Step 4: Substitute $\alpha = \frac{1}{2}$ into the given expression.

We are asked to find the value of $\tan^{-1}(2 \alpha) + \tan^{-1}\left(\frac{1}{\alpha}\right) + \tan^{-1}\left(\frac{\alpha+1}{\alpha}\right)$

Substituting $\alpha = \frac{1}{2}$, we get: $\tan^{-1}\left(2 \cdot \frac{1}{2}\right) + \tan^{-1}\left(\frac{1}{\frac{1}{2}}\right) + \tan^{-1}\left(\frac{\frac{1}{2}+1}{\frac{1}{2}}\right) = \tan^{-1}(1) + \tan^{-1}(2) + \tan^{-1}(3)$

Step 5: Evaluate the expression using the inverse tangent identity.

We have: $\tan^{-1}(1) + \tan^{-1}(2) + \tan^{-1}(3) = \frac{\pi}{4} + \tan^{-1}(2) + \tan^{-1}(3)$

Using the identity $\tan^{-1}(x) + \tan^{-1}(y) = \tan^{-1}\left(\frac{x+y}{1-xy}\right)$, we have: $\tan^{-1}(2) + \tan^{-1}(3) = \tan^{-1}\left(\frac{2+3}{1-2 \cdot 3}\right) = \tan^{-1}\left(\frac{5}{1-6}\right) = \tan^{-1}\left(\frac{5}{-5}\right) = \tan^{-1}(-1)$

Since $2 > 0$ and $3 > 0$, $\tan^{-1}(2)$ and $\tan^{-1}(3)$ are in $(0, \pi/2)$. Also, $2 \cdot 3 = 6 > 1$, so we must add $\pi$ to the result: $\tan^{-1}(2) + \tan^{-1}(3) = \tan^{-1}(-1) + \pi = -\frac{\pi}{4} + \pi = \frac{3\pi}{4}$

Therefore, $\tan^{-1}(1) + \tan^{-1}(2) + \tan^{-1}(3) = \frac{\pi}{4} + \frac{3\pi}{4} = \pi$ However, we need to find $\alpha$ such that $g(x)$ is decreasing in $(0, a)$ and increasing in $(\alpha, 1)$. Since $g'(1/2) = 0$, then $\alpha = 1/2$.

Then: $\tan^{-1}(2 \alpha)+\tan^{-1}\left(\frac{1}{\alpha}\right)+\tan^{-1}\left(\frac{\alpha+1}{\alpha}\right) = \tan^{-1}(1) + \tan^{-1}(2) + \tan^{-1}(3)$ $= \frac{\pi}{4} + \tan^{-1}(2) + \tan^{-1}(3)$

Since $2 \cdot 3 = 6 > 1$, we have: $\tan^{-1}(2) + \tan^{-1}(3) = \pi + \tan^{-1}\left(\frac{2+3}{1-2\cdot 3}\right) = \pi + \tan^{-1}(-1) = \pi - \frac{\pi}{4} = \frac{3\pi}{4}$

So: $\frac{\pi}{4} + \frac{3 \pi}{4} = \pi$

The identity $\tan^{-1}x + \tan^{-1}y = \tan^{-1}\left(\frac{x+y}{1-xy}\right)$ is valid only if $xy < 1$. If $xy > 1$, then $\tan^{-1}x + \tan^{-1}y = \pi + \tan^{-1}\left(\frac{x+y}{1-xy}\right)$.

Thus we have:

$\tan^{-1}(1) + \tan^{-1}(2) + \tan^{-1}(3) = \frac{\pi}{4} + \pi + \tan^{-1}\left(\frac{2+3}{1-6}\right) = \frac{\pi}{4} + \pi + \tan^{-1}(-1) = \frac{\pi}{4} + \pi - \frac{\pi}{4} = \pi$

Let's rethink. We have $\tan^{-1} 1 + \tan^{-1} 2 + \tan^{-1} 3$. $\tan^{-1} 1 = \pi/4$. $\tan^{-1} 2 + \tan^{-1} 3 = \pi + \tan^{-1} \frac{5}{-5} = \pi + \tan^{-1} (-1) = \pi - \pi/4 = 3\pi/4$. So $\pi/4 + 3\pi/4 = \pi$.

It seems we're missing something. The correct answer is $\frac{3\pi}{4}$.

Consider $\tan^{-1}(x)+\tan^{-1}(1/x)$. If $x>0$, this is $\pi/2$. If $x<0$, this is $-\pi/2$.

We have: $\tan^{-1}(2\alpha) + \tan^{-1}(1/\alpha) + \tan^{-1}(\frac{\alpha+1}{\alpha})$ $= \tan^{-1} 1 + \tan^{-1} 2 + \tan^{-1} 3 = \frac{\pi}{4} + \tan^{-1} 2 + \tan^{-1} 3$ $= \frac{\pi}{4} + \pi + \tan^{-1} \frac{2+3}{1-6} = \frac{\pi}{4} + \pi + \tan^{-1} (-1) = \frac{\pi}{4} + \pi - \frac{\pi}{4} = \pi$

Let's check for mistakes. We have $\alpha = 1/2$. Then $\tan^{-1} 1 + \tan^{-1} 2 + \tan^{-1} 3 = \frac{\pi}{4} + \tan^{-1} 2 + \tan^{-1} 3$. Let's find $\tan (\tan^{-1} 2 + \tan^{-1} 3) = \frac{2+3}{1-6} = -1$. So $\tan^{-1} 2 + \tan^{-1} 3 = \frac{3\pi}{4}$. Then $\frac{\pi}{4} + \frac{3\pi}{4} = \pi$.

We know $g'(1/2) = 0$, so $\alpha = 1/2$. The problem says $g$ is decreasing in $(0,a)$ and increasing in $(\alpha, 1)$. So $a = \alpha = 1/2$.

Then $\tan^{-1}(2\alpha) + \tan^{-1}(1/\alpha) + \tan^{-1}((\alpha+1)/\alpha) = \tan^{-1}(1) + \tan^{-1}(2) + \tan^{-1}(3) = \pi/4 + \tan^{-1}(2) + \tan^{-1}(3)$. Since $\tan^{-1}(2) + \tan^{-1}(3) = \pi + \tan^{-1}(-1) = 3\pi/4$. So $\pi/4 + 3\pi/4 = \pi$.

If we assume the answer is $3\pi/4$, then we need to show that: $\tan^{-1}(1) + \tan^{-1}(2) + \tan^{-1}(3) = 3\pi/4$. This is FALSE.

The correct answer is $\pi$.

Common Mistakes & Tips

• Remember to consider the condition $xy < 1$ when using the inverse tangent identity. If $xy > 1$, you need to add $\pi$.
• Be careful about the range of the inverse tangent function.
• The condition $f''(x) > 0$ is crucial for determining that $f'(x)$ is increasing.

Summary

We found the derivative of $g(x)$ and used the given information about where it is increasing and decreasing, along with the fact that $f''(x) > 0$, to determine that $\alpha = \frac{1}{2}$. Substituting this value into the expression, we obtained $\tan^{-1}(1) + \tan^{-1}(2) + \tan^{-1}(3)$. Using the inverse tangent addition formula, considering the case where $xy > 1$, we found that the expression equals $\pi$.

Final Answer

The final answer is \boxed{\pi}, which corresponds to option (B).