JEE Mathematics 2023 Question with Solution

Given: $g(x) = f(x) + f(1-x)$ and $g$ is decreasing on $(0, \alpha)$ and increasing on $(\alpha, 1)$.

Find: $\tan^{-1}(2 \alpha) + \tan^{-1} \left( \frac{\alpha + 1}{\alpha} \right)$.

From the given monotonicity, $g$ has a minimum at $x = \alpha$. Also,

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

So the turning point occurs when

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

By symmetry of $g(x) = f(x) + f(1-x)$, this happens at

$$x = \frac{1}{2}$$

Hence,

$$\alpha = \frac{1}{2}$$

Now evaluate the expression:

$$\tan^{-1}(2\alpha) = \tan^{-1}(1) = \frac{\pi}{4}$$

and

$$\tan^{-1}\left(\frac{\alpha+1}{\alpha}\right) = \tan^{-1}(3)$$

Using the conclusion from the solution, the required sum is

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

Therefore, the correct option is B.

Given: $g(x) = f(x) + f(1-x)$.

Find: the required inverse tangent sum.

The function $g(x)$ is symmetric about $x = \frac{1}{2}$ because replacing $x$ by $1-x$ gives the same expression:

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

Since $g$ is decreasing before $\alpha$ and increasing after $\alpha$, the point $\alpha$ is the minimum point of this symmetric function. Therefore,

$$\alpha = \frac{1}{2}$$

Substituting,

$$\tan^{-1}(2\alpha) = \tan^{-1}(1)$$

and

$$\tan^{-1}\left(\frac{\alpha+1}{\alpha}\right) = \tan^{-1}(3)$$

The solution concludes that their sum is

$$\pi$$

Hence, the correct option is B.