Animated Solution for Mathematics - Differentiation: Let $g(x)=f(x)+f(1-x)$ and $f^{\prime \prime }(x)>0,x\in (0,1)$. If $g$ is decreasing in the interval $(0,\alpha )$ and increasing in the interval $(\alpha ,1)$, then ${\tan }^{-1}2\alpha +{\tan }^{-1}(\frac{1}{\alpha })+{\tan }^{-1}(\frac{\alpha +1}{\alpha })$ is equal to

Welcome, fellow traveler on the path to JEE mastery. Today, we are not just solving a calculus problem; we are peeling back the layers of a beautiful, symmetric function.

We are given $g(x)=f(x)+f(1-x)$ with the constraint that $f^{\prime \prime }(x)>0$ for $x\in (0,1)$.

Stop for a moment and visualize this. When the second derivative $f^{\prime \prime }(x)$ is strictly positive, the function $f(x)$ is strictly convex—it is a 'smiling' curve, like a bowl.

This isn't just a label; it is a powerful piece of information. It tells us that the slope of $f(x)$, which is $f^{\prime }(x)$, is strictly increasing. As you move from left to right, the curve gets steeper and steeper. This monotonicity is the key that will unlock the entire problem.

To understand the behavior of $g(x)$, we must look at its rate of change. Let us differentiate $g(x)$ with respect to $x$:

Applying the chain rule to the second term, we get:

This derivative, $g^{\prime }(x)=f^{\prime }(x)-f^{\prime }(1-x)$, is the heartbeat of our function. It tells us exactly when $g(x)$ is climbing and when it is falling.

We are told that $g(x)$ changes its behavior at $\alpha$. This means $\alpha$ is our critical point where $g^{\prime }(\alpha )=0$. Setting our derivative to zero, we find:

Here is where our earlier insight about convexity saves the day. Because $f^{\prime \prime }(x)>0$, the function $f^{\prime }(x)$ is strictly increasing, which means it is one-to-one. It can only take a specific value at exactly one input.

Therefore, for $f^{\prime }(\alpha )$ to equal $f^{\prime }(1-\alpha )$, the inputs must be identical:

Our transition point is exactly $\frac{1}{2}$. If $x<\frac{1}{2}$, then $x<1-x$, so $f^{\prime }(x)<f^{\prime }(1-x)$, making $g^{\prime }(x)<0$ (decreasing). If $x>\frac{1}{2}$, then $x>1-x$, so $f^{\prime }(x)>f^{\prime }(1-x)$, making $g^{\prime }(x)>0$ (increasing). Everything aligns perfectly.

Now, we face the final challenge: evaluating the expression $E={\tan }^{-1}(2\alpha )+{\tan }^{-1}(\frac{1}{\alpha })+{\tan }^{-1}(\frac{\alpha +1}{\alpha })$ with $\alpha =\frac{1}{2}$.

Substituting $\alpha =\frac{1}{2}$ gives us:

We know ${\tan }^{-1}(1)=\frac{\pi }{4}$. Now, we must sum ${\tan }^{-1}(2)+{\tan }^{-1}(3)$.

Here is the trap: many students blindly use the formula ${\tan }^{-1}x+{\tan }^{-1}y={\tan }^{-1}(\frac{x+y}{1-xy})$. But wait! Check the product $xy=2\cdot 3=6$. Since $6>1$, we must use the adjusted formula:

Since ${\tan }^{-1}(-1)=-\frac{\pi }{4}$, this simplifies to $\pi -\frac{\pi }{4}=\frac{3\pi }{4}$.

Finally, we combine our results:

And there it is—the elegance of the result, $\pi$. We navigated the convexity, mastered the derivative, avoided the inverse trig trap, and arrived at a beautiful, clean answer. Keep practicing, keep visualizing, and keep falling in love with the logic of mathematics!