Animated Solution for Mathematics - Definite Integration: The total number of distinct $x\in [0,1]$ for which ${\int }_0^x\frac{t^2}{1+t^4}dt=2x-1$ is

The problem asks for the number of distinct solutions for $x\in [0,1]$ in the equation:

At first glance, this looks like a nightmare. However, in JEE Advanced, the most complex-looking problems often have the most elegant solutions if you change your perspective.

The first step is to stop treating this as an equality to be solved and start treating it as a function to be analyzed. We define a new function:

Finding the roots of $g(x)=0$ is mathematically equivalent to solving our original equation. By moving everything to one side, we have created a single entity whose behavior we can study.

Now, let us look at the boundaries of our interval, $[0,1]$. First, we evaluate $g(0)$:

Our function starts above the $x$-axis. Next, we evaluate $g(1)$:

We do not need the exact value of the integral; we only need its sign. We invoke the AM-GM inequality: $1+t^4\ge 2\sqrt{1\cdot t^4}=2t^2$. This implies:

Therefore, the integral satisfies:

Substituting this back into our expression for $g(1)$, we get $g(1)\le \frac{1}{2}-1=-0.5$. Since $g(1)$ is negative and $g(0)$ is positive, the Intermediate Value Theorem guarantees that our function must cross the $x$-axis at least once.

We know there is at least one root, but we must check if the function "wiggles." To answer this, we look at the derivative $g^{\prime }(x)$ using the Leibniz Rule:

This simplifies to:

We already established that $\frac{x^2}{1+x^4}\le \frac{1}{2}$. Therefore:

Because the derivative is strictly negative for all $x\in [0,1]$, the function is strictly decreasing. It is a one-way slide from positive to negative and cannot turn back.

We have proven two things: first, by the Intermediate Value Theorem, the function must cross the $x$-axis at least once. Second, by the strict monotonicity of the function, it cannot cross more than once.

Therefore, there is exactly one distinct root.