Animated Solution for Mathematics - Definite Integration: Let $f(x)={\int }_1^x\sqrt{2-t^2}dt$. Then the real roots of the equation $x^2-f^{\prime }(x)=0$ are

My dear student, welcome to a journey through one of the most elegant intersections of calculus and algebra. Today, we are dissecting the function $f(x)={\int }_1^x\sqrt{2-t^2}dt$.

At first glance, this might look like a daunting integral, but I want you to shift your perspective. This is an accumulation function; it is a dynamic process where the area under the curve $\sqrt{2-t^2}$ is being 'accumulated' as $x$ moves.

Before we touch any algebra, let us respect the domain. For the integrand $\sqrt{2-t^2}$ to exist in the real number system, we must have $2-t^2\ge 0$, which means $t\in [-\sqrt{2},\sqrt{2}]$. Keep this boundary in your mind; it is the fence within which our solution must live.

Now, we are tasked with finding the real roots of $x^2-f^{\prime }(x)=0$. Many students would immediately try to solve the integral, but that is a trap! We only need $f^{\prime }(x)$.

This is where the Newton-Leibniz Rule becomes our best friend. It tells us that the derivative of an integral with a variable upper limit $x$ is simply the integrand evaluated at $x$. Mathematically:

Since the derivative of $x$ is $1$ and the derivative of the constant $1$ is $0$, the entire expression collapses into the beautiful, simple result:

See how the complexity vanished? That is the power of knowing your theorems.

With $f^{\prime }(x)$ in our toolkit, the equation $x^2-f^{\prime }(x)=0$ transforms into $x^2=\sqrt{2-x^2}$. Now, we are in the realm of algebra.

To solve this, we must isolate the radical. Squaring both sides is the logical next step, but we must be vigilant. Squaring is a 'dangerous' operation because it can create extraneous roots.

When we square both sides, we get $(x^2{)}^2=(\sqrt{2-x^2}{)}^2$, which simplifies to $x^4=2-x^2$. Rearranging this, we arrive at the biquadratic equation:

This is a quadratic in disguise! By substituting $u=x^2$, we get $u^2+u-2=0$. Factoring this, we find $(u+2)(u-1)=0$. This gives us two potential paths: $x^2=-2$ or $x^2=1$.

We must now apply our filters. We already established that $x$ must be real, so $x^2=-2$ is impossible. We are left with $x^2=1$, which gives us $x=\pm 1$.

Finally, we check our domain constraint: $x^2\le 2$. Since $1\le 2$, both roots are valid.

Graphically, you are looking for the intersection of the parabola $y=x^2$ and the semicircle $y=\sqrt{2-x^2}$. They meet perfectly at $x=1$ and $x=-1$.

You have successfully navigated the trap, applied the theorem, and verified the result. This is the essence of JEE Advanced mathematics—not just calculation, but conceptual mastery.