Animated Solution for Mathematics - Differentiation: Let $f$ be a twice differentiable function such that $f^{\prime \prime }(x)=-f(x)$, and $f^{\prime }(x)=g(x),h(x)=[f(x){]}^2+[g(x){]}^2$. Find $h(10)$ if $h(5)=11$.

Imagine you are standing at the edge of a vast, tranquil lake. You throw a stone, and the ripples spread out in perfect, concentric circles.

In mathematics, we often encounter functions that seem complex at first glance, yet they possess a hidden, underlying symmetry that keeps them perfectly balanced. Today, we are going to explore one such problem—a problem that initially looks like a daunting task of calculus but reveals itself to be a beautiful demonstration of conservation.

Defining Our Players

We are given a twice-differentiable function $f(x)$ that satisfies the elegant differential equation $f^{\prime \prime }(x)=-f(x)$. This is the signature equation of Simple Harmonic Motion, the heartbeat of the physical universe.

We are also given a secondary function $g(x)=f^{\prime }(x)$, and a composite function $h(x)=[f(x){]}^2+[g(x){]}^2$. Our mission is to find $h(10)$ given that $h(5)=11$.

At first, you might be tempted to solve for $f(x)$ directly. You might think, "I need to find the exact form of $f(x)$ to plug in $x=10$."

But hold that thought. In the world of JEE Advanced, the most elegant path is rarely the one that requires the most brute force. Instead, let us ask a more profound question: How does $h(x)$ behave as $x$ changes?

To understand the behavior of $h(x)$, we must look at its rate of change. We differentiate $h(x)$ with respect to $x$ using the chain rule:

This expression is the key to the entire problem. We know $f^{\prime }(x)=g(x)$.

But what about $g^{\prime }(x)$? Since $g(x)=f^{\prime }(x)$, it follows that $g^{\prime }(x)=f^{\prime \prime }(x)$. And here is where the magic happens: we are given that $f^{\prime \prime }(x)=-f(x)$. Therefore, $g^{\prime }(x)=-f(x)$.

Now, let us substitute these pieces back into our derivative equation. Watch closely as the terms interact:

Look at that! We have $2f(x)g(x)-2f(x)g(x)$. The terms cancel out perfectly, leaving us with $h^{\prime }(x)=0$.

This is a monumental realization. If the derivative of a function is zero everywhere, the function itself must be a constant. It does not matter if $x$ is 5, 10, or 100; the value of $h(x)$ remains locked in time.

Because $h(x)$ is a constant, we can say $h(x)=C$. We were given the initial condition $h(5)=11$.

This tells us that our constant $C$ is exactly 11. Since the function never changes, it must be true that $h(10)=11$.

Think about what this means geometrically. The point $(f(x),g(x))$ is tracing a circle in the Cartesian plane. As $x$ increases, the point moves along the circumference, but its distance from the origin—represented by $\sqrt{h(x)}$—remains perfectly fixed.

You have just solved a problem that describes the conservation of energy in a harmonic system. You didn't just calculate a number; you uncovered a fundamental truth about the system. Keep this perspective, and you will find that even the most intimidating JEE problems are just stories waiting to be told.