Animated Solution for Mathematics - Definite Integration: If $f$ is a continuous function with ${\int }_0^{x}f(t)dt\to \infty$ as $|x|\to \infty$, then show that every line $y=mx$ intersects the curve $y^2+{\int }_0^{x}f(t)dt=2$

Proving the Unprovable

Welcome, future engineer. Today, we are not just solving an equation; we are embarking on a journey into the heart of calculus.

Often, in the JEE Advanced, you will be presented with problems that seem impossible because they lack specific numbers. You might look at the curve $y^2+{\int }_0^{x}f(t)dt=2$ and feel a sense of dread.

Where is the function? What is $f(t)$? How can I solve for $x$ if I don't know what $f(t)$ is?

Here is the secret: you don't need to know $f(t)$. You only need to understand its behavior. This is the difference between a calculator and a mathematician.

Let us peel back the layers of this problem together.

The Geometric Intuition

Imagine you are standing on a coordinate plane. You have a curve defined by the equation $y^2+{\int }_0^{x}f(t)dt=2$. It is a mysterious, shifting shape.

Then, you have a straight line, $y=mx$, passing through the origin. The question asks us to prove that these two entities must collide. They must intersect.

In geometry, an intersection is a point $(x,y)$ that satisfies both equations simultaneously. If we want to prove they intersect, we need to prove that there exists at least one pair $(x,y)$ that makes both equations true.

This is an existence proof. We are not looking for the coordinate; we are looking for the guarantee that the coordinate exists.

The Auxiliary Function

To bridge the gap between the line and the curve, we use a classic tool: the auxiliary function. We want to see if the line and the curve ever reach the same height.

Let us substitute the line's definition, $y=mx$, into the curve's equation. This gives us:

Now, let us move everything to one side. We define a new function, $F(x)$, which represents the 'gap' between the curve and the line:

If $F(x)=0$, then the gap is zero. If the gap is zero, the line and the curve are touching. Our mission is now crystal clear: we must prove that $F(x)$ takes the value zero at some point $x_0$.

The Power of Continuity

Here is where we invoke the Intermediate Value Theorem (IVT). The IVT is the bridge that connects the negative to the positive.

If a continuous function is negative at one point and positive at another, it must cross the x-axis somewhere in between. It has no choice; it cannot jump over the axis, it must pass through it.

Let us test our function $F(x)$ at the origin, $x=0$:

Since the integral from $0$ to $0$ is always $0$, we are left with $F(0)=-2$. We have our starting point: a negative value.

The Limit of Infinity

Now, let us look at the behavior as $|x|\to \infty$. This is where the problem gives us the key to the lock.

We are told that ${\int }_0^{x}f(t)dt\to \infty$ as $|x|\to \infty$. Look at our function $F(x)=m^2{x}^2+{\int }_0^{x}f(t)dt-2$.

As $x$ grows, $m^2{x}^2$ grows to positive infinity (assuming $m eq 0$). The integral also grows to positive infinity. We are adding two massive positive quantities and subtracting a measly $2$.

The result is undeniably positive infinity:

A Mathematical Certainty

We have established two facts: 1. At $x=0$, $F(x)=-2$ (Negative). 2. As $|x|\to \infty$, $F(x)\to \infty$ (Positive).

Because $f$ is a continuous function, the integral ${\int }_0^{x}f(t)dt$ is also continuous. Therefore, $F(x)$ is continuous.

By the Intermediate Value Theorem, there must exist some value $x_0$ such that $F(x_0)=0$. At this $x_0$, the line and the curve are at the exact same height. They intersect.

We have proven it without ever needing to know the exact form of $f(t)$. This is the elegance of calculus—using the properties of functions to reveal truths that are hidden from plain sight. You have mastered the logic; now, carry this confidence into your next problem.