Animated Solution for Mathematics - Differentiation: Comprehension Passage

Welcome, fellow traveler on the path to JEE mastery. Today, we are not just solving a problem; we are uncovering a hidden structure.

When you first look at the inequality $f^{\prime \prime }(x)-2f^{\prime }(x)+f(x)\ge e^x$, it might seem like a chaotic mess of derivatives. But in the world of advanced mathematics, chaos is often just order waiting to be discovered. Let us peel back the layers together.

The Integrating Factor

The expression $f^{\prime \prime }(x)-2f^{\prime }(x)+f(x)$ is not random. It is a signature. It is the expansion of the operator $(D-1{)}^2$, where $D$ is the derivative operator.

When you see this, your intuition should immediately scream, "Integrating Factor!" We need to multiply by $e^{-x}$.

Why? Because the derivative of $e^{-x}$ is $-e^{-x}$, which perfectly generates the alternating signs we see in our inequality. When we multiply both sides by $e^{-x}$, the right side becomes $e^{-x}\cdot e^x=1$.

The left side becomes:

This is the moment of clarity. This expression is exactly the second derivative of the product $e^{-x}f(x)$.

The Auxiliary Function $g(x)$

Let us define a new function, $g(x)=e^{-x}f(x)$. By calculating its derivatives, we find:

Suddenly, our terrifying inequality transforms into the elegant statement $g^{\prime \prime }(x)\ge 1$. This is the heart of the problem.

We now know that $g(x)$ is concave upward, like a smiling curve, because its second derivative is strictly positive. We also know the boundary conditions: since $f(0)=0$ and $f(1)=0$, it follows that $g(0)=0$ and $g(1)=0$.

The Geometric Insight

Imagine the graph of $g(x)$. It starts at the origin $(0,0)$, ends at $(1,0)$, and is concave upward everywhere in between.

There is only one way to draw such a curve: it must dip below the x-axis. Therefore, $g(x)<0$ for all $x\in (0,1)$.

Since $g(x)=e^{-x}f(x)$ and $e^{-x}$ is always positive, $f(x)$ must also be negative. This solves our first mystery: $f(x)$ is strictly negative in the interval $(0,1)$.

The Minimum and the Slope

Now, consider the second part of the problem. We are told that $g(x)$ reaches its minimum at $x=1/4$. At this point, the tangent is horizontal, so $g^{\prime }(1/4)=0$.

For $x<1/4$, the function is decreasing as it approaches the minimum, meaning $g^{\prime }(x)<0$. Substituting our expression for $g^{\prime }(x)$, we get:

Since $e^{-x}>0$, we conclude that $f^{\prime }(x)-f(x)<0$, or $f^{\prime }(x)<f(x)$.

This is the beauty of calculus. We started with a complex inequality and, through the lens of an auxiliary function, reduced it to a simple geometric argument about concavity and slopes. Never fear the differential equation; look for the symmetry, define your auxiliary function, and let the math reveal its own truth.