Animated Solution for Mathematics - Differentiation: Let $f:[2,4]\to \mathbb{R}$ be a differentiable function such that $(x{\log }_{e}x)f^{\prime }(x)+({\log }_{e}x)f(x)+f(x)\ge 1,x\in [2,4]$ with $f(2)=\frac{1}{2}$ and $f(4)=\frac{1}{2}$. Consider the following two statements: (A) $f(x)\le 1$, for all $x\in [2,4]$ (B) $f(x)\ge 1/8$, for all $x\in [2,4]$. Then,

Welcome, fellow traveler on the path to JEE mastery. Today, we are going to dismantle a problem that, at first glance, looks like a chaotic mess of calculus.

We are given a differentiable function $f:[2,4]\to \mathbb{R}$ and a terrifying-looking differential inequality:

It is easy to feel overwhelmed by such an expression. But in the world of JEE Advanced, intimidation is often the first trap. Let us breathe, look closer, and find the elegance hidden beneath the surface.

The Product Rule Revelation

Look at the Left Hand Side (LHS) again. We have $(x{\log }_{e}x)f^{\prime }(x)+({\log }_{e}x+1)f(x)$.

Does this structure feel familiar? Think back to the product rule of differentiation:

If we let $u(x)=x{\log }_{e}x$, then its derivative is $u^{\prime }(x)={\log }_{e}x+1$. Suddenly, the chaos organizes itself.

The entire LHS is simply the derivative of the product $x{\log }_{e}xf(x)$. We have transformed a complex differential inequality into the elegant form:

This is the turning point of the problem.

The Auxiliary Function

To make our lives easier, let us package this product into a new auxiliary function. Let $g(x)=x{\log }_{e}xf(x)$.

Now, our inequality is simply $g^{\prime }(x)\ge 1$. This is beautiful! We have moved from a complex differential equation to a simple statement about the slope of $g(x)$.

But how do we bound $f(x)$? We need to compare $g(x)$ to something we understand. Let us define $h(x)=g(x)-x$.

If we differentiate this, we get $h^{\prime }(x)=g^{\prime }(x)-1$. Since $g^{\prime }(x)\ge 1$, it follows that $h^{\prime }(x)\ge 0$. This tells us that $h(x)$ is a non-decreasing function on the interval $[2,4]$.

The Boundary Trap

Since $h(x)$ is non-decreasing, its behavior is strictly governed by its endpoints. The minimum value of $h(x)$ must be at $x=2$, and the maximum value must be at $x=4$.

Let us calculate these values using our boundary conditions $f(2)=\frac{1}{2}$ and $f(4)=\frac{1}{2}$.

At $x=2$:

At $x=4$:

Because $h(x)$ is non-decreasing, we have the inequality:

This is the 'squeeze' that traps our function $f(x)$.

The Final Verdict

Now, we isolate $f(x)$ from this double inequality. For the lower bound, we have:

Testing this against the claim $f(x)\ge \frac{1}{8}$, we find that even at the lowest points, the function stays well above $\frac{1}{8}$. Thus, Statement (B) is true.

For the upper bound, we have:

Testing this against the claim $f(x)\le 1$, we find that the upper ceiling of our function never reaches 1. Thus, Statement (A) is also true.

We have conquered the problem. By recognizing the hidden derivative and utilizing the power of monotonicity, we turned a daunting inequality into a clear, logical path. Keep this mindset—look for the structure, define your auxiliary functions, and trust the math. You are ready for the next challenge.