Animated Solution for Mathematics - Functions: Let $f(x)$ be defined for all $x>0$ and be continuous. Let $f(x)$ satisfy $f(x/y)=f(x)-f(y)$ for all $x,y$ and $f(e)=1$. Then

Decoding the Logarithm

Welcome, warriors of JEE Advanced. Today, we are not just solving a problem; we are performing a forensic analysis on a mathematical object.

We are given a function $f(x)$ defined for all $x>0$, and we are handed a cryptic clue:

This is a functional equation. It is the DNA of the function, telling us exactly how the function behaves without giving us the explicit formula. Our mission is to decode this DNA.

Finding the Anchor Point

Before we rush into complex derivations, let us find the anchor. Every function has a point of reference.

Let us test the simplest possible values. If we set $x=1$ and $y=1$, the equation becomes:

This simplifies beautifully to $f(1)=0$. This is our first breakthrough.

We now know that our function must pass through the point $(1,0)$. This is the hallmark of logarithmic functions, which always cross the x-axis at $x=1$.

The Logarithmic Revelation

Look closely at the property $f(\frac{x}{y})=f(x)-f(y)$. Does this ring a bell? This is the fundamental identity of logarithms.

Because the problem states that $f(x)$ is continuous, we can confidently assert that $f(x)$ must be a logarithmic function of the form $f(x)={\log }_{a}x$.

To find the base $a$, we use the second clue: $f(e)=1$. Substituting $x=e$ into our assumed form, we get:

By the definition of logarithms, this implies $a^1=e$, which means $a=e$. Thus, our function is revealed: $f(x)=\ln x$.

The Rigorous Verification

Now, we must be careful. In JEE Advanced, we never stop at finding the answer; we verify the options to ensure we haven't missed a subtle trap.

Option A: Is it bounded? We know $f(x)=\ln x$. As $x\to \infty$, $\ln x\to \infty$. As $x\to 0^{+}$, $\ln x\to -\infty$. The range is $(-\infty ,\infty )$, so it is clearly unbounded. Option A is incorrect.

Option B: The limit of $f(1/x)$? We evaluate:

As $x\to 0^{+}$, $\ln x\to -\infty$, so $-\ln x\to \infty$. The limit is not $0$. Option B is incorrect.

Option C: The limit of $xf(x)$? This is the trap. We need to evaluate ${\lim }_{x\to 0^{+}}x\ln x$. This is an indeterminate form of type $0\times (-\infty )$.

We rewrite it as:

Applying L'Hopital's Rule, we differentiate the numerator to get $\frac{1}{x}$ and the denominator to get $-\frac{1}{x^2}$. The expression becomes:

The limit is $0$, not $1$. Option C is incorrect.

We have systematically dismantled the problem. We identified the function, verified its properties, and debunked the distractors.

The only correct choice is $f(x)=\ln x$. Remember, in the exam hall, trust your derivation, but always verify your steps. You have the tools; now go forth and conquer.