Animated Solution for Mathematics - Differentiation: Let $f:(-1,\infty )\to \mathbb{R}$ be defined by $f(0)=1$ and $f(x)=\frac{1}{x}{\log }_e(1+x)$, $x\ne 0$. Then the function $f$:

A Journey into Monotonicity

Welcome, fellow traveler of the JEE Advanced journey. Today, we are not just solving a problem; we are exploring the landscape of a function.

We are looking at $f(x)=\frac{\ln (1+x)}{x}$ for $x eq 0$, with $f(0)=1$. This function is a classic, a beautiful blend of logarithmic growth and algebraic division. Our mission is to determine its monotonicity—is it climbing, or is it sliding down?

The Derivative Challenge

To understand the behavior of any function, we must look at its rate of change. We need the derivative, $f^{\prime }(x)$.

Since our function is a fraction, we invoke the quotient rule:

Here, $u=\ln (1+x)$ and $v=x$. Differentiating these pieces is straightforward: the derivative of $\ln (1+x)$ is $\frac{1}{1+x}$, and the derivative of $x$ is $1$.

When we assemble these into the quotient rule, we get:

This looks a bit messy, doesn't it? But don't panic. Let's simplify it. By taking a common denominator in the numerator, we arrive at:

The Auxiliary Function $g(x)$

Now, look closely at the expression for $f^{\prime }(x)$. The denominator $x^2(1+x)$ is always positive in our domain $x\in (-1,\infty )$.

This means the sign of the derivative—and thus the monotonicity of our function—depends entirely on the numerator. Let's isolate this numerator and call it:

This is the "trap catcher." Many students try to solve $g(x)<0$ by inspection, but that is a dangerous game. Instead, let's analyze $g(x)$ using its own derivative.

Analyzing the Behavior

We differentiate $g(x)$ to see how it moves. Using the product rule on $(1+x)\ln (1+x)$, we get:

The terms simplify beautifully:

Now, consider the behavior of $g^{\prime }(x)$. For $x>0$, $\ln (1+x)>0$, so $g^{\prime }(x)<0$. This means $g(x)$ is decreasing for $x>0$.

For $-1<x<0$, $\ln (1+x)<0$, so $g^{\prime }(x)>0$. This means $g(x)$ is increasing for $x<0$.

Since $g(0)=0-(1+0)\ln (1)=0$, we have a function that increases until it hits $0$ at $x=0$, and then decreases. This makes $x=0$ the absolute maximum of $g(x)$.

The Final Verdict

Because $g(0)=0$ is the maximum value, $g(x)$ must be strictly less than $0$ for all other $x$ in the domain.

Returning to our derivative $f^{\prime }(x)=\frac{g(x)}{x^2(1+x)}$, we see that we have a negative numerator divided by a positive denominator. Therefore, $f^{\prime }(x)<0$ for all $x \in (-1, \infty), x eq 0$.

The function is strictly decreasing everywhere. We have successfully navigated the landscape and found the truth. Keep this technique in your arsenal: when the derivative is complex, define an auxiliary function for the numerator and analyze its extrema. You are now one step closer to mastering the JEE.