Animated Solution for Mathematics - Functions: Given below are two statements: Statement I: The function $f:\mathbb{R}\to \mathbb{R}$ defined by $f(x)=\frac{x}{1+|x|}$ is one-one. Statement II: The function $f:\mathbb{R}\to \mathbb{R}$ defined by $f(x)=\frac{x^2+4x-30}{x^2-8x+18}$ is many-one. In the light of the above statements, choose the correct answer from the options given below :

A Journey into Injectivity

Welcome, student. Today, we are not just solving a problem; we are exploring the very anatomy of functions. In the JEE Advanced landscape, understanding the difference between a one-one (injective) function and a many-one function is not just about memorizing definitions—it is about developing a geometric intuition for how functions behave in the real number plane $\mathbb{R}$.

Before we touch the algebra, let us ground ourselves in the concept. What does it mean for a function to be one-one? Imagine a function as a machine. If you put in a unique input $x$, you must get a unique output $y$.

If two different inputs $x_1$ and $x_2$ produce the same output $y$, the machine is 'many-one'—it is collapsing different inputs into a single result.

Graphically, this is the Horizontal Line Test. If you draw a horizontal line anywhere on the graph and it hits the curve more than once, you have a many-one function.

But how do we prove this analytically? We look for monotonicity. If a function is strictly increasing ($f^{\prime }(x)>0$) or strictly decreasing ($f^{\prime }(x)<0$) for its entire domain, it can never 'turn back' to hit a previous $y$-value. It is a one-way street. This is our golden rule for today.

The Absolute Value Challenge

Let us analyze the first function: $f(x)=\frac{x}{1+|x|}$. Whenever you see an absolute value, do not panic. It is simply a signal to split the universe of $x$ into two distinct realities.

Case 1: $x\ge 0$ When $x$ is non-negative, $|x|=x$. Our function simplifies beautifully to:

To check for monotonicity, we calculate the derivative using the quotient rule. Let $u=x$ and $v=1+x$. Then $u^{\prime }=1$ and $v^{\prime }=1$. The derivative is:

Look at this result. The numerator is $1$, which is positive. The denominator is a square, which is always positive. Thus, $f^{\prime }(x)>0$. The function is strictly increasing here.

Case 2: $x<0$ When $x$ is negative, $|x|=-x$. Our function becomes:

Again, applying the quotient rule with $u=x$ and $v=1-x$, where $u^{\prime }=1$ and $v^{\prime }=-1$:

Again, $f^{\prime }(x)>0$. Since the function is strictly increasing for $x\ge 0$ and strictly increasing for $x<0$, and the function is continuous at $x=0$ (where $f(0)=0$), the entire function is strictly increasing on $\mathbb{R}$. It never turns back. Therefore, Statement I is True.

The Rational Function and the Turning Point

Now, let us turn our attention to the second function: $f(x)=\frac{x^2+4x-30}{x^2-8x+18}$. First, we check the domain. The denominator $x^2-8x+18$ can be written as $(x-4{)}^2+2$.

Since $(x-4{)}^2\ge 0$, the denominator is always at least $2$. It is never zero. The function is defined for all $x\in \mathbb{R}$.

To determine if it is one-one or many-one, we must find the derivative. Using the quotient rule $f^{\prime }(x)=\frac{u^{\prime }v-u{v}^{\prime }}{v^2}$: Let $u=x^2+4x-30$ and $v=x^2-8x+18$. Then $u^{\prime }=2x+4$ and $v^{\prime }=2x-8$.

The numerator of the derivative becomes:

Let us expand this carefully. The first part is $2x^3-12x^2+4x+72$. The second part is $2x^3-92x+240$.

Subtracting them yields:

We can factor out $-12$:

To find the critical points, we set $f^{\prime }(x)=0$. This implies $x^2-8x+14=0$. Using the quadratic formula:

Because we have two distinct real roots, the derivative $f^{\prime }(x)$ changes sign at these points. This means the function increases, then decreases, then increases again. It is not monotonic and fails the horizontal line test. Therefore, Statement II is True.

We have rigorously proven that Statement I describes a one-one function and Statement II describes a many-one function. Both statements are correct. Remember, in calculus, the derivative is your compass. It tells you exactly when a function is climbing, when it is falling, and when it is turning. Trust the math, stay calm through the algebra, and you will always find the truth.