Animated Solution for Mathematics - Functions: Let a function $f:(0,\infty )\to (0,\infty )$ be defined by $f(x)=|1-\frac{1}{x}|$. Then $f$ is:

Unveiling the Function

Welcome, future engineer. Today, we are not just solving a math problem; we are dissecting the behavior of a function.

When you look at $f(x)=|1-\frac{1}{x}|$, do not see just a collection of symbols. See a machine that takes a positive number, transforms it, and spits out a result. Our goal is to determine if this machine is 'Injective' (one-to-one) and 'Surjective' (onto).

Deconstructing the Modulus

The modulus function is the great divider. It forces us to look at the world in two ways: when the inside is positive, and when it is negative.

Our function $f(x)=|1-\frac{1}{x}|$ has a critical point where the inside expression $1-\frac{1}{x}$ equals zero. Solving $1-\frac{1}{x}=0$ gives us $x=1$. This is our pivot point.

For $x\in (0,1]$, the term $\frac{1}{x}$ is greater than or equal to $1$. Thus, $1-\frac{1}{x}$ is non-positive. The modulus flips the sign, giving us:

As $x$ approaches $0$ from the right, $\frac{1}{x}$ explodes to infinity. Our graph starts from the heavens!

For $x>1$, the term $\frac{1}{x}$ is less than $1$. Thus, $1-\frac{1}{x}$ is positive. The modulus drops away, leaving:

As $x$ grows towards infinity, $\frac{1}{x}$ shrinks to zero, and our function value approaches $1$. We have a horizontal asymptote at $y=1$.

The Hunt for Injectivity

Now, is this function injective? An injective function is like a perfect handshake: every input $x$ must have a unique output $y$.

If two different inputs map to the same output, the handshake is broken. Graphically, we use the Horizontal Line Test. Imagine drawing a horizontal line at $y=0.5$. Does it hit the graph twice? Yes!

Algebraically, we set $f(x)=0.5$. This leads to:

This splits into two cases: $1-\frac{1}{x}=0.5$ and $1-\frac{1}{x}=-0.5$. Solving these, we find $x=2$ and $x=\frac{2}{3}$.

Both inputs yield the same output of $0.5$. Because $f(2)=f(\frac{2}{3})=0.5$, the function is strictly not injective. It fails the test.

The Surjectivity Trap

Finally, let us look at surjectivity. A function is surjective if its Range covers the entire Codomain.

The problem defines our Codomain as $(0,\infty )$. We must find the Range. Looking at our graph, the lowest point is at $x=1$, where $f(1)=0$.

The function then rises to infinity. Thus, the Range is $[0,\infty )$.

Here lies the trap. The Range is $[0,\infty )$, which includes the value $0$. However, the Codomain is $(0,\infty )$, which strictly excludes $0$.

Because the Range is not equal to the Codomain, the function is not surjective.

We have analyzed the function, tested its injectivity, and scrutinized its surjectivity. It failed both.

It is neither injective nor surjective. Remember, in JEE Advanced, the beauty lies not just in the answer, but in the rigorous verification of every boundary condition. Keep practicing, keep questioning, and keep falling in love with the logic of mathematics!