Q.Let $f : \mathbb{R} \to \mathbb{R}$ be defined by $f(x) = \dfrac{1}{x}$, $\forall\, x \in \mathbb{R}$. Then $f$ is
(A) one-one
(B) onto
(C) bijective
(D) $f$ is not defined

The key idea is one-one (injective): a function is one-one if distinct inputs always give distinct outputs.

Step 1 — Check the domain. The function $f(x) = \frac{1}{x}$ is defined for all real numbers except $x = 0$. The problem states $f : \mathbb{R} \to \mathbb{R}$, but $f(0)$ is undefined.

Step 2 — Because $f$ is not defined at $x = 0$, it is not a function from $\mathbb{R}$ to $\mathbb{R}$ in the first place. …

The first thing to notice here is that the definition itself has a problem. The function is given as $f : \mathbb{R} \to \mathbb{R}$ with $f(x) = \frac{1}{x}$ for all $x \in \mathbb{R}$. But $\frac{1}{x}$ is not defined when $x = 0$ — division by zero is undefined in real numbers. So the domain cannot be all of $\mathbb{R}$; the function simply does not exist at $x = 0$.

This is not a subtle point — it is a fatal flaw. Before we even talk about one-one or onto, the function must be well-defined on its entire domain. Since $f(0)$ is not a real number, $f$ is not a function from $\mathbb{R}$ to $\mathbb{R}$.

Let’s go through the reasoning step by step.

1. Check the definition of a function.

   A function $f : A \to B$ must assign to every element of $A$ a unique element of $B$. Here $A = \mathbb{R}$ and $B = \mathbb{R}$. For $x = 0$, the expression $\frac{1}{0}$ is undefined in $\mathbb{R}$. So $f(0)$ does not exist. Therefore $f$ is not a function from $\mathbb{R}$ to $\mathbb{R}$.

2. What about the other options?

   Options (A), (B), and (C) all assume $f$ is a valid function. But since the function is not defined at $x = 0$, none of these properties can be meaningfully discussed for the given domain.

3. Could we restrict the domain? …

Method: Check a Map Is Well-Defined Before Classifying It

Use this for any 'is $f$ one-one / onto / bijective?' question — first confirm $f$ is even a valid function on the stated domain.

Steps

Step 1: Test that every element of the domain has an output.

A rule is a function from $A$ to $B$ only if it assigns exactly one value in $B$ to each $a\in A$. For a rational rule, find where the denominator is zero; for an even root, where the radicand is negative.

Step 2: Compare the trouble points with the stated domain. …

Common Mistakes

Mistake 1: Jumping straight to the one-one / onto test without checking the domain.

Why it's wrong: a rule like $1/x$ has no value at $x=0$, so it is not a function on all of $\mathbb{R}$; asking whether it is injective is premature. Correct approach: verify the rule produces an output at every point of the stated domain first.

Mistake 2: Silently shrinking the domain to make the rule work. …