Check if f(x) = x² is one-one and onto from R to R

Question

Check whether the function $f: \mathbb{R} \rightarrow \mathbb{R}$ defined by $f(x) = x^2$ is (a) one-one (injective) and (b) onto (surjective).

Solution — Step by Step

One-one (Injective): A function $f$ is one-one if distinct inputs give distinct outputs. Formally: $f(a) = f(b) \implies a = b$ for all $a, b \in$ domain.

Equivalently: if $a \neq b$ , then $f(a) \neq f(b)$ .

Onto (Surjective): A function $f: A \rightarrow B$ is onto if every element of the codomain $B$ is the image of at least one element from $A$ . Formally: For every $y \in B$ , there exists $x \in A$ such that $f(x) = y$ .

Take $f(a) = f(b)$ :

a^2 = b^2 \implies a^2 - b^2 = 0 \implies (a-b)(a+b) = 0

So either $a = b$ OR $a = -b$ .

Since $a = -b$ is possible (e.g., $a = 2, b = -2$ ), we can have $f(a) = f(b)$ with $a \neq b$ .

Counterexample: $f(2) = 4$ and $f(-2) = 4$ . So two different inputs (2 and -2) give the same output (4).

Conclusion: $f(x) = x^2$ is NOT one-one (not injective).

We need to check if every real number $y \in \mathbb{R}$ can be written as $y = x^2$ for some real $x$ .

Consider $y = -1$ (a negative real number). Is there any $x \in \mathbb{R}$ such that $x^2 = -1$ ?

x^2 = -1

There is no real solution — the square of any real number is non-negative. So $x^2 \geq 0$ for all $x \in \mathbb{R}$ .

This means no negative number is in the range of $f$ . The range of $f$ is $[0, \infty)$ , but the codomain is $\mathbb{R}$ (which includes negative numbers).

Since range $\subsetneq$ codomain, the function is NOT onto.

Conclusion: $f(x) = x^2$ is NOT onto (not surjective) as a function from $\mathbb{R}$ to $\mathbb{R}$ .

$f: \mathbb{R} \rightarrow \mathbb{R}$ , $f(x) = x^2$ :

NOT one-one: Because $f(2) = f(-2) = 4$ (two distinct inputs map to the same output)
NOT onto: Because negative real numbers (like $-1$ ) have no pre-image in $\mathbb{R}$ under $f$

Therefore, $f(x) = x^2$ is neither injective nor surjective as a function from $\mathbb{R}$ to $\mathbb{R}$ .

Why This Works

The issue with injectivity: $x^2$ is an even function — $f(-x) = f(x)$ . Any even function that maps both positive and negative values in the domain will fail the one-one test because symmetric inputs give the same output.

The issue with surjectivity: the codomain is $\mathbb{R}$ (all real numbers), but the function only produces non-negative values. The range is $[0, \infty) \subsetneq \mathbb{R}$ . A function is onto only when range = codomain.

Key insight: The same function with a DIFFERENT codomain can be onto. If we define $f: \mathbb{R} \rightarrow [0, \infty)$ , then $f(x) = x^2$ IS onto (every non-negative number is a square of some real number).

Similarly, $f: [0, \infty) \rightarrow [0, \infty)$ , $f(x) = x^2$ is both one-one AND onto — it becomes a bijection.

Alternative Method — Graph Analysis

The horizontal line test:

One-one test: If any horizontal line $y = c$ crosses the graph more than once → not one-one.
The graph of $y = x^2$ (parabola) has a horizontal line at $y = 4$ crossing at $x = 2$ and $x = -2$ → not one-one. ✓
Onto test: If there exists a horizontal line $y = c$ (for $c \in$ codomain) that doesn’t cross the graph → not onto.
For $y = -1$ , the horizontal line $y = -1$ is entirely below the parabola — no intersection → not onto. ✓

CBSE Class 12 and JEE Main both test this “check injectivity/surjectivity” question type. Always: (1) find a counterexample for “not one-one,” (2) find a value of $y$ with no $x$ -preimage for “not onto.” A concrete counterexample with specific numbers always earns full marks.

Common Mistake

Students sometimes say ” $f(x) = x^2$ is one-one because $f(2) \neq f(3)$ ” — showing that two specific values give different outputs doesn’t prove one-one. One-one requires ALL pairs of distinct inputs to give distinct outputs. You only need ONE counterexample (like $f(2) = f(-2) = 4$ ) to disprove it. For proofs, find the counterexample; don’t just verify specific cases.

Question

Solution — Step by Step

Why This Works

Alternative Method — Graph Analysis

Common Mistake

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