Let R be the set of real numbers. If f : R → R is a function defined by f(x) = x², then f is : (a) I

Q: Wait, I thought $f(x) = x^2$ is a function, so why is it not injective?

A function is simply a rule where every input has exactly one output. Injectivity is a much stricter condition: it requires that every output must have a *unique* input. The function $f(x) = x^2$ fails this because both $2$ and $-2$ map to $4$. Since two different inputs share the same output, it is 'many-to-one', not 'one-to-one'.

Q: Why do we say the range is $[0, \infty)$? Can't $x$ be negative?

You are absolutely right that $x$ can be negative! However, the range is the set of all *outputs* (the y-values). When you square any real number, the result is always non-negative. Therefore, the function can never produce a negative output. Since the codomain is all real numbers $\mathbb{R}$, but the outputs are only $[0, \infty)$, the function is not surjective.

Q: If the question said $f: [0, \infty) \to [0, \infty)$, would it be bijective?

That is a brilliant observation! By restricting the domain to $[0, \infty)$, we eliminate the negative inputs that caused the 'many-to-one' issue. By restricting the codomain to $[0, \infty)$, we ensure every output is covered. In that specific case, the function would indeed be bijective. This is how we 'fix' functions in higher mathematics!

Question

Let $R$ be the set of real numbers. If $f : R 	o R$ is a function defined by $f(x) = x^2$, then $f$ is :

Options:

(A) Injective but not surjective

(B) Surjective but not injective

(C) Bijective

(D) None of these

Sigma Percentile · Accepted Answer

Welcome, future engineers! Today, we are going to dissect one of the most fundamental yet misunderstood functions in the entire JEE syllabus: $f(x) = x^2$.

It seems simple, right? A basic parabola. But in the world of functions, simplicity is often where the most dangerous traps hide.

Let us embark on this journey to understand why this function is neither injective nor surjective.

### The Geometry of the Parabola

Imagine you are standing on a coordinate plane. The $x$-axis represents our domain, the set of all possible inputs, which is $\mathbb{R}$. The $y$-axis represents our codomain, the set of all potential outputs, also $\mathbb{R}$.

When we define $f(x) = x^2$, we are creating a mapping. For every $x$, we square it.

If you plot this, you get the classic, elegant, upward-opening parabola with its vertex at the origin $(0,0)$. This curve is the visual signature of our function.

### The Injective Test: The Horizontal Line

Now, let us talk about injectivity, or being 'one-to-one'. The definition is rigorous: $f(x_1) = f(x_2) \Rightarrow x_1 = x_2$.

But let us use the 'Horizontal Line Test' to visualize this. Imagine drawing a horizontal line at $y = 4$. Does it hit the graph once? No. It hits it at $x = 2$ and $x = -2$.

Mathematically, we see that:
$$f(2) = 2^2 = 4$$
$$f(-2) = (-2)^2 = 4$$

Because two distinct inputs ($2$ and $-2$) lead to the exact same output ($4$), the function fails the test of injectivity. It is 'many-to-one'. This is the first hurdle.

### The Surjective Trap: Range vs. Codomain

Next, we tackle surjectivity, or being 'onto'. This is where many students stumble.

A function is surjective if and only if its Range is equal to its Codomain. The problem explicitly states the codomain is $\mathbb{R}$, meaning the entire $y$-axis.

But look at our parabola. Does it ever dip below the $x$-axis? Never. For any real number $x$:
$$x^2 \ge 0$$

The range of our function is $[0, \infty)$. Because the range $[0, \infty)$ is a proper subset of the codomain $\mathbb{R}$, there are infinitely many values in the codomain (all the negative numbers) that have no pre-image.

There is no real number $x$ such that $x^2 = -5$. Thus, the function is not surjective.

### The Final Verdict

We have analyzed the function from both angles. It is not injective because it is 'many-to-one', and it is not surjective because it is 'into'.

Therefore, it is **neither**.

When you face such problems in the JEE, do not rush. Visualize the graph, check the horizontal line test for injectivity, and compare the range to the codomain for surjectivity. You have mastered the logic. Keep that curiosity alive, and keep solving!

Animated Solution for Mathematics - Functions: Let $R$ be the set of real numbers. If $f : R \to R$ is a function defined by $f (x) = x^{2}$ , then $f$ is :

Select Answer:

Visualized Solution

Understanding the Function $f (x) = x^{2}$

Graphing the Parabola

Defining Injectivity (One-One)

Applying the Horizontal Line Test

Analyzing Intersections

Mathematical Proof of Non-Injectivity

Defining Surjectivity (Onto)

Identifying the Codomain

Finding the Range

Comparing Range and Codomain

Final Conclusion

The Sigma Insight: Classification of Functions

The Geometry of the Parabola

The Injective Test

The Surjective Trap

The Final Verdict

Similar Questions

The function $f : R \to [- \frac{1}{2}, \frac{1}{2}]$ defined as $f (x) = \frac{x}{1 + x ^{2}}$ , is:

Let $A = {x \in R : x is not a positive integer}$ . Define a function $f : A \to R$ as $f (x) = \frac{2 x}{x - 1}$ then $f$ is

Let a function $f : (0, \infty) \to (0, \infty)$ be defined by $f (x) = ∣1 - \frac{1}{x} ∣$ . Then $f$ is:

Let function $f : R \to R$ be defined by $f (x) = 2 x + sin x$ for $x \in R$ , then $f$ is

The function $f : R \to R, f (x) = \frac{x ^{2} + 2 x - 15}{x ^{2} - 4 x + 9}, x \in R$ is

For real $x$ , let $f (x) = x^{3} + 5 x + 1$ , then

The function $f : (- \infty, \infty) \to (- \infty, 1)$ , defined by $f (x) = \frac{2 ^{x} - 2 ^{- x}}{2 ^{x} + 2 ^{- x}}$ is:

If the function $f : R \to R$ is defined by $f (x) = ∣ x ∣ (x - sin x)$ , then which of the following statements is TRUE?

Let a function $f : N \to N$ be defined by $f (n) = ⎩ ⎨ ⎧ 2 n, n - 1, \frac{n + 1}{2}, n = 2, 4, 6, 8, \dots n = 3, 7, 11, 15, \dots n = 1, 5, 9, 13, \dots$ then, $f$ is

A function $f$ from the set of natural numbers to integers defined by $f (n) = {\frac{n - 1}{2}, - \frac{n}{2}, when n is odd when n is even$ is

Animated Solution for Mathematics - Functions: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let R be the set of real numbers. If f:R→R is a function defined by f(x)=x2, then f is :