Q.Let $f: \mathbf{R} \rightarrow \mathbf{R}$ be defined as $f(x) = x^4$. Choose the correct answer. (A) $f$ is one-one onto (B) $f$ is many-one onto (C) $f$ is one-one but not onto (D) $f$ is neither one-one nor onto.

Concept: Function Properties (One-One and Onto)

A function is one-one (injective) if $f(a) = f(b)$ implies $a = b$.

Here, $f(x) = x^4$ gives $f(2) = 16$ and $f(-2) = 16$, so $2 \neq -2$ but outputs are equal — hence not one-one (it is many-one). …

The question asks us to classify $f(x)=x^4$ from $\mathbb{R}$ to $\mathbb{R}$ according to two properties: being one-one (injective) and being onto (surjective). Let’s understand what each means.

A function is one-one if $f(a)=f(b)$ forces $a=b$. Equivalently, different inputs must map to different outputs. A function is onto if every element in the codomain ($\mathbb{R}$ here) is actually hit by some input — that is, the range equals the codomain.

For $f(x)=x^4$, the first thing to notice is that squaring (or raising to an even power) destroys sign information: $(-2)^4 = 16$ and $(2)^4 = 16$. So two different inputs give the same output. That immediately kills injectivity.

For surjectivity: $x^4$ is always non-negative. The codomain is all real numbers, which includes negatives like $-1$. Since $x^4$ can never be negative, those numbers are never reached. So the function is not onto either.

Let’s walk through the reasoning step by step.

1. Check one-one (injectivity).

   Take $a=1$ and $b=-1$. Then $f(1)=1^4=1$ and $f(-1)=(-1)^4=1$. So $f(1)=f(-1)$ but $1 \neq -1$. This is a direct counterexample to the one-one property.

   More generally, for any $x \neq 0$, $f(x)=f(-x)$ with $x \neq -x$, so the function is many-one.

2. Check onto (surjectivity). …

Method: Classifying an even-power function on R (MCQ)

For $f(x)=x^{2n}$ on $\mathbf{R}\to\mathbf{R}$, decide one-one and onto with two quick observations, then match the option.

Steps

Step 1: One-one? — an even power satisfies $f(-x)=f(x)$, so any $x\neq 0$ produces a repeated output (e.g. $(-1)^4=1^4$). Hence it is many-one, not one-one. …

Common Mistakes

Mistake 1: Thinking $x^4$ is one-one because it increases for $x\ge 0$.

Why it's wrong: the domain is all of $\mathbf{R}$, and $f(-x)=f(x)$ (e.g. $(-1)^4=1^4=1$) makes it many-one over the full domain. Correct approach: test negative inputs too before judging injectivity.

Mistake 2: Assuming $x^4$ is onto $\mathbf{R}$. …