Q.Let $A = [-1, 1]$. Then, discuss whether the following functions defined on $A$ are one-one, onto or bijective:

Concept: One-One Function — A function is one-one (injective) if distinct inputs give distinct outputs: $f(x_1) = f(x_2) \implies x_1 = x_2$.

(i) $f(x) = \frac{x}{2}$ on $[-1,1]$.

Strictly increasing (derivative $1/2 > 0$), so one-one. Range is $[-1/2, 1/2] \subset A$, so not onto. Not bijective.

(ii) $g(x) = |x|$ on $[-1,1]$.

$g(-1) = g(1) = 1$, so not one-one. Range is $[0,1] \subset A$, so not onto. Not bijective.

(iii) $h(x) = x|x|$ on $[-1,1]$. …

We need to check each function for injectivity (one-one) and surjectivity (onto) on the domain $A=[-1,1]$. The codomain is also $A$ unless stated otherwise — here, each function maps $A$ to $\mathbb{R}$, but "onto" means the range equals $A$.

Let’s go function by function.

(i) $f(x) = \frac{x}{2}$

One-one?

If $f(x_1)=f(x_2)$, then $\frac{x_1}{2}=\frac{x_2}{2} \Rightarrow x_1=x_2$. So $f$ is injective.

Onto?

The range of $f$ on $[-1,1]$ is $[-\frac12, \frac12]$. This is a proper subset of $[-1,1]$. For example, $1 \in A$ has no preimage because $\frac{x}{2}=1 \Rightarrow x=2 \notin A$. So $f$ is not onto.

Conclusion: $f$ is one-one but not onto → not bijective.

(ii) $g(x) = |x|$

One-one?

$g(-0.5)=0.5$ and $g(0.5)=0.5$, so two distinct inputs give the same output. Hence $g$ is many-one (not injective).

Onto?

The range of $|x|$ on $[-1,1]$ is $[0,1]$. Negative numbers like $-0.5$ in $A$ are never attained. So $g$ is not onto.

Conclusion: $g$ is neither one-one nor onto → not bijective.

(iii) $h(x) = x|x|$

This is the interesting one. Let’s understand the function first.

For $x \ge 0$, $|x|=x$, so $h(x)=x \cdot x = x^2$.

For $x < 0$, $|x|=-x$, so $h(x)=x \cdot (-x) = -x^2$.

So $h(x) = \begin{cases} x^2, & x \ge 0 \\ -x^2, & x < 0 \end{cases}$.

One-one?

Suppose $h(x_1)=h(x_2)$.

• If both are non-negative, $x_1^2=x_2^2 \Rightarrow x_1=x_2$ (since both $\ge 0$).
• If both are negative, $-x_1^2=-x_2^2 \Rightarrow x_1^2=x_2^2 \Rightarrow x_1=x_2$ (both negative).
• If one is non-negative and the other negative, say $x_1 \ge 0$, $x_2<0$, then $h(x_1)=x_1^2 \ge 0$ and $h(x_2)=-x_2^2 < 0$. They can never be equal. So no cross-case equality.

Thus $h$ is injective.

Onto? …

Method: Testing one-one / onto on a restricted domain

Use this when several functions share a fixed domain (and codomain) — often a closed interval like $[-1,1]$ — and you must classify each as one-one, onto, or bijective.

Steps

Step 1: Fix what 'onto' means here.

When no separate codomain is stated, treat the given set as the codomain. So 'onto' means the range equals that set, not that the function hits all real numbers. Getting this wrong flips most answers.

Step 2: Test one-one via monotonicity or a collision.

A strictly increasing OR strictly decreasing function is automatically one-one. If the function is even or uses $|x|$, look for a collision $f(-a)=f(a)$ to show it is many-one.

Step 3: Test onto by finding the actual range on the domain. …

Common Mistakes

Mistake 1: Treating 'onto' as 'hits every real number' instead of every element of $A=[-1,1]$.

Why it's wrong: the codomain here is $[-1,1]$, so onto means the range equals $[-1,1]$; $f(x)=x/2$ has range $[-1/2,1/2]$, a proper subset, hence not onto. Correct approach: compare each range against $[-1,1]$, not against $\mathbb{R}$.

Mistake 2: Assuming $x|x|$ behaves like $x^2$ and is many-one. …