(a) Check whether $f : R - \{3\} \to R$ defined as $f(x) = \frac{x-2}{x-3}$ is onto or not. OR (b) Check whether $f : Z \times Z \to Z \times Z$ (where $Z$ is the set of integers) defined as $f(x, y) = (2y, 3x)$ is injective or not.

(a) Check whether f : R - \3\ → R defined as f(x) = (x- — CBSE Class 12 Mathematics 2026 PYQ

Question

f : R - \{3\} \to R

(OR)

(b) Check whether

f : Z \times Z \to Z \times Z

(where

Z

is the set of integers) defined as

f(x, y) = (2y, 3x)

is injective or not.

CBSECBSE Class XII Board 2026Subjective· 2mImportance★★★★★

🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

Part (a)Concept understanding — Onto Function Definition

Onto (Surjective) Functions

A function $f : A \to B$ is onto — also called surjective — if every element of the codomain $B$ is hit by at least one input from $A$ . Nothing in $B$ is left over.

Note

Picture posting letters into pigeonholes: the function is onto when every pigeonhole receives at least one letter. It doesn't matter if some hole gets two or three — only that none stays empty.

Precise statement

$f : A \to B$ is onto if

\text{for every } y \in B, \ \text{there exists } x \in A \text{ such that } f(x) = y.

Equivalently, the range equals the codomain: $f(A) = B$ .

How to check

Take an arbitrary $y$ in the codomain and try to solve $f(x) = y$ for some $x$ in the domain. If you can always produce such an $x$ , the function is onto.

Example. $f : \mathbb{R} \to \mathbb{R}$ , $f(x) = 2x + 1$ . For any $y$ , choose $x = \dfrac{y-1}{2} \in \mathbb{R}$ , and $f(x) = y$ . So $f$ is onto.

Non-example. $f : \mathbb{R} \to \mathbb{R}$ , $f(x) = x^2$ . No $x$ gives $f(x) = -1$ , because squares are never negative. The negative reals are never reached, so $f$ is not onto. …

Part (b)Concept understanding — One One Function

One-One (Injective) Function

Think of taking attendance by unique roll numbers: call a number and exactly one student responds — never two sharing a number. That is a one-one function: different inputs always land on different outputs.

The idea

A function is a machine turning inputs into outputs. It is one-one (or injective) if it never reuses an output — two different inputs can never produce the same result.

$f(x)=x+1$ is one-one: if $x_1\neq x_2$ then $x_1+1\neq x_2+1$ .
$g(x)=x^2$ on $\mathbb{R}$ is not one-one, because $g(2)=g(-2)=4$ .

Precise definition

$f:A\to B$ is one-one if for all $x_1,x_2\in A$ ,

x_1\neq x_2 \implies f(x_1)\neq f(x_2).

The contrapositive is usually easier in proofs:

f(x_1)=f(x_2)\implies x_1=x_2.

"If the outputs are equal, the inputs must have been equal."

How to check

Horizontal line test (graphs): if any horizontal line meets the graph more than once, the function is not one-one, because that line marks one output shared by several inputs.
Algebraic test: assume $f(x_1)=f(x_2)$ and try to deduce $x_1=x_2$ ; succeed and it is one-one, find a counterexample and it is not.

Tip

A strictly increasing or strictly decreasing function is automatically one-one. So a decreasing function like $f(x)=-x$ is one-one too — being one-one is about no repeated outputs, not about going up.

Why it matters …

Start your 14-day free trial to unlock the full solution →

(a) $f(x)=\frac{x-2}{x-3}$ has range $\mathbb{R}-\{1\}\neq\mathbb{R}$ , so it is not onto. (b) $f(x,y)=(2y,3x)$ is one-one because equal images force equal components.

Part (a)

A function $f:A\to B$ is onto (surjective) if its range equals the codomain $B$ , i.e. every $y\in B$ has some $x\in A$ with $f(x)=y$ .

Here $A=\mathbb{R}-\{3\}$ , $B=\mathbb{R}$ , $f(x)=\dfrac{x-2}{x-3}$ . Set $y=f(x)$ and solve for $x$ :

$y(x-3)=x-2\Rightarrow yx-3y=x-2\Rightarrow x(y-1)=3y-2\Rightarrow x=\frac{3y-2}{y-1}.$ …

Start your 14-day free trial to unlock the full solution →

Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

Showing the 12 most recent of 17 on this concept.

CBSE 2026Set A1 markMCQ
$f : A \to B$
(a) $f(A) \subset B$
(b) $f(A) = B$
(c) $f(A) \supset B$
(d) none of these
›Reveal solutionSolution

A function is into when its range does not fill the whole codomain: $f(A)\subset B$ .

For $f : A \to B$ , the range is $f(A) \subseteq B$ . The function is onto (surjective) when $f(A) = B$ . It is into when at least one element of $B$ is …

Start your 14-day free trial to see the full solution →
CBSE 2026Set ANNUAL1 mark
$f: X \to Y$
›Reveal solutionSolution

True - onto $\Leftrightarrow$ range $=Y$ .

A function $f:X\to Y$ is onto (surjective) if every element of $Y$ is an image of …

Start your 14-day free trial to see the full solution →
CBSE 2025Set E1 markMCQ
$f : A \to B$
(a) $f(A) \subset B$
(b) $f(A) = B$
(c) $f(A) \supset B$
(d) $f(A) \neq B$
›Reveal solutionSolution

$f$ is onto (surjective) precisely when the range $f(A)$ equals the whole codomain $B$ .

A function $f:A\to B$ is onto if every element of $B$ is the image of at least one element of $A$ ; that is, the range equals the codomain: …

Start your 14-day free trial to see the full solution →
CBSE 2025Set ANNUAL1 markMCQ
$f(a) = f(b) \Rightarrow a = b \; \forall a, b \in A$
(a) one-one
(b) constant
(c) onto
(d) many one
›Reveal solutionSolution

The statement given is the textbook definition of a one-one (injective) function.

A function f : A → B is called one-one (or injective) if distinct elements of A always map to distinct elements of B — equivalently, if f(a) = f(b) forces a = b (no two different inputs can share an output).

This is precisely the condition stated in the question, so f is one-one.

…

Start your 14-day free trial to see the full solution →
CBSE 2025Set ANNUAL1 mark
$f : \mathbb{R} \to \mathbb{R}$
›Reveal solutionSolution

Exhibit a counterexample: two distinct inputs with the same image.

A function $f$ is one-one (injective) if $f(x_1)=f(x_2)\Rightarrow x_1=x_2$ . To show $f(x)=x^2$ is not one-one, we produce two distinct inputs with equal outputs.

Take $x_1=1$ and $x_2=-1$ . These are distinct: $1\ne-1$ . But

$f(1)=1^2=1,\qquad f(-1)=(-1)^2=1,$

so $f(1)=f(-1)$ .

…

Start your 14-day free trial to see the full solution →
CBSE 2024Set ANNUAL1 markMCQ
$f : \mathbf{Z} \to \mathbf{Z}$
(a) Both A and R are correct and R is the correct explanation of A.
(b) Both A and R are correct but R is not the correct explanation of A.
(c) A is correct but R is incorrect.
(d) Both A and R are incorrect.
›Reveal solutionSolution

Check A (is $f$ one-one?) and R (is $f$ onto?) independently, then judge whether R explains A.

Checking Assertion (A): $f:\mathbf{Z}\to\mathbf{Z}$ , $f(x)=2x$ . If $f(x_1)=f(x_2)$ , then $2x_1=2x_2 \Rightarrow x_1=x_2$ . So $f$ is one-one. A is TRUE.

Checking Reason (R): The range of $f$ is $\{\ldots,-4,-2,0,2,4,\ldots\}$ , i.e. only even integers. Odd integers in the codomain $\mathbf{Z}$ (e.g. 1) have no pre-image. So $f$ is not onto. R is TRUE.

…

Start your 14-day free trial to see the full solution →
CBSE 2024Set D1 markMCQ
$f : A \to B$
(a) $f(A) \subset B$
(b) $f(A) = B$
(c) $f(A) \supset B$
(d) None of these
›Reveal solutionSolution

Onto $\Leftrightarrow f(A) = B$ .

A function $f : A \to B$ is called onto (surjective) if every element of the codomain $B$ is the image of at least one element of $A$ . Equivalently, the range (image) of $f$ equals the whole codomain:

$f(A) = B.$ …

Start your 14-day free trial to see the full solution →
CBSE 2024Set A1 mark
$f: X \to Y$
›Reveal solutionSolution

The stated condition is the wrong implication — it should read $f(x_1)=f(x_2)\Rightarrow x_1=x_2$ .

…

Start your 14-day free trial to see the full solution →
CBSE 2023Set A1 mark
$f: x \to y$
›Reveal solutionSolution

By definition, $f:X\to Y$ is onto exactly when every element of $Y$ is the image of some element of $X$ , i.e. range $=Y$ .

…

Start your 14-day free trial to see the full solution →
CBSE 2023Set ANNUAL1 mark
$f : R \to R$
›Reveal solutionSolution

For injectivity, show $f(x_1)=f(x_2)\Rightarrow x_1=x_2$ using uniqueness of real cube roots.

Let $f(x)=x^3$ and suppose $f(x_1)=f(x_2)$ for $x_1,x_2\in R$ . Then

$x_1^3=x_2^3 \Rightarrow x_1^3-x_2^3=0 \Rightarrow (x_1-x_2)(x_1^2+x_1x_2+x_2^2)=0.$

The quadratic factor $x_1^2+x_1x_2+x_2^2=\left(x_1+\tfrac{x_2}{2}\right)^2+\tfrac{3}{4}x_2^2\ge0$ , and it is zero only when $x_1=x_2=0$ . In every case we are forced to $x_1=x_2$ . Hence $f$ is one-one (injective).

…

Start your 14-day free trial to see the full solution →
CBSE 2022Set ANNUAL1 mark
$f(x) = x^2$
›Reveal solutionSolution

A function is many-one if two different inputs give the same output; $f(x)=x^2$ does this.

A function $f$ is one-one (injective) if distinct elements of the domain always map to distinct elements of the codomain; otherwise it is many-one.

Take $x_1=1$ and $x_2=-1$ in $\mathbf{R}$ . Then

$f(1)=1^2=1,\qquad f(-1)=(-1)^2=1$ …

Start your 14-day free trial to see the full solution →
CBSE 2022Set ANNUAL1 mark
$f:R\to R$
›Reveal solutionSolution

$f^{-1}(17)$ means: find all $x$ with $f(x)=17$ .

…

Start your 14-day free trial to see the full solution →

Q.(a) Check whether f:R−{3}→Rf : R - \{3\} \to Rf:R−{3}→R defined as f(x)=x−2x−3f(x) = \frac{x-2}{x-3}f(x)=x−3x−2​ is onto or not.

Part (a)

🎓Unlock everything free for 14 days

Q.(a) Check whether $f : R - \{3\} \to R$ defined as $f(x) = \frac{x-2}{x-3}$ is onto or not.

Unlock everything free for 14 days