Show that a function $f : R \to R$ defined by $f(x) = \dfrac{2x}{1 + x^2}$ is neither one-one nor onto. Further, find set $A$ so that the given function $f : R \to A$ becomes an onto function. OR A relation $R$ is defined on $N \times N$ (where $N$ is the set of natural numbers) as: $(a, b)\, R\, (c, d) \Leftrightarrow a - c = b - d$ Show that $R$ is an equivalence relation.

Show that a function f : R → R defined by f(x) = (2x)/(1 + x²) is neither one-one nor onto. Further, find set

Question

f : R \to R

(OR)

R

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Part (a)Concept understanding — One One Onto

One-One Onto (Bijective) Functions

Picture seating students on chairs so that every student gets a chair, every chair is used, no two students share one and none is left empty. A function that manages this perfect pairing between its domain and codomain is one-one onto, or bijective.

One-one (injective)

$f$ is one-one if different inputs always give different outputs — no two students on one chair. Formally, $f(x_1)=f(x_2)\implies x_1=x_2$ (equivalently $x_1\neq x_2\implies f(x_1)\neq f(x_2)$ ).

$f(x)=2x$ on $\mathbb{R}$ is one-one, since $2a=2b\Rightarrow a=b$ . But $f(x)=x^2$ is not: $f(2)=f(-2)=4$ while $2\neq-2$ .

Onto (surjective)

$f$ is onto if every element of the codomain is actually hit — no chair left empty. Formally, for every $y$ in the codomain there is some $x$ with $f(x)=y$ . Here $f(x)=2x$ is onto (take $x=y/2$ ), whereas $f:\mathbb{R}\to\mathbb{R},\ f(x)=x^2$ is not, since negative values are never outputs.

Both together — bijective

A function that is one-one and onto is bijective: a one-to-one correspondence in which the two sets match up exactly.

Important

One-one and onto are independent properties. $f(x)=e^x$ (from $\mathbb{R}$ to $\mathbb{R}$ ) is one-one but not onto; $f(x)=x^3-x$ is onto but not one-one. You must verify both.

Why it matters

Only a bijection has a genuine inverse function: because each output comes from exactly one input (one-one) and every codomain element is used (onto), the map can be reversed unambiguously. …

Part (b)Concept understanding — Equivalence Relation Proof

Proving a Relation is an Equivalence Relation

A relation $R$ on a set $A$ is an equivalence relation when it satisfies exactly three properties: it is reflexive, symmetric, and transitive. To prove a given relation is an equivalence relation, you check these three — in this order — one at a time.

Note

Antisymmetry plays no role here; that property belongs to partial orders. For an equivalence relation you need only reflexive, symmetric, transitive.

The three checks

Reflexive — show $(a, a) \in R$ for every $a \in A$ .
Symmetric — assume $(a, b) \in R$ and deduce $(b, a) \in R$ .
Transitive — assume $(a, b) \in R$ and $(b, c) \in R$ , and deduce $(a, c) \in R$ .

If all three hold, $R$ is an equivalence relation. If even one fails, produce a single counterexample and you are done.

A worked template

Let $R$ be defined on $\mathbb{Z}$ by $a\,R\,b \iff a - b$ is divisible by $5$ .

Reflexive: $a - a = 0$ , and $0$ is divisible by $5$ , so $a\,R\,a$ for every integer $a$ . ✓

Symmetric: if $a\,R\,b$ , then $a - b = 5k$ for some integer $k$ . Then $b - a = -5k = 5(-k)$ , also a multiple of $5$ , so $b\,R\,a$ . ✓

Transitive: if $a\,R\,b$ and $b\,R\,c$ , then $a - b = 5k$ and $b - c = 5m$ . Adding, $a - c = 5(k + m)$ , a multiple of $5$ , so $a\,R\,c$ . ✓

All three hold, so $R$ is an equivalence relation. …

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Part (a): $f(x)=\dfrac{2x}{1+x^2}$ is neither one-one nor onto (range $[-1,1]$ ); choosing $A=[-1,1]$ makes it onto. Part (b): the relation $(a,b)R(c,d)\Leftrightarrow a-c=b-d$ is reflexive, symmetric and transitive, so it is an equivalence relation.

Part (a)

Concept

Injective = distinct inputs give distinct outputs; surjective = every codomain value is attained. Find the range via the discriminant, then restrict the codomain.

Steps

Not one-one. For $x\ne0$ ,

f\!\Big(\tfrac1x\Big)=\frac{2/x}{1+1/x^2}=\frac{2x}{x^2+1}=f(x).

E.g. $f(2)=\tfrac45=f(\tfrac12)$ with $2\ne\tfrac12$ . Not one-one.

2. Not onto (find range). Set $y=\dfrac{2x}{1+x^2}\Rightarrow yx^2-2x+y=0$ . Real $x$ requires

\Delta=(-2)^2-4y\cdot y=4-4y^2\ge0\Rightarrow y^2\le1\Rightarrow y\in[-1,1].

So the range is $[-1,1]\ne\mathbb R$ ; e.g. $y=2$ has no preimage. Not onto. …

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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 58 on this concept.

CBSE 2026Set CX1 markMCQ
$f(x)=2x$
(a) one-one but not onto
(b) one-one and onto
(c) many-one and onto
(d) many-one but not onto
›Reveal solutionSolution

$f(x)=2x$ on $\mathbb{R}$ is a bijection — both one-one and onto — option (b).

One-one: If $f(x_1)=f(x_2)$ then $2x_1=2x_2\Rightarrow x_1=x_2$ . So $f$ is injective.

…

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CBSE 2026Set ANNUAL1 markMCQ
$f: R \to R$
(a) one-one onto
(b) onto only
(c) neither one-one nor onto
(d) none of these
›Reveal solutionSolution

$f(x)=3-4x$ is a linear function with non-zero slope, so it is both one-one and onto.

Given $f:\mathbb{R}\to\mathbb{R}$ , $f(x)=3-4x$ .

One-one: Let $f(x_1)=f(x_2)$ . Then $3-4x_1=3-4x_2 \Rightarrow x_1=x_2$ . So $f$ is injective.

…

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CBSE 2026Set ANNUAL1 markMCQ
$Q. If A = {0, 1, 4, 9, 16, 25, ......} then function defined by f: Z \to A, f(x) = x² is:$
(a) one-one but not onto
(b) onto but not one-one
(c) one-one and onto
(d) neither one-one nor onto
›Reveal solutionSolution

Two different integers with the same absolute value (like $2$ and $-2$ ) give the same square, so $f$ is not one-one; but every element of $A$ is a perfect square that some integer squares to, so $f$ is onto.

$f:\mathbb{Z}\to A$ is defined by $f(x) = x^2$ , where $A=\{0,1,4,9,16,25,\dots\}$ is the set of all perfect squares of non-negative integers.

One-one check: Take $x=2$ and $x=-2$ . Both are in $\mathbb{Z}$ and $f(2) = 4 = f(-2)$ , but $2\ne -2$ . Different inputs give the same output — $f$ is not one-one.

…

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CBSE 2026Set ANNUAL1 markMCQ
$Q. Let f : R \to R be defined by f(x) = 3x, choose the correct answer:$
(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
›Reveal solutionSolution

$f(x)=3x$ is a straight-line map with non-zero slope, so it is both injective and surjective on $\mathbb{R}$ — a bijection.

Checking one-one (injective):

Let $f(x_1)=f(x_2)$ . Then $3x_1=3x_2 \Rightarrow x_1=x_2$ . So distinct inputs never share an output — $f$ is one-one.

Checking onto (surjective): …

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CBSE 2026Set ANNUAL1 mark
$f : \mathbb{R} \to \mathbb{R}$
›Reveal solutionSolution

Prove injectivity by showing $f(x_1)=f(x_2)$ forces $x_1=x_2$ , and prove surjectivity by exhibiting a pre-image for an arbitrary $y\in\mathbb{R}$ .

Given $f:\mathbb{R}\to\mathbb{R}$ , $f(x)=2x$ .

Step 1: One-one (injective).

Let $x_1,x_2\in\mathbb{R}$ such that $f(x_1)=f(x_2)$ .

$2x_1 = 2x_2 \implies x_1 = x_2$

So $f(x_1)=f(x_2) \implies x_1=x_2$ , hence $f$ is one-one.

Step 2: Onto (surjective).

Let $y\in\mathbb{R}$ be arbitrary (any element of the codomain). We need $x\in\mathbb{R}$ (the domain) such that $f(x)=y$ .

…

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CBSE 2026Set ANNUAL1 markMCQ
$f : \mathbb{R} \to \mathbb{R}$
(a) $f$ is one-one and onto
(b) $f$ is many-one and onto
(c) $f$ is one-one but not onto
(d) $f$ is neither one-one nor onto
›Reveal solutionSolution

Find a counter-example pair with equal outputs to disprove one-one, and note the range excludes negative numbers to disprove onto.

Given $f:\mathbb{R}\to\mathbb{R}$ , $f(x)=x^4$ .

One-one? Consider $x_1=1$ and $x_2=-1$ :

$f(1) = 1^4 = 1, \qquad f(-1) = (-1)^4 = 1$

$f(1)=f(-1)=1$ but $1\ne-1$ . So $f$ is not one-one (it is many-one).

…

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CBSE 2026Set ANNUAL1 mark
$Q. Write the answer in one word/sentence: Define equivalence relation.$
›Reveal solutionSolution

Equivalence relation = reflexive + symmetric + transitive.

…

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CBSE 2025Set 65/4/11 markMCQ
$x$
›Reveal solutionSolution
The function $f(x)=x^3+5x+1$ is strictly increasing (since $f'(x)=3x^2+5>0$ for all real $x$ ), so it is one-one. As a cubic with odd degree and positive leading coefficient, its range is all real numbers, so it is onto $\mathbb{R}$ . Hence the correct option is (C).

Understanding one-one (injective) — why the derivative tells the story

A function is one-one if different inputs give different outputs. For a differentiable function, a sufficient condition is that the derivative never changes sign — that is, the function is strictly monotonic (always increasing or always decreasing).

Here, $f'(x) = 3x^2 + 5$ . Since $x^2 \ge 0$ for all real $x$ , we have $3x^2 \ge 0$ , so $3x^2 + 5 \ge 5 > 0$ . The derivative is always positive.

Therefore $f$ is strictly increasing on $\mathbb{R}$ . A strictly increasing function is automatically one-one: if $x_1 < x_2$ , then $f(x_1) < f(x_2)$ , so no two distinct $x$ 's can map to the same $y$ .

Understanding onto (surjective) — why the range is all reals

A function $f: \mathbb{R} \to \mathbb{R}$ is onto if every real number appears as an output. For a polynomial of odd degree with a positive leading coefficient, the end behaviour guarantees this:

As $x \to -\infty$ , $x^3 \to -\infty$ , so $f(x) \to -\infty$ .

As $x \to +\infty$ , $x^3 \to +\infty$ , so $f(x) \to +\infty$ . Since $f$ is continuous (every polynomial is continuous), by the Intermediate Value Theorem it takes every value between $-\infty$ and $+\infty$ . That is, the range is $\mathbb{R}$ . …
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CBSE 2025Set 65/4/11 markMCQ
$f : N \to W$
›Reveal solutionSolution
The function maps all odd naturals to 0 and each even natural to half its value, so it is surjective onto $W$ (every whole number is hit) but not injective (many inputs give the same output). The correct option is (B).

The core idea here is to understand what the function does to its domain, $N$ (the set of natural numbers, typically $\{1,2,3,\dots\}$ ), and where it lands, $W$ (the set of whole numbers, $\{0,1,2,3,\dots\}$ ). The definition splits the domain into two clear cases: odd numbers and even numbers.

For every odd natural number — 1, 3, 5, 7, … — the output is 0. That means infinitely many inputs all map to the single output 0. That alone kills injectivity: a function is injective (one-to-one) only if different inputs always give different outputs. Here, $f(1)=0$ , $f(3)=0$ , $f(5)=0$ , and so on, so it is clearly not injective.

For every even natural number — 2, 4, 6, 8, … — the output is half of that number. So $f(2)=1$ , $f(4)=2$ , $f(6)=3$ , $f(8)=4$ , and so on. This gives us every positive whole number exactly once. And the odd numbers already cover 0. So every whole number — 0, 1, 2, 3, … — appears as an output at least once. That makes the function surjective (onto).

Let’s walk through it step by step.

Check injectivity (one-to-one)

Take two different inputs, say $n=1$ and $n=3$ . Both are odd, so $f(1)=0$ and $f(3)=0$ . Since $1 \neq 3$ but $f(1)=f(3)$ , the function is not injective.

Watch out

A common mistake is to only check the even case and think the function looks one-to-one. But the odd case collapses everything to 0 — that’s the trap.

Check surjectivity (onto)

We need to see if every element of $W$ (the codomain) is actually hit by some $n$ in $N$ . …
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CBSE 2025Set ANNUAL1 markMCQ
$f : \mathbb{R} \to \mathbb{R}$
(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
›Reveal solutionSolution

$f(x)=x^2$ on $\mathbb{R}\to\mathbb{R}$ is neither one-one nor onto.

Not one-one: $f(-1)=1=f(1)$ but $-1\neq 1$ , so two different inputs give the same output.

…

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CBSE 2025Set IX1 markMCQ
$f: R \to R^{+}$
(a) one-one and onto
(b) many-one and onto
(c) one-one but not onto
(d) neither one-one nor onto
›Reveal solutionSolution

$f(x)=|x|$ is many-one (as $f(2)=f(-2)$ ) and onto $R^{+}$ ; option (b).

Concept. A function is one-one if different inputs give different outputs, and onto if every element of the codomain is actually attained.

One-one? Take $x=2$ and $x=-2$ : $f(2)=|2|=2$ and $f(-2)=|-2|=2$ . Two different inputs share the same image, so $f$ is many-one.

…

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CBSE 2025Set A1 markMCQ
$f: R \to R$
(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.
›Reveal solutionSolution

A linear function $f(x) = 3x$ with non-zero slope is always a bijection on $\mathbb{R}$ .

Checking one-one (injective): Suppose $f(x_1) = f(x_2)$ for $x_1, x_2 \in R$ .

$3x_1 = 3x_2 \implies x_1 = x_2$

So distinct inputs always give distinct outputs — $f$ is one-one.

…

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Q.Show that a function f:R→Rf : R \to Rf:R→R defined by f(x)=2x1+x2f(x) = \dfrac{2x}{1 + x^2}f(x)=1+x22x​ is neither one-one nor onto. Further, find set AAA so that the given function f:R→Af : R \to Af:R→A becomes an onto function.

Part (a)

Concept

Steps

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Q.Show that a function $f : R \to R$ defined by $f(x) = \dfrac{2x}{1 + x^2}$ is neither one-one nor onto. Further, find set $A$ so that the given function $f : R \to A$ becomes an onto function.

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