(a) If f : R^+ → R is defined as f(x) = loga x (a > 0 a — CBSE Class 12 Mathematics 2025 PYQ

Question

f : R^+ \to R

(OR)

(b) Let

A = \{1, 2, 3\}

and

B = \{4, 5, 6\}

. A relation R from A to B is defined as

R = \{(x, y) : x + y = 6, x \in A, y \in B\}

(i) Write all elements of R.

(ii) Is R a function ? Justify.

(iii) Determine domain and range of R.

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Part (a)Concept understanding — Bijection Proof

Proving a Function is a Bijection

A bijection is a function that is both one-one (injective) and onto (surjective). To prove that a given function $f:A\to B$ is a bijection, you establish these two things separately — there is no shortcut that does both at once.

Step 1 — prove it is one-one

Assume $f(x_1)=f(x_2)$ for $x_1,x_2\in A$ and derive $x_1=x_2$ .

Note

Injectivity means no two different inputs share an output.

Step 2 — prove it is onto

Take an arbitrary $y\in B$ and produce an $x\in A$ — usually by solving $f(x)=y$ for $x$ — such that $f(x)=y$ , checking that this $x$ really lies in $A$ .

Note

Surjectivity means every element of the codomain is used.

Once both parts hold, $f$ is a bijection.

Watch out

Both parts are compulsory. A function can be one-one but not onto (e.g. $f:\mathbb{N}\to\mathbb{N},\ f(x)=2x$ misses the odd numbers) or onto but not one-one. Proving only one property does not prove a bijection.

A worked template

To show $f:\mathbb{R}\to\mathbb{R},\ f(x)=2x+3$ is a bijection:

One-one: $2x_1+3=2x_2+3 \Rightarrow 2x_1=2x_2 \Rightarrow x_1=x_2.$
Onto: given any $y\in\mathbb{R}$ , set $x=\dfrac{y-3}{2}\in\mathbb{R}$ ; then $f(x)=2\!\left(\dfrac{y-3}{2}\right)+3=y.$

So $f$ is a bijection, with inverse $f^{-1}(y)=\dfrac{y-3}{2}$ .

Why it matters …

Part (b)Concept understanding — Relation Properties

Properties of a Relation

A relation $R$ on a set $A$ pairs elements of $A$ with one another. Some relations behave in regular, predictable ways, and we name these behaviours properties. Three matter most for CBSE Class 12 — reflexive, symmetric, transitive (together they build an equivalence relation); a fourth, antisymmetric, is worth knowing for order relations.

Reflexive — everything relates to itself

$R$ is reflexive if $a\,R\,a$ for every $a\in A$ . "Has the same age as" is reflexive; "is taller than" is not. If even one element misses its self-pair, reflexivity fails: on $\{1,2,3\}$ , $\{(1,1),(2,2)\}$ is not reflexive because $(3,3)$ is absent.

Symmetric — the relation runs both ways

$R$ is symmetric if $a\,R\,b \implies b\,R\,a$ . "Is married to" is symmetric; "is taller than" is not. Symmetry does not demand that every pair be related — only that any pair which appears also appears reversed. So $\{(1,2),(2,1),(3,3)\}$ is symmetric, but $\{(1,2),(2,1),(1,3)\}$ is not, since $(3,1)$ is missing.

Transitive — relations chain

$R$ is transitive if $a\,R\,b$ and $b\,R\,c$ together force $a\,R\,c$ . "Is an ancestor of" is transitive; "is a friend of" is not. A single broken chain breaks the property: $\{(1,2),(2,3)\}$ is not transitive because $(1,3)$ is missing.

Antisymmetric — two-way ties force equality

$R$ is antisymmetric if $a\,R\,b$ and $b\,R\,a$ together force $a=b$ . The order relation $\le$ is antisymmetric: $a\le b$ and $b\le a$ give $a=b$ . It does not ban self-pairs like $(1,1)$ ; it only rules out distinct elements related both ways.

Tip

Test the properties in order of ease — reflexivity first, then symmetry, transitivity. A single counterexample is enough to disprove any of them.

Property	Condition
Reflexive	$\forall a,\ a\,R\,a$

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(a) $f(x)=\log_a x$ is one-one and onto $\Rightarrow$ bijection. (b) $R=\{(1,5),(2,4)\}$ ; not a function; domain $\{1,2\}$ , range $\{4,5\}$ .

Part (a)

Let $f:\mathbb{R}^+\to\mathbb{R}$ , $f(x)=\log_a x$ , with $a>0,\ a\neq1$ .

Injective (one-one): Suppose $f(x_1)=f(x_2)$ for $x_1,x_2\in\mathbb{R}^+$ . Then $\log_a x_1=\log_a x_2$ . Raising $a$ to both sides, $a^{\log_a x_1}=a^{\log_a x_2}$ , i.e. $x_1=x_2$ . Hence $f$ is one-one.

Surjective (onto): Let $y\in\mathbb{R}$ be arbitrary. Choose $x=a^y$ ; since $a>0$ , $x=a^y>0$ so $x\in\mathbb{R}^+$ . Then

f(x)=\log_a(a^y)=y.

Thus every $y\in\mathbb{R}$ has a pre-image, so $f$ is 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 45 on this concept.

CBSE 2026Set V11 markMCQ
$R$
(a) symmetric but not transitive
(b) transitive but not symmetric
(c) symmetric and transitive
(d) neither symmetric nor transitive
›Reveal solutionSolution

$R=\{(1,1),(2,2)\}$ is both symmetric and transitive, so the answer is (c).

Symmetry: whenever $(a,b)\in R$ we need $(b,a)\in R$ . Here the only pairs are $(1,1)$ and $(2,2)$ , and each is its own reverse, so symmetry holds. …
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CBSE 2026Set A1 markMCQ
$Q. What type of a relation is "less than" in the set of real numbers?$
(a) Only symmetric
(b) Only transitive
(c) Only reflexive
(d) Equivalence
›Reveal solutionSolution
" $<$ " on $\mathbb{R}$ is transitive only.

Consider the relation " $a < b$ " on $\mathbb{R}$ :

Reflexive? $a < a$ is false for every $a$ , so NOT reflexive.

Symmetric? $a < b$ does not imply $b < a$ , so NOT symmetric. …
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CBSE 2026Set ANNUAL1 markMCQ
$R$
(a) Reflexive but not symmetric
(b) Reflexive but not transitive
(c) Symmetric and transitive
(d) Neither symmetric nor transitive
›Reveal solutionSolution

$R$ is reflexive (it contains every $(x,x)$ ) but not symmetric, since $(1,2)\in R$ while $(2,1)\notin R$ .

Given: $R=\{(1,1),(2,2),(3,3),(1,2),(2,3),(1,3)\}$ on $A=\{1,2,3\}$ .

Reflexivity: $R$ is reflexive if $(x,x)\in R$ for every $x\in A$ . Here $(1,1),(2,2),(3,3)$ are all present, so $R$ is reflexive.

Symmetry: $R$ is symmetric if $(x,y)\in R\Rightarrow(y,x)\in R$ . Here $(1,2)\in R$ but $(2,1)\notin R$ . So $R$ is not symmetric.

…

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CBSE 2026Set ANNUAL1 markMCQ
$R$
(a) $(2, 4) \in R$
(b) $(3, 8) \in R$
(c) $(6, 8) \in R$
(d) $(8, 6) \in R$
›Reveal solutionSolution
Check each pair against both conditions $a=b-2$ and $b>6$ .

$R=\{(a,b): a=b-2,\ b>6\}$

$(2,4)$ : $a=b-2 \Rightarrow 2=2$ ✓, but $b>6 \Rightarrow 4>6$ ✗. Not in $R$ .

$(3,8)$ : $a=b-2 \Rightarrow 3=6$ ✗. Not in $R$ . …
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CBSE 2026Set ANNUAL1 markMCQ
$R$
(a) $R$ is reflexive and symmetric but not transitive
(b) $R$ is reflexive and transitive but not symmetric
(c) $R$ is symmetric and transitive but not reflexive
(d) $R$ is an equivalence relation
›Reveal solutionSolution
Test $R$ against the three defining properties (reflexive, symmetric, transitive) one at a time, directly from its listed ordered pairs.

$R=\{(1,2),(2,2),(1,1),(4,4),(1,3),(3,3),(3,2)\}$ on $\{1,2,3,4\}$ .

Reflexive? Need $(a,a)\in R$ for every $a\in\{1,2,3,4\}$ : $(1,1)$ ✓, $(2,2)$ ✓, $(3,3)$ ✓, $(4,4)$ ✓. All present — $R$ is reflexive.

Symmetric? Need: whenever $(a,b)\in R$ , also $(b,a)\in R$ . Take $(1,2)\in R$ : is $(2,1)\in R$ ? It is not in the list. So $R$ is not symmetric.

Transitive? Need: whenever $(a,b)\in R$ and $(b,c)\in R$ , also $(a,c)\in R$ . Checking every chain:

$(1,1),(1,2)\Rightarrow(1,2)$ ✓

$(1,1),(1,3)\Rightarrow(1,3)$ ✓

$(1,2),(2,2)\Rightarrow(1,2)$ ✓

$(1,3),(3,3)\Rightarrow(1,3)$ ✓

$(1,3),(3,2)\Rightarrow(1,2)$ ✓ …
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CBSE 2026Set ANNUAL1 markMCQ
$Q.Assertion (A): Let Z be the set of integers. A function f: Z -> Z defined as f(x)=2x-3, \forall x \in Z is bijective. Reason (R): A function is a bijective if it is both injective and surjective.$
(a)
(i) Both A and R are correct and R is the correct explanation of A.
(b)
(ii) Both A and R are correct but R is not the correct explanation of A.
(c)
(iii) A is correct but R is incorrect.
(d)
(iv) Both A and R are incorrect.
›Reveal solutionSolution
$f$ is injective but not surjective onto $\mathbb{Z}$ , so it is not bijective — A is false; R (the definition) is true. This corresponds to the standard assertion–reason option "A is incorrect but R is correct."

Concept. A function is bijective iff it is both injective (one-one) and surjective (onto). For $f:\mathbb{Z}\to\mathbb{Z}$ , onto means every integer must be attained.

Steps.

Injective: if $f(x_1)=f(x_2)$ then $2x_1-3=2x_2-3\Rightarrow x_1=x_2$ . So $f$ is one-one. ✓

Surjective? For $f(x)=n$ we need $2x-3=n\Rightarrow x=\dfrac{n+3}{2}$ . This is an integer only when $n$ is odd. For an even integer such as $n=0$ , $x=\dfrac{3}{2}\notin\mathbb{Z}$ , so $0$ has no pre-image in $\mathbb{Z}$ . Hence $f$ is not onto $\mathbb{Z}$ . ✗

Therefore $f$ is not bijective: Assertion (A) is false.

Reason (R) states the correct definition of a bijection, so R is true. …
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CBSE 2026Set ANNUAL1 markMCQ
$R$
(a) $(2, 4) \in R$
(b) $(3, 8) \in R$
(c) $(6, 8) \in R$
(d) $(8, 7) \in R$
›Reveal solutionSolution

Only $(6,8)$ satisfies $a=b-2$ with $b>6$ .

…

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CBSE 20251 mark
$Z$
›Reveal solutionSolution
The key idea is to check whether $f(x)=3x-5$ is both injective (one-one) and surjective (onto) over the integers. It is one-one, but not onto because not every integer can be written as $3x-5$ for integer $x$ . So Assertion (A) is false, Reason (R) is true.

Let’s unpack this carefully. The question gives us an Assertion (A) and a Reason (R), and we need to decide if each is true, and if (R) correctly explains (A).

The function is $f: \mathbb{Z} \to \mathbb{Z}$ , $f(x) = 3x - 5$ . The domain and codomain are both the set of all integers. The Reason (R) simply states the definition of a bijection — that’s a standard fact, so it’s true. The real work is in checking whether $f$ is actually bijective.

Why the bijection proof approach works

To test if $f$ is bijective, we check two properties separately:

One-one (injective): If $f(a) = f(b)$ , does it force $a = b$ ?

Onto (surjective): For every integer $y$ in the codomain, does there exist an integer $x$ such that $f(x) = y$ ?

If either fails, the function is not bijective. Let’s go step by step.

Step-by-step reasoning

Check one-one (injectivity) Suppose $f(a) = f(b)$ for integers $a, b$ . Then:

$3a - 5 = 3b - 5$
Adding 5 to both sides: $3a = 3b$ . Dividing by 3 (which is allowed over integers): $a = b$ .

So $f$ is one-one. No integer can map to the same output from two different inputs.

Check onto (surjectivity) We need: for any integer $y$ , can we find an integer $x$ such that $3x - 5 = y$ ? Solve for $x$ :

$3x = y + 5 \quad \Rightarrow \quad x = \frac{y + 5}{3}$
For $x$ to be an integer, $y+5$ must be divisible by 3. That means $y \equiv 1 \pmod{3}$ (since $y+5 \equiv 0 \pmod{3}$ implies $y \equiv -5 \equiv 1 \pmod{3}$ ).

So only integers of the form $y = 3k + 1$ (for integer $k$ ) are hit. For example, $y = 0$ gives $x = \frac{5}{3}$ , not an integer — so 0 is not in the range. …
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CBSE 2025Set 65/1/11 markMCQ
$\mathbb{Z}$
›Reveal solutionSolution
The function $f(x) = 3x - 5$ from integers to integers is injective but not surjective, so it is not bijective. The reason states the correct definition of a bijective function. Therefore, Assertion (A) is false, and Reason (R) is true. The correct option is (D).

To determine if a function is bijective, we must check two properties: injectivity (one-to-one) and surjectivity (onto). A function is bijective if and only if it possesses both these properties.

Evaluate Reason (R):

Reason (R) states: "A function is a bijective if it is both surjective and injective."

This statement is the fundamental definition of a bijective function. A function that is both injective and surjective is indeed called a bijection.

Therefore, Reason (R) is true.

Evaluate Assertion (A):

Assertion (A) states: "Let $\mathbb{Z}$ be the set of integers. A function $f : \mathbb{Z} \to \mathbb{Z}$ defined as $f(x) = 3x - 5$ , $\forall x \in \mathbb{Z}$ is a bijective."

We need to check if $f(x) = 3x - 5$ is both injective and surjective.

Check for Injectivity (One-to-one): A function $f: A \to B$ is injective if for any $x_1, x_2 \in A$ , $f(x_1) = f(x_2)$ implies $x_1 = x_2$ . Let $x_1, x_2 \in \mathbb{Z}$ such that $f(x_1) = f(x_2)$ .

$3x_1 - 5 = 3x_2 - 5$
Add 5 to both sides:
$3x_1 = 3x_2$
Divide by 3:
$x_1 = x_2$
Since $f(x_1) = f(x_2)$ implies $x_1 = x_2$, the function $f$ is injective. * **Check for Surjectivity (Onto):** A function $f: A \to B$ is surjective if for every element $y$ in the codomain $B$, there exists at least one element $x$ in the domain $A$ such that $f(x) = y$. Here, the codomain is $\mathbb{Z}$. We need to check if for every $y \in \mathbb{Z}$, there exists an $x \in \mathbb{Z}$ such that $f(x) = y$. Set $f(x) = y$:
$3x - 5 = y$
Solve for $x$: …
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CBSE 2025Set X11 markMCQ
$R$
(a) $(a, a) \in R$ for all $a \in A$
(b) $(a, a) \in R$ for atleast one $a \in A$
(c) $(a, b) \in R$ implies $(b, a) \in R$
(d) $(a, b) \in R$ and $(b, c) \in R$ implies $(a, c) \in R$
›Reveal solutionSolution

Tests the definition of a reflexive relation — correct option is (a).

A relation $R$ on a set $A$ is reflexive when every element is related to itself. Formally, $(a,a)\in R$ must hold for all $a\in A$ — not just for at least one element. Option (c) states symmetry ( $(a,b)\in R\Rightarrow(b,a)\in R$ ) and option (d) states transitivit …
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CBSE 2025Set IX1 markMCQ
$R = \{(a, b) : a = b - 1,\ b \geq 3\}$
(a) $(2, 4) \in R$
(b) $(4, 5) \in R$
(c) $(4, 6) \in R$
(d) $(1, 3) \in R$
›Reveal solutionSolution
Only $(4,5)$ satisfies $a=b-1$ with $b\ge 3$ ; option (b).

Concept. A pair $(a,b)$ belongs to $R$ only if it satisfies both conditions: $a=b-1$ and $b\ge 3$ .

$(2,4)$ : $b-1=3\ne 2$ . ✗ …
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CBSE 2025Set ANNUAL1 markMCQ
$A=\{1,2,3,4\}$
(i) Reflexive
(ii) Symmetric
(iii) Transitive
(iv) Equivalent
›Reveal solutionSolution

Check each property directly: $R$ turns out to be symmetric only.

$A=\{1,2,3,4\}$ , $R=\{(a,b): a+b \text{ is odd}\}$ .

Reflexive? $(a,a)$ needs $a+a=2a$ to be odd — but $2a$ is always even. So $R$ is not reflexive.

Symmetric? If $a+b$ is odd, then $b+a=a+b$ is the same sum, also odd. So $(a,b)\in R \Rightarrow (b,a)\in R$ . $R$ is symmetric.

…

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Q.(a) If f:R+→Rf : R^+ \to Rf:R+→R is defined as f(x)=log⁡axf(x) = \log_a xf(x)=loga​x (a>0a > 0a>0 and a≠1a \neq 1a=1), prove that fff is a bijection. (R+R^+R+ is a set of all positive real numbers.)

Part (a)

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Q.(a) If $f : R^+ \to R$ is defined as $f(x) = \log_a x$ ( $a > 0$ and $a \neq 1$ ), prove that $f$ is a bijection. ( $R^+$ is a set of all positive real numbers.)

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