Class 12 Mathematics Chapter 1 Relations and Functions

Key Concept: Function Identification

c) $\{(1, 2), (2, 3), (3, 4)\}$
[Solution Description] A function maps every element of the domain to exactly one element in the co-domain. The set $\{(1, 2), (2, 3), (3, 4)\}$ satisfies this requirement.

Key Concept: Real Number Functions

b) No
[Solution Description] For the exponential function $f(x) = e^x$, the range is $(0, \infty)$, which means it does not cover all real numbers, particularly negative numbers and zero. Therefore, it is not onto.

Key Concept: Function Composition

c) It is only one-one.
[Solution Description] If both $f$ and $g$ are injective, then for any two elements $x_1, x_2 \in A$, if $g(f(x_1)) = g(f(x_2))$, then $f(x_1) = f(x_2)$ because $g$ is injective. Since $f$ is also injective, we have $x_1 = x_2$.

Therefore, the composition $g \circ f$ is also injective (one-one).

Key Concept: Symmetric Property

a) Yes
[Solution Description] A relation is symmetric if whenever $(a_1, a_2) \in R$, then $(a_2, a_1) \in R$. Here, since both $(1, 2)$ and its reverse $(2, 1)$ are included, $R$ is symmetric.

Key concept: Composition Basics

d) $2x + 2$

[Solution Description] To find the composition $gof(x)$, substitute $f(x)$ into $g(x)$. So $g(f(x)) = g(x+1) = 2(x+1) = 2x + 2$.

Key Concept: Simple Mapping

b) $\{1, 3, 5\}$
[Solution Description] To find the range, calculate $g(0) = 2(0) + 1 = 1$, $g(1) = 2(1) + 1 = 3$, and $g(2) = 2(2) + 1 = 5$. So, the range of $g(x)$ is $\{1, 3, 5\}$.

Key Concept: Reflexive Example

a) {(x, x), (y, y)}
[Solution Description] A relation is reflexive if it contains the pair (a, a) for each element a in set E. Thus, the relation {(x, x), (y, y)} is reflexive.

Key Concept: Combined Function Analysis

b) No, it is not injective.
[Solution Description] To determine if the function $(f \circ g)(x) = f(g(x)) = f(x^2) = 3x^2 + 1$ is injective, we need to check if $3x_1^2 + 1 = 3x_2^2 + 1$ implies $x_1 = x_2$. Simplifying gives $3x_1^2 = 3x_2^2$ or $x_1^2 = x_2^2$. This implies $x_1 = x_2$ or $x_1 = -x_2$. Therefore, $(f \circ g)(x)$ is not injective.

Key Concept: Basic Identification

b) $\{(x, x), (x, y), (y, x), (y, y)\}$
[Solution Description] The universal relation for the set $B = \{x, y\}$ will include all possible pairs of elements from $B$. Hence, it includes $(x, x), (x, y), (y, x), (y, y)$.

Key Concept: Identify Relation Type

a) Empty relation
[Solution Description] In this case, for any elements $a$ and $b$ in set $A$, there are no elements such that $a – b = 5$. Hence, $R$ contains no elements and is an empty relation.

Key Concept: Advanced Equivalence Class, Complex Real-World Application

c) Assertion is true, but Reason is false.
[Solution Description] To determine whether the assertion and reason are true, we need to analyze the properties of the given relation $R$:

1. **Reflexivity**: For any real number $a$, $a – a = 0$, which is clearly a multiple of 5. Thus, $aRa$ for all $a \in \mathbb{R}$, proving reflexivity.

2. **Symmetry**: Assume $aRb$ holds, meaning $(a-b)$ is a multiple of 5. Then $(b-a) = -(a-b)$ is also a multiple of 5, hence $bRa$. This proves symmetry.

3. **Transitivity**: Suppose $aRb$ and $bRc$; thus $(a-b)$ and $(b-c)$ are multiples of 5. Therefore, $(a-c) = (a-b) + (b-c)$ is also a multiple of 5, confirming transitivity.

Hence, $R$ is an equivalence relation on $\mathbb{R}$. It partitions $\mathbb{R}$ into equivalence classes such as $\{x + 5k \mid k \in \mathbb{Z}\}$.

Now analyzing the reason:

The relation $R$ does group numbers that differ by integer multiples of 5, but it doesn’t necessarily mean they have identical fractional parts. Consider two numbers $2.3$ and $7.3$: they belong to different equivalence classes under $R$, but have the same fractional part. Thus, while the reason describes grouping based on integer differences, it inaccurately connects this to fractional parts, which makes it false in this context.

Therefore, both the assertion and reason do not correctly align regarding equivalence class formation based on fractional parts.

Key concept: Onto Function

b) No

[Solution Description] A function is onto if every real number in the codomain has a pre-image in the domain. The function $f(x) = x^2$ only maps non-negative values since the square of any real number is non-negative. Therefore, negative numbers do not have any pre-images, so the function is not onto.

Key Concept: Composition of Functions

a) $5x + 13$
[Solution Description] Compute $(h \circ j)(x) = h(j(x))$. Start by finding $j(x) = x + 3$. Plug this into $h(x)$:
$ h(j(x)) = 5(x + 3) – 2 = 5x + 15 – 2 = 5x + 13. $
Thus, $(h \circ j)(x) = 5x + 13$.

Key concept: Real-World Function Application, Binary Operations

a) 0

[Solution Description] Evaluate each operation step-by-step:

– Compute $2 * 3$:

$2 * 3 = (2 \times 3) \mod 4 = 6 \mod 4 = 2$

– Compute $1 * 2$:

$1 * 2 = (1 \times 2) \mod 4 = 2 \mod 4 = 2$

– Finally, compute $(2 * 3) * (1 * 2)$:

$2 * 2 = (2 \times 2) \mod 4 = 4 \mod 4 = 0$

Therefore, the result is 0.

Key Concept: Transitive Check

a) Yes, it is transitive.
[Solution Description] A relation is transitive if whenever $(a, b)$ and $(b, c)$ are in $R$, then $(a, c)$ must also be in $R$. In this case, $(1, 2)$ and $(2, 3)$ imply $(1, 3)$ which is present in $R$. Hence, the relation is transitive.

Key Concept: Function Mapping

c) 3
[Solution Description] In a one-one function, each element in the domain maps to a unique element in the co-domain. Since the number of elements in the domain is 3, all 3 elements of the co-domain will have to be mapped uniquely.

Key Concept: Empty Relation

c) $\{(a, b): a > b \text{ where } a = b \}$
[Solution Description] An empty relation means there are no pairs $(a, b)$ that satisfy the condition within the given set. For option $(c)$, since no natural number is greater than itself, it forms an empty relation.

Key Concept: Symmetric Definition

b) If $(a, b) \in R$, then $(b, a) \in R$
[Solution Description] According to the definition, a relation is symmetric if for all $a, b \in A$, $(a, b) \in R$ implies $(b, a) \in R$.

Key Concept: Basic Example Identification

a) $R = \{(a, b): a – b = 10\}$
[Solution Description] To identify an empty relation, we look for a relation that results in no valid pairs from the given set. For $A = \{1, 2, 3, 4\}$, the relation $R = \{(a, b): a – b = 10\}$ is empty because neither $1, 2, 3,$ nor $4$ subtracted by another gives $10$.

Key Concept: One-One Function

a) Yes
[Solution Description] A function is one-one if for $f(x_1) = f(x_2)$, we must have $x_1 = x_2$.

Assume $f(x_1) = 3x_1 + 4$ and $f(x_2) = 3x_2 + 4$.

If $f(x_1) = f(x_2)$ then:

$
3x_1 + 4 = 3x_2 + 4
$

Subtracting $4$ from both sides gives:

$
3x_1 = 3x_2
$

Dividing both sides by $3$ yields:

$
x_1 = x_2
$

Therefore, the function is one-one.

Key Concept: Onto Check

b) No, not onto $\mathbb{R}$
[Solution Description] The function $f(x) = x^2$ maps real numbers to non-negative real numbers. Hence, it is not onto $\mathbb{R}$ as it cannot produce negative outputs.

Key Concept: Composition Basics

a) $4x^2 + 4x + 1$
[Solution Description] To find the composition $gof(x)$, we substitute $f(x)$ into $g(x)$:
$
gof(x) = g(f(x)) = g(2x + 1) = (2x + 1)^2
$
Simplifying further:
$
(2x + 1)^2 = 4x^2 + 4x + 1
$

Key Concept: Transitive Property Application

a) $ (x, z) \in R $
[Solution Description] Applying the transitive property, if $(x, y) \in R$ and $(y, z) \in R$, for the relation $R$ to be transitive, we must also have $(x, z) \in R$. This ensures the continuation of the sequence is preserved under transitivity.

Key Concept: Symmetric and Transitive

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description] The assertion states that the relation $R = \{(a, b) : 3 \text{ divides } a – b\}$ is symmetric and transitive.

– **Symmetry**: A relation $R$ is symmetric if whenever $(a, b) \in R$, then $(b, a) \in R$. Here, if $3$ divides $a – b$, it implies $a – b = 3k$ for some integer $k$. Hence, $b – a = -3k$, which shows $3$ divides $b – a$. Thus, symmetry holds.

– **Transitivity**: A relation $R$ is transitive if whenever $(a, b) \in R$ and $(b, c) \in R$, then $(a, c) \in R$. Here, if $3$ divides $a – b$, then $a – b = 3m$ for some integer $m$, and if $3$ divides $b – c$, then $b – c = 3n$ for some integer $n$. Adding these gives $a – c = (a – b) + (b – c) = 3m + 3n = 3(m+n)$, showing $3$ divides $a – c$. Thus, transitivity holds.

Therefore, the Assertion is true because the given relation is indeed symmetric and transitive.

The reason provided correctly explains why the relation is symmetric and transitive since it verifies the condition for symmetry and transitivity using standard definitions of equivalence relations involving divisibility by $3$.

Hence, both the assertion and reason are true, and the reason is the correct explanation of the assertion.

Key Concept: Function Type Identification

d) Both
[Solution Description] To classify the function: – One-One: Check if different inputs always result in different outputs. For $ f(x) = x^3 $, this is true as no two different real numbers have the same cube.
– Onto: Check if every real number output can be achieved by some input. For $ f(x) = x^3 $, every real number can indeed be represented as $ x^3 $ for some real $ x $.

Therefore, $ f(x) = x^3 $ is both one-one and onto, but since the option “both” is not available, we’ll mark it as neither.

Key Concept: Domain and Range

c) $\textbf{R}$
[Solution Description] The range of a function $gof$ is determined by applying $g$ to the range of $f$. Since both $f(x) = 2x$ and $g(x) = x + 3$ map all real numbers to all real numbers, the range of $gof$ is also $\textbf{R}$.