Class 11 Mathematics Chapter 2 Relations and Functions

Key Concept: Relation Basics

c) A relation where each element in the domain is related to exactly one element in the codomain

[Solution Description] A function is a special type of relation where each element in the domain is related to exactly one element in the codomain. This means for every input, there is only one corresponding output.

Key Concept: Ordered Pair Definition

a) $(7, x)$

[Solution Description] An ordered pair from sets $U$ and $V$ is denoted as $(u, v)$ where $u \in U$ and $v \in V$. Considering the options:

– Option a: $(7, x)$ follows the correct format for an ordered pair.

– Options b, c, d: Do not follow the definition of an ordered pair.

Key Concept: Basic Definition

a) $\alpha f(x)$

[Solution Description] The multiplication of a function $f(x)$ by a scalar $\alpha$ is defined as $$(\alpha f)(x) = \alpha f(x)$$, which means for any value of $x$, the function value is multiplied by the scalar.

Key Concept: Visual Representation

b) A function

[Solution Description] An arrow diagram showing that every element in set $A$ points to exactly one element in set $B$ represents a function. This is because each input (element of $A)$ has exactly one output (image in $B)$.

Key Concept: Range of a Function

b) $[0, \infty)$

[Solution Description] The absolute value function $f(x) = |x|$ outputs non-negative values for all real numbers. Thus, the range is $[0, \infty)$ because it includes all positive numbers and zero.

Key Concept: Domain Identification

c) All real numbers except $x = -1$

[Solution Description] The domain of a rational function like $$\frac{P(x)}{Q(x)}$$ is all real numbers except those that make $Q(x)$ zero. For $f(x) = \frac{x-2}{x+1}$, set $x+1=0$ to find restrictions:

$x + 1 = 0$

$x = -1$

Thus, $x=-1$ is excluded from the domain. Hence, the domain is $$\mathbb{R} \setminus \{-1\}$$, meaning all real numbers except $-1$.

Key Concept: Subset Identification

a) $\{(1, x), (2, y)\}$

[Solution Description] The Cartesian product $A \times B$ consists of all ordered pairs where the first element is from set A and the second element is from set B. Hence, $$A \times B = \{(1, x), (1, y), (1, z), (2, x), (2, y), (2, z)\}$$. Any subset of this can be a relation.

Key Concept: Definition Check

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

[Solution Description] According to the definition of a function, every element in the domain must have exactly one image in the codomain. Option (a) satisfies this condition as each element in the first set has only one corresponding element in the second set.

Key Concept: Basic Definition

b) $g(x) \neq 0$

[Solution Description] The quotient $$\frac{f(x)}{g(x)}$$ is defined as long as the denominator $g(x)$ is not equal to zero, ensuring the division is valid.

Key Concept: Graph Interpretation, Linear Function Analysis

a) Value at $x=0$: $-5$; Intercepts: $(0,-5), (2.5,0), (-3,0)$

[Solution Description] At $x = 0$, we use the definition of the piecewise function for $x \geq 0$: $f(x) = 2x – 5$. Substituting $x = 0$, we get:

$$f(0) = 2 \times 0 – 5 = -5$$

Now, interpreting the intersection points:

– Intersection with the y-axis occurs at $x = 0$, so it’s $(0, -5)$.

– For the positive part $(x \geq 0)$, solving $2x – 5 = 0$ gives $x = 2.5$, hence another axis intercept is $(2.5, 0)$.

– For the negative part $(x < 0)$, solving $-x + 3 = 0$ gives $x = 3$, hence the last axis intercept is $(-3, 0)$.

Key Concept: Intersection of Cartesian Products

b) $\{(x,2), (y,2)\}$

[Solution Description] We first determine $A \times B$ and $A \times C$:

$$A \times B = \{(x,1), (x,2), (y,1), (y,2)\}$$

$$A \times C = \{(x,2), (x,3), (y,2), (y,3)\}$$

The intersection $$(A \times B) \cap (A \times C)$$ consists of pairs present in both products:

$$(A \times B) \cap (A \times C) = \{(x,2), (y,2)\}$$

Thus, the result is $$\{(x,2), (y,2)\}$$.

Key Concept: Domain and Range

b) $[0, \infty)$

[Solution Description] The function $f(x) = x^2$ is a parabola that opens upwards, so its lowest point is at $x = 0$. Therefore, the range starts from 0 and goes to infinity.

Key Concept: Function Identification

b) No, it is not a function

[Solution Description] The relation $R$ is not a function because for an element in the domain, there should be exactly one image in the codomain. Here, the element 3 in set $A$ has two images: 4 and 5.

Key Concept: Function Identification

c) No, because one element maps to multiple outputs.

[Solution Description] To determine if the given relation R is a function, we must check that every element in the domain has only one image in the codomain. The domain of R is $\{1, 2, 3\}$. We notice that the element $2$ maps to both $4$ and $6$. This means it does not have a unique mapping. Hence, this relation is not a function.

Key Concept: Function from Relations

b) No, $R$ is not a function because $3$ has multiple images.

[Solution Description] A relation is a function if every element in the domain has a unique image in the codomain. Here, the elements in the domain are $\{1, 3, 5\}$. The element $3$ is associated with both $4$ and $9$, which means it does not have a unique image. Therefore, this relation is not a function.

Key Concept: Identity Function

d) A straight line through the origin

[Solution Description] The graph of an identity function $f(x) = x$ is a straight line that passes through the origin (0,0) with a slope of 1, implying that it makes a 45-degree angle with each axis.

Key Concept: Definition Check

b) $(2,3)$

[Solution Description] The Cartesian product $A \times B$ is given by $\{(a, b): a \in A, b \in B\}$. Therefore, the possible ordered pairs are $(1,3)$, $(1,4)$, $(2,3)$, and $(2,4)$.

Key Concept: Basic Inequality

b) $12 < 15$ [Solution Description] Let's evaluate each option: a) $7 > 10$: False, because $7$ is less than $10$.

b) $12 < 15$: True, because $12$ is indeed less than $15$. c) $20 = 25$: False, because $20$ does not equal $25$. d) $6 > 8$: False, because $6$ is less than $8$.

Hence, the correct answer is $12 < 15$.

Key Concept: Set-Builder Basics

a) $R = \{(x, y): x = 2y, x \in P, y \in Q\}$

[Solution Description] The set-builder representation specifies that x should be twice y. So, $$R = \{(x, y): x = 2y, x \in P, y \in Q\}$$

Key Concept: Set Product Formation

a) $\{(a, 1), (a, 2), (b, 1), (b, 2)\}$

[Solution Description] The Cartesian product $P \times Q$ consists of all possible ordered pairs $(p, q)$ such that $p \in P$ and $q \in Q$. We list them as follows:

$$P \times Q = \{(a, 1), (a, 2), (b, 1), (b, 2)\}$$

Thus, $$P \times Q = \{(a, 1), (a, 2), (b, 1), (b, 2)\}$$.

Key Concept: Basic Definition

b) 4

[Solution Description] The greatest integer less than or equal to 4.3 is 4. Therefore, $[4.3] = 4$.

Key Concept: Domain Identification

b) $\{2, 3, 4\}$

[Solution Description] The domain of a relation is the set of all first elements in each ordered pair. Therefore, for relation $R = \{(2,5), (3,8), (4,11)\}$, the domain is $\{2, 3, 4\}$.

Key Concept: Subtraction Formula

a) $x^2 – 3x + 1$

[Solution Description] To find $(f – g)(x)$, we subtract $g(x)$ from $f(x)$:

$$f(x) = x^2 + 5$$

$$g(x) = 3x + 4$$

Therefore, $$(f-g)(x) = f(x) – g(x) = (x^2 + 5) – (3x + 4) = x^2 – 3x + 1$$.

Key Concept: Identify Scalar

c) $6$

[Solution Description] The scalar in an expression of the form $(\alpha f)(x)$ is represented by the number that multiplies the function $f(x)$. In this case, the given expression is $(6f)(x)$, where $6$ is the scalar.

Key Concept: Identify Subtraction Result

b) $x^3 – 2x – 1$

[Solution Description] Applying the subtraction, we have:

$$(f-g)(x) = f(x) – g(x) = x^3 – (2x + 1)$$

This simplifies to:

$$x^3 – 2x – 1$$

Key Concept: Function Identification

a) $6x^2$

[Solution Description] The multiplication of $f(x) = 2x$ and $g(x) = 3x$ is given by:

$$(fg)(x) = f(x) \cdot g(x) = 2x \cdot 3x = 6x^2$$

Key Concept: Real Number Product

c) Points in 3D space

[Solution Description] In mathematics, the Cartesian product $R \times R \times R$ refers to the set of all possible ordered triples involving real numbers. Each dimension in this product represents one axis in space.

Hence, $R \times R \times R$ corresponds to points in 3D space, as each point is determined by three coordinates (x, y, z).

Consequently, $R \times R \times R$ represents a three-dimensional space.

Key Concept: Arrow Diagram

a) Diagram with arrows from 2 to 3 and 4 to 5

[Solution Description] In an arrow diagram for a relation $R$, we draw arrows from each element in set $A$ to its corresponding element in set $B$. For $R = \{(2, 3), (4, 5)\}$, there should be arrows from $2$ to $3$ and from $4$ to $5$.

Key Concept: Relation Properties

d) Both Reflexive and Transitive

[Solution Description] A relation is reflexive if every element is related to itself. Here, both $(1,1)$ and $(2,2)$ are present, making $S$ reflexive. A relation is symmetric if for any pair $(a, b)$ in the relation, $(b, a)$ is also in the relation. In $S$, only $(1,2)$ is present but not $(2,1)$, so it is not symmetric. A relation is transitive if whenever $(a,b)$ and $(b,c)$ are in the relation, $(a,c)$ is also in the relation. Here, there’s no such sequence, so it’s trivially transitive.

Key Concept: Basic Properties

c) It is both one-to-one and onto

[Solution Description] The identity function is both one-to-one and onto because it maps every element of the domain to a unique element in the range and covers all elements in the range.

Key Concept: Identity Function

d) $f(x) = x$

[Solution Description] The identity function is defined as $f(x) = x$ for each $x in R$. This means that the output value is exactly the same as the input value.

Key Concept: Relation to Function

c) No, because $4$ has two different outputs

[Solution Description] A relation is a function if every element in the domain corresponds to exactly one element in the range. In the given relation, the domain elements are $2, 3,$ and $4$. Observe that $4$ maps to both $5$ and $7$, meaning there is more than one output for a single input value (4). Hence, this relation is not a function.

Key Concept: Real Number Representation

c) Coordinates of all points in three-dimensional space

[Solution Description] The Cartesian product $\mathbb{R} \times \mathbb{R} \times \mathbb{R}$ represents a three-dimensional space. Each element is an ordered triplet $(x, y, z)$ corresponding to coordinates in this space. This is analogous to how $\mathbb{R} \times \mathbb{R}$ represents two-dimensional space.

Key Concept: Element Count

d) 0

[Solution Description] When either set in a Cartesian product is empty, the resultant set is also empty, having zero elements. Here, since $P = \emptyset$, $P \times Q$ will have 0 elements.

Key Concept: Modulus Graph

c) 3

[Solution Description] The modulus function $f(x) = |x|$ returns the absolute value of $x$. For any positive number or zero, it remains unchanged. Thus, when $x = 3$, the calculation is as follows:

$f(3) = |3| = 3$

Key Concept: Identify Components

a) $f(x) = 3x – 4$, $g(x) = 2x – 3$

[Solution Description] We need to determine the individual components that sum up to form the given function $(f + g)(x) = 5x – 7$. One possible pair could be $f(x) = 3x – 4$, $g(x) = 2x – 3$ because adding them gives $3x – 4 + 2x – 3 = 5x – 7$.

Key Concept: Domain and Range

c) $R$

[Solution Description] The modulus function $f(x) = |x|$ is defined for all real numbers. Therefore, the domain of the modulus function is $R$, the set of all real numbers.

Key Concept: Domain and Range Identification

b) $\{3, 4, 5\}$

[Solution Description] The domain of a relation is the set of all first elements from each ordered pair in the relation.

For the relation $R = \{(3,7), (4,8), (5,9)\}$, the first elements (or x-values) are 3, 4, and 5. So, the domain of the relation is $\{3, 4, 5\}$.

Key Concept: Properties of Relations

a) Reflexivity

[Solution Description] A relation is reflexive if every element is related to itself. Here, each element in $S$ is mapped to itself in $R$: $(a, a)$, $(b, b)$, $(c, c)$. Hence, the relation $R$ is reflexive.

Key Concept: Domain Identification

b) $R$

[Solution Description] The domain of a constant function like $f(x) = -4$ is all real numbers, represented as $R$. Since a constant function outputs the same value for any real number input, there are no restrictions on its domain.

Key Concept: Graph Shape

b) Hyperbola

[Solution Description] The function $$f(x) = \frac{1}{x^2}$$ has a graph shaped like a hyperbola. It approaches the x-axis and y-axis asymptotically but never touches them.

Key Concept: Square Roots

c) 6

[Solution Description] We need to find a number that when multiplied by itself gives 36. The possible numbers are 6 and -6 because $6 \times 6 = 36$ and $(-6) \times (-6) = 36$. But typically in such questions, we take the positive root, so the answer is 6.

Key Concept: Order Importance

b) $(\text{DL}, 01)$ is different from $(01, \text{DL})$

[Solution Description] In a Cartesian product, the order of the pairs matters. Thus, $(\text{DL}, 01)$ is not the same as $(01, \text{DL})$.

Key Concept: Linear Function

b) $f(x) = 5x + 3$

[Solution Description] A linear function has the form $$f(x) = mx + c$$. Among the options, $$f(x) = 5x + 3$$ fits this form with $m = 5$ and $c = 3$.

Key Concept: Cartesian Product Formation

b) No, they are not equal.

[Solution Description] The set $$M \times N = \{(x, 10), (y, 10)\}$$ and $$N \times M = \{(10, x), (10, y)\}$$. These two sets are not equal because the order of elements in ordered pairs matters, and the pairs are different.

Key Concept: Domain Identification

b) $\{a, b, c\}$

[Solution Description] The domain of a relation consists of all the first elements from each pair in the relation. For $R = \{(a,1), (b,2), (c,3)\}$, the first elements are $a$, $b$, and $c$. Therefore, the domain is $\{a, b, c\}$.

Key Concept: Scalar Multiplication

a) $8x – 2$

[Solution Description] To find $(\alpha f)(x)$, use the scalar multiplication formula:

$$(\alpha f)(x) = \alpha f(x)$$

Substitute the given function and scalar:

$$(\alpha f)(x) = 2(4x – 1) = 8x – 2$$

Key Concept: Domain Identification

d) All real numbers

[Solution Description] The given definition states that the domain of the signum function is all real numbers (R). This means any real number can be used as an input for the function.

Key Concept: Function vs. Non-Function

b) $\{(2, 3), (3, 4)\}$

[Solution Description] According to the definition, a relation $R$ is a function if every element in the domain corresponds to exactly one element in the range. We examine each option:

Option a: The pairs are $$\{(1, 2), (1, 3)\}$$ where ‘1’ is related to both ‘2’ and ‘3’, violating the rule of a function.

Option b: The pairs are $$\{(2, 3), (3, 4)\}$$ where each input has only one output.

Option c: The pairs are $$\{(4, 5), (4, 6)\}$$ where ‘4’ is related to both ‘5’ and ‘6’, which violates the rule.

Option d: The pairs are $$\{(7, 8), (8, 9)\}$$ where each input has only one output.

Therefore, options b and d satisfy the conditions of a function. However, since we choose one, option b fits better due to being more commonly in textbook examples with consecutive numbers.

Key Concept: Reflexive Set

a) $\emptyset$

[Solution Description] To determine the largest set on which this relation is reflexive, we need each element $n$ in any proposed set to have $(n, n)$ in $R$. However, $R$ contains ordered pairs like $(n, n+1)$, not $(n,n)$. As $(n, n)$ never appears for any $n$, the relation cannot be reflexive on any non-empty set. Thus, the largest set on which it would be reflexive is the empty set.

Key Concept: Constant Function

b) Range is $\{c\}$

[Solution Description] A constant function $f(x) = c$ has a domain of all real numbers (R) and a range as the single value set $\{c\}$.

Key Concept: Basic Identification

b) $\{4, 5\}$

[Solution Description] The range of a relation is the set of all second elements in the ordered pairs. In the given relation $R = \{(1, 4), (2, 5), (3, 5)\}$, the second elements are 4, 5, and 5. Therefore, the unique elements form the range: $\{4, 5\}$.