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CBSE - Class 11 Mathematics Relations and Functions Worksheet
Miscellaneous Exercise on Chapter 2
Worksheet Answers
Solution:
The relation $R$ is defined on the set of natural numbers $\mathbb{N}$ as $R = \{(a, b) : a, b \in \mathbb{N} \text{ and } a = b^2\}$.
The statement to evaluate is the transitivity property: $(a, b) \in R$ and $(b, c) \in R \implies (a, c) \in R$.
Let $(a, b) \in R$ and $(b, c) \in R$. By the definition provided in the problem statement, we have:
We substitute the expression for $b$ from the second equation into the first equation to determine the relationship between $a$ and $c$:
$a = (c^2)^2$
$a = c^4$ [By the Law of Indices, $(x^m)^n = x^{m \cdot n}$]
For the condition $(a, c) \in R$ to be true, it must satisfy the definition $a = c^2$. However, we have derived $a = c^4$.
To disprove the statement, we must find a set of values $(a, b, c) \in \mathbb{N}$ that satisfy the antecedent but fail the consequent.
| Variable | Value |
|---|---|
| $c$ | $2$ |
| $b$ | $b = c^2 = 2^2 = 4$ |
| $a$ | $a = b^2 = 4^2 = 16$ |
Testing the condition $(a, c) \in R$ with these values:
Since $a = 16$ and $c = 2$, $a \neq c^2$. Therefore, $(16, 2) \notin R$.
Final Solution: The statement is False. Since $a=b^2$ and $b=c^2$ implies $a=c^4$, and $c^4 \neq c^2$ for $c \in \mathbb{N} \setminus \{1\}$, the relation is not transitive. A counter-example is $(16, 4) \in R$ and $(4, 2) \in R$, but $(16, 2) \notin R$.
Solution:
We are given the set $A = \{9, 10, 11, 12, 13\}$. The function $f: A \to \mathbb{N}$ is defined such that $f(n)$ maps each element $n \in A$ to its highest prime factor.
To determine the highest prime factor for each element, we perform prime factorization for each $n \in A$:
We represent the function $f$ as a set of ordered pairs $(n, f(n))$:
| $n$ | $f(n)$ |
|---|---|
| $9$ | $3$ |
| $10$ | $5$ |
| $11$ | $11$ |
| $12$ | $3$ |
| $13$ | $13$ |
The range of a function $f$ is the set of all unique images $f(n)$ for all $n$ in the domain $A$. [Per the definition of the range of a function, $\text{Range}(f) = \{f(n) : n \in A\}$.]
Collecting the values obtained in Step 2:
$\text{Range}(f) = \{3, 5, 11, 3, 13\}$
By removing duplicate elements to satisfy the properties of a set:
$\text{Range}(f) = \{3, 5, 11, 13\}$
Final Solution: The range of $f$ is $\{3, 5, 11, 13\}$.
Solution:
The relation $R$ is defined on the set of natural numbers $\mathbb{N}$ as follows:
$R = \{(a, b) : a, b \in \mathbb{N} \text{ and } a = b^2\}$
The question asks to evaluate the truth value of the following implication: if $(a, b) \in R$, then $(b, a) \in R$.
In formal logic, this property is known as Symmetry. For the relation to be symmetric, the condition $(a, b) \in R \implies (b, a) \in R$ must hold for all $a, b \in \mathbb{N}$.
To disprove the statement, we seek a pair $(a, b)$ such that $(a, b) \in R$ but $(b, a) \notin R$.
1. Choose a value for $b$ such that $b \in \mathbb{N}$ and $b > 1$. Let $b = 2$.
2. Calculate $a$ based on the relation definition $a = b^2$:
$a = 2^2 = 4$
3. Thus, the ordered pair $(4, 2) \in R$ because $4 = 2^2$ [Per the definition of the relation $R$].
4. Now, check if the converse pair $(2, 4)$ belongs to $R$. For $(2, 4) \in R$, the condition $a = b^2$ must hold, where $a=2$ and $b=4$:
$2 = 4^2 \implies 2 = 16$
Since $2 \neq 16$, the pair $(2, 4) \notin R$ [By the definition of equality in arithmetic].
Since we have found at least one instance where $(a, b) \in R$ but $(b, a) \notin R$, the universal statement " $(a, b) \in R \implies (b, a) \in R$ " is false.
| Condition | Status |
|---|---|
| $(4, 2) \in R$ | True ($4 = 2^2$) |
| $(2, 4) \in R$ | False ($2 \neq 4^2$) |
Final Solution: The statement is False. The relation $R$ is not symmetric because for elements $a, b \in \mathbb{N}$, $a = b^2$ does not imply $b = a^2$ unless $a = b = 1$.
Solution:
The domain of a rational function $f(x) = \frac{P(x)}{Q(x)}$ consists of all real numbers $x \in \mathbb{R}$ such that the denominator $Q(x) \neq 0$. Any value of $x$ that results in $Q(x) = 0$ is excluded from the domain to avoid division by zero, which is undefined in the set of real numbers [Per the definition of functions in $\mathbb{R}$].
Given $f(x) = \frac{x^2 + 2x + 1}{x^2 - 8x + 12}$, the denominator is defined as the quadratic polynomial:
$Q(x) = x^2 - 8x + 12$
To find the excluded values, we set $Q(x) = 0$ and solve for $x$ using the factorization method [By the Zero Product Property]:
$x^2 - 8x + 12 = 0$
We seek two numbers that multiply to $12$ and add to $-8$. These numbers are $-6$ and $-2$.
$x^2 - 6x - 2x + 12 = 0$
$x(x - 6) - 2(x - 6) = 0$
$(x - 2)(x - 6) = 0$
Therefore, $x - 2 = 0 \implies x = 2$ or $x - 6 = 0 \implies x = 6$.
The function $f(x)$ is defined for all real numbers except for the set of points where the denominator vanishes. Thus, the domain is the set of all real numbers excluding $\{2, 6\}$.
Expressing the result in standard set notation:
$Domain(f) = \mathbb{R} \setminus \{2, 6\}$
Alternatively, in interval notation:
$Domain(f) = (-\infty, 2) \cup (2, 6) \cup (6, \infty)$
Final Solution: The domain of the function is the set of all real numbers except $2$ and $6$, represented as $\mathbb{R} - \{2, 6\}$.
Solution:
The function is defined as $f(x) = x^2$. We are tasked with evaluating the difference quotient:
$\frac{f(1.1) - f(1)}{1.1 - 1}$
[Per the definition of the derivative/average rate of change, this expression represents the slope of the secant line passing through points $(1, f(1))$ and $(1.1, f(1.1))$.]
We calculate $f(x)$ for the specified points in the domain:
Performing the arithmetic multiplication $(1.1 \times 1.1) = 1.21$. [By the law of exponents where $(a \cdot b)^n = a^n \cdot b^n$ is not applicable, rather simple decimal multiplication $11 \times 11 = 121$ with two decimal places].
Substitute the evaluated values into the expression:
$\frac{1.21 - 1}{1.1 - 1}$
Perform the subtraction in the numerator and denominator:
$\text{Numerator: } 1.21 - 1.00 = 0.21$
$\text{Denominator: } 1.1 - 1.0 = 0.1$
$\frac{0.21}{0.1}$
Multiply both numerator and denominator by $10$ to remove the decimal fraction: [Property of Equivalent Fractions: $\frac{a}{b} = \frac{a \cdot k}{b \cdot k}$ for $k \neq 0$]
$\frac{0.21 \times 10}{0.1 \times 10} = \frac{2.1}{1} = 2.1$
Final Solution: 2.1
Solution:
Given the functions $f: \mathbb{R} \to \mathbb{R}$ and $g: \mathbb{R} \to \mathbb{R}$ defined by:
By the definition of the sum of two functions, $(f + g)(x) = f(x) + g(x)$ [Per the algebraic definition of function addition].
Substituting the given expressions:
$(f + g)(x) = (x + 1) + (2x - 3)$
Grouping like terms:
$(f + g)(x) = (x + 2x) + (1 - 3)$
$(f + g)(x) = 3x - 2$
By the definition of the difference of two functions, $(f - g)(x) = f(x) - g(x)$ [Per the algebraic definition of function subtraction].
Substituting the given expressions:
$(f - g)(x) = (x + 1) - (2x - 3)$
Distributing the negative sign:
$(f - g)(x) = x + 1 - 2x + 3$
$(f - g)(x) = -x + 4$
By the definition of the quotient of two functions, $\left(\frac{f}{g}\right)(x) = \frac{f(x)}{g(x)}$, provided $g(x) \neq 0$ [Per the algebraic definition of function division].
Setting the denominator to zero to find the domain restriction:
$2x - 3 = 0 \implies 2x = 3 \implies x = \frac{3}{2}$
Thus, $\left(\frac{f}{g}\right)(x) = \frac{x + 1}{2x - 3}$, where $x \in \mathbb{R} \setminus \{\frac{3}{2}\}$.
Final Solution:
Solution:
A relation $f: A \to B$ is defined as a function if every element $x \in A$ has one and only one image $f(x) \in B$. [Per the Uniqueness Axiom of Functions]. If a relation yields two different outputs for the same input, it fails the definition of a function.
The relation $f$ is defined as:
$f(x) = \begin{cases} x^2, & 0 \le x \le 3 \\ 3x, & 3 \le x \le 10 \end{cases}$The only point of contention is the boundary value $x = 3$, where the definitions overlap. We evaluate the image of $x=3$ for both branches:
[By the Principle of Consistency], since $9 = 9$, the relation assigns a unique value to $x=3$. For all other values in the domain, the relation is clearly defined by a single expression. Thus, $f$ is a function.
The relation $g$ is defined as:
$g(x) = \begin{cases} x^2, & 0 \le x \le 2 \\ 3x, & 2 \le x \le 10 \end{cases}$We evaluate the image of the boundary value $x = 2$ for both branches:
[Per the Uniqueness Axiom], for a relation to be a function, each input must map to exactly one output. Here, $g(2) = 4$ and $g(2) = 6$. Since $4 \neq 6$, the input $x=2$ corresponds to two distinct images. Therefore, $g$ is not a function.
Final Solution: The relation $f$ is a function because it maps the boundary element $x=3$ to a unique value ($9$). The relation $g$ is not a function because it maps the boundary element $x=2$ to two distinct values ($4$ and $6$), violating the uniqueness property required for a function.
Solution:
The function is defined as $f(x) = \sqrt{x - 1}$. In the set of real numbers $\mathbb{R}$, the square root operation $\sqrt{u}$ is defined if and only if the radicand $u \geq 0$.
Setting the condition for the function $f(x)$:
$x - 1 \geq 0$ [By the definition of real-valued functions requiring non-negative radicands for square roots]
$x \geq 1$ [By the Addition Property of Inequality]
Therefore, the domain of $f$ is the interval $[1, \infty)$.
Let $y = f(x) = \sqrt{x - 1}$. By the definition of the principal square root function, the output $y$ must always be non-negative, regardless of the value of $x$ within the domain.
Since $x - 1 \geq 0$, it follows that $\sqrt{x - 1} \geq 0$. Consequently, $y \geq 0$ [By the definition of the range of the principal square root function].
The range of $f$ is the interval $[0, \infty)$.
| Feature | Interval Notation |
|---|---|
| Domain | $[1, \infty)$ |
| Range | $[0, \infty)$ |
Final Solution: The domain of the function $f(x) = \sqrt{x - 1}$ is $[1, \infty)$ and the range is $[0, \infty)$.
Solution:
The function is defined as $f: \mathbb{R} \to \mathbb{R}$ where for every $x \in \mathbb{R}$, the image is given by:
$y = f(x) = \frac{x^2}{1+x^2}$
To determine the range, we identify all possible values of $y$ such that $y = \frac{x^2}{1+x^2}$ for some $x \in \mathbb{R}$.
Multiply both sides by $(1+x^2)$:
$y(1+x^2) = x^2$
$y + yx^2 = x^2$
$y = x^2 - yx^2$
$y = x^2(1 - y)$
Assuming $y \neq 1$, we can write:
$x^2 = \frac{y}{1 - y}$ [By Algebraic Rearrangement]
Since $x \in \mathbb{R}$, it must follow that $x^2 \geq 0$. Therefore:
$\frac{y}{1 - y} \geq 0$
To solve this inequality, we analyze the signs of the numerator and the denominator:
For the fraction to be $\geq 0$, the signs of the numerator and denominator must either be both positive or both negative.
Case 1: $y \geq 0$ and $1 - y > 0 \implies 0 \leq y < 1$.
Case 2: $y \leq 0$ and $1 - y < 0 \implies y \leq 0$ and $y > 1$ (This is a contradiction, no solution).
At $x = 0$, $y = \frac{0}{1+0} = 0$.
As $x \to \infty$ or $x \to -\infty$, $y = \frac{x^2}{1+x^2} = \frac{1}{\frac{1}{x^2} + 1} \to \frac{1}{0+1} = 1$.
Since $1+x^2 > x^2$ for all $x \in \mathbb{R}$, it follows that $0 \leq \frac{x^2}{1+x^2} < 1$.
The set of all possible values of $y$ is the interval $[0, 1)$.
Final Solution: The range of the function $f$ is $[0, 1)$.
Solution:
By definition, a relation $f$ from a set $A$ to a set $B$ is a subset of the Cartesian product $A \times B$. That is, $f \subseteq A \times B$. [Per the Set-Theoretic definition of a binary relation].
Given sets:
The provided set $f$ is: $f = \{(1,5), (2,9), (3,1), (4,5), (2,11)\}$.
To verify if $f \subseteq A \times B$, we check every ordered pair in $f$ against the elements of $A \times B$:
| Ordered Pair $(x, y) \in f$ | Is $x \in A$? | Is $y \in B$? | In $A \times B$? |
|---|---|---|---|
| $(1, 5)$ | Yes | Yes | Yes |
| $(2, 9)$ | Yes | Yes | Yes |
| $(3, 1)$ | Yes | Yes | Yes |
| $(4, 5)$ | Yes | Yes | Yes |
| $(2, 11)$ | Yes | Yes | Yes |
Since every element $(x, y)$ in the set $f$ is also an element of the Cartesian product $A \times B$, it follows logically that $f \subseteq A \times B$. [By the subset criterion in set theory].
Final Solution: Yes, the statement is true. $f$ is a relation from $A$ to $B$ because $f$ is a subset of $A \times B$.
Solution:
The function $f(x) = |x - 1|$ is a modulus (absolute value) function. By the fundamental definition of the absolute value operator, the function maps any input $x$ to its non-negative magnitude, defined piecewise as:
$f(x) = \begin{cases} -(x - 1), & \text{if } x - 1 < 0 \\ (x - 1), & \text{if } x - 1 \ge 0 \end{cases}$This simplifies to:
$f(x) = \begin{cases} 1 - x, & \text{if } x < 1 \\ x - 1, & \text{if } x \ge 1 \end{cases}$The domain of a real-valued function is the set of all possible real values of $x$ for which $f(x)$ is defined. Since the expression $(x - 1)$ is a polynomial, it is defined for all real numbers $x \in \mathbb{R}$. There are no values of $x$ that lead to division by zero, square roots of negative numbers, or undefined logarithmic arguments. [By the property of polynomial functions in $\mathbb{R}$]
Domain: $\text{Dom}(f) = (-\infty, \infty)$ or $\mathbb{R}$.
The range is the set of all resulting values $f(x)$ for all $x \in \text{Domain}$. By the definition of the modulus function, $|x - 1| \ge 0$ for all $x \in \mathbb{R}$. [By the non-negativity axiom of the absolute value].
As $x$ varies from $-\infty$ to $\infty$, $x-1$ also varies from $-\infty$ to $\infty$. The absolute value of any real number in $(-\infty, \infty)$ yields the set of all non-negative real numbers $[0, \infty)$.
Range: $\text{Ran}(f) = [0, \infty)$.
| Property | Result |
|---|---|
| Domain | $(-\infty, \infty)$ |
| Range | $[0, \infty)$ |
| Vertex Point | $(1, 0)$ |
Final Solution: The domain of the function $f(x) = |x - 1|$ is $\mathbb{R}$ (all real numbers), and the range is $[0, \infty)$ (all non-negative real numbers).
Solution:
The relation $R$ is defined on the set of natural numbers $\mathbb{N} = \{1, 2, 3, \dots\}$. The condition for any ordered pair $(a, b)$ to be an element of $R$ is given by the predicate:
$R = \{ (a, b) : a, b \in \mathbb{N}, a = b^2 \}$
For the statement $(a, a) \in R$ to be true for all $a \in \mathbb{N}$, the condition $a = b^2$ must hold when $b$ is replaced by $a$. Substituting $b = a$ into the defining predicate, we obtain:
$a = a^2$
[By substitution property of equality]
We test the validity of $a = a^2$ for all $a \in \mathbb{N}$:
A universal quantifier ("for all $a \in \mathbb{N}$") is falsified if at least one counterexample exists within the domain [Per the Principle of Universal Instantiation].
Let $a = 2$. Since $2 \in \mathbb{N}$, we examine if $(2, 2) \in R$.
According to the relation definition, $(2, 2) \in R$ if and only if $2 = 2^2$.
Since $2 \neq 4$, $(2, 2) \notin R$.
Since the condition $a = a^2$ is not satisfied for all $a \in \mathbb{N}$ (specifically for any $a > 1$), the statement $(a, a) \in R$ for all $a \in \mathbb{N}$ is logically invalid.
Final Solution: The statement is False. As demonstrated by the counterexample $a = 2$, $(2, 2) \notin R$ because $2 \neq 2^2$.
Solution:
A relation $f: A \to B$ is classified as a function if and only if it satisfies two fundamental criteria:
In the given problem, the relation $f$ is defined as $f = \{(ab, a + b) : a, b \in \mathbb{Z}\}$. Here, the "input" corresponds to the first component $x = ab$ and the "output" corresponds to the second component $y = a + b$.
To determine if $f$ is a function, we must test if the same input $x$ can produce different outputs $y$ for different pairs $(a, b)$.
Consider the case where $x = 6$. We search for integer pairs $(a, b)$ such that $ab = 6$:
| Pair $(a, b)$ | Input $x = ab$ | Output $y = a + b$ |
|---|---|---|
| $(2, 3)$ | $2 \times 3 = 6$ | $2 + 3 = 5$ |
| $(1, 6)$ | $1 \times 6 = 6$ | $1 + 6 = 7$ |
From the table above, we observe the following mapping within the relation $f$:
Since the same input $x = 6$ is associated with two distinct outputs $y = 5$ and $y = 7$, the condition of uniqueness is violated. Specifically, $f(6) = 5$ and $f(6) = 7$, which implies $5 \neq 7$.
A relation is a function if and only if each input maps to exactly one unique output. Because there exists at least one input ($x=6$) that maps to multiple distinct outputs ($y=5, 7$), the relation fails the vertical line test/uniqueness condition.
Final Solution: No, $f$ is not a function from $\mathbb{Z}$ to $\mathbb{Z}$ because the relation is not well-defined; the same input $ab$ can result in different values of $a+b$, violating the uniqueness requirement of a function.
Solution:
Given the function $f(x) = ax + b$, where $a, b \in \mathbb{Z}$, we can substitute the ordered pairs $(x, f(x))$ from the set $f = \{(1,1), (2,3), (0,–1), (–1, –3)\}$ into the functional form to generate a system of equations.
Using the point $(0, -1)$:
$f(0) = a(0) + b = -1$
$\implies 0 + b = -1$
$\implies b = -1$ [By the Identity Property of Multiplication and Additive Inverse property]
Now, using the point $(1, 1)$ and the value of $b$ determined in Step 1:
$f(1) = a(1) + b = 1$
$a(1) + (-1) = 1$ [By substitution method]
$a - 1 = 1$
$a = 1 + 1$ [By transposition property]
$a = 2$
To ensure the function $f(x) = 2x - 1$ is consistent across the entire domain provided in the set, we check the remaining ordered pairs $(2, 3)$ and $(-1, -3)$.
| Point $(x, y)$ | Calculation: $f(x) = 2x - 1$ | Result |
|---|---|---|
| $(2, 3)$ | $f(2) = 2(2) - 1 = 4 - 1$ | $3$ |
| $(-1, -3)$ | $f(-1) = 2(-1) - 1 = -2 - 1$ | $-3$ |
Since the values obtained match the output values in the given set, the parameters $a = 2$ and $b = -1$ are verified as the unique integers satisfying the condition for all elements in $f$.
Final Solution: $a = 2, b = -1$
Solution:
By the fundamental definition of a function in set theory, a relation $f$ from set $A$ to set $B$ is a function if and only if:
[Reference: Definition of a Function, Cartesian Product $A \times B$]
The relation $f$ is defined as the set of ordered pairs: $f = \{(1, 5), (2, 9), (3, 1), (4, 5), (2, 11)\}$.
To determine if $f$ is a function, we examine the mapping of elements from the domain $A = \{1, 2, 3, 4\}$:
| Element ($x \in A$) | Associated Images ($y \in B$) |
|---|---|
| $1$ | $5$ |
| $2$ | $9$ and $11$ |
| $3$ | $1$ |
| $4$ | $5$ |
Per the uniqueness property of a function, for each $x \in A$, there must exist a unique $y \in B$ such that $(x, y) \in f$.
Observe the element $2 \in A$:
Since $9 \neq 11$, the element $2$ is mapped to two distinct elements in the codomain $B$. This constitutes a "one-to-many" relation, which violates the axiom of uniqueness required for a function. [Per the Vertical Line Test/Function Definition Axiom].
Because the element $2$ in the domain $A$ is associated with two different images ($9$ and $11$) in the codomain $B$, the relation $f$ fails the definition of a function.
Final Solution: The statement is False. $f$ is not a function from $A$ to $B$ because the element $2 \in A$ has two distinct images in $B$.
