Class 9-10 Math Solution Exercise 2.2

Exercise 2.2: Sets and Functions

1. Which one is the set of factors of 8?

1) $\{8, 16, 24, \dots\}$

2) $\{1, 2, 4, 8\}$

3) $\{2, 4, 8\}$

4) $\{1, 2\}$

Solution:

The numbers that divide 8 without leaving a remainder are the factors of 8.

The factors of 8 are: $1, 2, 4, 8$

Answer: 2) $\{1, 2, 4, 8\}$

2. If $R$ is a relation from the set $C$ to the set $B$, then which one of the following is true?

1) $R \subseteq C$

2 $R \subseteq B$

3) $R \subseteq C \times B$

4) $C \times B \subseteq R$

Solution:

By definition of a relation, if $R$ is a relation from set $C$ to set $B$, then $R$ will be a subset of the Cartesian product $C \times B$. That is, $R \subseteq C \times B$.

Answer: 3) $R \subseteq C \times B$

3. If $A = \{1, 2\}$ and $B = \{2, 5\}$, which one of the following is the number of elements of $P(A \cap B)$?

1) $1$

2) $2$

3) $3$

4) $8$

Solution:

First, $A \cap B = \{1, 2\} \cap \{2, 5\} = \{2\}$

Here, the number of elements in set $A \cap B$ is $n = 1$.

According to the formula $2^n$ for finding the number of elements in a power set,

the number of elements in $P(A \cap B) = 2^1 = 2$.

Answer: 2) $2$

4. Which one of the following expresses the set $\{x \in \mathbb{N} : 13 < x < 17 \text{ and } x \text{ is a prime number}\}$ in tabular method?

1) $\emptyset$

2) $\{\emptyset\}$

3) $\{13, 17\}$

4) $\{14, 15, 16\}$

Solution: The natural numbers greater than 13 and less than 17 are: $14, 15, 16$.

None of these are prime numbers ($14 = 2 \times 7$, $15 = 3 \times 5$, $16 = 2 \times 8$).

Since no value satisfies the given condition, it is an empty set $\emptyset$.

Answer: 1) $\emptyset$

5. If $A = \{3, 4\}$, $B = \{2, 4\}$, $x \in A$, and $y \in B$, then determine the relation $x > y$ in $A$ and $B$.

Solution: Given, $A = \{3, 4\}$ and $B = \{2, 4\}$

Cartesian product of $A$ and $B$:

$A \times B = \{3, 4\} \times \{2, 4\} = \{(3, 2), (3, 4), (4, 2), (4, 4)\}$

According to the question, the required relation is $R = \{(x, y) \in A \times B : x > y\}$

Checking the ordered pairs based on the condition:

• For $(3, 2)$: $3 > 2$ (satisfies condition)
• For $(3, 4)$: $3 \ngtr 4$
• For $(4, 2)$: $4 > 2$ (satisfies condition)
• For $(4, 4)$: $4 \ngtr 4$

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

Answer:$\{(3, 2), (4, 2)\}$

6. If $C = \{2, 5\}$, $D = \{4, 6, 7\}$, $x \in C$, and $y \in D$, then determine the relation $x + 1 < y$ in $C$ and $D$.

Solution: Given, $C = \{2, 5\}$ and $D = \{4, 6, 7\}$

$C \times D = \{2, 5\} \times \{4, 6, 7\} = \{(2, 4), (2, 6), (2, 7), (5, 4), (5, 6), (5, 7)\}$

According to the question, the required relation is $R = \{(x, y) \in C \times D : x + 1 < y\}$

Checking according to the condition ($x + 1 < y$):

• For $(2, 4)$: $2 + 1 = 3 < 4$ (satisfies condition)
• For $(2, 6)$: $2 + 1 = 3 < 6$ (satisfies condition)
• For $(2, 7)$: $2 + 1 = 3 < 7$ (satisfies condition)
• For $(5, 4)$: $5 + 1 = 6 \nless 4$
• For $(5, 6)$: $5 + 1 = 6 \nless 6$
• For $(5, 7)$: $5 + 1 = 6 < 7$ (satisfies condition)

$\therefore R = \{(2, 4), (2, 6), (2, 7), (5, 7)\}$

Answer:$\{(2, 4), (2, 6), (2, 7), (5, 7)\}$

7. If $f(x) = x^4 + 5x – 3$, then determine $f(-1)$, $f(2)$, and $f\left(\frac{1}{2}\right)$.

Solution: Given, $f(x) = x^4 + 5x – 3$

$\therefore f(-1) = (-1)^4 + 5(-1) – 3 = 1 – 5 – 3 = -7$

$\therefore f(2) = 2^4 + 5(2) – 3 = 16 + 10 – 3 = 23$

$\therefore f\left(\frac{1}{2}\right) = \left(\frac{1}{2}\right)^4 + 5\left(\frac{1}{2}\right) – 3$

$= \frac{1}{16} + \frac{5}{2} – 3$

$= \frac{1 + 40 – 48}{16} = \frac{-7}{16}$

Answer:$f(-1) = -7$, $f(2) = 23$, and $f\left(\frac{1}{2}\right) = -\frac{7}{16}$

8. If $f(y) = y^3 + ky^2 – 4y – 8$, then for what value of $k$ will $f(-2) = 0$?

Solution: Given, $f(y) = y^3 + ky^2 – 4y – 8$

Substituting $y = -2$:

$f(-2) = (-2)^3 + k(-2)^2 – 4(-2) – 8$

$\implies f(-2) = -8 + 4k + 8 – 8$

$\implies f(-2) = 4k – 8$

According to the question,

$f(-2) = 0$

$\implies 4k – 8 = 0$

$\implies 4k = 8$

$\implies k = 2$

Answer:$k = 2$

9. If $f(x) = x^3 – 6x^2 + 11x – 6$, then for what values of $x$ is $f(x) = 0$?

Solution: Given, $f(x) = x^3 – 6x^2 + 11x – 6$

According to the question,

$f(x) = 0$

$\implies x^3 – 6x^2 + 11x – 6 = 0$

$\implies x^3 – x^2 – 5x^2 + 5x + 6x – 6 = 0$

$\implies x^2(x – 1) – 5x(x – 1) + 6(x – 1) = 0$

$\implies (x – 1)(x^2 – 5x + 6) = 0$

$\implies (x – 1)(x^2 – 2x – 3x + 6) = 0$

$\implies (x – 1)[x(x – 2) – 3(x – 2)] = 0$

$\implies (x – 1)(x – 2)(x – 3) = 0$

Either,

$x – 1 = 0 \implies x = 1$

or,

$x – 2 = 0 \implies x = 2$

or,

$x – 3 = 0 \implies x = 3$

Answer:$x = 1, 2, 3$

10. If $f(x) = \frac{2x + 1}{2x – 1}$, then determine the value of $\frac{f\left(\frac{1}{x^2}\right) + 1}{f\left(\frac{1}{x^2}\right) – 1}$.

Solution: Given, $f(x) = \frac{2x + 1}{2x – 1}$

Now,

$f\left(\frac{1}{x^2}\right) = \frac{2\left(\frac{1}{x^2}\right) + 1}{2\left(\frac{1}{x^2}\right) – 1} = \frac{\frac{2 + x^2}{x^2}}{\frac{2 – x^2}{x^2}} = \frac{2 + x^2}{2 – x^2}$

Then,

$\frac{f\left(\frac{1}{x^2}\right) + 1}{f\left(\frac{1}{x^2}\right) – 1} = \frac{\frac{2 + x^2}{2 – x^2} + 1}{\frac{2 + x^2}{2 – x^2} – 1}$

$= \frac{\frac{(2 + x^2) + (2 – x^2)}{2 – x^2}}{\frac{(2 + x^2) – (2 – x^2)}{2 – x^2}}$

$= \frac{2 + x^2 + 2 – x^2}{2 + x^2 – 2 + x^2}$

$= \frac{4}{2x^2} = \frac{2}{x^2}$

Answer:$\frac{2}{x^2}$

11. If $g(x) = \frac{1 + x^2 + x^4}{x^2}$, prove that $g\left(\frac{1}{x^2}\right) = g(x^2)$.

Solution: Given, $g(x) = \frac{1 + x^2 + x^4}{x^2}$

L.H.S.:

$g\left(\frac{1}{x^2}\right) = \frac{1 + \left(\frac{1}{x^2}\right)^2 + \left(\frac{1}{x^2}\right)^4}{\left(\frac{1}{x^2}\right)^2}$

$= \frac{1 + \frac{1}{x^4} + \frac{1}{x^8}}{\frac{1}{x^4}}$

$= \frac{\frac{x^8 + x^4 + 1}{x^8}}{\frac{1}{x^4}}$

$= \frac{x^8 + x^4 + 1}{x^8} \times x^4$

$= \frac{1 + x^4 + x^8}{x^4}$

R.H.S.:

$g(x^2) = \frac{1 + (x^2)^2 + (x^2)^4}{(x^2)^2}$

$= \frac{1 + x^4 + x^8}{x^4}$

$\therefore g\left(\frac{1}{x^2}\right) = g(x^2)$ (Shown)

12. Determine domain and range from the following relations:

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

Solution:

The set of first elements of the ordered pairs in a relation is called the Domain, and the set of second elements is called the Range.

Here,

Domain $R = \{2\}$

Range $R = \{1, 2, 3\}$

Answer: Domain $R = \{2\}$, Range $R = \{1, 2, 3\}$

(2) $S = \{(-2, 4), (-1, 1), (0, 0), (1, 1), (2, 4)\}$

Solution:

First elements: $-2, -1, 0, 1, 2$

Second elements: $4, 1, 0, 1, 4$

Domain $S = \{-2, -1, 0, 1, 2\}$

Range $S = \{0, 1, 4\}$

Answer: Domain $S = \{-2, -1, 0, 1, 2\}$, Range $S = \{0, 1, 4\}$

(3) $F = \left\{\left(\frac{1}{2}, 0\right), (1, 1), (1, -1), \left(\frac{5}{2}, 2\right), \left(\frac{5}{2}, -2\right)\right\}$

Solution:

First elements: $\frac{1}{2}, 1, 1, \frac{5}{2}, \frac{5}{2}$

Second elements: $0, 1, -1, 2, -2$

Domain $F = \left\{\frac{1}{2}, 1, \frac{5}{2}\right\}$

Range $F = \{-2, -1, 0, 1, 2\}$

Answer: Domain $F = \left\{\frac{1}{2}, 1, \frac{5}{2}\right\}$, Range $F = \{-2, -1, 0, 1, 2\}$

13. Express the following relations in tabular method and determine domain and range for each:

(1) $R = \{(x, y) : x \in A, y \in A \text{ and } x + y = 1\}$ where $A = \{-2, -1, 0, 1, 2\}$

Solution: Given, $A = \{-2, -1, 0, 1, 2\}$

Condition: $x + y = 1 \implies y = 1 – x$

Now for each $x \in A$, we calculate the value of $y = 1 – x$:

• For $x = -2$: $y = 1 – (-2) = 3 \notin A$
• For $x = -1$: $y = 1 – (-1) = 2 \in A$
• For $x = 0$: $y = 1 – 0 = 1 \in A$
• For $x = 1$: $y = 1 – 1 = 0 \in A$
• For $x = 2$: $y = 1 – 2 = -1 \in A$

$\therefore R = \{(-1, 2), (0, 1), (1, 0), (2, -1)\}$

Domain $R = \{-1, 0, 1, 2\}$

Range $R = \{-1, 0, 1, 2\}$

Answer:

Roster method: $R = \{(-1, 2), (0, 1), (1, 0), (2, -1)\}$

Domain $R = \{-1, 0, 1, 2\}$, Range $R = \{-1, 0, 1, 2\}$

(2) $F = \{(x, y) : x \in C, y \in C \text{ and } y = 2x\}$ where $C = \{-1, 0, 1, 2, 3\}$

Solution: Given, $C = \{-1, 0, 1, 2, 3\}$

Condition: $y = 2x$

Now for each $x \in C$, we calculate the value of $y = 2x$:

• For $x = -1$: $y = 2(-1) = -2 \notin C$
• For $x = 0$: $y = 2(0) = 0 \in C$
• For $x = 1$: $y = 2(1) = 2 \in C$
• For $x = 2$: $y = 2(2) = 4 \notin C$
• For $x = 3$: $y = 2(3) = 6 \notin C$

$\therefore F = \{(0, 0), (1, 2)\}$

Domain $F = \{0, 1\}$

Range $F = \{0, 2\}$

Answer:

Roster method: $F = \{(0, 0), (1, 2)\}$

Domain $F = \{0, 1\}$, Range $F = \{0, 2\}$

14. Draw the points $(-3, 2), (0, -5), \left(\frac{1}{2}, -\frac{5}{6}\right)$ on graph paper.

Solution (Drawing Instructions):

1. Draw two mutually perpendicular lines $XOX’$ (X-axis) and $YOY’$ (Y-axis) on a graph paper intersecting at the origin $O$.
2. Taking 1 small square = 1 unit:
   • Point $(-3, 2)$: From the origin $O$, move 3 squares to the left along the X-axis and 2 squares upward along the Y-axis to plot the point.
   • Point $(0, -5)$: From the origin $O$, move 5 squares downward along the Y-axis to plot the point.
   • Point $\left(\frac{1}{2}, -\frac{5}{6}\right)$: From the origin $O$, move half a square ($0.5$) to the right along the X-axis and slightly less than 1 square (approximately $0.83$ squares) downward along the Y-axis to plot the point.

15. Draw the points $(1, 2), (-1, 1), (11, 7)$ on the graph paper and show that all three points are on the same straight line.

Solution:

Algebraic Proof: Let the three points be $A(1, 2)$, $B(-1, 1)$, and $C(11, 7)$.

Slope of line segment $AB = \frac{1 – 2}{-1 – 1} = \frac{-1}{-2} = \frac{1}{2}$

Slope of line segment $BC = \frac{7 – 1}{11 – (-1)} = \frac{6}{12} = \frac{1}{2}$

Since the slopes of $AB$ and $BC$ are equal and point $B$ is common to both segments, the three points are collinear (lie on the same straight line).

Plotting and Observation on Graph Paper:

1. Draw X and Y axes on a graph paper and mark the origin.
2. Plot the three points $A(1, 2)$, $B(-1, 1)$, and $C(11, 7)$.
3. Connecting the three points using a ruler shows that all three points lie on a single straight line.
   (Shown)

16. Draw the graphs for the function as below:

(1) $y = 3x + 5$

Solution: Given equation: $y = 3x + 5$

Taking some values of $x$, we find the corresponding values of $y$:

x	−2	−1	0	1
$y$	$-1$	$2$	$5$	$8$

Drawing the graph:

Plot the points $A(-2, -1)$, $B(-1, 2)$, $C(0, 5)$, and $D(1, 8)$ on a graph paper and connect them to get a straight line. This line is the required graph of the function.

(2) $x + y = 2$

Solution:

Given equation: $x + y = 2 \implies y = 2 – x$

Taking some values of $x$, we find the corresponding values of $y$:

x	−1	0	1	2
$y$	$3$	$2$	$1$	$0$

Drawing the graph:

Plot the points $P(-1, 3)$, $Q(0, 2)$, $R(1, 1)$, and $S(2, 0)$ on a graph paper and connect them to get a straight line. This line is the required graph of the function.

Sample Questions (Multiple Choice)

1. If $A = \{2, 3, 4\}$ and $B = \{1, 2\}$, how many elements are in the set $(A \times B)$?

a) 16

b) 8

c) 6

d) 5

Solution:

Number of elements in set $A$: $n(A) = 3$

Number of elements in set $B$: $n(B) = 2$

Number of elements in Cartesian product $(A \times B) = n(A) \times n(B) = 3 \times 2 = 6$.

Answer: c) 6

2. If $A \cup B = \{a, b, c\}$, then—

(i) $A = \{a, b\}$, $B = \{a, b, c\}$

(ii) Number of elements in $P(A \cup B)$ is 8.

(iii) $A$ is a finite set.

Which of the following is correct?

a) i & ii

b) i & iii

c) ii & iii

d) i, ii & iii

Solution:

(i) If $A = \{a, b\}$ and $B = \{a, b, c\}$, then $A \cup B = \{a, b, c\}$, which is correct.

(ii) The number of elements in $A \cup B$ is 3, so elements in power set $= 2^3 = 8$, which is correct.

(iii) The elements of set $A$ can be counted, so it is a finite set, which is correct.

All three statements are correct.

Answer: d) i, ii & iii

If $A = \{6, 7, 8, 9, 10, 11, 12, 13\}$, then answer the questions no. 3 and 4:

3. Which one is the correct expression for set $A$?

a) $\{x \in \mathbb{N} : 6 < x < 13\}$

b) $\{x \in \mathbb{N} : 6 \le x < 13\}$

c) $\{x \in \mathbb{N} : 6 \le x \le 13\}$

d) $\{x \in \mathbb{N} : 6 < x \le 13\}$

Solution:

The elements of set $A$ are natural numbers ($\mathbb{N}$) that are greater than or equal to 6 and less than or equal to 13.

That is, $6 \le x \le 13$.

Answer: c) $\{x \in \mathbb{N} : 6 \le x \le 13\}$

4. Which one is the set of primes in $A$?

a) $\{6, 8, 10, 12\}$

b) $\{7, 9, 11, 13\}$

c) $\{7, 11, 13\}$

d) $\{9, 12\}$

Solution:

Among the elements of set $A = \{6, 7, 8, 9, 10, 11, 12, 13\}$, the prime numbers are: $7, 11, 13$.

Answer: c) $\{7, 11, 13\}$

Creative Question:

5. $y = f(x) = \frac{4x – 7}{2x – 4}$ is a function.

(a) Determine the value of $f\left(-\frac{1}{2}\right)$.

Solution: Given, $f(x) = \frac{4x – 7}{2x – 4}$

Substituting $-\frac{1}{2}$ in place of $x$:

$f\left(-\frac{1}{2}\right) = \frac{4\left(-\frac{1}{2}\right) – 7}{2\left(-\frac{1}{2}\right) – 4} = \frac{-2 – 7}{-1 – 4} = \frac{-9}{-5} = \frac{9}{5}$

Answer:$\frac{9}{5}$

(b) Determine the value of $\frac{f(x) + 2}{f(x) – 1}$.

Solution: Given, $f(x) = \frac{4x – 7}{2x – 4}$

Now,

$\frac{f(x) + 2}{f(x) – 1} = \frac{\frac{4x – 7}{2x – 4} + 2}{\frac{4x – 7}{2x – 4} – 1}$

$= \frac{\frac{(4x – 7) + 2(2x – 4)}{2x – 4}}{\frac{(4x – 7) – 1(2x – 4)}{2x – 4}}$

$= \frac{4x – 7 + 4x – 8}{4x – 7 – 2x + 4}$

$= \frac{8x – 15}{2x – 3}$

Answer:$\frac{8x – 15}{2x – 3}$

(c) Prove that $f(y) = x$.

Solution:

From the given stem, $y = \frac{4x – 7}{2x – 4}$ and $f(x) = \frac{4x – 7}{2x – 4}$

To find $f(y)$, substitute $y$ in place of $x$ in $f(x)$:

$f(y) = \frac{4y – 7}{2y – 4}$

Now substitute the value of $y$, we get

$f(y) = \frac{4\left(\frac{4x – 7}{2x – 4}\right) – 7}{2\left(\frac{4x – 7}{2x – 4}\right) – 4}$

$= \frac{\frac{4(4x – 7) – 7(2x – 4)}{2x – 4}}{\frac{2(4x – 7) – 4(2x – 4)}{2x – 4}}$

$= \frac{16x – 28 – 14x + 28}{8x – 14 – 8x + 16}$

$= \frac{2x}{2} = x$

$\therefore f(y) = x$ (Shown)

Short-Answer Questions:

6. (a) If $A$ and $B$ are subsets of a universal set $U$, use a Venn diagram to represent $(A \cup B)’$ .

Solution (Description of Venn Diagram):

1. Draw a rectangle representing the universal set $U$.
2. Inside the rectangle, draw two intersecting circles representing sets $A$ and $B$.
3. $A \cup B$ represents the entire region inside both circles $A$ and $B$.
4. Therefore, $(A \cup B)’ = U \setminus (A \cup B)$ is the region outside circles $A$ and $B$, but inside the rectangle $U$.
   (In a diagram, shade/color the empty region inside the rectangle, i.e., outside the two circles.)

(b) If the sets of prime factors of 10 and 21 are $A$ and $B$ respectively, show that $(A \cap B)$ is a null set.

Solution:

Prime factors of 10: $2, 5$

$\therefore A = \{2, 5\}$

Prime factors of 21: $3, 7$

$\therefore B = \{3, 7\}$

Now,

$A \cap B = \{2, 5\} \cap \{3, 7\} = \emptyset$

Since $A \cap B = \emptyset$ (empty set), sets $A$ and $B$ are disjoint sets.

(Shown)

(c) If $f(x) = x^3 – 3x^2 + 3x – 1$ and $f(a) = 0$, find the value of $a$.

Solution: Given, $f(x) = x^3 – 3x^2 + 3x – 1$

$\therefore f(a) = a^3 – 3a^2 + 3a – 1$

According to the question,

$f(a) = 0$

$\implies a^3 – 3 \cdot a^2 \cdot 1 + 3 \cdot a \cdot 1^2 – 1^3 = 0$

$\implies (a – 1)^3 = 0$

$\implies a – 1 = 0$

$\implies a = 1$

Answer:$a = 1$

(d) Determine all subsets of the set $A = \{a, b, c\}$ and then find the proper subsets.

Solution: Given, $A = \{a, b, c\}$

The subsets of set $A$ are:

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

Subsets having fewer elements than the original set are called proper subsets (i.e., excluding the original set itself).

$\therefore$ The proper subsets of set $A$ are:

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

Answer:

Subsets: $\{a, b, c\}, \{a, b\}, \{a, c\}, \{b, c\}, \{a\}, \{b\}, \{c\}, \emptyset$

Proper Subsets: $\{a, b\}, \{a, c\}, \{b, c\}, \{a\}, \{b\}, \{c\}, \emptyset$