Q1
If \( \left( \frac{x}{3} + 1, y - \frac{2}{3} \right) = \left( \frac{5}{3}, \frac{1}{3} \right) \), find the values of \( x \) and \( y \).
Answer:
Since the ordered pairs are equal, their corresponding elements are equal.
\( \frac{x}{3} + 1 = \frac{5}{3} \Rightarrow \frac{x}{3} = \frac{5}{3} - 1 = \frac{2}{3} \Rightarrow x = 2 \)
\( y - \frac{2}{3} = \frac{1}{3} \Rightarrow y = \frac{1}{3} + \frac{2}{3} = \frac{3}{3} = 1 \)
Result: \( x = 2, y = 1 \)
Q2
If the set A has 3 elements and the set B = \( \{3, 4, 5\} \), then find the number of elements in \( (A \times B) \).
Answer:
Number of elements in A, \( n(A) = 3 \).
Number of elements in B, \( n(B) = 3 \).
Number of elements in \( A \times B = n(A) \times n(B) = 3 \times 3 = 9 \).
Q3
If \( G = \{7, 8\} \) and \( H = \{5, 4, 2\} \), find \( G \times H \) and \( H \times G \).
Answer:
\( G \times H = \{ (7,5), (7,4), (7,2), (8,5), (8,4), (8,2) \} \)
\( H \times G = \{ (5,7), (5,8), (4,7), (4,8), (2,7), (2,8) \} \)
Q4
State whether each of the following statements are true or false. If the statement is false, rewrite the given statement correctly.- If \( P = \{m, n\} \) and \( Q = \{n, m\} \), then \( P \times Q = \{(m, n), (n, m)\} \).
- If A and B are non-empty sets, then \( A \times B \) is a non-empty set of ordered pairs \( (x, y) \) such that \( x \in A \) and \( y \in B \).
- If \( A = \{1, 2\}, B = \{3, 4\} \), then \( A \times (B \cap \phi) = \phi \).
- If \( P = \{m, n\} \) and \( Q = \{n, m\} \), then \( P \times Q = \{(m, n), (n, m)\} \).
- If A and B are non-empty sets, then \( A \times B \) is a non-empty set of ordered pairs \( (x, y) \) such that \( x \in A \) and \( y \in B \).
- If \( A = \{1, 2\}, B = \{3, 4\} \), then \( A \times (B \cap \phi) = \phi \).
Answer:
- False. Correct statement: If \( P = \{m, n\} \) and \( Q = \{n, m\} \), then \( P \times Q = \{(m, n), (m, m), (n, n), (n, m)\} \).
- True. By definition of Cartesian product.
- True. \( B \cap \phi = \phi \), and the Cartesian product of any set with the null set is the null set.
Q5
If \( A = \{-1, 1\} \), find \( A \times A \times A \).
Answer:
\( A \times A = \{ (-1, -1), (-1, 1), (1, -1), (1, 1) \} \)
\( A \times A \times A = \{ (-1, -1, -1), (-1, -1, 1), (-1, 1, -1), (-1, 1, 1), (1, -1, -1), (1, -1, 1), (1, 1, -1), (1, 1, 1) \} \)
Q6
If \( A \times B = \{ (a, x), (a, y), (b, x), (b, y) \} \). Find A and B.
Answer:
A is the set of all first components: \( A = \{a, b\} \).
B is the set of all second components: \( B = \{x, y\} \).
Q7
Let \( A = \{1, 2\}, B = \{1, 2, 3, 4\}, C = \{5, 6\} \) and \( D = \{5, 6, 7, 8\} \). Verify that
(i) \( A \times (B \cap C) = (A \times B) \cap (A \times C) \)
(ii) \( A \times C \) is a subset of \( B \times D \)
(i) \( A \times (B \cap C) = (A \times B) \cap (A \times C) \)
(ii) \( A \times C \) is a subset of \( B \times D \)
Answer:
(i) \( B \cap C = \phi \). So LHS \( = A \times \phi = \phi \).
\( A \times B \) and \( A \times C \) share no common elements because B and C are disjoint. So RHS \( = \phi \). Hence verified.
(ii) \( A \times C = \{ (1, 5), (1, 6), (2, 5), (2, 6) \} \).
Since all elements of A are in B and all elements of C are in D, every pair in \( A \times C \) is present in \( B \times D \). Hence verified.
Q8
Let \( A = \{1, 2\} \) and \( B = \{3, 4\} \). Write \( A \times B \). How many subsets will \( A \times B \) have? List them.
Answer:
\( A \times B = \{ (1, 3), (1, 4), (2, 3), (2, 4) \} \).
Number of elements \( n = 4 \).
Number of subsets \( = 2^4 = 16 \).
Subsets:
\( \phi \)
\( \{(1, 3)\}, \{(1, 4)\}, \{(2, 3)\}, \{(2, 4)\} \)
\( \{(1, 3), (1, 4)\}, \{(1, 3), (2, 3)\}, \dots \) (and so on combining 2, 3, and 4 elements).
Q9
Let A and B be two sets such that \( n(A) = 3 \) and \( n(B) = 2 \). If \( (x, 1), (y, 2), (z, 1) \) are in \( A \times B \), find A and B, where \( x, y \) and \( z \) are distinct elements.
Answer:
A contains the first components: \( x, y, z \). Since \( n(A) = 3 \), \( A = \{x, y, z\} \).
B contains the second components: \( 1, 2, 1 \). Since \( n(B) = 2 \), \( B = \{1, 2\} \).
Q10
The Cartesian product \( A \times A \) has 9 elements among which are found \( (-1, 0) \) and \( (0, 1) \). Find the set A and the remaining elements of \( A \times A \).
Answer:
Since \( n(A \times A) = 9 \), \( n(A) = 3 \).
The elements in the pairs are -1, 0, 1. Thus, \( A = \{-1, 0, 1\} \).
Remaining elements of \( A \times A \):
\( (-1, -1), (-1, 1), (0, -1), (0, 0), (1, -1), (1, 0), (1, 1) \).
Q1
Let \( A = \{1, 2, 3, \dots, 14\} \). Define a relation R from A to A by \( R = \{(x, y) : 3x - y = 0, \text{ where } x, y \in A\} \). Write down its domain, codomain and range.
Answer:
Relation \( 3x - y = 0 \Rightarrow y = 3x \).
Pairs: \( (1, 3), (2, 6), (3, 9), (4, 12) \). (For \( x=5, y=15 \notin A \)).
Domain: \( \{1, 2, 3, 4\} \)
Range: \( \{3, 6, 9, 12\} \)
Codomain: Set A \( \{1, 2, \dots, 14\} \)
Q2
Define a relation R on the set N of natural numbers by \( R = \{(x, y) : y = x + 5, x \text{ is a natural number less than 4}; x, y \in N\} \). Depict this relationship using roster form. Write down the domain and the range.
Answer:
Possible values for \( x \): 1, 2, 3.
If \( x = 1, y = 6 \). Pair: \( (1, 6) \).
If \( x = 2, y = 7 \). Pair: \( (2, 7) \).
If \( x = 3, y = 8 \). Pair: \( (3, 8) \).
Roster form: \( \{(1, 6), (2, 7), (3, 8)\} \)
Domain: \( \{1, 2, 3\} \)
Range: \( \{6, 7, 8\} \)
Q3
\( A = \{1, 2, 3, 5\} \) and \( B = \{4, 6, 9\} \). Define a relation R from A to B by \( R = \{(x, y): \text{the difference between } x \text{ and } y \text{ is odd}; x \in A, y \in B\} \). Write R in roster form.
Answer:
Difference is odd if one number is even and the other is odd.
- \( 1 \in A \) (odd): Pairs with even in B: \( (1, 4), (1, 6) \).
- \( 2 \in A \) (even): Pairs with odd in B: \( (2, 9) \).
- \( 3 \in A \) (odd): Pairs with even in B: \( (3, 4), (3, 6) \).
- \( 5 \in A \) (odd): Pairs with even in B: \( (5, 4), (5, 6) \).
Roster form: \( \{(1, 4), (1, 6), (2, 9), (3, 4), (3, 6), (5, 4), (5, 6)\} \)
Q4
The Fig 2.7 shows a relationship between the sets P and Q. Write this relation
(i) in set-builder form
(ii) roster form.
What is its domain and range?
(i) in set-builder form
(ii) roster form.
What is its domain and range?
Answer:
From the figure (assuming standard NCERT figure): \( 5 \to 3, 6 \to 4, 7 \to 5 \).
The relationship is \( y = x - 2 \).
(i) Set-builder form: \( R = \{(x, y) : y = x - 2, x \in P, y \in Q\} \)
(ii) Roster form: \( \{(5, 3), (6, 4), (7, 5)\} \)
Domain: \( \{5, 6, 7\} \)
Range: \( \{3, 4, 5\} \)
Q5
Let \( A = \{1, 2, 3, 4, 6\} \). Let R be the relation on A defined by \( \{(a, b) : a, b \in A, b \text{ is exactly divisible by } a\} \).
(i) Write R in roster form
(ii) Find the domain of R
(iii) Find the range of R.
(i) Write R in roster form
(ii) Find the domain of R
(iii) Find the range of R.
Answer:
(i) Roster form:
\( \{(1,1), (1,2), (1,3), (1,4), (1,6), (2,2), (2,4), (2,6), (3,3), (3,6), (4,4), (6,6)\} \)
(ii) Domain: \( \{1, 2, 3, 4, 6\} \)
(iii) Range: \( \{1, 2, 3, 4, 6\} \)
Q6
Determine the domain and range of the relation R defined by \( R = \{(x, x+5) : x \in \{0, 1, 2, 3, 4, 5\}\} \).
Answer:
Values of \( x \): 0, 1, 2, 3, 4, 5.
Corresponding values of \( y (x+5) \): 5, 6, 7, 8, 9, 10.
Domain: \( \{0, 1, 2, 3, 4, 5\} \)
Range: \( \{5, 6, 7, 8, 9, 10\} \)
Q7
Write the relation \( R = \{(x, x^3) : x \text{ is a prime number less than 10}\} \) in roster form.
Answer:
Primes less than 10 are 2, 3, 5, 7.
Cubes: \( 2^3=8, 3^3=27, 5^3=125, 7^3=343 \).
Roster form: \( \{(2, 8), (3, 27), (5, 125), (7, 343)\} \)
Q8
Let \( A = \{x, y, z\} \) and \( B = \{1, 2\} \). Find the number of relations from A to B.
Answer:
\( n(A) = 3, n(B) = 2 \).
\( n(A \times B) = 3 \times 2 = 6 \).
Number of relations = Number of subsets of \( A \times B = 2^6 = 64 \).
Q9
Let R be the relation on Z defined by \( R = \{(a, b) : a, b \in Z, a - b \text{ is an integer}\} \). Find the domain and range of R.
Answer:
The difference of any two integers is always an integer.
Thus, for any \( a \in Z \), we can pick any \( b \in Z \).
Domain: Z (Integers)
Range: Z (Integers)
Q1
Which of the following relations are functions? Give reasons. If it is a function, determine its domain and range.- \( \{(2,1), (5,1), (8,1), (11,1), (14,1), (17,1)\} \)
- \( \{(2,1), (4,2), (6,3), (8,4), (10,5), (12,6), (14,7)\} \)
- \( \{(1,3), (1,5), (2,5)\} \)
- \( \{(2,1), (5,1), (8,1), (11,1), (14,1), (17,1)\} \)
- \( \{(2,1), (4,2), (6,3), (8,4), (10,5), (12,6), (14,7)\} \)
- \( \{(1,3), (1,5), (2,5)\} \)
Answer:
- Function. Every element in the domain has a unique image (1).
Domain: \( \{2, 5, 8, 11, 14, 17\} \), Range: \( \{1\} \). - Function. Every first element has a unique second element.
Domain: \( \{2, 4, 6, 8, 10, 12, 14\} \), Range: \( \{1, 2, 3, 4, 5, 6, 7\} \). - Not a Function. The element 1 has two different images (3 and 5).
Q2
Find the domain and range of the following real functions:
(i) \( f(x) = -|x| \)
(ii) \( f(x) = \sqrt{9 - x^2} \)
(i) \( f(x) = -|x| \)
(ii) \( f(x) = \sqrt{9 - x^2} \)
Answer:
- \( f(x) = -|x| \) is defined for all real numbers.
Domain: R
Since \( |x| \ge 0 \), \( -|x| \le 0 \).
Range: \( (-\infty, 0] \) - \( f(x) = \sqrt{9 - x^2} \) is defined when \( 9 - x^2 \ge 0 \Rightarrow x^2 \le 9 \Rightarrow -3 \le x \le 3 \).
Domain: \( [-3, 3] \)
The value of square root is non-negative and max value is \( \sqrt{9}=3 \).
Range: \( [0, 3] \)
Q3
A function \( f \) is defined by \( f(x) = 2x - 5 \). Write down the values of
(i) \( f(0) \)
(ii) \( f(7) \)
(iii) \( f(-3) \)
(i) \( f(0) \)
(ii) \( f(7) \)
(iii) \( f(-3) \)
Answer:
- \( f(0) = 2(0) - 5 = -5 \)
- \( f(7) = 2(7) - 5 = 14 - 5 = 9 \)
- \( f(-3) = 2(-3) - 5 = -6 - 5 = -11 \)
Q4
The function 't' which maps temperature in degree Celsius into temperature in degree Fahrenheit is defined by \( t(C) = \frac{9C}{5} + 32 \).
Find (i) \( t(0) \) (ii) \( t(28) \) (iii) \( t(-10) \) (iv) The value of C when \( t(C) = 212 \).
Find (i) \( t(0) \) (ii) \( t(28) \) (iii) \( t(-10) \) (iv) The value of C when \( t(C) = 212 \).
Answer:
- \( t(0) = 0 + 32 = 32 \)
- \( t(28) = \frac{9 \times 28}{5} + 32 = \frac{252}{5} + 32 = 50.4 + 32 = 82.4 \)
- \( t(-10) = \frac{9 \times (-10)}{5} + 32 = -18 + 32 = 14 \)
- \( 212 = \frac{9C}{5} + 32 \Rightarrow 180 = \frac{9C}{5} \Rightarrow C = \frac{180 \times 5}{9} = 100 \).
Q5
Find the range of each of the following functions.- \( f(x) = 2 - 3x, x \in R, x > 0 \).
- \( f(x) = x^2 + 2, x \) is a real number.
- \( f(x) = x, x \) is a real number.
- \( f(x) = 2 - 3x, x \in R, x > 0 \).
- \( f(x) = x^2 + 2, x \) is a real number.
- \( f(x) = x, x \) is a real number.
Answer:
- \( x > 0 \Rightarrow 3x > 0 \Rightarrow -3x < 0 \Rightarrow 2 - 3x < 2 \).
Range: \( (-\infty, 2) \) - \( x^2 \ge 0 \Rightarrow x^2 + 2 \ge 2 \).
Range: \( [2, \infty) \) - \( f(x) = x \) takes all real values.
Range: R
Q1
The relation \( f \) is defined by \( f(x) = \begin{cases} x^2, & 0 \le x \le 3 \\ 3x, & 3 \le x \le 10 \end{cases} \). The relation \( g \) is defined by \( g(x) = \begin{cases} x^2, & 0 \le x \le 2 \\ 3x, & 2 \le x \le 10 \end{cases} \). Show that \( f \) is a function and \( g \) is not a function.
Answer:
For \( f(x) \): At \( x = 3 \), \( x^2 = 9 \) and \( 3x = 9 \). Since values match, \( f \) is consistent and has a unique image for every \( x \). Thus, \( f \) is a function.
For \( g(x) \): At \( x = 2 \), \( x^2 = 4 \) and \( 3x = 6 \). Since \( 4 \ne 6 \), the element 2 has two distinct images. Thus, \( g \) is not a function.
Q2
If \( f(x) = x^2 \), find \( \frac{f(1.1) - f(1)}{1.1 - 1} \).
Answer:
\( f(1.1) = (1.1)^2 = 1.21 \)
\( f(1) = 1^2 = 1 \)
\( \frac{1.21 - 1}{0.1} = \frac{0.21}{0.1} = 2.1 \)
Q3
Find the domain of the function \( f(x) = \frac{x^2 + 2x + 1}{x^2 - 8x + 12} \).
Answer:
The function is undefined when the denominator is zero.
\( x^2 - 8x + 12 = 0 \Rightarrow (x - 6)(x - 2) = 0 \)
So \( x = 6, 2 \).
Domain: \( R - \{2, 6\} \)
Q4
Find the domain and the range of the real function \( f \) defined by \( f(x) = \sqrt{x - 1} \).
Answer:
Domain: \( x - 1 \ge 0 \Rightarrow x \ge 1 \). So, \( [1, \infty) \).
Range: The square root yields non-negative values. So, \( [0, \infty) \).
Q5
Find the domain and the range of the real function \( f \) defined by \( f(x) = |x - 1| \).
Answer:
Domain: Defined for all real numbers. R.
Range: Absolute value is always non-negative. \( [0, \infty) \).
Q6
Let \( f = \left\{ \left( x, \frac{x^2}{1+x^2} \right) : x \in R \right\} \) be a function from R into R. Determine the range of \( f \).
Answer:
Let \( y = \frac{x^2}{1+x^2} \).
Since \( x^2 \ge 0 \) and \( 1+x^2 > x^2 \), \( y \) is always positive and less than 1.
Also at \( x=0, y=0 \).
Range: \( [0, 1) \).
Q7
Let \( f, g : R \to R \) be defined, respectively by \( f(x) = x + 1, g(x) = 2x - 3 \). Find \( f + g, f - g \) and \( f/g \).
Answer:
\( (f + g)(x) = (x + 1) + (2x - 3) = 3x - 2 \)
\( (f - g)(x) = (x + 1) - (2x - 3) = -x + 4 \)
\( \left( \frac{f}{g} \right)(x) = \frac{x+1}{2x-3}, x \ne \frac{3}{2} \)
Q8
Let \( f = \{(1, 1), (2, 3), (0, -1), (-1, -3)\} \) be a function from Z to Z defined by \( f(x) = ax + b \), for some integers \( a, b \). Determine \( a, b \).
Answer:
Using point \( (1, 1) \): \( 1 = a(1) + b \Rightarrow a + b = 1 \).
Using point \( (0, -1) \): \( -1 = a(0) + b \Rightarrow b = -1 \).
Substitute \( b = -1 \) in first eqn: \( a - 1 = 1 \Rightarrow a = 2 \).
Result: \( a = 2, b = -1 \)
Q9
Let R be a relation from N to N defined by \( R = \{(a, b) : a, b \in N \text{ and } a = b^2\} \). Are the following true?
(i) \( (a, a) \in R \) for all \( a \in N \)
(ii) \( (a, b) \in R \implies (b, a) \in R \)
(iii) \( (a, b) \in R, (b, c) \in R \implies (a, c) \in R \)
(i) \( (a, a) \in R \) for all \( a \in N \)
(ii) \( (a, b) \in R \implies (b, a) \in R \)
(iii) \( (a, b) \in R, (b, c) \in R \implies (a, c) \in R \)
Answer:
(i) False. \( 2 \ne 2^2 \).
(ii) False. \( 4 = 2^2 \) but \( 2 \ne 4^2 \).
(iii) False. \( 16 = 4^2 \) and \( 4 = 2^2 \), but \( 16 \ne 2^2 \).
Q10
Let \( A = \{1, 2, 3, 4\}, B = \{1, 5, 9, 11, 15, 16\} \) and \( f = \{(1, 5), (2, 9), (3, 1), (4, 5), (2, 11)\} \). Are the following true?
(i) \( f \) is a relation from A to B
(ii) \( f \) is a function from A to B.
(i) \( f \) is a relation from A to B
(ii) \( f \) is a function from A to B.
Answer:
(i) True. Every pair in \( f \) has first element in A and second in B.
(ii) False. Element 2 has two images: 9 and 11.
Q11
Let \( f \) be the subset of \( Z \times Z \) defined by \( f = \{(ab, a+b) : a, b \in Z\} \). Is \( f \) a function from Z to Z? Justify your answer.
Answer:
No. Consider \( a=1, b=4 \Rightarrow (4, 5) \). Consider \( a=2, b=2 \Rightarrow (4, 4) \).
The element 4 has two different images 5 and 4. Thus, it is not a function.
Q12
Let \( A = \{9, 10, 11, 12, 13\} \) and let \( f: A \to N \) be defined by \( f(n) = \) the highest prime factor of \( n \). Find the range of \( f \).
Answer:
\( f(9) = 3 \) (Factors 1, 3, 9)
\( f(10) = 5 \) (Factors 1, 2, 5, 10)
\( f(11) = 11 \) (Prime)
\( f(12) = 3 \) (Factors 1, 2, 3, 4, 6, 12)
\( f(13) = 13 \) (Prime)
Range: \( \{3, 5, 11, 13\} \)