Exercise 2.1

Question 1:

 If (x/3 + 1, y – 2/3) = (5/3, 1/3), find the values of x and y.

Answer:

It is given that

(x/3 + 1, y – 2/3) = (5/3, 1/3)

Since the ordered pairs are equal, the corresponding elements will also be equal.

Therefore,

=> x/3 + 1 = 5/3 and y – 2/3 = 1/3

=> x/3 = 5/3 - 1 and y = 1/3 + 2/3

=> x/3 = 2/3 and y = 3/3

=> x/3 = 2/3 and y = 1

=> x = 2 and y = 1

Question 2:

If the set A has 3 elements and the set B = {3, 4, 5}, then find the number of elements in (A * B)?

Answer:

It is given that set A has 3 elements and the elements of set B are 3, 4, and 5.

Number of elements in set B = 3

Number of elements in (A * B) = (Number of elements in A) * (Number of elements in B)

                                                       = 3 * 3 = 9

Thus, the number of elements in (A * B) is 9

Question 3:

If G = {7, 8} and H = {5, 4, 2}, find G * H and H * G.

Answer:

G = {7, 8} and H = {5, 4, 2}

We know that the Cartesian product P * Q of two non-empty sets P and Q is defined as

P * Q = {(p, q): p ∈ P, q ∈ Q}

So, G * H = {(7, 5), (7, 4), (7, 2), (8, 5), (8, 4), (8, 2)}

and H * G = {(5, 7), (5, 8), (4, 7), (4, 8), (2, 7), (2, 8)}

Question 4:

State whether each of the following statements are true or false. If the statement is false, rewrite the given statement correctly.

(i) If P = {m, n} and Q = {n, m}, then P * Q = {(m, n), (n, m)}.

(ii) If A and B are non-empty sets, then A * B is a non-empty set of ordered pairs (x, y) such that x ∈ A and y ∈ B.

(iii) If A = {1, 2}, B = {3, 4}, then A * (B ∩ φ) = φ.

Answer:

(i) False

If P = {m, n} and Q = {n, m}, then

P * Q = {(m, m), (m, n), (n, m), (n, n)}

(ii) True

(iii) True

Question 5:

If A = {–1, 1}, find A * A * A.

Answer:

It is known that for any non-empty set A, A * A * A is defined as

A * A * A = {(a, b, c): a, b, c ∈ A}

It is given that A = {–1, 1}

Now, A * A * 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)}

Question 6:

If A * B = {(a, x), (a , y), (b, x), (b, y)}. Find A and B.

Answer:

It is given that A * B = {(a, x), (a, y), (b, x), (b, y)}

We know that the Cartesian product of two non-empty sets P and Q is defined as

P * Q = {(p, q): p ∈ P, q ∈ Q}

So, A is the set of all first elements and B is the set of all second elements.

Thus, A = {a, b} and B = {x, y}

Question 7:

Let A = {1, 2}, B = {1, 2, 3, 4}, C = {5, 6} and D = {5, 6, 7, 8}. Verify that

(i) A * (B ∩ C) = (A * B) ∩ (A * C).              (ii) A * C is a subset of B * D.

Answer:

(i) To verify: A * (B ∩ C) = (A * B) ∩ (A * C)

We have B ∩ C = {1, 2, 3, 4} ∩ {5, 6} = ф

L.H.S. = A * (B ∩ C) = A * ф = ф

A * B = {(1, 1), (1, 2), (1, 3), (1, 4), (2, 1), (2, 2), (2, 3), (2, 4)}

A * C = {(1, 5), (1, 6), (2, 5), (2, 6)}

R.H.S. = (A * B) ∩ (A * C) = ф

Since L.H.S. = R.H.S

Hence, A * (B ∩ C) = (A * B) ∩ (A * C)

(ii) To verify: A * C is a subset of B * D

A * C = {(1, 5), (1, 6), (2, 5), (2, 6)}

A * D = {(1, 5), (1, 6), (1, 7), (1, 8), (2, 5), (2, 6), (2, 7), (2, 8), (3, 5), (3, 6), (3, 7), (3, 8), (4, 5),           

              (4, 6), (4, 7), (4, 8)}

We can observe that all the elements of set A * C are the elements of set B * D.

Therefore, A * C is a subset of B * D.

Question 8:

Let A = {1, 2} and B = {3, 4}. Write A * B. How many subsets will A * B have? List them.

Answer:

A = {1, 2} and B = {3, 4}

So, A * B = {(1, 3), (1, 4), (2, 3), (2, 4)}

=> n(A * B) = 4

We know that if C is a set with n(C) = m, then n[P(C)] = 2^m

Therefore, the set A * B has 2⁴ = 16 subsets. These are

ф, {(1, 3)}, {(1, 4)}, {(2, 3)}, {(2, 4)}, {(1, 3), (1, 4)}, {(1, 3), (2, 3)}, {(1, 3), (2, 4)}, {(1, 4), (2, 3)},    

{(1, 4), (2, 4)}, {(2, 3), (2, 4)}, {(1, 3), (1, 4), (2, 3)}, {(1, 3), (1, 4), (2, 4)}, {(1, 3), (2, 3), (2, 4)},

{(1, 4), (2, 3), (2, 4)}, {(1, 3), (1, 4), (2, 3), (2, 4)}

Question 9:

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 * B, find A and B, where x, y and z are distinct elements.

Answer:

It is given that n(A) =3 and n(B) =2 and (x, 1), (y, 2), (z, 1) are in A * B.

We know that

A = Set of first elements of the ordered pair elements of A * B

B = Set of second elements of the ordered pair elements of A * B.

So, x, y and z are the elements of A and 1 and 2 are the elements of B.

Since n(A) = 3 and n(B) = 2

It is clear that A = {x, y, z} and B = {1, 2}.

Question 10:

The Cartesian product A * A has 9 elements among which are found (–1, 0) and (0, 1). Find the set A and the remaining elements of A * A.

Answer:

We know that if n(A) = p and n(B) = q, then n(A * B) = pq.

So, n(A * A) = n(A) * n(A)

It is given that n(A * A) = 9

Hence, n(A) * n(A) = 9

=> n(A) = 3

The ordered pairs (–1, 0) and (0, 1) are two of the nine elements of A * A.

We know that A * A = {(a, a): a ∈ A}. Therefore, –1, 0, and 1 are elements of A.

Since n(A) = 3, it is clear that A = {–1, 0, 1}.

The remaining elements of set A * A are

(–1, –1), (–1, 1), (0, –1), (0, 0), (1, –1), (1, 0), and (1, 1).

 

                                                                        Exercise 2.2

Question 1:

Let A = {1, 2, 3,...,14}. Define a relation R from A to A by R = {(x, y): 3x – y = 0, where x, y ∈ A}.

Write down its domain, co-domain and range.

Answer:

The relation R from A to A is given as

R = {(x, y): 3x – y = 0, where x, y ∈ A}

i.e. R = {(x, y): 3x = y, where x, y ∈ A}

So, R = {(1, 3), (2, 6), (3, 9), (4, 12)}

The domain of R is the set of all first elements of the ordered pairs in the relation.

So, Domain of R = {1, 2, 3, 4}

The whole set A is the co-domain of the relation R.

So, Co-domain of R = A = {1, 2, 3... 14}

The range of R is the set of all second elements of the ordered pairs in the relation.

Hence, Range of R = {3, 6, 9, 12}

Question 2:

Define a relation R on the set N of natural numbers by R = {(x, y): y = x + 5, x is a natural number less than 4; x, y ∈N}.

Depict this relationship using roster form. Write down the domain and the range.

Answer:

R = {(x, y): y = x + 5, x is a natural number less than 4, x, y ∈ N}

The natural numbers less than 4 are 1, 2, and 3.

So, R = {(1, 6), (2, 7), (3, 8)}

The domain of R is the set of all first elements of the ordered pairs in the relation.

So, Domain of R = {1, 2, 3}

The range of R is the set of all second elements of the ordered pairs in the relation.

Hence, Range of R = {6, 7, 8}

Question 3:

A = {1, 2, 3, 5} and B = {4, 6, 9}. Define a relation R from A to B by R = {(x, y): the difference between x and y is odd; x ∈ A, y ∈ B}.

Write R in roster form.

Answer:

Given, A = {1, 2, 3, 5} and B = {4, 6, 9}

R = {(x, y): the difference between x and y is odd; x ∈ A, y ∈ B}

Hence, R = {(1, 4), (1, 6), (2, 9), (3, 4), (3, 6), (5, 4), (5, 6)}

Question 4:

The Fig2.7 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?

 

Answer:

According to the given figure, P = {5, 6, 7}, Q = {3, 4, 5}

(i) R = {(x, y): y = x – 2; x ∈ P} or R = {(x, y): y = x – 2 for x = 5, 6, 7}

(ii) R = {(5, 3), (6, 4), (7, 5)}

Now, Domain of R = {5, 6, 7}

and Range of R = {3, 4, 5}

 

Question 5:

Let A = {1, 2, 3, 4, 6}. Let R be the relation on A defined by

{(a, b): a, b ∈ A, b is exactly divisible by a}.

(i) Write R in roster form

(ii) Find the domain of R

(iii) Find the range of R.

Answer:

Given, A = {1, 2, 3, 4, 6}, R = {(a, b): a, b ∈ A, b is exactly divisible by a}

(i) R = {(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 of R = {1, 2, 3, 4, 6}

(iii) Range of R = {1, 2, 3, 4, 6}

Question 6:

Determine the domain and range of the relation R defined by

R = {(x, x + 5): x ∈ {0, 1, 2, 3, 4, 5}}.

Answer:

Given, R = {(x, x + 5): x ∈ {0, 1, 2, 3, 4, 5}}

So, R = {(0, 5), (1, 6), (2, 7), (3, 8), (4, 9), (5, 10)}

Now, Domain of R = {0, 1, 2, 3, 4, 5}

and Range of R = {5, 6, 7, 8, 9, 10}

Question 7:

Write the relation R = {(x, x³): x is a prime number less than 10} in roster form.

Answer:

Given, R = {(x, x³): x is a prime number less than 10}

The prime numbers less than 10 are 2, 3, 5, and 7.

So, R = {(2, 8), (3, 27), (5, 125), (7, 343)}

Question 8:

Let A = {x, y, z} and B = {1, 2}. Find the number of relations from A to B.

Answer:

It is given that A = {x, y, z} and B = {1, 2}.

Now, A * B = {(x, 1), (x, 2), (y, 1), (y, 2), (z, 1), (z, 2)}

Since n(A * B) = 6, the number of subsets of A * B is 2⁶.

Therefore, the number of relations from A to B is 2⁶.

Question 9:

Let R be the relation on Z defined by R = {(a, b): a, b ∈ Z, a – b is an integer}. Find the domain and range of R.

Answer:

R = {(a, b): a, b ∈ Z, a – b is an integer}

It is known that the difference between any two integers is always an integer.

So, Domain of R = Z

and Range of R = Z

 

                                                  Exercise 2.3

Question 1:

1. Which of the following relations are functions? Give reasons. If it is a function, determine its domain and range.

(i) {(2, 1), (5, 1), (8, 1), (11, 1), (14, 1), (17, 1)}

(ii) {(2, 1), (4, 2), (6, 3), (8, 4), (10, 5), (12, 6), (14, 7)}

(iii) {(1, 3), (1, 5), (2, 5)}.

Answer:

(i) {(2, 1), (5, 1), (8, 1), (11, 1), (14, 1), (17, 1)}

Since 2, 5, 8, 11, 14, and 17 are the elements of the domain of the given relation having their

unique images, this relation is a function.

Here, domain = {2, 5, 8, 11, 14, 17} and range = {1}

(ii) {(2, 1), (4, 2), (6, 3), (8, 4), (10, 5), (12, 6), (14, 7)}

Since 2, 4, 6, 8, 10, 12, and 14 are the elements of the domain of the given relation having

their unique images, this relation is a function.

Here, domain = {2, 4, 6, 8, 10, 12, 14} and range = {1, 2, 3, 4, 5, 6, 7}

(iii) {(1, 3), (1, 5), (2, 5)}

Since the same first element i.e., 1 corresponds to two different images i.e., 3 and 5, this

relation is not a function.

Question 2:

Find the domain and range of the following real functions:

(i) f(x) = -|x|                  (ii) f(x) = √(9 – x²)

Answer:

(i) f(x) = –|x|, x ∈ R

We know that |x| =    x, if x ≥ 0

                                      −x, if x < 0

So, f(x) = -|x| =    −x, if x ≥ 0

                                 x, if x < 0

Since f(x) is defined for x ∈ R, the domain of f is R.

It can be observed that the range of f(x) = –|x| is all real numbers except positive real

numbers.

So, the range of f is (−∞, 0].

(ii) f(x) = √(9 − x²)

Since √(9 − x²) is defined for all real numbers that are greater than or equal to –3 and less than

or equal to 3, the domain of f(x) is {x : –3 ≤ x ≤ 3} or [–3, 3].

For any value of x such that –3 ≤ x ≤ 3, the value of f(x) will lie between 0 and 3.

So, the range of f(x) is {x: 0 ≤ x ≤ 3} or [0, 3].

Question 3:

A function f is defined by f(x) = 2x –5. Write down the values of

(i) f(0)                                            (ii) f(7)                                          (iii) f(-3).

Answer:

The given function is f(x) = 2x – 5

Therefore,

(i) f(0) = 2 * 0 – 5 = 0 – 5 = –5

(ii) f(7) = 2 * 7 – 5 = 14 – 5 = 9

(iii) f(-3) = 2 * (-3) – 5 = – 6 – 5 = –11

 

Question 4:

The function ‘t’ which maps temperature in degree Celsius into temperature in degree Fahrenheit is defined by t(C) = 9C/5 + 32. Find

(i) t(0)                   (ii) t(28)                (iii) t(–10)                  (iv) The value of C, when t(C) = 212.

Answer:

The given function is f(C) = 9C/5 + 32

Therefore,

(i) t(0) = (9 * 0)/5 + 32 = 0 + 32 = 32          

(ii) t(28) = (9 * 28)/5 + 32 = 252/5 + 32 = (252 + 32 * 5)/5 = (252 + 160)/5 = 412/5       

(iii) t(–10) = {9 * (-10}/5 + 32 = 9 * (-2) + 32 = -18 + 32 = 14                

(iv) It is given that t(C) = 212

=> 212 = 9C/5 + 32

=> 9C/5 = 212 – 32

=> 9C/5 = 180

=> 9C = 180 * 5

=> 9C = 900

=> C = 900/9

=> C = 100

Hence, the value of t, when t(C) = 212, is 100

Question 5:

Find the range of each of the following functions.

(i) f(x) = 2 – 3x, x ∈ R, x > 0.

(ii) f(x) = x² + 2, x is a real number.

(iii) f (x) = x, x is a real number.

Answer:

Let x > 0

=> 3x > 0

=> -3x < 0

=> 2 –3x < 2

=> f(x) < 2

Hence, Range of f = (-∞, 2)

(ii) Let x be any real number.

Now, x² ≥ 0

=> x² + 2 ≥ 0 + 2

=> x² + 2 ≥ 2

=> f(x) ≥ 2

Hence, Range of f = [2, ∞)

(iii) Given, f(x) = x, x is a real number

It is clear that the range of f is the set of all real numbers.

Hence, Range of f = R

 

                                                       Miscellaneous Exercise on Chapter 2

Question 1:

The relation f is defined by

f(x) =      x², 0 ≤ x ≤ 3

              3x, 3 ≤ x ≤ 10

The relation g is defined by

g(x) =     x², 0 ≤ x ≤ 2

              3x, 2 ≤ x ≤ 10

Show that f is a function and g is not a function.

Answer:

The relation f is defined as

f(x) =      x², 0 ≤ x ≤ 3

              3x, 3 ≤ x ≤ 10

It is observed that for

0 ≤ x < 3, f(x) = x²

3 ≤ x ≤ 10, f(x) = 3x

Also, at x = 3, f(x) = 3² = 9 or f(x) = 3 * 3 = 9

i.e., at x = 3, f(x) = 9

Therefore, for 0 ≤ x ≤ 10, the images of f(x) are unique. Thus, the given relation is a function.

The relation g is defined as

g(x) =     x², 0 ≤ x ≤ 2

              3x, 2 ≤ x ≤ 10

It can be observed that for x = 2, g(x) = 2² = 4 and g(x) = 3 * 2 = 6

Hence, element 2 of the domain of the relation g corresponds to two different images

i.e. 4 and 6.

Hence, this relation is not a function.

Question 2:

If f(x) = x², find. {f(1.1) – f(1)}/(1.1 - 1)

Answer:

Given, f(x) = x²

Now, {f(1.1) – f(1)}/(1.1 - 1) =  {(1.1)² – (1)²}/(1.1 - 1)

                                                   = (1.21 - 1)/0.1

                                                   = 0.21/0.1

                                                   = 2.1   

Question 3:

Find the domain of the function f(x) = (x² + 2x + 1)/(x² – 8x + 12)

Answer:

Given function is

      f(x) = (x² + 2x + 1)/(x² – 8x + 12)

=> f(x) = (x² + 2x + 1)/{(x - 6)(x - 2)}

It can be seen that function f is defined for all real numbers except at x = 6 and x = 2.

Hence, the domain of f is R – {2, 6}.

Question 4:

Find the domain and the range of the real function f defined by

f(x) = √(x − 1)

 

Answer:

The given real function is f(x) = √(x − 1)

It can be seen that √(x − 1) is defined for x ≥ 1

Therefore, the domain of f is the set of all real numbers greater than or equal to 1 i.e., the

domain of f = [1, ∞)

As x ≥ 1

=> (x – 1) ≥ 0

=> √(x − 1) ≥ 0

Therefore, the range of f is the set of all real numbers greater than or equal to 0 i.e.,

 the range of f = [0, ∞)

Question 5:

Find the domain and the range of the real function f defined by

f (x) = |x – 1|

Answer:

The given real function is f (x) = |x – 1|

It is clear that |x – 1| is defined for all real numbers.

So, Domain of f = R

Also, for x ∈ R, |x – 1| assumes all real numbers.

Hence, the range of f is the set of all non-negative real numbers.

Question 6:

Let f = {(x, x²/(1 + x²)): x ∈ R} be a function from R into R. Determine the range of f.

Answer:

Given, f = {(x, x²/(1 + x²)): x ∈ R}

               = {(0, 0), (±0.5, 1/5), (±1, 1/2), (±1.5, 9/13), (±2, 4/5), (3, 9/10), (4, 16/17), ………} 

The range of f is the set of all second elements. It can be observed that all these elements are

greater than or equal to 0 but less than 1. [Denominator is greater numerator]

Thus, range of f = [0, 1)

Question 7:

Let f, g: R → R be defined, respectively by f(x) = x + 1, g(x) = 2x – 3. Find f + g, f – g and f/g.

Answer:

f, g: R → R is defined as f(x) = x + 1, g(x) = 2x – 3

(f + g) (x) = f(x) + g(x)

                 = (x + 1) + (2x – 3)

                 = 3x – 2

So, (f + g) (x) = 3x – 2

(f – g) (x) = f(x) – g(x)

                 = (x + 1) – (2x – 3)

                 = x + 1 – 2x + 3

                 = -x + 4

So, (f – g) (x) = –x + 4

(f/g)(x) = f(x)/g(x), g(x) ≠ 0

(f/g)(x) = (x + 1)/(2x + 3), 2x - 3 ≠ 0 or 2x ≠ 3

(f/g)(x) = (x + 1)/(2x + 3), x ≠ 2/3

Question 8:

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:

f = {(1, 1), (2, 3), (0, –1), (–1, –3)} and f(x) = ax + b

Now, (1, 1) ∈ f

=> f(1) = 1

=> a * 1 + b = 1

=> a + b = 1

Again, (0, –1) ∈ f

=> f(0) = –1

=> a * 0 + b = –1

=> b = –1

On substituting b = –1 in a + b = 1,

We obtain

     a + (–1) = 1

=> a = 1 + 1 = 2

Thus, the respective values of a and b are 2 and –1

Question 9:

Let R be a relation from N to N defined by R = {(a, b): a, b ∈ N and a = b²}. Are the following true?

(i) (a, a) ∈ R, for all a ∈ N

(ii) (a, b) ∈ R, implies (b, a) ∈ R

(iii) (a, b) ∈ R, (b, c) ∈ R implies (a, c) ∈ R.

Justify your answer in each case.

Answer:

R = {(a, b): a, b ∈ N and a = b²}

(i) It can be seen that 2 ∈ N; however, 2 ≠ 2² = 4

Therefore, the statement “(a, a) ∈ R, for all a ∈ N” is not true.

(ii) It can be seen that (9, 3) ∈ N because 9, 3 ∈ N and 9 = 3²

Now, 3 ≠ 9² = 81; therefore, (3, 9) ∉ N

Therefore, the statement “(a, b) ∈ R, implies (b, a) ∈ R” is not true.

(iii) It can be seen that (9, 3) ∈ R, (16, 4) ∈ R because 9, 3, 16, 4 ∈ N and 9 = 3² and 16 = 4²

Now, 9 ≠ 4² = 16; therefore, (9, 4) ∉ N

Therefore, the statement “(a, b) ∈ R, (b, c) ∈ R implies (a, c) ∈ R” is not true.

Question 10:

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.

Justify your answer in each case.

Answer:

A = {1, 2, 3, 4} and B = {1, 5, 9, 11, 15, 16}

Now, A * B = {(1, 1), (1, 5), (1, 9), (1, 11), (1, 15), (1, 16), (2, 1), (2, 5), (2, 9), (2, 11), (2, 15),

                          (2, 16), (3, 1), (3, 5), (3, 9), (3, 11), (3, 15), (3, 16), (4, 1), (4, 5), (4, 9), (4,11),

                         (4, 15), (4, 16)}

It is given that f = {(1, 5), (2, 9), (3, 1), (4, 5), (2, 11)}

(i) A relation from a non-empty set A to a non-empty set B is a subset of the Cartesian product

A * B.

It is observed that f is a subset of A * B.

Thus, f is a relation from A to B.

(ii) Since the same first element i.e., 2 corresponds to two different images i.e., 9 and 11,

relation f is not a function.

Question 11:

Let f be the subset of Z * Z defined by f = {(ab, a + b): a, b ∈ Z}. Is f a function from Z to Z:     justify your answer.

Answer:

The relation f is defined as f = {(ab, a + b): a, b ∈ Z}

We know that a relation f from a set A to a set B is said to be a function if every element of set

A has unique images in set B.

Since 2, 6, –2, –6 ∈ Z, (2 × 6, 2 + 6), (–2 × –6, –2 + (–6)) ∈ f i.e., (12, 8), (12, –8) ∈ f

It can be seen that the same first element i.e., 12 corresponds to two different images

i.e. 8 and -8.

Thus, relation f is not a function.

Question 12:

Let A = {9, 10, 11, 12, 13} and let f: A → N be defined by f(n) = the highest prime factor of n. Find the range of f.

Answer:

A = {9, 10, 11, 12, 13} f: A → N is defined as f(n) = The highest prime factor of n

Prime factor of 9 = 3

Prime factors of 10 = 2, 5

Prime factor of 11 = 11

Prime factors of 12 = 2, 3

Prime factor of 13 = 13

So, f(9) = The highest prime factor of 9 = 3

       f(10) = The highest prime factor of 10 = 5

       f(11) = The highest prime factor of 11 = 11

       f(12) = The highest prime factor of 12 = 3

       f(13) = The highest prime factor of 13 = 13

The range of f is the set of all f(n), where n ∈ A.

Hence, Range of f = {3, 5, 11, 13}