PRACTICE WORKSHEET RELATIONS AND FUNCTIONS WITH ANSWERS

Question 1 :

If the ordered pairs (x− 3x, y2 + 4y) and (-2,5) are equal, then find x and y.

Question 2 :

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

Question 3 :

Given that f(x)  =  

Question 4 :

Let A= {9, 10, 11, 12, 13, 14, 15, 16, 17} and let f : A-> N be defined by f (n) = the highest prime factor of n  A. Write f as a set of ordered pairs and find the range of f.

Detailed Answer Key

Question 1 :

If the ordered pairs (x− 3x, y2 + 4y) and (-2,5) are equal, then find x and y.

Answer :

x2 - 3x  =  -2

x2 - 3x + 2  =  0

(x - 1)(x - 2)  =  0

x = 1 and x = 2

y2 + 4y  =  5

y2 + 4y - 5  =  0

(y + 5)(y - 1)  =  0

y = -5 and y = 1

Question 2 :

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

Answer :

So, in this question since A X A has 9 elements, so the number of elements in the set  A must obviously be 3.

Since the given elements of the cross product have -1, 0 and 1 as part of the entries, clearly these only must be the elements of the set A.

So A = {-1, 0, 1}

A X A

  =  {(-1, -1),(-1, 0),(-1, 1),(0, -1),(0, 0),(0, 1),(1, -1),(1, 0),(1, 1)}

Question 3 :

Given that f(x)  =  

(i) f (0) (ii) f (3) (iii) f (a+1) in terms of a.(Given that  0)

Answer :

(i)  f(0)

Instead of x, we have 0, it is less than 1

f(0)  =  4

(ii)  f(3)

Instead of x, we have 3, it is greater than 1

f(3)  =  √(3 - 1)

f(3)  =  √2

(iii) f (a+1) 

Instead of x, we have a+1, it is greater than 1

f(a + 1)  =  √(a + 1 - 1)

f(a + 1)  =  √a

Question 4 :

Let A= {9, 10, 11, 12, 13, 14, 15, 16, 17} and let f : A-> N be defined by f (n) = the highest prime factor of n  A. Write f as a set of ordered pairs and find the range of f.

Answer :

Elements of A

9

10

11

12

13

14

15

16

17

Highest prime factor

3

5

11

3

13

7

5

2

17

Set ordered pairs :

=  {(9,3) (10, 5)(11, 11)(12, 3)(13, 13) (14, 7) (15, 5) (16,2) (17, 17)}

Range  =  {2, 3, 5, 7,  11, 13, 17}

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