In this page, you can find the complete solutions of the first exercise of Relations, Functions, and Graphs chapter from Basic Mathematics Grade XI book published and distributed by Sukunda Pustak Bhawan.

In the above-mentioned book, the Relations, Functions, and Graphs is the 2nd chapter and has five exercises only. Out of which, this is the solution of the first exercise.

Check: Basic Mathematics Grade 11 (Sukunda Publication) Guide:
Grade 11 Basic Mathematics by Sukunda Pustak Vawan Notes and Solutions | Nepal

 

Disclaimer:

Answers mentioned here are not solved by teachers. These are the solutions written by a student of Grade 11. Answers are all correct. However, the language or process of solving the questions might be informal and in examinations, you might have to add little more language and show more calculations than what has been done here. So, we highly encourage you to view these solutions as guide rather than just copying everything mentioned here. Few questions have been typed while most of them have been updated as pictures.
1. Identify which of the following pairs are equal:

A set of ordered pairs are said to be equal if their corresponding components are equal. For example: (a,b) and (c,d) are equal if a = c and b = d.

a) (1,3) and (3,1): Not equal
b) (a,b) and (a,b): Equal
c) (1,a) and (1,x): Not equal
d) (x,x) and (y,y): Not equal

2. Find x and y, if:

A set of ordered pairs are said to be equal if their corresponding components are equal. For example: (a,b) and (c,d) are equal if a = c and b = d.

a) (x+y,3) = (1,x-y)

Solution: From equality of ordered pairs,

$x + y = 1$ - (i)
$3 = x - y \implies x - y = 3$ - (ii)

On solving equations (i) and (ii), we get, x = 2 and y = -1

b) (x + 2y, 3) = (-1, 2x - y)

Solution: From equality of ordered pairs,

$x + 2y = -1$ - (i)
$3 = 2x - y \implies 2x - y = 3$ - (ii)

On solving equations (i) and (ii), we get, x = 1 and y = -1

c) (x -2, y + 1) = (1,0)

Solution: From equality of ordered pairs,

$x - 2 = 1$ - (i)
$y + 1 = 0$ - (ii)
On solving equations (i) and (ii), we get, x = 3 and y = -1

d) (2x - 1, -3) = (1, y + 3)

Solution: From equality of ordered pairs,

$2x -1 = 1$ - (i)
$-3 =  y + 3$ - (ii)

On solving equations (i) and (ii), we get, x = 1 and y = -6


3. If A = {1,2,3} and B = {a,b}, find A x A, B x B and B x A.

Solution:

A x B is read as A cross B. It is the set all possible ordered pairs from set A to set B.

Here,

A = {1,2,3}

B = {a,b}

Now,

A x A = {1,2,3} x {1,2,3}

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

B x B = {a,b} x {a,b}

= { (a,a), (a,b), (b,a), (b,b)}

A x B = {1,2,3} x {a,b}

= { (1,a), (2,a), (3,a), (1,b), (2,b), (3,b)}


4. Let A = {a,b}, B = {b,c} and C = {c,d}. Find:

Solution:

Here,

A = {a,b}

B = {b,c}

a) A x (B $\cup$ C)

$\implies$ A x [ {b,c} $\cup$ {c,d} ]

= A x {b,c,d}

= {a,b} x {b,c,d}

= {(a,b), (a,c), (a,d), (b,a), (b,c), (b,d)}

b) A x (B $\cap$ C)

$\implies$ A x [ {b,c} $\cap$ {c,d}]

= A x {c}

= {a,b} x {c}

= {(a,c), (b,c)}

c) (A x B) $\cup$ (A x C)

$\implies [ {a,b} x {b,c} ] $\cup$ [ {a,b} x {c,d}]$

= {(a,b), (a,c), (b,b), (b,c)} $\cup$ {(a,c), (a,d), (b,c), (b,d)}

= {(a,b), (a,c), (a,d), (b,b), (b,c), (b,d)}

d) (A x B) $\cap$ (A xC)

$\implies [ {a,b} x {b,c} ] $\cap$ [ {a,b} x {c,d}]$

= {(a,b), (a,c), (b,b), (b,c)} $\cap$ {(a,c), (a,d), (b,c), (b,d)}

= {(a,c) (b,c)}

5. Let A = {a,b}, B = {c,d} and C = {d,e}. Verify that:

a) A x (B $\cup$ C) = (A x B) $\cup$ (A x C)

Solution:

LHS

= A x (B $\cup$ C)

= A x [ {c,d} $\cup$ {d,e} ]

= {a,b} x {c,d,e}

= { (a,c), (a,d), (a,e), (b,c), (b,d), (b,e) }

RHS

= (A x B) $\cup$ (A x C)

= [ {a,b} x {c,d} ] $\cup$ [ {a,b} x {d,e} ]

= { (a,c), (a,d), (b,c), (b,d) } $\cup$ { (a,d), (a,e), (b,d), (b,e) }

= { (a,c), (a,d), (a,e), (b,c), (b,d), (b,e) }

Hence, it is verified.


b) A x (B $\cap$ C) = (A x B) $\cap$ (A x C)

Solution:

LHS

= A x (B $\cap$ C)

= A x [ {c,d} $\cap$ {d,e} ]

= {a,b} x { d}

= { (a,d), (b,d) }

RHS

= (A x B) $\cap$ (A x C)

= [ {a,b} x {c,d} ] $\cap$ [ {a,b} x {d,e} ]

= { (a,c), (a,d), (b,c), (b,d) } $\cap$ { (a,d), (a,e), (b,d), (b,e) }

= { (a,d), (b,d) }

Hence, it is verfied.


6. Find the carerstian product A x B of the following sets:

a) A = {x:x = 1,2,3}, B = {y:y = 3-x}

Solution:

Here,

A = {1,2,3}

For x = 1, y = 3 - 1 = 2
For x = 2, y = 3 - 2 = 1
For x = 3, y = 3 - 3 = 0

B = {2,1,0}

Now,

A x B = {1,2,3} x {2,1,0}

= { (1,2), (1,1), (1,0), (2,2), (2,1), (2,0), (3,2), (3,1), (3,0) }


b) A = {x: x = 0,1,2} and B = {y: y = x$^2$}

Solution:

Here,

A = {0,1,2}

For x = 0, y = 0^2 = 0
For x = 1, y = 1^2 = 1
For x = 2, y = 2^2 = 4

B = {0,1,4}

Now,

AxB = {0,1,2} x {0,1,4}

= { (0,0), (0,1), (0,4), (1,0), (1,1), (1,4), (2,0), (2,1), (2,4) }


About the Textbook:

Name: Basic Mathematics Grade XI
Author(s): D.R. Bajracharya | R.M. Shresththa | M.B. Singh | Y.R. Sthapit | B.C. Bajracharya
Publisher: Sukunda Pustak Bhawan (Bhotahity, Kathmandu)
Telephone: 5320379, 5353537
Price: 695 /- (2078 BS)