Relation, Function and Graphs Exercise: 2.1 Class 11 Basic Mathematics Solution [NEB UPDATED]

Exercise 2.1

1) Identify which of the following pairs are equal.

Solution:

Here,

a. (1, 3) and (3, 1) are unequal i.e. (1, 3) ≠ (3, 1). 

b. (a, b) and (a, b) are equal i.e. (a, b) = (a, b). 

c. (1, a) and (1, x) are unequal i.e. (1, a) ≠ (1, x). 

d. (x, x) and (y, y) are unequal i.e. (x, x) ≠ (y, y).

 

2) Find x and y, if

Solution:

a) (x + y, 3) = (1, x – y)

(x + y, 3) = (1, x – y)

So, x + y = 1 …(i)

And, x – y = 3 …(ii)

Adding (i) and (ii) we get,

2x = 4

So, x = 2

Putting x= 2 in (i) we get,

2 + y = 1

So, y = –1.

Hence, (x, y) = (2, –1) 

b) (x + 2y, 3) = (–1, 2x – y)

(x + 2y, 3) = (–1, 2x – y)

So, x + 2y = –1 …(i)

And 2x – y = 3

So, 4x – 2y = 6 …(ii)          [multiplication by 2]

Adding (i) and (ii) we have,

5x = 5

So, x = 1

So, Putting x = 1 in (i) we get,

1 + 2y = –1

Or, 2y = –2

So, y = –1

Hence, x = 1, y = –1.

c) (x – 2, y + 1) = (1, 0)

(x – 2, y + 1) = (1, 0)

So, x – 2 = 1

So, x = 3

And y + 1 = 0

So, y = –1

Hence, x = 3 and y = –1. 

d) (2x – 1, –3) = (1, y + 3)

(2x – 1, –3) = (1, y + 3)

So, 2x – 1 = 1

Or, 2x = 2

So, x =1.

And y + 3 = –3

So, y = –6

Hence, 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 = {1, 2, 3} and B = {a, b}

So, 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)}

Or, B x B = {a, b} x {a, b} = {(a, a), (a, b), (b, a), (b, b)}

And B x A = {a, b) x {1, 2, 3}

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

 

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

Solution:

A = {a, b}, B = {b, c}, C = {c, d}

Now,

a) A x  (BUC)

BUC = {b, c} U {c, d} = {b, c, d}

So, A x  (BUC) = {a, b} x  {b, c, d} = {(a, b), (a, c), (a, d), (b, b), (b, c), (b, d)} 

b) A x  (B∩C)

B ∩ C = {b, c) ∩ {c, d} = {c}

So, A x  (B∩C) = {a, b} x {c} = {(a, c), (b, c)} 

c) (A x B) U (A x C)

A x B = {a, b} x  {b, c} = {(a, b), (a, c), (b, b), (b, c)}

A x C = {a, b} x {c, d} = {(a, c), (a, d), (b, c), (b, d)}

So, (A x B) U (A x C) = {(a, b), (a, c), (b, b), (b, c)} U {(a, c), (a, d), (b, c), (b, d)}

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

d) (A x  B) ∩ (A x  C)

(A x  B) ∩ (A x  C) = {(a, b), (a, c), (b, b), (b, c)} ∩ {(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

Solution:

A = {a, b}, B = {c, d}, C = {d, e}

Now, B U C = {c, d} U {d, e} = {c, d, e}

Now, B ∩ C = {c, d} ∩ {d, e} = {d}

A x  B = {a, b} x  {c, d} = {(a, c), (a, d), (b, c), (b, d)}

A x  C = {a, b} x  {d, e} = {(a, d), (a, e), (b, d), (b, e)}

Now,

a) A x (B U C) = (A x  B) U (A x C)

A x (B U C) = {a, b} x  {c, d, e}

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

And (A x  B) U (A x  C) = {(a, c), (a, d), (b, c), (b, d)} U {(a, d), (a, e), (b, d), (b, e)}

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

Hence, A x  (B U C) = (A x  B) U (A x C). Verified. 

b) A x  (B ∩ C) = (A x  B) ∩ (A x C)

A x  (B ∩ C) = {a, b} ∩ {d}

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

And (A x  B) ∩ (A x  C) = {(a, c), (a, d), (b, c), (b, d)} ∩ {(a, d), (a, e), (b, d), (b, e)}

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

Hence, A x  (B ∩ C) = (A x  B) ∩ (A x C). Verified.

 

6) Find the Cartesian product A xB of the following sets:

Solution:

a) A = {x:x = 1, 2, 3} = {1, 2, 3}, B = {y:y = 3 – x} = {0, 1, 2}

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

And B = {y:y = 3 – x} = {0, 1, 2}

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

So, A x B = {(1, 0), (1, 1), (1, 2), (2, 0), (2, 1), (2, 2), (3, 0), (3, 1), (3, 2)}

b) A = {x:x = 0, 1, 2} = {0, 1, 2}, B = {y:y = x²} = {0, 1, 4}

A = {x:x = 0, 1, 2} = {0, 1, 2}

B = {y:y = x²} = {0, 1, 4}

Now, A x B = {0, 1, 2} x {0, 1, 4}

So, A x B = {(0, 0), (0, 1), (0, 4), (1, 0), (1, 1), (1, 4), (2, 0), (2, 1), (2, 4)}

 

7. a) Let A= {1,2,3,4} and B={1,3,5}. Find the relation R from set A to set B determined by the condition.

Solution:

Here, A x B = {1, 2, 3, 4} x {1, 3, 5}

So, A x B = {(1, 1), (1, 3), (1, 5), (2, 1), (2, 3), (2, 5), (3, 1), (3, 3), (3, 5), (4, 1), (4, 3), (4, 5)}

Now,

i) x>y

R = {(x, y): x>y}

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

ii) x < y

 R = {(x, y): x < y}

= {(1, 1), (1, 3), (1, 5), (2, 3), (2, 5), (3, 3), (3, 5), (4, 5)} 

iii) y = x²

R = {(x, y) : y = x²} = {(1, 1)} 

b. Let A = {1, 2, 3, 4}. Find the relation on A satisfying the condition

Here, A = {1, 2, 3, 4}

So, A x A = {1, 2, 3, 4} x {1, 2, 3, 4}

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

Now,

i) y = 2x

R = {(x, y) : y = 2x}

So, R = {(1, 2), (2, 4)} 

ii) x + y = 6

R = {(x, y): x + y = 6}

So, R ={(2, 4), (3, 3), (4, 2)} 

iii) x + y ≤ 4

R = {(x, y): x + y ≤ 4}

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

iv) x – y ≥ 1

R = {(x, y): x – y ≥ 1}

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

 

8. a) R₁={(1,2),(2,3),(3,4),(4,5)}

Solution: here,

So, D(R₁) = {1, 2, 3, 4}

R(R₁) = {2, 3, 4, 5}

And R₁^–1 = {(2, 1), (3, 2), (4, 3), (5, 4)} 

b) R₂={(1,3),(2,5),(3,7),(4,9)}

Solution:

So, D(R₂) = {1, 2, 3, 4}

R(R₂) = {3, 5, 7, 9}

And R₂^–1 = {(3, 1), (5, 2), (7, 3), (9, 4)} 

c) R₃={(a,b),(b,a),(b,c),(c,b),(c,a),(a,c)}

Solution:

So, D(R₃) = {a, b, c}

R(R₃) = {a, b, c}

And R₃^–1 = {(b, a), (a, b), (c, b), (b, c), (a, c), (c, a)} 

d) R₄= {(a,b), (c,b), (d, b), (e, b)}

Solution:

So, D(R₄) = {a, c, d, e}

R(R₄) = {b}

And R₄^–1 = {(b, a), (b, c), (b, d), (b, e)} 

 

9. a) Let A= {2,3,4} and B={2,3,4,6}. Find a relation from set A to set B determined by the condition that x divides y. Also, find the domain and range of the function.

Solution:

Here, A = {2, 3, 4} and B = {2, 3, 4, 6}

So, A x B = {2, 3, 4) x {2, 3, 4, 6}

So, A x B = {(2, 2), (2, 3), (2, 4), (2, 6), (3, 2), (3, 3), (3, 4), (3, 6), (4, 2), (4, 3), (4, 4), (4, 6)}

Let R be the required relation on A x B.

Where, R = {(x.y): x divides y}

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

Domain R = {2, 3, 4}

And range R = {2, 3, 4, 6} 

b) Let A= {3, 4, 5, 6} and the relation is defined as R= {(x,y): x,y ԑ A and x + y = 7}. Express R as a set of ordered pairs. Find the domain, range and R^-1.

Solution:

Here, A = {3, 4, 5, 6}

So, A x A = {3, 4, 5, 6} x {3, 4, 5, 6}

So, A x A = {(3, 3), (3, 4), (3, 5), (3, 6), (4, 3), (4, 4), (4, 5), (4, 6), (5, 3), (5, 4), (5, 5), (5, 6), (6, 3), (6, 4), (6, 5), (6, 6)}

Now, R = {(x, y): x, y ԑ A and x + y = 7}

So, R = {(3, 4), (4, 3)}

Domain R = {3, 4}

Range R = {3, 4}

And R^–1 = {(4, 3), (3, 4)}