State |
Tamil Nadu |

Board |
State Board [Samacheer Kalvi] |

Class |
10th/SSLC |

Subject |
Maths [English Medium] |

Download |
Guide/ Solution/ Book Answers |

Chapter |
Chapter 1 Relations and Functions |

Exercise |
Exercise 1.1 |

## Samacheer Kalvi Class 10th Maths Chapter 1 Guide

Download Tamil Nadu Samacheer Kalvi Class 10th English Medium Maths Chapter 1 Relations and Functions Exercise 1.1 Guide/ Solution/ Book Answers.

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## Samacheer Kalvi Class 10th Maths Chapter 1 Ex 1.1 Book Answers

Samacheer Kalvi Class 10th Maths Chapter 1 Relations and Functions Exercise 1.1 Answers

**Question 1 :**

Find A × B, A × A and B × A

(i) A = {2,-2,3} and B = {1,-4}

(ii) A = B = {p,q]

(iii) A= {m,n} ; B = (Φ)

**Solution**:

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

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

A × A = {(2, 2), (2,-2), (2, 3), (-2, 2), (-2, -2), (-2, 3), (3, 2), (3, -2), (3,3) }

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

(ii) A = B = {(p,q)]

A × B = {(p, p), {p, q), (q, p), (q, q)}

A × A = {(p, p), (p, q), (q, p), (q, q)}

B × A = {(p,p), {p, q), (q, p), (q, q)}

(iii) A = {m,n} × Φ

A × B = { }

A × A = {(m, m), (m, n), (n, m), (n, n)}

B × A = { }

**———————————————————————————————————————————————————————————————————————————————————————————————————————————————-**

**Question 2:
**Let A= {1,2,3} and B = {× | x is a prime number less than 10}. Find A × B and B × A.

**Answer**:

A = {1,2,3}, B = {2, 3, 5, 7}

A × B = {1,2,3} × {2, 3, 5, 7}

= {(1, 2) (1, 3) (1, 5) (1, 7) (2, 2)

(2, 3) (2, 5) (2, 7)(3, 2) (3, 3) (3, 5) (3, 7)}

B × A = {2, 3, 5, 7} × {1,2,3}

= {(2, 1)(2, 2)(2, 3)(3, 1)(3, 2)(3, 3) (5, 1)(5, 2)(5, 3) (7, 1) (7,2)(7, 3)}

**———————————————————————————————————————————————————————————————————————————————————————————————————————————————-**

**Question 3:**

If B × A = {(-2, 3),(-2, 4),(0, 3),(0, 4),(3, 3), (3, 4)} find A and B.

**Solution**:

B × A ={(-2, 3), (-2, 4), (0, 3), (0, 4), (3, 3), (3, 4)}

A = {3, 4), B = { -2, 0, 3}

**———————————————————————————————————————————————————————————————————————————————————————————————————————————————-**

**Question 4:**

If A= {5, 6}, B = {4, 5 ,6}, C = {5, 6, 7}, Show that A × A = (B × B) ∩ (C × C)

**Answer**:

A ={5,6}, B = {4,5,6}, C = {5, 6,7}

A × A = {5, 6} × {5,6}

= {(5, 5) (5, 6) (6, 5) (6, 6)} ….(1)

B × B = {4, 5, 6} × {4, 5, 6}

= {(4, 4)(4, 5)(4, 6)(5, 4)(5, 5) (5, 6) (6, 4)(6, 5) (6, 6)}

C × C = {5,6,7} × {5,6,7}

= {(5, 5)(5, 6)(5, 7)(6, 5)(6, 6) (6, 7)(7, 5)(7, 6) (7, 7)}

(B × B) ∩ (C × C) = {(5, 5)(5, 6)(6, 5)(6, 6)} ….(2)

From (1) and (2) we get

A × A = (B × B) ∩ (C × C)

**Question 5:**

Given A ={1, 2, 3}, B = {2, 3, 5}, C = {3, 4} and D = {1, 3, 5}, check if (A ∩ C) x (B ∩ D) = (A × B) ∩ (C × D) is true?

**Solution**:

LHS = {(A∩C) × (B∩D)

A ∩C = {3}

B ∩D = {3, 5}

(A ∩ C) × (B ∩ D) = {(3, 3) (3, 5)} ………….. (1)

RHS = (A × B) ∩ (C × D)

A × B = {(1, 2), (1, 3), (1, 5), (2, 2), (2, 3), (2, 5), (3, 2), (3, 3), (3, 5)}

C × D = {(3, 1), (3, 3), (3, 5), (4, 1), (4, 3), (4, 5)}

(A × B) ∩ (C × D) = {(3, 3), (3, 5)} …(2)

∴ (1) = (2) ∴ It is true.

**Question 6:**

Let A = {x ∈ W | x < 2},

B = {x ∈ N | 1 < 1 < × < 4} and

C = {3,5}. Verify that

(i) A × (B ∪ C) = (A × B) ∪ (A × C)

(ii) A × (B ∩ C) = (A × B) ∩ (A × C)

(iii) (A ∪ B) × C = (A × C) ∪ (B × C)

**Answer**:

(i) A = {0, 1}

B = {2,3,4}

C = {3,5}

(i) A × (B ∪ C) = (A × B) ∪ (A × c)

B ∪ C = {2, 3,4} ∪ {3,5}

= {2, 3, 4, 5}

A × (B ∪ C) = {0, 1} × {2, 3, 4, 5}

= {(0, 2) (0, 3) (0, 4) (0, 5) (1, 2) (1, 3)(1, 4)(1, 5)} ….(1)

A × B = {0, 1} × {2,3,4}

= {(0,2) (0,3) (0,4) (1,2) (1,3) (1,4) }

A × C = {0, 1} × {3, 5}

{(0, 3) (0, 5) (1,3) (1,5)}

(A × B) ∪ (A × C) = {(0, 2) (0, 3) (0, 4) (0, 5) (1, 2)(1, 3)(1, 4)(1, 5)} ….(2)

From (1) and (2) we get

A × (B ∪ C) = (A × B) ∪ (A × C)

(ii) A × (B n C) = (A × B) n (A × C)

B ∩ C = {2,3,4} ∩ {3,5}

= {3}

A × (B ∩ C) = {0, 1} × {3}

= {(0,3) (1,3)} ….(1)

A × B = {0,1} × {2,3,4}

= {(0, 2) (0, 3) (0, 4) (1,2) (1,3) (1,4)}

A × C = {0,1} × {3,5}

{(0, 3) (0, 5) (1,3) (1,5)}

(A × B) n (A × C) = {(0, 3) (1, 3)} ….(2)

From (1) and (2) we get

A × ( B n C) = (A × B) n (A × C)

(iii) (A ∪ B) × C = (A × C) ∪ (B × C)

A ∪ B = {0, 1} ∪ {2,3,4}

= {0,1, 2, 3, 4}

(A ∪ B) × C = {0, 1,2, 3,4} × {3,5}

= {(0, 3) (0, 5) (1, 3) (1, 5)(2, 3) (2, 5) (3, 3)(3, 5) (4, 3)(4, 5)} ….(1)

A × C = {0, 1} × {3,5}

= {(0,3) (0,5) (1,3) (1,5)}

B × C = {2,3,4} × {3,5}

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

(A × C) ∪ (B × C) = {(0, 3) (0, 5) (1, 3) (1, 5) (2, 3)(2, 5) (3, 3) (3, 5) (4, 3) (4, 5)} ….(2)

From (1) and (2) we get

(A ∪ B) × C = (A × C) ∪ (B × C)

** Question 7:**

Let A = The set of all natural numbers less than 8, B = The set of all prime numbers less than 8, C = The set of even prime number. Verify that

(i) (A ∩ B) × c = (A × C) ∩ (B × C)

(ii) A × (B – C ) = (A × B) – (A × C)

A = {1, 2, 3, 4, 5, 6, 7}

B = {2, 3, 5, 7}

C = {2}

**Solution**:

(i)(A ∩ B) × C = (A × c) ∩ (B × C)

LHS = (A ∩ B) × C

A ∩ B = {2, 3, 5, 7}

(A ∩ B) × C = {(2, 2), (3, 2), (5, 2), (7, 2)} ………… (1)

RHS = (A × C) ∩ (B × C)

(A × C) = {(1, 2), (2, 2), (3, 2), (4, 2), (5, 2), (6, 2), (7, 2)}

(B × C) = {2, 2), (3, 2), (5, 2), (7, 2)}

(A × C) ∩ (B × C) = {(2, 2), (3, 2), (5, 2), (7, 2)} ……….. (2)

(1) = (2)

∴ LHS = RHS. Hence it is verified.

(ii) A × (B – C) = (A × B) – (A × C)

LHS = A × (B – C)

(B – C) = {3,5,7}

A × (B – C) = {(1, 3), (1, 5), (1, 7), (2, 3), (2, 5), (2, 7) , (3, 3), (3, 5), (3, 7), (4, 3), (4, 5), (4, 7), (5, 3), (5, 5), (5, 7), (6, 3) , (6, 5), (6, 7), (7, 3), (7, 5), (7, 7)} …………. (1)

RHS = (A × B) – (A × C)

(A × B) = {(1,2), (1,3), (1,5), (1,7),

(2, 2), (2, 3), (2, 5), (2, 7),

(3, 2), (3, 3), (3, 5), (3, 7),

(4, 2), (4, 3), (4, 5), (4, 7),

(5, 2), (5, 3), (5, 5), (5, 7),

(6, 2), (6, 3), (6, 5), (6, 7),

(7, 2), (7, 3), (7, 5), (7,7)}

(A × C) = {(1, 2), (2, 2),(3, 2),(4, 2), (5, 2), (6, 2), (7, 2)}

(A × B) – (A × C) = {(1, 3), (1, 5), (1, 7), (2, 3), (2, 5), (2, 7), (3, 3), (3, 5), (3, 7), (4, 3), (4, 5), (4, 7), (5, 3), (5, 5), (5, 7), (6, 3), (6, 5), (6, 7), (7, 3), (7, 5), (7,7) } ………….. (2)

(1) = (2) ⇒ LHS = RHS.

Hence it is verified.

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