Chapter 1: Relations and Functions (உறவுகளும் சார்புகளும்)

பாடம் 1: உறவுகளும் சார்புகளும் — Tamil Nadu Samacheer Kalvi Book Back Answers & Solutions

Complete exercise-wise book-back solutions from Exercise 1.1 to 1.6 covering Cartesian Product, Relations, Functions, Types of Functions, and Function Compositions with step-by-step mathematical working.

கார்டீசியன் பெருக்கல், உறவுகள், சார்புகள், சார்புகளின் வகைகள் மற்றும் சார்புகளின் சேர்ப்பு ஆகிய அனைத்து பயிற்சிகளுக்குமான (பயிற்சி 1.1 முதல் 1.6 வரை) முழுமையான படிபடியான கணிதத் தீர்வுகள்.

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📐 Important Formulas & Laws Quick Revision

  • Cartesian Product Cardinality (கார்டீசியன் பெருக்கலின் உறுப்புகளின் எண்ணிக்கை): n(A × B) = n(A) × n(B) If n(A) = p and n(B) = q, then n(A × B) = pq. Note that A × B ≠ B × A in general, but n(A × B) = n(B × A).
  • Total Number of Relations from A to B (A லிருந்து B க்கான மொத்த உறவுகளின் எண்ணிக்கை): Total Relations = 2^(n(A) × n(B)) = 2^(pq) A relation R from A to B is any subset of A × B. The empty set ∅ and A × B itself are extreme relations.
  • Composition of Functions (சார்புகளின் சேர்ப்பு): (f ∘ g)(x) = f(g(x)) and (g ∘ f)(x) = g(f(x)) In general, f ∘ g ≠ g ∘ f (not commutative). But composition is associative: (f ∘ g) ∘ h = f ∘ (g ∘ h).
  • Linear Function / Constant Function (நேரியல் சார்பு / மாறிலி சார்பு): f(x) = mx + c (Linear), f(x) = c (Constant) The graph of a linear function is always a straight line.

Part I: Choose the Correct Answer (1 Mark)

1. If n(A × B) = 6 and A = {1, 3}, then n(B) is ________.

n(A × B) = 6 மற்றும் A = {1, 3} எனில், n(B) ஆனது ________ ஆகும்.

  • (a) 1
  • (b) 2
  • (c) 3 ✓ (Correct Answer)
  • (d) 6
Explanation: Formula: n(A × B) = n(A) × n(B). Here A = {1, 3} ⟹ n(A) = 2. Therefore, 6 = 2 × n(B) ⟹ n(B) = 6 / 2 = 3.

2. A = {a, b, p}, B = {2, 3}, C = {p, q, r, s} then n[(A ∪ C) × B] is ________.

A = {a, b, p}, B = {2, 3}, C = {p, q, r, s} எனில் n[(A ∪ C) × B] ஆனது ________ ஆகும்.

  • (a) 8
  • (b) 20
  • (c) 12 ✓ (Correct Answer)
  • (d) 16
Explanation: Step 1: A ∪ C = {a, b, p} ∪ {p, q, r, s} = {a, b, p, q, r, s} (p is common, written once). So n(A ∪ C) = 6. Step 2: n(B) = 2. Step 3: n[(A ∪ C) × B] = n(A ∪ C) × n(B) = 6 × 2 = 12.

3. If A = {1, 2}, B = {1, 2, 3, 4}, C = {5, 6} and D = {5, 6, 7, 8}, then state which of the following statement is true.

A = {1, 2}, B = {1, 2, 3, 4}, C = {5, 6} மற்றும் D = {5, 6, 7, 8} எனில் கீழ்க்கண்டவற்றுள் எது சரியான கூற்று?

  • (a) (A × C) ⊂ (B × D) ✓ (Correct Answer)
  • (b) (B × D) ⊂ (A × C)
  • (c) (A × B) ⊂ (A × D)
  • (d) (D × A) ⊂ (B × A)
Explanation: Here A ⊂ B and C ⊂ D. By subset Cartesian product property: If A ⊂ B and C ⊂ D, then (A × C) ⊂ (B × D).

4. If there are 1024 relations from a set A = {1, 2, 3, 4, 5} to a set B, then the number of elements in B is ________.

A = {1, 2, 3, 4, 5} என்ற கணத்திலிருந்து B-க்கு 1024 உறவுகள் உள்ளது எனில் B-ல் உள்ள உறுப்புகளின் எண்ணிக்கை ________.

  • (a) 3
  • (b) 2 ✓ (Correct Answer)
  • (c) 4
  • (d) 8
Explanation: Total relations = 2^(n(A) × n(B)). Here n(A) = 5. Total relations = 1024 = 2^10. Therefore, 5 × n(B) = 10 ⟹ n(B) = 10 / 5 = 2.

5. The range of the relation R = {(x, x²) | x is a prime number less than 13} is ________.

R = {(x, x²) | x என்பது 13-ஐ விடக் குறைந்த பகா எண்கள்} என்ற உறவின் வீச்சகமானது ________.

  • (a) {2, 3, 5, 7}
  • (b) {2, 3, 5, 7, 11}
  • (c) {4, 9, 25, 49, 121} ✓ (Correct Answer)
  • (d) {1, 4, 9, 25, 49, 121}
Explanation: Prime numbers less than 13 are 2, 3, 5, 11. Their squares are: 2² = 4, 3² = 9, 5² = 25, 7² = 49, 11² = 121. Range = {4, 9, 25, 49, 121}.

Part VI: Short Answer Questions — 2 Marks (சுருக்கமான விடையளி)

Q1. Define a Relation from a set A to a set B.

கணம் A-லிருந்து கணம் B-க்கான உறவை வரையறுக்க.

Answer:
  • Let A and B be any two non-empty sets.
  • A Relation R from A to B is defined as a subset of the Cartesian product A × B.
  • Symbolically: R ⊆ (A × B).
  • If an element (a, b) belongs to R, we write a R b, read as 'a is related to b by R'.

Q2. State the Vertical Line Test used to determine whether a curve represents a function.

ஒரு வளைவரை சார்பினைக் குறிக்கிறதா என்பதைத் தீர்மானிக்கும் குத்துக்கோட்டுச் சோதனையைக் கூறுக.

Answer:
  • A curve drawn in a plane represents a function if and only if every vertical line intersects the curve in at most one point.
  • If any vertical line intersects the curve at two or more points, then the curve does NOT represent a function.

Part VII: Long Answer Questions — 5 / 7 Marks (விரிவான விடையளி)

Q1. Let A = {x ∈ W | x < 2}, B = {x ∈ N | 1 < x ≤ 4} and C = {3, 5}. Verify that A × (B ∪ C) = (A × B) ∪ (A × C).

A = {x ∈ W | x < 2}, B = {x ∈ N | 1 < x ≤ 4} மற்றும் C = {3, 5} எனில், A × (B ∪ C) = (A × B) ∪ (A × C) என்பதைச் சரிபார்க்கவும்.

Step 1: Express Sets in Roster Form (1 Mark)
  • Given W = {0, 1, 2, ...} is the set of Whole numbers. Since x < 2: A = {0, 1}.
  • Given N = {1, 2, 3, ...} is the set of Natural numbers. Since 1 < x ≤ 4: B = {2, 3, 4}.
  • Given set C = {3, 5}.
Step 2: Evaluate Left Hand Side (LHS) (2 Marks)
  • First find (B ∪ C): B ∪ C = {2, 3, 4} ∪ {3, 5} = {2, 3, 4, 5}.
  • Now compute A × (B ∪ C):
  • 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)} ---- (Equation 1)
Step 3: Evaluate Right Hand Side (RHS) (1.5 Marks)
  • First compute (A × B): A × B = {(0, 2), (0, 3), (0, 4), (1, 2), (1, 3), (1, 4)}
  • Next compute (A × C): A × C = {(0, 3), (0, 5), (1, 3), (1, 5)}
  • Now take union (A × B) ∪ (A × C):
  • = {(0, 2), (0, 3), (0, 4), (0, 5), (1, 2), (1, 3), (1, 4), (1, 5)} ---- (Equation 2)
Step 4: Conclusion (0.5 Mark)
  • Comparing (Equation 1) and (Equation 2), every ordered pair on LHS is identically present on RHS.
  • Therefore, LHS = RHS.
  • Hence, A × (B ∪ C) = (A × B) ∪ (A × C) is verified.

Part VIII: Solved Problems Corner / Numerical Calculations

Problem 1: Find A × B, A × A and B × A for: (i) A = {2, -2, 3} and B = {1, -4} (ii) A = B = {p, q} (iii) A = {m, n}, B = ∅

பின்வருவனவற்றிற்கு A × B, A × A மற்றும் B × A ஆகியவற்றைக் காண்க: (i) A = {2, -2, 3} மற்றும் B = {1, -4} (ii) A = B = {p, q} (iii) A = {m, n}, B = ∅

Step-by-Step Working:
Subdivision (i):
A × B = {2, -2, 3} × {1, -4} = {(2, 1), (2, -4), (-2, 1), (-2, -4), (3, 1), (3, -4)}
A × A = {2, -2, 3} × {2, -2, 3} = {(2, 2), (2, -2), (2, 3), (-2, 2), (-2, -2), (-2, 3), (3, 2), (3, -2), (3, 3)}
B × A = {1, -4} × {2, -2, 3} = {(1, 2), (1, -2), (1, 3), (-4, 2), (-4, -2), (-4, 3)}
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Subdivision (ii): Since A = B = {p, q}:
A × B = {p, q} × {p, q} = {(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)}
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Subdivision (iii): Since B is an empty set (∅):
A × B = {m, n} × ∅ = ∅
A × A = {m, n} × {m, n} = {(m, m), (m, n), (n, m), (n, n)}
B × A = ∅ × {m, n} = ∅
Boxed Final Answer: (i) A × B has 6 elements, A × A has 9 elements, B × A has 6 elements. (ii) A × B = A × A = B × A = {(p, p), (p, q), (q, p), (q, q)}. (iii) A × B = ∅, A × A = {(m, m), (m, n), (n, m), (n, n)}, B × A = ∅.

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

A = {1, 2, 3} மற்றும் B = {x | x என்பது 10-ஐ விடக் குறைந்த பகா எண்} எனில், A × B மற்றும் B × A ஆகியவற்றைக் காண்க.

Step-by-Step Working:
Step 1: Write set B in roster form:
Prime numbers less than 10 are 2, 3, 5, 7. (Note: 1 is neither prime nor composite!).
So, B = {2, 3, 5, 7}.
Here n(A) = 3 and n(B) = 4, so n(A × B) must be 3 × 4 = 12 elements.
Step 2: Compute A × B:
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)}
Step 3: Compute B × A:
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)}
Boxed Final Answer: 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)} B × A = {(2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (3, 3), (5, 1), (5, 2), (5, 3), (7, 1), (7, 2), (7, 3)}

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

B × A = {(-2, 3), (-2, 4), (0, 3), (0, 4), (3, 3), (3, 4)} எனில், A மற்றும் B ஆகியவற்றைக் காண்க.

Step-by-Step Working:
Step 1: Understand the definition of Cartesian product B × A:
In B × A, the FIRST coordinate of every ordered pair belongs to B.
The SECOND coordinate of every ordered pair belongs to A.
Step 2: Collect all first coordinates for set B:
First coordinates are: -2, -2, 0, 0, 3, 3.
Removing duplicate elements (since sets contain unique elements):
B = {-2, 0, 3}
Step 3: Collect all second coordinates for set A:
Second coordinates are: 3, 4, 3, 4, 3, 4.
Removing duplicate elements:
A = {3, 4}
Boxed Final Answer: A = {3, 4} B = {-2, 0, 3}

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

A = {5, 6}, B = {4, 5, 6}, C = {5, 6, 7} எனில், A × A = (B × B) ∩ (C × C) எனக் காட்டுக.

Step-by-Step Working:
Step 1: Compute Left Hand Side (LHS) = A × A:
A × A = {5, 6} × {5, 6}
A × A = {(5, 5), (5, 6), (6, 5), (6, 6)} ---- (Equation 1)
Step 2: Compute B × B:
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)}
Step 3: Compute C × C:
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)}
Step 4: Compute Right Hand Side (RHS) = (B × B) ∩ (C × C):
Find common ordered pairs between B × B and C × C:
(B × B) ∩ (C × C) = {(5, 5), (5, 6), (6, 5), (6, 6)} ---- (Equation 2)
Step 5: Compare Equations (1) and (2):
LHS = RHS. Hence, A × A = (B × B) ∩ (C × C) is proved.
Boxed Final Answer: From (1) and (2), LHS = RHS. Hence proved.

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

A = {1, 2, 3}, B = {2, 3, 5}, C = {3, 4} மற்றும் D = {1, 3, 5} எனில், (A ∩ C) × (B ∩ D) = (A × B) ∩ (C × D) என்பது உண்மையா என சோதிக்கவும்.

Step-by-Step Working:
Step 1: Compute LHS = (A ∩ C) × (B ∩ D):
A ∩ C = {1, 2, 3} ∩ {3, 4} = {3}
B ∩ D = {2, 3, 5} ∩ {1, 3, 5} = {3, 5}
(A ∩ C) × (B ∩ D) = {3} × {3, 5} = {(3, 3), (3, 5)} ---- (Equation 1)
Step 2: Compute RHS = (A × B) ∩ (C × D):
A × B = {1, 2, 3} × {2, 3, 5} = {(1, 2), (1, 3), (1, 5), (2, 2), (2, 3), (2, 5), (3, 2), (3, 3), (3, 5)}
C × D = {3, 4} × {1, 3, 5} = {(3, 1), (3, 3), (3, 5), (4, 1), (4, 3), (4, 5)}
(A × B) ∩ (C × D) = {(3, 3), (3, 5)} ---- (Equation 2)
Step 3: Comparing (1) and (2):
LHS = RHS = {(3, 3), (3, 5)}.
Therefore, (A ∩ C) × (B ∩ D) = (A × B) ∩ (C × D) is TRUE.
Boxed Final Answer: Yes, it is TRUE.

Problem 6: Let A = {x ∈ W | x < 2}, B = {x ∈ N | 1 < x ≤ 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)

A = {x ∈ W | x < 2}, B = {x ∈ N | 1 < x ≤ 4} மற்றும் C = {3, 5} எனில் கீழ்க்கண்டவற்றைச் சரிபார்க்க: (i) A × (B ∪ C) = (A × B) ∪ (A × C) (ii) A × (B ∩ C) = (A × B) ∩ (A × C) (iii) (A ∪ B) × C = (A × C) ∪ (B × C)

Step-by-Step Working:
Step 1: Write sets A, B, C clearly in roster form:
A = {0, 1}, B = {2, 3, 4}, C = {3, 5}
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Subdivision (i): Verify A × (B ∪ C) = (A × B) ∪ (A × C):
LHS: 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)
RHS: A × B = {(0, 2), (0, 3), (0, 4), (1, 2), (1, 3), (1, 4)}
A × C = {(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), LHS = RHS. Hence (i) is verified.
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Subdivision (ii): Verify A × (B ∩ C) = (A × B) ∩ (A × C):
LHS: B ∩ C = {2, 3, 4} ∩ {3, 5} = {3}
A × (B ∩ C) = {0, 1} × {3} = {(0, 3), (1, 3)} ---- (3)
RHS: (A × B) ∩ (A × C) from above:
Common pairs in A × B and A × C are {(0, 3), (1, 3)} ---- (4)
From (3) and (4), LHS = RHS. Hence (ii) is verified.
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Subdivision (iii): Verify (A ∪ B) × C = (A × C) ∪ (B × C):
LHS: 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)} ---- (5)
RHS: A × C = {(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)} ---- (6)
From (5) and (6), LHS = RHS. Hence (iii) is verified.
Boxed Final Answer: All three distributive identities verified successfully. LHS = RHS.

Problem 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 என்பது 8-ஐ விடக் குறைவான இயல் எண்களின் கணம், B என்பது 8-ஐ விடக் குறைவான பகா எண்களின் கணம், C என்பது இரட்டைப்படை பகா எண்களின் கணம் எனில் கீழ்க்கண்டவற்றைச் சரிபார்க்க: (i) (A ∩ B) × C = (A × C) ∩ (B × C) (ii) A × (B - C) = (A × B) - (A × C)

Step-by-Step Working:
Step 1: Write sets A, B, C:
A = {1, 2, 3, 4, 5, 6, 7}
B = {2, 3, 5, 7}
C = {2}
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Subdivision (i): Verify (A ∩ B) × C = (A × C) ∩ (B × C):
LHS: A ∩ B = {2, 3, 5, 7}
(A ∩ B) × C = {2, 3, 5, 7} × {2} = {(2, 2), (3, 2), (5, 2), (7, 2)} ---- (1)
RHS: 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)
From (1) and (2), LHS = RHS. Hence (i) is verified.
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Subdivision (ii): Verify A × (B - C) = (A × B) - (A × C):
LHS: B - C = {2, 3, 5, 7} - {2} = {3, 5, 7}
A × (B - C) = {1, 2, 3, 4, 5, 6, 7} × {3, 5, 7}
= {(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)} ---- (3)
RHS: Compute (A × B) - (A × C):
A × B contains all pairs with second coordinate in {2, 3, 5, 7}.
A × C contains all pairs with second coordinate 2.
Removing (A × C) from (A × B) eliminates pairs ending in 2, leaving exactly the 21 pairs ending in 3, 5, 7 ---- (4)
From (3) and (4), LHS = RHS. Hence (ii) is verified.
Boxed Final Answer: Both subdivisions verified. LHS = RHS.

Problem 1: Let A = {1, 2, 3, 7} and B = {3, 0, -1, 7}, which of the following are relation from A to B? (i) R₁ = {(2, 1), (7, 1)} (ii) R₂ = {(-1, 1)} (iii) R₃ = {(2, -1), (7, 7), (1, 3)} (iv) R₄ = {(7, -1), (0, 3), (3, 3), (0, 7)}

A = {1, 2, 3, 7} மற்றும் B = {3, 0, -1, 7} எனில், பின்வருவனவற்றில் எவை A-லிருந்து B-க்கான உறவுகள் ஆகும்? (i) R₁ = {(2, 1), (7, 1)} (ii) R₂ = {(-1, 1)} (iii) R₃ = {(2, -1), (7, 7), (1, 3)} (iv) R₄ = {(7, -1), (0, 3), (3, 3), (0, 7)}

Step-by-Step Working:
Test condition: Every pair (a, b) in R must have a ∈ A and b ∈ B.
--------------------------------------------------
(i) In R₁ = {(2, 1), (7, 1)}:
Here 1 ∉ B. So (2, 1) ∉ A × B and (7, 1) ∉ A × B.
Hence, R₁ is NOT a relation from A to B.
--------------------------------------------------
(ii) In R₂ = {(-1, 1)}:
Here -1 ∉ A and 1 ∉ B.
Hence, R₂ is NOT a relation from A to B.
--------------------------------------------------
(iii) In R₃ = {(2, -1), (7, 7), (1, 3)}:
2 ∈ A and -1 ∈ B ⟹ (2, -1) ∈ A × B
7 ∈ A and 7 ∈ B ⟹ (7, 7) ∈ A × B
1 ∈ A and 3 ∈ B ⟹ (1, 3) ∈ A × B
Since R₃ ⊆ A × B, R₃ IS a relation from A to B.
--------------------------------------------------
(iv) In R₄ = {(7, -1), (0, 3), (3, 3), (0, 7)}:
Here 0 ∉ A. So (0, 3) ∉ A × B and (0, 7) ∉ A × B.
Hence, R₄ is NOT a relation from A to B.
Boxed Final Answer: Only R₃ is a relation from A to B. R₁, R₂ and R₄ are NOT relations.

Problem 2: Let A = {1, 2, 3, 4, ..., 45} and R be the relation defined as 'is square of a number' on A. Write R as a subset of A × A. Also, find the domain and range of R.

A = {1, 2, 3, 4, ..., 45} மற்றும் R என்ற உறவு 'A-ன் மீது ஓர் எண்ணின் வர்க்கம்' என வரையறுக்கப்பட்டால், R-ஐ A × A-ன் உட்கணமாக எழுதுக. மேலும் R-க்கான மதிப்பகம் மற்றும் வீச்சகத்தைக் காண்க.

Step-by-Step Working:
Step 1: Check squares of numbers in A that do not exceed 45:
1² = 1 ∈ A
2² = 4 ∈ A
3² = 9 ∈ A
4² = 16 ∈ A
5² = 25 ∈ A
6² = 36 ∈ A
7² = 49 ∉ A (exceeds 45, so 7 cannot be in the domain)
Step 2: Write relation R in roster form:
R = {(1, 1), (2, 4), (3, 9), (4, 16), (5, 25), (6, 36)}
Step 3: Domain of R (set of all first coordinates):
Domain = {1, 2, 3, 4, 5, 6}
Step 4: Range of R (set of all second coordinates):
Range = {1, 4, 9, 16, 25, 36}
Boxed Final Answer: R = {(1, 1), (2, 4), (3, 9), (4, 16), (5, 25), (6, 36)} Domain = {1, 2, 3, 4, 5, 6} Range = {1, 4, 9, 16, 25, 36}

Problem 3: A Relation R is given by the set {(x, y) / y = x + 3, x ∈ {0, 1, 2, 3, 4, 5}}. Determine its domain and range.

R என்ற ஓர் உறவு {(x, y) / y = x + 3, x ∈ {0, 1, 2, 3, 4, 5}} எனக் கொடுக்கப்பட்டுள்ளது. இதன் மதிப்பகம் மற்றும் வீச்சகத்தைக் காண்க.

Step-by-Step Working:
Substitute each value of x to find y:
When x = 0 ⟹ y = 0 + 3 = 3
When x = 1 ⟹ y = 1 + 3 = 4
When x = 2 ⟹ y = 2 + 3 = 5
When x = 3 ⟹ y = 3 + 3 = 6
When x = 4 ⟹ y = 4 + 3 = 7
When x = 5 ⟹ y = 5 + 3 = 8
Therefore, relation in roster form:
R = {(0, 3), (1, 4), (2, 5), (3, 6), (4, 7), (5, 8)}
Domain of R = {0, 1, 2, 3, 4, 5}
Range of R = {3, 4, 5, 6, 7, 8}
Boxed Final Answer: Domain = {0, 1, 2, 3, 4, 5} Range = {3, 4, 5, 6, 7, 8}

Problem 4: Represent each of the given relations by (a) an arrow diagram, (b) a graph, and (c) a set in roster form, wherever possible: (i) {(x, y) | x = 2y, x ∈ {2, 3, 4, 5}, y ∈ {1, 2, 3, 4}} (ii) {(x, y) | y = x + 3, x, y are natural numbers < 10}

கொடுக்கப்பட்ட உறவுகள் ஒவ்வொன்றையும் (a) அம்புக்குறி படம், (b) வரைபடம், (c) பட்டியல் முறையில் குறிக்க: (i) {(x, y) | x = 2y, x ∈ {2, 3, 4, 5}, y ∈ {1, 2, 3, 4}} (ii) {(x, y) | y = x + 3, x, y ஆகியவை < 10 என்ற இயல் எண்கள்}

Step-by-Step Working:
Subdivision (i):
When x = 2 ⟹ y = 2/2 = 1 ∈ y
When x = 3 ⟹ y = 3/2 = 1.5 ∉ y
When x = 4 ⟹ y = 4/2 = 2 ∈ y
When x = 5 ⟹ y = 5/2 = 2.5 ∉ y
(c) Roster form: R = {(2, 1), (4, 2)}
(a) Arrow Diagram: Set X = {2, 3, 4, 5}, Set Y = {1, 2, 3, 4}. Draw arrows from 2 ⟶ 1 and 4 ⟶ 2.
(b) Graph: On X-Y plane, plot points (2, 1) and (4, 2).
--------------------------------------------------
Subdivision (ii): Natural numbers < 10 are {1, 2, 3, 4, 5, 6, 7, 8, 9}:
x = 1 ⟹ y = 4
x = 2 ⟹ y = 5
x = 3 ⟹ y = 6
x = 4 ⟹ y = 7
x = 5 ⟹ y = 8
x = 6 ⟹ y = 9
(Note: for x = 7, y = 10 ∉ natural numbers < 10)
(c) Roster form: R = {(1, 4), (2, 5), (3, 6), (4, 7), (5, 8), (6, 9)}
(a) Arrow Diagram: Draw two ovals with numbers 1 to 9, connect 1 ⟶ 4, 2 ⟶ 5, 3 ⟶ 6, 4 ⟶ 7, 5 ⟶ 8, 6 ⟶ 9.
(b) Graph: Plot points (1, 4), (2, 5), (3, 6), (4, 7), (5, 8), (6, 9) on coordinate axes.
Boxed Final Answer: (i) R = {(2, 1), (4, 2)} (ii) R = {(1, 4), (2, 5), (3, 6), (4, 7), (5, 8), (6, 9)}

Problem 5: A company has four categories of employees: Assistants (A), Clerks (C), Managers (M) and Executive Officer (E). The company provides ₹10,000, ₹25,000, ₹50,000 and ₹1,00,000 as salaries to categories A, C, M and E respectively. If A₁, A₂, A₃, A₄, A₅ were Assistants; C₁, C₂, C₃, C₄ were Clerks; M₁, M₂, M₃ were Managers and E₁, E₂ were Executive Officers and if relation R is defined by x R y, where x is the salary given to person y, express the relation R through an ordered pair and an arrow diagram.

ஒரு நிறுவனத்தில் உதவியாளர்கள் (A), எழுத்தர்கள் (C), மேலாளர்கள் (M) மற்றும் நிர்வாக அலுவலர்கள் (E) ஆகிய நான்கு பிரிவுகளில் பணியாளர்கள் உள்ளனர். அவர்களுக்கு முறையே ₹10,000, ₹25,000, ₹50,000 மற்றும் ₹1,00,000 ஊதியம் வழங்கப்படுகிறது. A₁, A₂, A₃, A₄, A₅ உதவியாளர்கள்; C₁, C₂, C₃, C₄ எழுத்தர்கள்; M₁, M₂, M₃ மேலாளர்கள் மற்றும் E₁, E₂ நிர்வாக அலுவலர்கள். x R y என்ற உறவில் x என்பது y என்பவருக்கு வழங்கப்படும் ஊதியம் எனில், R-ஐ வரிசைச் சோடிகள் மற்றும் அம்புக்குறி படம் மூலம் குறிக்க.

Step-by-Step Working:
Note: The relation is defined as x R y, meaning FIRST element is SALARY, SECOND element is PERSON.
1. Set of Ordered Pairs:
R = {
(10000, A₁), (10000, A₂), (10000, A₃), (10000, A₄), (10000, A₅),
(25000, C₁), (25000, C₂), (25000, C₃), (25000, C₄),
(50000, M₁), (50000, M₂), (50000, M₃),
(100000, E₁), (100000, E₂)
}
2. Arrow Diagram:
Draw Oval X containing {10000, 25000, 50000, 100000}.
Draw Oval Y containing {A₁, A₂, A₃, A₄, A₅, C₁, C₂, C₃, C₄, M₁, M₂, M₃, E₁, E₂}.
Draw arrows from 10000 to all A₁...A₅, 25000 to C₁...C₄, 50000 to M₁...M₃, and 100000 to E₁, E₂.
Boxed Final Answer: Relation R written with 14 ordered pairs and represented via arrow diagram.

Problem 1: Let f = {(x, y) | x, y ∈ N and y = 2x} be a relation on N. Find the domain, co-domain and range. Is this relation a function?

f = {(x, y) | x, y ∈ N மற்றும் y = 2x} ஆனது N-ன் மீதான ஓர் உறவு என்க. மதிப்பகம், துணை மதிப்பகம் மற்றும் வீச்சகத்தைக் காண்க. இந்த உறவு ஒரு சார்பாகுமா?

Step-by-Step Working:
Step 1: Determine relation pairs for x ∈ N:
x = 1 ⟹ y = 2(1) = 2
x = 2 ⟹ y = 2(2) = 4
x = 3 ⟹ y = 2(3) = 6
x = 4 ⟹ y = 2(4) = 8
So f = {(1, 2), (2, 4), (3, 6), (4, 8), ...}
Step 2: Domain = All elements x in N = {1, 2, 3, 4, ...} = N
Step 3: Co-domain = Second set N = {1, 2, 3, 4, ...} = N
Step 4: Range = Set of all even natural numbers = {2, 4, 6, 8, ...}
Step 5: Function check: Every element in N has a unique and distinct image in N. Therefore, YES, this relation is a FUNCTION.
Boxed Final Answer: Domain = N Co-domain = N Range = {2, 4, 6, 8, ...} Yes, it is a function.

Problem 2: Let X = {3, 4, 6, 8}. Determine whether the relation R = {(x, f(x)) | x ∈ X, f(x) = x² + 1} is a function from X to N?

X = {3, 4, 6, 8} என்க. R = {(x, f(x)) | x ∈ X, f(x) = x² + 1} என்ற உறவானது X-லிருந்து N-க்கான சார்பாகுமா?

Step-by-Step Working:
Compute f(x) for each x ∈ X:
When x = 3 ⟹ f(3) = 3² + 1 = 9 + 1 = 10 ∈ N
When x = 4 ⟹ f(4) = 4² + 1 = 16 + 1 = 17 ∈ N
When x = 6 ⟹ f(6) = 6² + 1 = 36 + 1 = 37 ∈ N
When x = 8 ⟹ f(8) = 8² + 1 = 64 + 1 = 65 ∈ N
Relation in roster form:
R = {(3, 10), (4, 17), (6, 37), (8, 65)}
Observation: Each element in X has a unique image in N.
Therefore, R is a function from X to N.
Boxed Final Answer: R = {(3, 10), (4, 17), (6, 37), (8, 65)}. Yes, R is a function from X to N.

Problem 3: Given the function f : x ↦ x² - 5x + 6, evaluate: (i) f(-1) (ii) f(2a) (iii) f(2) (iv) f(x - 1)

f : x ↦ x² - 5x + 6 எனக் கொடுக்கப்பட்டால், பின்வருவனவற்றைக் காண்க: (i) f(-1) (ii) f(2a) (iii) f(2) (iv) f(x - 1)

Step-by-Step Working:
(i) Substitute x = -1:
f(-1) = (-1)² - 5(-1) + 6 = 1 + 5 + 6 = 12
--------------------------------------------------
(ii) Substitute x = 2a:
f(2a) = (2a)² - 5(2a) + 6 = 4a² - 10a + 6
--------------------------------------------------
(iii) Substitute x = 2:
f(2) = (2)² - 5(2) + 6 = 4 - 10 + 6 = 10 - 10 = 0
--------------------------------------------------
(iv) Substitute x = x - 1:
f(x - 1) = (x - 1)² - 5(x - 1) + 6
= (x² - 2x + 1) - 5x + 5 + 6
= x² - 7x + 12
Boxed Final Answer: (i) f(-1) = 12 (ii) f(2a) = 4a² - 10a + 6 (iii) f(2) = 0 (iv) f(x - 1) = x² - 7x + 12

Problem 5: Let f(x) = 2x + 5. If x ≠ 0, then find [f(x + 2) - f(2)] / x.

f(x) = 2x + 5 என்க. x ≠ 0 எனில், [f(x + 2) - f(2)] / x -ஐக் காண்க.

Step-by-Step Working:
Step 1: Compute f(x + 2):
f(x + 2) = 2(x + 2) + 5 = 2x + 4 + 5 = 2x + 9
Step 2: Compute f(2):
f(2) = 2(2) + 5 = 4 + 5 = 9
Step 3: Substitute into the given expression:
[f(x + 2) - f(2)] / x = [(2x + 9) - 9] / x
= 2x / x
Since x ≠ 0, we can cancel x from numerator and denominator:
= 2
Boxed Final Answer: [f(x + 2) - f(2)] / x = 2

Problem 2: Let f : A ⟶ B be a function defined by f(x) = x/2 - 1, where A = {2, 4, 6, 10, 12}, B = {0, 1, 2, 4, 5, 9}. Represent f by: (i) set of ordered pairs (ii) a table (iii) an arrow diagram (iv) a graph

f : A ⟶ B என்ற சார்பானது f(x) = x/2 - 1 என வரையறுக்கப்படுகிறது. இங்கு A = {2, 4, 6, 10, 12}, B = {0, 1, 2, 4, 5, 9} எனில், f-ஐ: (i) வரிசைச் சோடிகளின் கணம் (ii) அட்டவணை (iii) அம்புக்குறி படம் (iv) வரைபடம் ஆகியவற்றின் மூலம் குறிக்க.

Step-by-Step Working:
Step 1: Compute f(x) for each element in A:
f(2) = 2/2 - 1 = 1 - 1 = 0 ∈ B
f(4) = 4/2 - 1 = 2 - 1 = 1 ∈ B
f(6) = 6/2 - 1 = 3 - 1 = 2 ∈ B
f(10) = 10/2 - 1 = 5 - 1 = 4 ∈ B
f(12) = 12/2 - 1 = 6 - 1 = 5 ∈ B
--------------------------------------------------
(i) Set of Ordered Pairs:
f = {(2, 0), (4, 1), (6, 2), (10, 4), (12, 5)}
--------------------------------------------------
(ii) Table Form:
| x | 2 | 4 | 6 | 10 | 12 |
| f(x) | 0 | 1 | 2 | 4 | 5 |
--------------------------------------------------
(iii) Arrow Diagram:
Draw Oval A with elements {2, 4, 6, 10, 12}.
Draw Oval B with elements {0, 1, 2, 4, 5, 9}.
Connect: 2 ⟶ 0, 4 ⟶ 1, 6 ⟶ 2, 10 ⟶ 4, 12 ⟶ 5.
(Note: 9 in B has no pre-image).
--------------------------------------------------
(iv) Graph:
On coordinate Cartesian plane, mark points: (2, 0), (4, 1), (6, 2), (10, 4), (12, 5) and label them.
Boxed Final Answer: Function represented in all 4 required forms.

Problem 4: Show that the function f : N ⟶ N defined by f(x) = 2x - 1 is one-one but not onto.

f : N ⟶ N என்ற சார்பு f(x) = 2x - 1 என வரையறுக்கப்பட்டால், அது ஒன்றுக்கொன்றான சார்பு ஆனால் மேல்சார்பு இல்லை எனக் காட்டுக.

Step-by-Step Working:
Part A: Prove One-One (Injective):
Method: Let f(x₁) = f(x₂).
2x₁ - 1 = 2x₂ - 1
Adding 1 to both sides: 2x₁ = 2x₂
Dividing by 2: x₁ = x₂.
Since f(x₁) = f(x₂) implies x₁ = x₂, f is a ONE-ONE FUNCTION.
--------------------------------------------------
Part B: Prove Not Onto (Surjective):
Let us check the range of f for natural numbers N:
f(1) = 2(1) - 1 = 1
f(2) = 2(2) - 1 = 3
f(3) = 2(3) - 1 = 5
Range of f = {1, 3, 5, 7, ...} = Set of odd natural numbers.
Co-domain of f = N = {1, 2, 3, 4, 5, ...}.
Here Range ≠ Co-domain (all even numbers 2, 4, 6, 8... have no pre-images in domain N).
Therefore, f is NOT an onto function.
Hence, f is one-one but not onto.
Boxed Final Answer: Hence proved: f is one-one but not onto.

Problem 1: Using the functions f and g given below, find f ∘ g and g ∘ f. Check whether f ∘ g = g ∘ f: (i) f(x) = x - 6, g(x) = x² (ii) f(x) = 2/x, g(x) = 2x² - 1 (iii) f(x) = (x + 6)/3, g(x) = 3 - x

f மற்றும் g ஆகிய சார்புகளைப் பயன்படுத்தி f ∘ g மற்றும் g ∘ f ஆகியவற்றைக் காண்க. மேலும் f ∘ g = g ∘ f என்பது உண்மையா என சோதிக்க: (i) f(x) = x - 6, g(x) = x² (ii) f(x) = 2/x, g(x) = 2x² - 1 (iii) f(x) = (x + 6)/3, g(x) = 3 - x

Step-by-Step Working:
Subdivision (i): f(x) = x - 6, g(x) = x²:
(f ∘ g)(x) = f(g(x)) = f(x²) = x² - 6 ---- (1)
(g ∘ f)(x) = g(f(x)) = g(x - 6) = (x - 6)² = x² - 12x + 36 ---- (2)
Comparing (1) and (2): x² - 6 ≠ x² - 12x + 36.
Therefore, f ∘ g ≠ g ∘ f.
--------------------------------------------------
Subdivision (ii): f(x) = 2/x, g(x) = 2x² - 1:
(f ∘ g)(x) = f(g(x)) = f(2x² - 1) = 2 / (2x² - 1) ---- (3)
(g ∘ f)(x) = g(f(x)) = g(2/x) = 2(2/x)² - 1 = 2(4/x²) - 1 = 8/x² - 1 ---- (4)
Comparing (3) and (4): 2 / (2x² - 1) ≠ 8/x² - 1.
Therefore, f ∘ g ≠ g ∘ f.
--------------------------------------------------
Subdivision (iii): f(x) = (x + 6)/3, g(x) = 3 - x:
(f ∘ g)(x) = f(3 - x) = [(3 - x) + 6] / 3 = (9 - x) / 3 ---- (5)
(g ∘ f)(x) = g((x + 6)/3) = 3 - [(x + 6) / 3] = (9 - (x + 6)) / 3 = (3 - x) / 3 ---- (6)
Comparing (5) and (6): (9 - x)/3 ≠ (3 - x)/3.
Therefore, f ∘ g ≠ g ∘ f.
Boxed Final Answer: In all three cases, f ∘ g ≠ g ∘ f.

Problem 2: Find the value of k, such that f ∘ g = g ∘ f: (i) f(x) = 3x + 2, g(x) = 6x - k (ii) f(x) = 2x - k, g(x) = 4x + 5

f ∘ g = g ∘ f எனில் k-ன் மதிப்பைக் காண்க: (i) f(x) = 3x + 2, g(x) = 6x - k (ii) f(x) = 2x - k, g(x) = 4x + 5

Step-by-Step Working:
Subdivision (i): f(x) = 3x + 2, g(x) = 6x - k:
Find (f ∘ g)(x) = f(6x - k) = 3(6x - k) + 2 = 18x - 3k + 2
Find (g ∘ f)(x) = g(3x + 2) = 6(3x + 2) - k = 18x + 12 - k
Given that f ∘ g = g ∘ f:
18x - 3k + 2 = 18x + 12 - k
Cancel 18x from both sides:
-3k + 2 = 12 - k
Rearrange k terms:
-3k + k = 12 - 2
-2k = 10
k = 10 / -2 = -5
--------------------------------------------------
Subdivision (ii): f(x) = 2x - k, g(x) = 4x + 5:
Find (f ∘ g)(x) = f(4x + 5) = 2(4x + 5) - k = 8x + 10 - k
Find (g ∘ f)(x) = g(2x - k) = 4(2x - k) + 5 = 8x - 4k + 5
Equating both since f ∘ g = g ∘ f:
8x + 10 - k = 8x - 4k + 5
10 - k = -4k + 5
-k + 4k = 5 - 10
3k = -5
k = -5/3
Boxed Final Answer: (i) k = -5 (ii) k = -5/3

Problem 8: Consider the functions f(x), g(x), h(x) as given below. Show that (f ∘ g) ∘ h = f ∘ (g ∘ h) in each case: (i) f(x) = x - 1, g(x) = 3x + 1 and h(x) = x²

கீழ்க்காணும் சார்புகளில் (f ∘ g) ∘ h = f ∘ (g ∘ h) எனக்காட்டுக: (i) f(x) = x - 1, g(x) = 3x + 1 மற்றும் h(x) = x²

Step-by-Step Working:
Step 1: Compute LHS = (f ∘ g) ∘ h:
First compute (f ∘ g)(x):
(f ∘ g)(x) = f(g(x)) = f(3x + 1) = (3x + 1) - 1 = 3x
Now compose with h(x):
((f ∘ g) ∘ h)(x) = (f ∘ g)(h(x)) = (f ∘ g)(x²) = 3(x²) = 3x² ---- (Equation 1)
Step 2: Compute RHS = f ∘ (g ∘ h):
First compute (g ∘ h)(x):
(g ∘ h)(x) = g(h(x)) = g(x²) = 3(x²) + 1 = 3x² + 1
Now compose with f(x):
(f ∘ (g ∘ h))(x) = f(3x² + 1) = (3x² + 1) - 1 = 3x² ---- (Equation 2)
Step 3: Conclusion:
From (1) and (2), LHS = RHS = 3x².
Hence, (f ∘ g) ∘ h = f ∘ (g ∘ h) is proved.
Boxed Final Answer: From (1) and (2), LHS = RHS = 3x². Hence verified.