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Study Material Chapter 1 – Relations and Functions

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Chapter 1 - Relations and Functions

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Definition: A set is a collection of well defined objects. If 𝐴 and 𝐵 are two non-empty sets, then the set of all ordered pairs (𝑎, 𝑏) such that 𝑎 ∈ 𝐴, 𝑏 ∈ 𝐵 is

called the Cartesian Product of 𝑨 and 𝑩, and is denoted by 𝐴 × 𝐵. Thus 𝐴 × 𝐵 = {(𝑎, 𝑏)|𝑎 ∈ 𝐴, 𝑏 ∈ 𝐵} Note: 𝐴 × 𝐵 is the set of all possible ordered pairs between the elements of 𝐴 and 𝑩 such that the first

coordinate is an element of 𝐴 and the second coordinate is an element of 𝐵. 𝐵 × 𝐴 is the set of all possible ordered pairs between the element of 𝐴 and 𝐵 such that the first

coordinate is an element of 𝐵 and the second coordinate is an element of 𝐴 If 𝑎 = 𝑏, then (𝑎, 𝑏) = (𝑏, 𝑎). The “Cartesian product” is also referred as “cross product” In general 𝐴 × 𝐵 ≠ 𝐵 × 𝐴, but 𝑛(𝐴 × 𝐵) = 𝑛(𝐵 × 𝐴) 𝐴 × 𝐵 = ∅ if and only if 𝐴 = ∅ or 𝐵 = ∅ If 𝑛(𝐴) = 𝑝 and 𝑛(𝐵) = 𝑞 then 𝑛(𝐴 × 𝐵) = 𝑝𝑞 The set of all points in the Cartesian plane can be viewed as the set of all ordered pairs (𝑥, 𝑦)

where 𝑥, 𝑦 are real numbers. In fact ℝ × ℝ is the set of all points which we call as the Cartesian plane.

Distributive property of Cartesian product: (i) 𝐴 × (𝐵 ∪ 𝐶) = (𝐴 × 𝐵) ∪ (𝐴 × 𝐶) (ii) 𝐴 × (𝐵 ∩ 𝐶) = (𝐴 × 𝐵) ∩ (𝐴 × 𝐶)

𝐴 × 𝐵 represent a shape in two dimensions and 𝐴 × 𝐵 × 𝐶 represent an object in three dimensions.

Text Book Page Number : 3

1. For any two non-empty sets 𝐴 and 𝐵, 𝐴 × 𝐵 is called as cartesian product. 2. If 𝑛 (𝐴 × 𝐵) = 20 and 𝑛(𝐴) = 5 then 𝑛(𝐵) is 𝟒. 3. If 𝐴 = {−1, 1} and 𝐵 = {−1, 1} then geometrically describe the set points of 𝐴 × 𝐵. 𝐴 = {−1, 1} , 𝐵 = {−1, 1} 𝐴 × 𝐵 = {−1,1} × {−1,1} = {(−1, −1)(−1,1)(1, −1)(1,1)} 4. If 𝐴, 𝐵 are the line segments given by the intervals (−4, 3) and (−2, 3) Respectively, represent the cartesian product of 𝐴 and 𝐵. 𝐴 × 𝐵 = {−4,3} × {−2,3} = {(−4, −2), (−4,3), (3, −2), (3,3)}

1. Relations and Functions

Concept corner

Introduction for Exercise 1.1


Way to Success - 10th Maths 6


When will 𝐴 × 𝐵 be equal to 𝐵 × 𝐴 ? 𝐴 × 𝐵 = 𝐵 × 𝐴 Only when 𝐴 and 𝐵 are equal sets.

1. Find 𝑨 × 𝑩, 𝑨 × 𝑨 and 𝑩 × 𝑨

(i) 𝑨 = {𝟐, −𝟐, 𝟑} and 𝑩 = {𝟏, −𝟒}

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)} (ii) 𝑨 = 𝑩 = {𝒑, 𝒒}

𝐴 × 𝐵 = {𝑝, 𝑞} × {𝑝, 𝑞}

= {(𝒑, 𝒑), (𝒑, 𝒒), (𝒒, 𝒑), (𝒒, 𝒒)}

𝐴 × 𝐴 = {𝑝, 𝑞} × {𝑝, 𝑞}

= {(𝒑, 𝒑), (𝒑, 𝒒), (𝒒, 𝒑), (𝒒, 𝒒)}

𝐵 × 𝐴 = {𝑝, 𝑞} × {𝑝, 𝑞}

= {(𝒑, 𝒑), (𝒑, 𝒒), (𝒒, 𝒑), (𝒒, 𝒒)}

(iii) 𝑨 = {𝒎, 𝒏}; 𝑩 = ∅

𝐴 × 𝐵 = { } 𝐴 × 𝐴 = {𝑚, 𝑛} × {𝑚, 𝑛}

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

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} 𝐵 = {𝑥|𝑥 is a prime number less than 10} = {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)}

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

𝐵 = set of all first co-ordinates of elements of 𝐵 × 𝐴

𝑩 = {−𝟐, 𝟎, 𝟑}

𝐴 = set of all second co-ordinates of elements of 𝐵 × 𝐴

𝑨 = {𝟑, 𝟒}

PTA-1

Exercise 1.1

Try Your Self… 1. Find 𝐴 × 𝐵, 𝐵 × 𝐴 and 𝐵 × 𝐵. i) 𝐴 = {𝑎, 𝑏, 𝑐} and 𝐵 = {𝑎, 𝑒}

ii) 𝐴 = 𝐵 = {1, 3} iii) 𝐴 = { } and 𝐵 = {2, 5}

Ans: (i) 𝐴 × 𝐵 = {(𝑎, 𝑎) (𝑎, 𝑒) (𝑏, 𝑎) (𝑏, 𝑒) (𝑐, 𝑎) (𝑐, 𝑒)}

𝐵 × 𝐴 = (𝑎, 𝑎), (𝑎, 𝑏), (𝑎, 𝑐), (𝑒, 𝑎), (𝑒, 𝑏), (𝑒, 𝑐)

𝐵 × 𝐵 = {(𝑎, 𝑎), (𝑎, 𝑒), (𝑒, 𝑎)(𝑒, 𝑒)}

(ii) 𝐴 × 𝐵 = {(1,1), (1,3), (3,1), (3,3)}

𝐵 × 𝐴 = {(1,1), (1,3), (3,1), (3,3)}

𝐵 × 𝐵 = {(1,1), (1,3), (3,1), (3,3)}

(iii) we can’t find Cartesian product of A and B.

Because A is an empty set.

Try Your Self… 2. Let 𝐴 = {1,2,5} and 𝐵 =

{𝑥/𝑥 is a even prime

number less than 10} find

𝐴 × 𝐵 and 𝐵 × 𝐴.

Ans: 𝐴 × 𝐵 = {(1,2), (2,2), (5,2)},

𝐵 × 𝐴 = {(2,1), (2,2), (2,5)}

Try Your Self…

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

find 𝐴 and 𝐵 Ans: 𝐴 = {2, 3}, 𝐵 = {1, 3, 5}

𝑨 = {𝟐, −𝟐, 𝟑} and 𝑩 = {𝟏, −𝟒}

𝑨 × 𝑩 =?

𝐴 × 𝐵 𝐵

1 −4 𝐴 2 (2,1) (2, −4)

−2 (−2,1) (−2, −4) 3 (3,1) (3, −4)

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

Shortcut corner



7

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

𝐴 × 𝐴 = (𝐵 × 𝐵) ∩ (𝐶 × 𝐶)

LHS: 𝐴 × 𝐴 = {5,6} × {5,6}

= {(𝟓, 𝟓), (𝟓, 𝟔), (𝟔, 𝟓), (𝟔, 𝟔)} ……….(1)

RHS:

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), 𝑨 × 𝑨 = (𝑩 × 𝑩) ∩ (𝑪 × 𝑪)

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 ∩ 𝐶 = {1,2, 𝟑} ∩ {2, 𝟑, 5} = {3}

𝐵 ∩ 𝐷 = {2, 𝟑, 𝟓} ∩ {1, 𝟑, 𝟓} = {3,5}

LHS:

(𝐴 ∩ 𝐶) × (𝐵 ∩ 𝐷) = {3} × {3,5}

= {(3,3), (3,5)} ………….(1)

RHS: 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)} ……….(2)

From (1) and (2), (𝑨 ∩ 𝑪) × (𝑩 ∩ 𝑫) = (𝑨 × 𝑩) ∩ (𝑪 × 𝑫) is true.

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)

𝐴 = {𝑥 ∈ 𝕎|𝑥 < 2} = {0,1}, 𝐵 = {𝑥 ∈ ℕ|1 < 𝑥 ≤ 4} = {2,3,4}, 𝐶 = {3,5}

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 = {0,1} × {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), (1,2), (1,3), (1,4)} ∪ {(0,3), (0,5), (1,3), (1,5)}

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

From (1) and (2), 𝑨 × (𝑩 ∪ 𝑪) = (𝑨 × 𝑩) ∪ (𝑨 × 𝑪)

PTA-2

Try Your Self…

4. If 𝐴 = {1, 2} 𝐵 = {1, 2, 3} 𝐶 = {1, 2, 4} show that

𝐴 × 𝐴 = (𝐵 × 𝐵) ∩ (𝐶 × 𝐶).

Ans: 𝐴 × 𝐴 = {(1, 1), (1, 2), (2, 1), (2,2)}

𝐵 × 𝐵 = {(1, 1), (1, 2), (1, 3), (2, 1),

(2, 2) , (2, 3), (3, 1) , (3, 2) , (3, 3)}

𝐶 × 𝐶 = {(1, 1), (1, 2), (1, 4), (2, 1),

(2, 2), (2, 4), (4, 1), (4, 2), (4, 4)}

(𝐵 × 𝐵) ∩ (𝐶 × 𝐶) = {(1,1), (1,2), (2,1), (2,2)}

Try Your Self…

5. Given 𝐴 = {𝑎, 𝑏, 𝑐}, 𝐵 = {𝑏, 𝑐, 𝑑}, 𝐶 = {𝑐, 𝑑} and

𝐷 = {𝑎, 𝑐, 𝑑} check if

(𝐴 ∩ 𝐶) × (𝐵 ∩ 𝐷) = (𝐴 × 𝐵) ∩ (𝐶 × 𝐷) is true?

Ans: 𝐴 ∩ 𝐶 = {𝑐}, 𝐵 ∩ 𝐷 = {𝑐, 𝑑}

(𝐴 ∩ 𝐶) × (𝐵 ∩ 𝐷) = {(𝑐, 𝑐), (𝑐, 𝑑)}

𝐴 × 𝐵 = {(𝑎, 𝑏), (𝑎, 𝑐), (𝑎, 𝑑), (𝑏, 𝑏), (𝑏, 𝑐)

(𝑏, 𝑑), (𝑐, 𝑏), (𝑐, 𝑐), (𝑐, 𝑑)}

𝐶 × 𝐷 = {(𝑐, 𝑎), (𝑐, 𝑐), (𝑐, 𝑑), (𝑑, 𝑎), (𝑑, 𝑐), (𝑑, 𝑑)}

(𝐴 × 𝐵) ∩ (𝐶 × 𝐷) = {(𝑐, 𝑐) (𝑐, 𝑑)}



(ii) 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)} …………..(1)

RHS: 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), (1,2), (𝟏, 𝟑), (1,4)} ∩ {(𝟎, 𝟑), (0,5), (𝟏, 𝟑), (1,5)}

= {(0,3), (1,3)} ……………(2)

From (1) and (2),

𝑨 × (𝑩 ∩ 𝑪) = (𝑨 × 𝑩) ∩ (𝑨 × 𝑪)

(iii) (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)} ………..(1)

RHS:

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)}

= {(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), (𝑨 ∪ 𝑩) × 𝑪 = (𝑨 × 𝑪) ∪ (𝑩 × 𝑪)

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

𝐴 = The set of all natural numbers less than 8 = {1,2,3,4,56,7}

𝐵 = The set of all prime numbers less than 8 = {2,3,5,7} 𝐶 = The set of even prime number = {2}

(i) (𝑨 ∩ 𝑩) × 𝑪 = (𝑨 × 𝑪) ∩ (𝑩 × 𝑪) LHS: 𝐴 ∩ 𝐵 = {1,2,3,4,5, 6,7} ∩ {2,3,5,7}

= {2,3,5,7}

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

RHS:

𝐴 × 𝐶 = {1,2,3,4,56,7} × {2}

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

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

PTA-5



9

(ii) 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)}…………(1)

RHS: 𝐴 × 𝐵 = {1,2,3,4,5,6,7} × {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), (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 = {1,2,3,4,5,6,7} × {2} = {(1,2), (2,2), (3,2), (4,2), (5,2), (6,2), (7,2)}

(𝐴 × 𝐵) − (𝐴 × 𝐶)

= {(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)}

− {(1,2), (2,2), (3,2), (4,2), (5,2), (6,2), (7,2)}

= {(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)

From (1) and (2)

𝑨 × (𝑩 − 𝑪) = (𝑨 × 𝑩) − (𝑨 × 𝑪)

1. Let 𝑨 = {𝒙 ∈ 𝑾/𝟎 < 𝒙 < 𝟓}, 𝑩 = {𝒙 ∈ 𝑾/𝟎 ≤ 𝒙 ≤ 𝟐}, 𝑪 = {𝒙 ∈ 𝑾/𝒙 < 𝟑} then verify that

𝑨 × (𝑩 ∩ 𝑪) = (𝑨 × 𝑩) ∩ (𝑨 × 𝑪)

𝐴 = {𝑥 ∈ 𝑊|0 < 𝑥 < 5} = {1, 2, 3, 4}, 𝐵 = {𝑥 ∈ 𝑊|0 ≤ 𝑥 ≤ 2} = {0, 1, 2},

𝐶 = {𝑥 ∈ 𝑊|𝑥 < 3} = {0, 1, 2}

𝐵 ∩ 𝐶 = {0, 1, 2} ∩ {0, 1, 2} = {0, 1, 2}

𝐴 × (𝐵 ∩ 𝐶) = {1, 2, 3, 4} × {0, 1, 2}

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

𝐴 × 𝐵 = {1, 2, 3, 4} × {0, 1, 2}

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

𝐴 × 𝐶 = {1, 2, 3, 4} × {0, 1, 2}

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

(𝐴 × 𝐵) ∩ (𝐴 × 𝐶)

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

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

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

From (1) and (2), 𝐴 × (𝐵 ∩ 𝐶) = (𝐴 × 𝐵) ∩ (𝐴 × 𝐶) is verified.

2. If 𝑨 × 𝑩 = {(𝟏, 𝟑) , (𝟏, 𝟒), (𝟐, 𝟑), (𝟐, 𝟒)} then find A and B.

𝐴 × 𝐵 = {(1,3) (1,4) (2,3) (2,4)}

𝐴 = {set of all first co-ordinates of elements of 𝐴 × 𝐵} ⇒ 𝐴 = {1, 2}

𝐵 = {set of all second co-ordinates of elements of 𝐴 × 𝐵} ⇒ 𝐵 = {3,4}

Thus 𝐴 = {1, 2} and 𝐵 = {3, 4}

Creative Questions

PTA-3



3. Let 𝑨 = {𝒙 ∈ 𝑵/𝒙 < 𝟓} B= The set of all prime numbers less than 6 𝑪 = {𝟐, 𝟑}. verify that i) (𝑨 ∩ 𝑩) × 𝑪 = (𝑨 × 𝑪) ∩ (𝑩 × 𝑪) ii) 𝑨 × (𝑩 − 𝑪) = (𝑨 × 𝑩) − (𝑨 × 𝑪) i) (𝐴 ∩ 𝐵) × 𝐶 = (𝐴 × 𝐶) ∩ (𝐵 × 𝐶)

𝐴 ∩ 𝐵 = {1,2,3,4} ∩ {2,3,5} = {2,3}

(𝐴 ∩ 𝐵) × 𝐶 = {2,3} × {2,3}

= {(2,2), (2,3), (3,2), (3,3)} …… (1)

𝐴 × 𝐶 = {1,2,3,4} × {2,3}

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

(3,2), (3,3), (4,2), (4,3)}

𝐵 × 𝐶 = {2,3,5} × {2,3}

= {(2,2), (2,3), (3,2),

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

(𝐴 × 𝐶) ∩ (𝐵 × 𝐶)

= {(2,2), (2,3), (3,2), (3,3)} ……….. (2)

From (1) and (2),

(𝐴 ∩ 𝐵) × 𝐶 = (𝐴 × 𝐶) ∩ (𝐵 × 𝐶) is verified.

ii) 𝐴 × (𝐵 − 𝐶) = (𝐴 × 𝐵) − (𝐴 × 𝐶)

𝐵 − 𝐶 = {2,3,5} − {2,3} = {5}

𝐴 × (𝐵 − 𝐶) = {1,2,3,4} × {5}

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

𝐴 × 𝐵 = {1,2,3,4} × {2,3,5}

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

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

𝐴 × 𝐶 = {1,2,3,4} × {2,3}

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

(3,2), (3,3), (4,2), (4,3)}

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

From (1) and (2)

𝐴 × (𝐵 − 𝐶) = (𝐴 × 𝐵) − (𝐴 × 𝐶) is Verified.

4. If 𝑨 = {𝟐, 𝟒, 𝟔} and 𝑩 = {𝟑, 𝟒} then i) Find 𝑨 × 𝑩 and 𝑩 × 𝑨.

ii) Is 𝑨 × 𝑩 = 𝑩 × 𝑨 if not why? iii) Show that 𝒏(𝑨 × 𝑩) = 𝒏(𝑩 × 𝑨)

= 𝒏(𝑨) × 𝒏(𝑩). Given that 𝐴 = {2,4,6} and 𝐵 = {3,4}

i) 𝐴 × 𝐵 = {2,4,6} × {3,4}

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

(4,4), (6,3), (6,4)} …… (1)

𝐵 × 𝐴 = {3,4} × {2,4,6}

= {(3,2), (3,4), (3,6),

(4,2), (4,4), (4,6)} … … (2)

ii) From (1) and (2), 𝐴 × 𝐵 ≠ 𝐵 × 𝐴

Because (2,3) ≠ (3,2) and

(2,4) ≠ (4,2), etc.

iii) 𝑛(𝐴) = 3; 𝑛(𝐵) = 2

From (1) and (2) ⇒

𝑛(𝐴 × 𝐵) = 𝑛(𝐴) × 𝑛(𝐵)

= 3 × 2 = 6

𝑛(𝐵 × 𝐴) = 𝑛(𝐵) × 𝑛(𝐴)

= 2 × 3

= 6

𝑛(𝐴) × 𝑛(𝐵) = 3 × 2

= 6

∴ 𝑛(𝐴 × 𝐵) = 𝑛(𝐵 × 𝐴)

= 𝑛(𝐴) × 𝑛(𝐵) = 6

5. Let 𝑨 = {𝒙 ∈ 𝑵/ 𝟐 < 𝒙 ≤ 𝟒} , 𝑩 = {𝒙 ∈ 𝑾/ 𝒙 < 𝟐} , 𝑪 = {𝒙 ∈ 𝑵/ 𝒙 ≤ 𝟐} Then verify that

i) 𝑨 × (𝑩 ∪ 𝑪) = (𝑨 × 𝑩) ∪ (𝑨 × 𝑪) ii) 𝑨 × (𝑩 ∩ 𝑪) = (𝑨 × 𝑩) ∩ (𝑨 × 𝑪)

𝐴 = {𝑥 ∈ 𝑁/2 < 𝑥 ≤ 4} = {3, 4}, 𝐵 = {𝑥 ∈ 𝑊/𝑥 < 2} = {0, 1}

𝐶 = {𝑥 ∈ 𝑁/𝑥 ≤ 2} = {1, 2}

i) 𝐴 × (𝐵 ∪ 𝐶) = (𝐴 × 𝐵) ∪ (𝐴 × 𝐶)

𝐵 ∪ 𝐶 = {0, 1, 2}

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

𝐴 × 𝐵 = {3, 4} × {0, 1} = {(3, 0), (3, 1), (4, 0), (4,1)}

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

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

From (1) and (2), 𝐴 × (𝐵 ∪ 𝐶) = (𝐴 × 𝐵) ∪ (𝐴 × 𝐶) is verified.



11

ii) 𝐴 × (𝐵 × 𝐶) = (𝐴 × 𝐵) ∩ (𝐴 × 𝐶)

𝐵 ∩ 𝐶 = {0,1} ∩ {1,2} = {1}

𝐴 × (𝐵 ∩ 𝐶 ) = {3,4} × {1} = {(3, 1) (4, 1)}………… (1)

𝐴 × 𝐵 = {(3, 0), (3, 1), (4, 0), (4, 1)}

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

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

From (1) and (2), 𝐴 × (𝐵 ∩ 𝐶) = (𝐴 × 𝐵) ∩ (𝐴 × 𝐶) is verified.

Definition: Let 𝐴 and 𝐵 be any two non-empty sets. A relation (R) from 𝐴 to 𝐵 is a subset of 𝐴 × 𝐵 satisfying some specified conditions. If 𝑥 ∈ 𝐴 is related to 𝑦 ∈ 𝐵 through R, then we write it as 𝑥𝑅𝑦.

𝑥𝑅𝑦 if and only if (𝑥, 𝑦) ∈ 𝑅

The domain of the relation 𝑅 = {𝑥 ∈ 𝐴|𝑥𝑅𝑦 , for some 𝑦 ∈ 𝐵} The co-domain of the relation 𝑅 is 𝐵 The range of the relation 𝑅 = {𝑦 ∈ 𝐵|𝑥𝑅𝑦, for some 𝑥 ∈ 𝐴}

Note:

A relation may be represented algebraically either by the roster method or by the set builder method. An arrow diagram is a visual representation of a relation. A relation which contains no element is called a ‘Null relation’. If 𝑛(𝐴) = 𝑝, 𝑛(𝐵) = 𝑞 then the total number of relations that exist from 𝐴 to 𝐵 is 2𝑝𝑞 . A relation which contains no elements is called a “Null relation”


Let 𝐴 = {1, 2, 3, 4} and 𝐵 = {𝑎, 𝑏, 𝑐}

1. Which of the following are relations from 𝐴 to 𝐵?

2. Which of the following are relations from 𝐵 to 𝐴?

(i) {(1, 𝑏), (1, 𝑐), (3, 𝑎), (4, 𝑏)} (i) {(𝑐, 𝑎), (𝑐, 𝑏), (𝑐, 1)} (ii) {(1, 𝑎), (𝑏, 4), (𝑐, 3)} (ii) {(𝑐, 1), (𝑐, 2), (𝑐, 3), (𝑐, 4)} (iii) {(1, 𝑎), (𝑎, 1), (2, 𝑏), (𝑏, 2)} (iii) {(𝑎, 4), (𝑏, 3), (𝑐, 2)}

1. Which of the following are relations from A to B?

𝐴 × 𝐵 = {1,2,3,4} × {𝑎, 𝑏, 𝑐} = {(1, 𝑎), (1, 𝑏), (1, 𝑐), (2, 𝑎), (2, 𝑏), (2, 𝑐)(3, 𝑎), (3, 𝑏), (3, 𝑐), (4, 𝑎), (4, 𝑏), (4, 𝑏)

}

(i) {(1, 𝑏)(1, 𝑐)(3, 𝑎)(4, 𝑏)} yes This is a relation from A to B (ii) {(1, 𝑎)(𝑏, 4)(𝑐, 3)} No. This is not a relation from A to B (iii) {(1, 𝑎)(𝑎, 1)(2, 𝑏)(𝑏, 2)} No. This is not a relation from A to B

2. Which of the following are relations from B to A?

𝐵 × 𝐴 = {𝑎, 𝑏, 𝑐} × {1,2,3,4} = {(𝑎, 1)(𝑎, 2)(𝑎, 3)(𝑎, 4)(𝑏, 1)(𝑏, 2)(𝑏, 3)(𝑏, 4)(𝑐, 1)(𝑐, 2)(𝑐, 3)(𝑐, 4)

}

(i) {(𝑐, 𝑎)(𝑐, 𝑏)(𝑐, 1)} No. This is not a relation from B to A (ii) {(𝑐, 1)(𝑐, 2)(𝑐, 3)(𝑐, 4)} yes. This is a relation from B to A

(iii) {(𝑎, 4)(𝑏, 3)(𝑐, 2)} Yes. This is a relation from B to A

Concept corner




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)}

𝐴 × 𝐵 = {1,2,3,7} × {3,0, −1,7}

= {(1,3), (1,0), (1, −1), (1,7), (2,3), (2,0), (2, −1), (2,7), (3,3), (3,0),

(3, −1), (3,7), (7,3), (7,0), (7, −1), (7,7)}

We know that,

(2,1) and (7,1)∈ 𝑅1 but (2,1), (7,1) ∉ 𝐴 × 𝐵

So, 𝑅1 is not a relation from 𝐴 to 𝐵

(ii) 𝑅2 = {(−1,1)}

Here (−1,1) ∈ 𝑅2 but (−1,1) ∉ 𝐴 × 𝐵

So 𝑅2 is not a relation from 𝐴 to 𝐵

(iii) 𝑅3 = {(2, −1), (7,7), (1,3)}

Here 𝑅3 ⊆ 𝐴 × 𝐵

Hence 𝑅3 is a relation from 𝐴 to 𝐵

(iv) 𝑅4 = {(7, −1), (0,3), (3,3), (0,7)}

Here (0,3) and (0,7) ∈ 𝑅4 but (0,3) and (0,7) ∉ (𝐴 × 𝐵)

So, 𝑅4 is not a relation from 𝐴 to 𝐵

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.

Given 𝐴 = {1,2,3,4, … ,45}

𝐴 × 𝐴 = {(1,1), (1,2), (1,3), (1,4) … … (45,45)}

Then, 𝑅 be the relation defined as is “square of a number ” on 𝐴.

Hence,𝑅 = {(1,1), (2,4), (3,9), (4,16), (5,25), (6,36)}

So 𝑅 ⊆ 𝐴 × 𝐴

The domain of 𝑅 = {𝟏, 𝟐, 𝟑, 𝟒, 𝟓, 𝟔}

The range of 𝑅 = {𝟏, 𝟒, 𝟗, 𝟏𝟔, 𝟐𝟓, 𝟑𝟔}

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 = {(𝑥, 𝑦)/𝑦 = 𝑥 + 3, 𝑥 ∈ {0,1,2,3,4,5}}

Here domain (𝑥) = {0,1,2,3,4,5}

Co-domain (𝑦) = 𝑥 + 3

𝑦0 = 0 + 3 = 3 , 𝑦1 = 1 + 3 = 4

𝑦2 = 2 + 3 = 5 , 𝑦3 = 3 + 3 = 6

𝑦4 = 4 + 3 = 7

𝑦5 = 5 + 3 = 8

𝑅 = {(0,3), (1,4), (2,5), (3,6), (4,7), (5,8)}

Domain = {𝟎, 𝟏, 𝟐, 𝟑, 𝟒, 𝟓}

Range = {𝟑, 𝟒, 𝟓, 𝟔, 𝟕, 𝟖}

PTA-5

Exercise 1.2

Try Your Self…

1. Let 𝐴 = {1,2,3,4 … … … … .25} and 𝑅 be

the relation difined as “is multiple of 5”

on 𝐴. write 𝑅 as a subset of 𝐴 × 𝐴.

Also, find the domain and range of 𝑅.

Ans: 𝐴 × 𝐴 = {(1,1) (1,2) (1,3) (1,4)… . . . . . . . . . . . . . . . . . (25,25)}

𝑅 = {(1,5) (2,10) (3,15) (4,20) (5,25)}

𝑅 ⊆ 𝐴 × 𝐴.

The domain 𝐷 = {1,2,3,4,5}

The range of 𝑅 = {5,10,15,20,25}

Try Your Self…

2. Let 𝐴 = {1,2,3,4} and 𝐵 = {−1,2,3,4,5,6,7,9,10}. A

relation 𝑅 = {(1,3) (2,6) (3,10) (4,9)}. Find its range.

Ans: The range = {3,6,10,9}



13

4. Represent each of the given relation 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}} (a) An arrow diagram

Given, 𝑥 = 2𝑦 If 𝑦 = 1 ⇒ 𝑥 = 2 If 𝑦 = 2 ⇒ 𝑥 = 4

(b) a graph

(c) a set in roster form 𝑅 = {(2,1), (4,2)}

(ii) {(𝒙, 𝒚)|𝒚 = 𝒙 + 𝟑, 𝒙, 𝒚 are natural numbers



1. A Relation 𝑹 is given by the set {(𝒙, 𝒚)/𝒚 = 𝒙𝟐 + 𝟑, 𝒙 ∈ {, 𝟏, 𝟐, 𝟑, 𝟒, 𝟓}} Determine its domain and range 𝑅 = {(𝑥, 𝑦)/𝑦 = 𝑥2 + 3, 𝑥 ∈ {0, 1, 2, 3, 4, 5}}

Here domain (𝑥) = {0, 1, 2, 3, 4, 5}, Co-domain (𝑦) = 𝑥2 + 3 𝑦0 = 0 + 3 = 3 𝑦1 = 1 + 3 = 4 𝑦2 = 4 + 3 = 7 𝑦3 = 9 + 3 = 12 𝑦4 = 16 + 3 = 19 𝑦5 = 25 + 3 = 28 𝑅 = {(0,3), (1,4), (2,7), (3,12), (4,19), (5,28)} Domain = {0,1,2,3,4,5} Range = {3,4,7,12,19,28}

2. Let 𝑨 = {𝟏, 𝟐, 𝟑, 𝟒, 𝟓}, 𝑩 = {𝟏, 𝟑, 𝟓, 𝟕, 𝟗} which of the following sets are relations from 𝑨 to 𝑩?

i) 𝑹𝟏 = {(𝟏, 𝟏), (𝟐, 𝟏), (𝟑, 𝟑), (𝟒, 𝟑), (𝟓, 𝟓)} ii) 𝑹𝟐 = {(𝟑, 𝟏), (𝟑, 𝟕), (𝟒, 𝟗), (𝟓, 𝟏𝟎)} iii) 𝑹𝟑 = {(𝟏, 𝟑), (𝟐, 𝟓), (𝟒, 𝟕), (𝟓, 𝟗), (𝟑, 𝟏)} 𝐴 × 𝐵 = {(1,1), (1,3), (1,5), (1,7), (1,9), (2,1),

(2,3), (2,5), (2,7), (2,9), (3,1), (3,3), (3,5), (3,7) , (3,9), (4,1), (4,3),

(4,5), (4,7), (4,9)(5,1) , (5,3), (5,5), (5,7), (5,9)}

i) we note that, 𝑅1 ⊆ 𝐴 × 𝐵 Hence, 𝑅1 is a relation from 𝐴 to 𝐵. ii) Here (5,10) ∈ 𝑅2 but (5,10) ∉ (𝐴 × 𝐵) So, 𝑅2 is not a relation from 𝐴 to 𝐵 iii) we note that 𝑅3 ⊆ 𝐴 × 𝐵 So, 𝑅3 is a relation from 𝐴 to 𝐵.

3. The arrow diagram shows a relationship between the sets 𝑷 and 𝑸

write the relation in i) set builder form ii) Roster form iii) What is the domain and Range of 𝑹 i) set builder form of 𝑅 𝑅 = {(𝑥, 𝑦)/ 𝑦 = 𝑥2, 𝑥 ∈ 𝑃, 𝑦 ∈ 𝑄}

ii) Roster form of 𝑅 𝑅 = {(1,1), (2,2), (3,9), (4,16)}

iii) Domain of 𝑅 = {1,2,3,4} and range of 𝑅 = {1,4,9,16}

4. Let 𝑿 = {𝟏, 𝟐, 𝟑, 𝟒} which of the following are relation from 𝑿 to 𝑿?

i) 𝑹𝟏 = {(𝟐, 𝟑), (𝟏, 𝟒) , (𝟐, 𝟏), (𝟑, 𝟐), (𝟒, 𝟒)} ii) 𝑹𝟐 = {(𝟑, 𝟏), (𝟒, 𝟐), (𝟐, 𝟏)} iii) 𝑹𝟑 = {(𝟐, 𝟏), (𝟑, 𝟓), (𝟏, 𝟒), (𝟓, 𝟑)} 𝑋 × 𝑋 = {(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)}

i) 𝑅1 = {(2,3), (1,4), (2,1), (3,2), (4,4)}

we observe that, 𝑅1 ⊆ 𝑋 × 𝑋

Thus 𝑅1 is a relation from 𝑋 to 𝑋

ii) 𝑅2 = {(3,1) (4,2) (2,1)}

we observe that, 𝑅2 ⊆ 𝑋 × 𝑋

Thus 𝑅2 is a relation from 𝑋 to 𝑋

iii) 𝑅3 = {(2,1), (3,5), (1,4), (5,3)}

Here (3,5) and (5,3) ∈ 𝑅3 but

(3,5) and (5,3) ∉ 𝑋 × 𝑋.

So 𝑅3 is not a relation from 𝑋 to 𝑋.

5. A cell phone store sells three different types of cell phones and we call them 𝑪𝟏, 𝑪𝟐 and 𝑪𝟑. Let us also suppose that the price of 𝑪𝟏 is C𝟏𝟐𝟎𝟎, price of 𝑪𝟐 is C 𝟐𝟓𝟎𝟎 and price of 𝑪𝟑 is C 𝟑𝟎𝟎𝟎. and the relation 𝑹 is defined by 𝒙𝑹𝒚 , where 𝒙 is the price of cell phone y. Express the relation 𝑹 through an ordered pair and an arrow diagram.

Here 𝑅 is defined by 𝑥𝑅𝑦. cost of cell phones (𝐴)

= {1200, 2500, 3000} Types of cell phones (𝐵) = {𝐶1, 𝐶2, 𝐶3} a) order of pairs 𝑅 = {(1200, 𝐶1) (2500, 𝐶2) (3000, 𝐶3)}

b) Arrow diagram

Creative Questions

PTA-2



15


1. Relation are subsets of cartesion product. Functions are subsets of relations . 2. True or False: All the elements of a relation should have images. False

[Example: Let 𝑓: 𝐴 → 𝐵 be a any relation from set A to set B. If relation 𝑓 is not a function then it is possible that its domain is not equal to the set A and hence all elements of set A not have a image in set B.]

3. True or False: All the elements of a function should have images. True [Example: A function is a relation from a set A to set B in which each elements of a set A is mapped to a unique element of set B. Hence all the elements of a function should have images.]

4. True or False: If 𝑅: 𝐴 → 𝐵 is a relation then the domain of 𝑅 = 𝐴. False 5. If 𝑓: ℕ → ℕ is defined as 𝑓(𝑥) = 𝑥2 the pre-image(s) of 1 and 2 are 1 and no pre-image

𝑓(𝑥) = 𝑥2 𝑓(1) = 1 𝑓(2) = 4 𝑓(3) = 9 𝑓(4) = 16

6. The difference between relation and function is _______. When every input has unique output then it is a function, otherwise relation. 7. Let 𝐴 and 𝐵 be two non-empty finite sets. Then which one among the following two

collection is large? (i) The number of relations between 𝐴 and 𝐵. (ii) The number of relations between 𝑨 and 𝑩. [ The number of relations between A and B is large because funtions are subsets of relations.]

Definition: A relations 𝑓 between two non-empty sets 𝑋 and 𝑌 is called a function from 𝑋 to 𝑌 if, for each 𝑥 ∈ 𝑋 there exists only one 𝑦 ∈ 𝑌 such that (𝑥, 𝑦) ∈ 𝑓.That is, 𝑓 = {(𝑥, 𝑦)/ for all 𝑥 ∈ 𝑋, 𝑦 ∈ 𝑌} Note: If 𝑓: 𝑋 → 𝑌 is a function then, the set 𝑋 is called the domain, 𝑓 and the set 𝑌 is called its co-domain. A function is also called as a mapping or transformation. 𝑓: 𝑋 → 𝑌 is a function only if i) every element in the domain of 𝑓 has an image. ii) the image is unique. If 𝐴 and 𝐵 are finite sets such that 𝑛(𝐴) = 𝑝, 𝑛(𝐵) = 𝑞 then the total number of functions

that exist between 𝐴 and 𝐵 is 𝑞𝑝 If 𝑓(𝑎) = 𝑏, then 𝑏 is called image of a under 𝑓 and 𝑎 is called a pre-image of 𝑏. The set of all images of the elements 𝑋 under 𝑓 is called the range of 𝑓. Describing domain of a function

(i) Let 𝑓(𝑥) =1

1+𝑥 . If 𝑥 = −1 then 𝑓(−1) is not defined. Hence 𝑓 is defined for all real

numbers except at 𝑥 = −1. So, domain of 𝑓 is ℝ − {−1}

(ii) Let 𝑓(𝑥) =1

𝑥2−5𝑥+6 , if 𝑥 = 2,3 then 𝑓(2) and 𝑓(3) are not defined. Hence 𝑓 is defined

for all real numbers except at 𝑥 = 2 and 3. So domain of 𝑓 = ℝ − {2,3}


Concept corner




Is the relation representing the association between planets and their respective moons a function? No, it is not a function. Because mercury and Venus have no moons. Earth has one moon. The other planets have more than one moon. Here, the relation between planets and their respective moon is not a function.

1. Let 𝒇 = {(𝒙, 𝒚)|𝒙, 𝒚 ∈ ℕ 𝐚𝐧𝐝 𝒚 = 𝟐𝒙} be a relation on 𝑵. Find the domain, codomain and range. Is

this relation a function?

𝑦 = 𝑓(𝑥) = 2𝑥

𝑓(1) = 2(1) = 2

𝑓(2) = 2(2) = 4

𝑓(3) = 2(3) = 6

𝑓(4) = 2(4) = 8

⋮

𝑓 be a relation on 𝑁

Domain of 𝑓 = {1,2,3,4, … }

Codomain of 𝑓 = {1,2,3,4, … } , Range of 𝑓 = {2,4,6,8, … }

From the arrow diagram of 𝑓, for each 𝑥 ∈ 𝐴 there exists only one 𝑦 ∈ 𝐵. Yes, 𝑓 is a function.

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?

Given 𝑋 = {3,4,6,8}

𝑌 = {1,2,3,4, … . }

𝑅 = {(𝑥, 𝑓(𝑥)/𝑥 ∈ 𝑋, 𝑓(𝑥) = 𝑥2 + 1)}

Let 𝑦 = 𝑓(𝑥) = 𝑥2 + 1

𝑓(3) = 32 + 1 = 10 , 𝑓(4) = 42 + 1 = 17 , 𝑓(6) = 62 + 1 = 37 , 𝑓(8) = 82 + 1 = 65

𝑅 = {(3,10), (4,17), (6,37), (8,65)}

We note that each element is the domains of 𝑋 has a unique image in 𝑁.

Yes, it is a function from 𝑋 to 𝑁

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 (i) 𝑓(−1) 𝑓(−1) = (−1)2 − 5(−1) + 6 = 1 + 5 + 6 = 𝟏𝟐 (ii) 𝑓(2𝑎) 𝑓(2𝑎) = (2𝑎)2 − 5(2𝑎) + 6 = 𝟒𝒂𝟐 − 𝟏𝟎𝒂 + 𝟔 (iii) 𝑓(2) 𝑓(2) = 22 − 5(2) + 6 = 4 − 10 + 6 = 𝟎 (iv) 𝑓(𝑥 − 1) 𝑓(𝑥 − 1) = (𝑥 − 1)2 − 5(𝑥 − 1) + 6 = 𝑥2 − 2𝑥 + 1 − 5𝑥 + 5 + 6 = 𝒙𝟐 − 𝟕𝒙 + 𝟏𝟐

Exercise 1.3

Try Your Self…

1. Let 𝐴 = {1,2,3,4,5}, 𝐵 = 𝑁 be a

relation defined by 𝑦 = 𝑥2.

Find the co-domain and range?

Ans: Range of 𝑓 = {1,4,9,16,25}

Co-domain = {1,2,3, . . . . . . . . . . . }

Try Your Self…

2. Let 𝐴 = {0,1,2,3} and 𝐵 = {1,3,5,7,9} Determine

whether the relation 𝑅 = {(𝑥, 𝑓(𝑥)/𝑥 ∈ 𝐴, 𝑓(𝑥) = 2𝑥 + 1}

is a function from 𝐴 to 𝐵?

Ans: Each element in 𝐴 is associated with a unique element

in 𝐵. It is a function from 𝐴 to 𝐵.

Try Your Self…

3. Given the function 𝑓(𝑥) = 𝑥2 + 2𝑥 + 1, evaluate

i) 𝑓(1) ii) 𝑓(−3) iii) 𝑓(2𝑎) iv) 𝑓(𝑥 + 1)

Ans: i) 𝑓(1) = 4

ii) 𝑓(−3) = 4

iii) 𝑓(2𝑎) = 4𝑎2 + 4𝑎 + 1

iv) 𝑓(𝑥 + 1) = 𝑥2 + 4𝑥 + 4



17

4. A graph representing the function 𝒇(𝒙) is given in adjacent figure.

It is clear that 𝒇(𝟗) = 𝟐

(i)Find the following values of the function

(a) 𝒇(𝟎) (b) 𝒇(𝟕) (c) 𝒇(𝟐) (d) 𝒇(𝟏𝟎)

(ii) for What value of 𝒙 is 𝒇(𝒙) = 𝟏

(iii) Describe the following (i)Domain (ii) Range

(iv) What is the image of 6 under 𝒇?

(i) From the figure,

(a) 𝑓(0) = 𝟗 (b) 𝑓(7) = 𝟔 (c) 𝑓(2) = 𝟔 (d) 𝑓(10) = 𝟎

(ii) 𝑓(𝑥) = 1 ⇒ 𝑥 = 𝟗. 𝟓

(iii) (a) Domain of 𝑓 = {𝟎, 𝟏, 𝟐, 𝟑, 𝟒, 𝟓, 𝟔, 𝟕, 𝟖, 𝟗, 𝟏𝟎} or

𝑓 = {𝒙/𝟎 ≤ 𝒙 ≤ 𝟏𝟎, 𝒙 ∈ 𝑹}

(b) Range of 𝑓 = {𝟎, 𝟏, 𝟐, 𝟑, 𝟒, 𝟓, 𝟔, 𝟕, 𝟖, 𝟗} or

𝑓 = {𝒙/𝟎 ≤ 𝒙 ≤ 𝟗, 𝒙 ∈ 𝑹}

(iv) The image of 6 under 𝑓 = 𝟓

5. Let 𝒇(𝒙) = 𝟐𝒙 + 𝟓. If 𝒙 ≠ 𝟎 then find 𝒇(𝒙+𝟐)−𝒇(𝟐)

𝒙

Given, 𝑓(𝑥) = 2𝑥 + 5, 𝑥 ≠ 0

𝑓(𝑥 + 2) = 2(𝑥 + 2) + 5 = 2𝑥 + 4 + 5 = 2𝑥 + 9

𝑓(2) = 2(2) + 5 = 4 + 5 = 9

Then 𝑓(𝑥+2)−𝑓(2)

𝑥=

2𝑥+9−9

𝑥

=2𝑥

𝑥= 𝟐

6. A function 𝒇 is defined by 𝒇(𝒙) = 𝟐𝒙 − 𝟑

(i) find 𝒇(𝟎)+𝒇(𝟏)

𝟐 (ii)find 𝒙 such that 𝒇(𝒙) = 𝟎 (iii) find 𝒙 sucht that 𝒇(𝒙) = 𝒙

(iv) find 𝒙 such that 𝒇(𝒙) = 𝒇(𝟏 − 𝒙)

(i) 𝑓(0)+𝑓(1)

2

𝑓(0) = 2(0) − 3 = −3

𝑓(1) = 2(1) − 3 = −1 𝑓(0)+𝑓(1)

2=

−3−1

2= −

4

2= −𝟐

(ii) 𝑓(𝑥) = 0

2𝑥 − 3 = 0

2𝑥 = 3

𝒙 =𝟑

𝟐

(iii) 𝑓(𝑥) = 𝑥

2𝑥 − 3 = 𝑥

2𝑥 − 𝑥 = 3

𝒙 = 𝟑

(iv) 𝑓(𝑥) = 𝑓(1 − 𝑥)

2𝑥 − 3 = 2(1 − 𝑥) − 3 = 2 − 2𝑥 − 3 = −2𝑥 − 1

2𝑥 + 2𝑥 = 3 − 1

4𝑥 = 2

𝒙 =𝟏

𝟐

Try Your Self…

4. Let 𝑓(𝑥) = 2𝑥 − 1, {2 ≤ 𝑥 ≤ 4, 𝑥 ∈ 𝑅}

then find 𝑓(2)+𝑓(4)

𝑓(3).

Ans: 2



7. An open box is to be made from a square piece of material, 24 cm on a side, by

cutting equal squares from the corners and turning up the sides as shown

figure. Express the volume 𝑽 of the box as a function of 𝒙.

Side of the square material = 24cm

Side of cutting equal = 𝑥 cm

Square from the corners

Now, it is to be made an open box.

Side of the open box = 24 − 2𝑥 cm

Height = 𝑥 cm

Volume of the open box 𝑉 = 𝑙𝑏ℎ cubic units.

= (24 − 2𝑥)(24 − 2𝑥)(𝑥)

= (24 − 2𝑥)2(𝑥)

= (576 − 96𝑥 + 4𝑥2)𝑥

= 𝟒𝒙𝟑 − 𝟗𝟔𝒙𝟐 + 𝟓𝟕𝟔𝒙

8. A function 𝒇 is defined by 𝒇(𝒙) = 𝟑 − 𝟐𝒙. Find

𝒙 such that 𝒇(𝒙𝟐) = (𝒇(𝒙))𝟐

Given, 𝑓(𝑥) = 3 − 2𝑥

To find 𝑥, 𝑓(𝑥2) = (𝑓(𝑥))2

3 − 2𝑥2 = (3 − 2𝑥)2

3 − 2𝑥2 = 9 − 12𝑥 + 4𝑥2

4𝑥2 + 2𝑥2 − 12𝑥 + 6 = 0

. 6𝑥2 − 12𝑥 + 6 = 0

𝑥2 − 2𝑥 + 1 = 0

(𝑥 − 1)2 = 0

𝑥 − 1 = 0

𝒙 = 𝟏

9. A plane is flying at a speed of 500 km per hour. Express the distance 𝒅 travelled by the plane as function of time 𝒕 in hours. Given , speed (𝑆 )= 500 km/hr

Speed =Distance

Time

𝑆 =𝑑

𝑡

𝒅 = 𝟓𝟎𝟎𝒕

10. The data in the adjacent table depicts the length of a woman’s

forehand and her corresponding height. Based on this data, a

student finds a relationship between the height (𝒚) and the

forehand length (𝒙) as 𝒚 = 𝒂𝒙 + 𝒃, where 𝒂, 𝒃 are constants.

(i) Check if this relation is a function. (ii) Find 𝒂 and 𝒃 (iii)

Find the height of a woman whose forehand length is 40cm (iv)

Find the length of forehand of a woman if her height is 53.3

inches.

Given 𝑦 = 𝑎𝑥 + 𝑏

(i) Arrow diagram

Each element in 𝑥 is associated with a unique element in 𝑦

Yes, this relation is a function

Length 𝒙 of forehand (in cm)

Height ‘𝒚’ (in inches)

35 56 45 65 50 69.5 55 74

PTA-4

Try Your Self…

5. A bus travels at a speed of 50 𝑘𝑚 per

hour. Express the distance 𝑑 travelled

by the bus as function of time t hours.

Ans: 𝑑 = 50 𝑡



19

(ii) find 𝑎 and 𝑏

From the table

35𝑎 + 𝑏 = 56 ……………(1)

45𝑎 + 𝑏 = 65 …………..(2) (−) (−) (−)

−10𝑎 = −9

𝑎 =9

10= 0.9

𝑎 = 0.9 substitute in (1)

35(0.9) + 𝑏 = 56

31.5 + 𝑏 = 56

𝑏 = 56 − 31.5 = 24.5

𝒂 = 𝟎. 𝟗 and 𝒃 = 𝟐𝟒. 𝟓

(iii) Length = 40cm, 𝑎 = 0.9, 𝑏 = 24.5

𝑦 = 𝑎𝑥 + 𝑏

= (0.9)(40) + 24.5

= 60.5

The height of a woman whose forehand length is 40 cm = 60.5 inches.

(iv) Height = 53.3 inches

𝑦 = 𝑎𝑥 + 𝑏

53.3 = (0.9)𝑥 + 24.5

= 0.9𝑥 + 24.5

0.9𝑥 = 53.3 − 24.5

= 28.8

𝑥 =28.8

0.9

𝑥 = 32 cm

The length of forehand of a women = 32 cm

1. An open box is to be made from a rectangle piece of material, 𝟏𝟐𝒄𝒎 × 𝟓𝒄𝒎, by cutting equal squres from the corners and turning up the the sides as shown. Express the volume 𝑽 of the box as a function of 𝒙. Length of the Rectangle material = 12 𝑐𝑚 Breadth of the Rectangle material = 5𝑐𝑚 side of cutting equal square from the corners = 𝑥 𝑐𝑚 Now It is to be made an open box length of the open box = 12 − 2𝑥 𝑐𝑚 Breadth of the open box = 5 − 2𝑥 𝑐𝑚 height = 𝑥 𝑐𝑚 volume of the open box 𝑉 = 𝑙𝑏ℎ = (12 − 2𝑥) (5 − 2𝑥) (𝑥) = (60 − 24𝑥 − 10𝑥 + 4𝑥2) (𝑥) = 4𝑥3 − 34𝑥2 + 60𝑥

2. Given 𝒇(𝒙) = 𝟐𝒙 − 𝟏, Find i) 𝒇(𝟏) ii) 𝒇(𝒙 + 𝟏) iii) 𝒇(𝒙) + 𝒇(𝟏)

i) 𝑓(1) = 2(1) − 1 = 2 − 1 = 1

ii) 𝑓(𝑥 + 1) = 2(𝑥 + 1) − 1

= 2𝑥 + 2 − 1 = 2𝑥 + 1

iii) 𝑓(𝑥) + 𝑓(1) = (2𝑥 − 1) + (1) = 2𝑥 − 1 + 1 = 2𝑥

Try Your Self…

6.

𝑥 1 4 9 16

𝑦 2 5 10 17

𝑦 = 𝑎𝑥 + 𝑏, where 𝑎, 𝑏 are constants.

i) check if this relation is a function.

ii) Find 𝑎 and 𝑏 iii) Find 𝑦 when 𝑥 = 2

Ans: i) 𝑓 is a function.

ii) 𝑎 = 1, 𝑏 = 1

iii) when 𝑥 = 2 ⇒ 𝑦 = 3

Creative Questions



3. A function 𝒇 is defined by 𝒇(𝒙) = 𝟐𝒙 + 𝟑

i) Find 𝒇(𝟎)+𝒇(𝟏)

𝟐 ii)Find 𝒙 such that 𝒇(𝒙) = 𝟎

iii) Find 𝒙 such that 𝒇(𝒙) = 𝒙 iv) Find 𝒙 such that 𝒇(𝒙) = 𝒇(𝟏 − 𝒙) Given, 𝑓(𝑥) = 2𝑥 + 3

i) 𝑓(0)+𝑓(1)

2

𝑓(0) = 2(0) + 3 = 3

𝑓(1) = 2(1) + 3 = 2 + 3 = 5

𝑓(0)+𝑓(1)

2=

3+5

2=

8

2 = 4

ii) 𝑓(𝑥) = 0

2𝑥 + 3 = 0

2𝑥 = −3

𝑥 =−3

2

iii) 𝑓(𝑥) = 𝑥

2𝑥 + 3 = 𝑥

2𝑥 − 𝑥 = −3

𝑥 = −3

iv) 𝑓(𝑥) = 𝑓(1 − 𝑥)

2𝑥 + 3 = 2(1 − 𝑥) + 3

= 2 − 2𝑥 + 3

2𝑥 + 2𝑥 = 5 − 3

4𝑥 = 2 ⇒ 𝑥 =2

4=

1

2

4. A function 𝒇 is defined by 𝒇(𝒙) = 𝟏 − 𝟐𝒙.

Find 𝒙 such that 𝒇(𝒙𝟐) = (𝒇(𝒙))𝟐

Given 𝑓(𝑥) = 1 − 2𝑥

𝑓(𝑥2) = (𝑓(𝑥))2

1 − 2𝑥2 = (1 − 2𝑥)2 1 − 2𝑥2 = 1 − 4𝑥 + 4𝑥2 4𝑥2 + 2𝑥2 − 4𝑥 + 1 − 1 = 0 6𝑥2 − 4𝑥 + 0 = 0 6𝑥2 − 4𝑥 = 0 𝑥(6𝑥 − 4) = 0

𝑥 = 0 (or) 6𝑥 − 4 = 0

𝑥 = 0 (or) 𝑥 =4

6

𝑥 = 0 (or) 𝑥 =2

3

𝑥 = 0,2

3

5. A relation 𝒇 is defined by 𝒇(𝒙) = 𝟐𝒙 − 𝟏 where 𝒙 ∈ {𝟐, 𝟑, 𝟒, 𝟓}

i) List the elements of 𝒇

ii) Is 𝒇 a function?

Given, 𝑓(𝑥) = 2𝑥 − 1 where 𝑥 ∈ {2,3,4,5}

i) 𝑓(2) = 2(2) − 1

= 4 − 1

= 3

𝑓(3) = 2(3) − 1

= 6 − 1

= 5

𝑓(4) = 2(4) − 1

= 8 − 1

= 7

𝑓(5) = 2(5) − 1

= 10 − 1

= 9

𝑓 = {(2,3), (3,5), (4,7), (5,9)}

ii) we observe that each element in the domain of 𝑓 has a unique image. So 𝒇 is a function.

6. Let 𝑨 = {𝟏, 𝟐, 𝟑, 𝟒} and 𝑩 = {𝟏, 𝟒, 𝟗, 𝟏𝟔, 𝟐𝟓} and 𝑹 = {(𝟏, 𝟏), (𝟐, 𝟒), (𝟑, 𝟗), (𝟒, 𝟏𝟔)} show that 𝑹 is a function and find its domain , co-domain and range?

Given, 𝐴 = {1,2,3,4} 𝐵 = {1,4,9,16,25} 𝑅 = {(1,1), (2,4),

(3,9), (4,16)}

From the arrow diagram, we see that for each 𝑥 ∈ 𝐴, there exists only one 𝑦 ∈ 𝐵. Thus all elements in 𝐴 have only one image in 𝐵. ∴ 𝑅 is a function. Domain 𝐴 = {1,2,3,4} Co-domain 𝐵 = {1,4,9,16,25}

Range of 𝑓 = {1,4,9,16}



21

Note: Any equation represented in a graph is usually called a curve. Representation of functions

a) a set of ordered pairs b) a table form c) An arrow diagram d) a graphical form.

Vertical line test: A curve drawn in a graph represents a function, if every vertical line intersects the curve in at most one point.

Horizontal Line Test: A function represented in a graph in one – one, if every horizontal line intersects the curve in at most one point.

Every function can be represented by a curve in a graph. But not every curve drawn in a graph will represent a function.

If 𝑓: 𝐴 − 𝐵 is an onto function then, the range of 𝑓 = 𝐵

Note: A one-one and onto function is also called a one-one correspondence.

Types of functions

Sl.No Name Definition Mapping Example

1 One-One function

(Injection)

A function 𝑓: 𝐴 → 𝐵 is called one-one function if distinct elements of 𝐴 have distinct images in 𝐵.

2 Many-one

function A function 𝑓: 𝐴 → 𝐵 is called many-one function if two or more elements of 𝐴 have same image in 𝐵

3 Onto function

(Surjection)

A function 𝑓: 𝐴 → 𝐵 is said to be onto function if the range of 𝑓 is equal to the co-domain of 𝑓.

4 Into function A function 𝑓: 𝐴 → 𝐵 is called an into

function if there exists at least one element in 𝐵 which is not the image of any element of 𝐴

5 Constant

function A function 𝑓: 𝐴 → 𝐵 is called a constant function if the range of 𝑓 contains only one element. That is, 𝑓(𝑥) = 𝑐 for all 𝑥 ∈ 𝐴 and for some fixed 𝑐 ∈ 𝐵.

6 Identity function

Let 𝐴 be anon-empty set. Then the function 𝑓: 𝐴 → 𝐴 defined by 𝑓(𝑥) = 𝑥 for all 𝑥 ∈ 𝐴 is called an identity function on 𝐴 and is denoted by 𝐼𝐴.

Concept corner




7 Bijection If a function 𝑓: 𝐴 → 𝐵 is both one-one and onto, then 𝑓 is called a bijection from 𝐴 to 𝐵

8 Real – Valued

function A function 𝑓: 𝐴 → 𝐵 is called a real valued function if the range of 𝑓 is a subset of the set of all real numbers 𝑅. That is 𝑓(𝐴) ⊆ 𝑅


State True or False. 1. All one – one function are onto functions.

False Example:

The above function 𝑓: 𝐴 → 𝐵 is called one – one but not onto. Because the range

of 𝑓 is not equal to the co-domain of 𝑓.

2. There will be no one – one function from A to

B when 𝑛(𝐴) = 4, 𝑛(𝐵) = 3. True If 𝑓 is called one – one

function, distinct

elements of A have

distinct images in B.

But 2 and 4 have the

same image. Here 𝑛(𝐴) = 4, 𝑛(𝐵) = 3

3. All onto function are one – one functions.

False Here, the range of 𝑓 is B. Here onto function.

But ‘b’ and ‘d’ have same image. Hence not one – one function.

4. There will be no onto function from A to B when 𝑛(𝐴) = 4, 𝑛(𝐵) = 5. True The range of f is not equal to B.

That is 𝑛(𝐴) ≠ 𝑛(𝐵)

Here, 𝑛(𝐴) = 4, 𝑛(𝐵) = 5

5. If 𝑓 is a bijection from A to B then 𝑛(𝐴) = 𝑛(𝐵). True

𝑓: 𝐴 → 𝐵 is both one – one and

onto (bijection)

Hence, 𝑛(𝐴) = 3 𝑛(𝐵) = 3

6. If 𝑛(𝐴) = 𝑛(𝐵), then 𝑓 is a bijection from A to B.

True 𝑓: 𝐴 → 𝐵 is both one – one and

onto (bijection)

Hence, 𝑛(𝐴) = 3 𝑛(𝐵) = 3

7. All constant function are bijection. False

Here, distinct elements do not have distinct images and 𝑛(𝐴) ≠ 𝑛(𝐵)

Pg.No 20 Can there be a one to many function?

As per definition, it is Not possible. 23 Is an identity function one – one function?

Yes. An identity function is one-one funciton.



23

1. Determine whether the graph given below represent functions. Give reason for your answers

concerning each graph.

Using vertical line test, the curve does not represent a function as the vertical line meets the curve in two points P and Q.

The curve represent a function as the vertical line meets the curve in at most one point.

The curve does not represent a function as the vertical line meets the curve in three points P, Q and R.

The graph represent a function as the vertical line meets the line is at most one point P.

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 by (i) set of ordered pairs (ii) a table (iii) an arrow diagram (iv) a graph

Given, 𝐴 = {2,4,6,10,12}, 𝐵 = {0,1,2,4,5,9}

Now 𝑓(𝑥) =𝑥

2− 1

Thus, 𝑓(2) =2

2− 1 = 1 − 1 = 𝟎

𝑓(4) =4

2− 1

= 2 − 1 = 𝟏

𝑓(6) =6

2− 1

= 3 − 1 = 𝟐

𝑓(10) =10

2− 1

= 5 − 1

= 𝟒

𝑓(12) =12

2− 1

= 6 − 1 = 𝟓

(i) set of ordered pairs,

𝒇 = {(𝟐, 𝟎), (𝟒, 𝟏), (𝟔, 𝟐), (𝟏𝟎, 𝟒), (𝟏𝟐, 𝟓)}

(ii) Table

𝑥 2 4 6 10 12

𝑓(𝑥) 0 1 2 4 5

Exercise 1.4

Try Your Self…

1. Let 𝐴 = {6,9,15,18,21}; 𝐵 = {1,2,4,5,6} and 𝑓: 𝐴 → 𝐵 be defined

by 𝑓(𝑥) =𝑥−3

3. Represent 𝑓 by i) an arrow diagram ii) a set of

ordered pairs iii) a table iv) a graph.

Ans: i) An arrow diagram

ii) Set of ordered pairs

𝑓 = {(6,1) (9,2) (15,4) (18,5) (21,6)}

iii) Table

𝑥 6 9 15 18 21

𝑓(𝑥) 1 2 4 5 6

iv) Graph



(iii) An arrow diagram

(iv) Graph

3. Represent the function 𝒇 = {(𝟏, 𝟐), (𝟐, 𝟐), (𝟑, 𝟐), (𝟒, 𝟑), (𝟓, 𝟒)} through (i) an arrow diagram (ii)

a table form (iii) a graph

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

(i) an arrow diagram

(ii) a table form

𝑥 1 2 3 4 5

𝑓(𝑥) 2 2 2 3 4

(iii) a graph

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

The function 𝑓: ℕ → ℕ defined by 𝑓(𝑥) = 2𝑥 − 1

If 𝑥 = 1, 𝑓(1) = 2(1) − 1 = 1

If 𝑥 = 2, 𝑓(2) = 2(2) − 1 = 3

If 𝑥 = 3, 𝑓(3) = 2(3) − 1 = 5

Arrow diagram:

Then 𝑓 is a function from 𝑁 to 𝑁 and for different elements in 𝑁 , there are different images in 𝑁 . Hence 𝑓 one-one function.

But the even numbers in the co-domain do not have any pre-images of the domain. Hence 𝑓 is not onto, So 𝑓 is one-one but not onto function.

5. Show that the function 𝒇: ℕ → ℕ defined by 𝒇(𝒎) = 𝒎𝟐 + 𝒎 + 𝟑 is one – one function

The function 𝑓: ℕ → ℕ defined by

𝑓(𝑚) = 𝑚2 + 𝑚 + 3

𝑚 = 1, 𝑓(1) = (1)2 + 1 + 3 = 1 + 1 + 3 = 5

𝑚 = 2, 𝑓(2) = (2)2 + 2 + 3 = 4 + 2 + 3 = 9

𝑚 = 3, 𝑓(3) = (3)2 + 3 + 3 = 9 + 3 + 3 = 15

𝑚 = 4, 𝑓(4) = (4)2 + 4 + 3 = 16 + 4 + 3 = 23

Since different elements of 𝑁 have different images in the codomain the function of 𝒇 is one-one

function.

Try Your Self…

2. Represent the function 𝑓 = {(−1,2) (−3,1) (−5,6) (−4,3)}

as i) a table ii) an arrow diagram

Ans: i) Table ii) Arrow diagram

𝑥 −1 −3 −5 −4

𝑓(𝑥) 2 1 6 3

Try Your Self… 3. Let 𝑓 be function 𝑓: 𝑁 → 𝑁 be defined by

𝑓(𝑥) = 3𝑥 − 2 is one-one but not onto.

Ans: Since different elements of 𝑁 have different

images in the co-domain, the function 𝑓 is

one-one function. 𝑓 is one-one but not onto

function.

Try Your Self…

4. Show that the function the function 𝑓: 𝑁 → 𝑁 defined by

𝑓(𝑥) = 𝑥2 + 𝑥 + 1 is one-one function.

Ans: Since different elements of 𝑁 have different image in

the co-domain the function 𝑓 is one-one function.



25

6. Let 𝑨 = {𝟏, 𝟐, 𝟑, 𝟒} and 𝑩 = ℕ. Let 𝒇: 𝑨 → 𝑩 be defined by 𝒇(𝒙) = 𝒙𝟑 then, (i) find the range of 𝒇

(ii) identify the type of function

Now 𝐴 = {1,2,3,4} , 𝐵 = {1,2,3, … }

Given 𝑓: 𝐴 → 𝐵 and 𝑓(𝑥) = 𝑥3

𝑓(1) = 13 = 1, 𝑓(3) = 33 = 27

𝑓(2) = 23 = 8, 𝑓(4) = 43 = 64

(i) Range of 𝒇 = {𝟏, 𝟖, 𝟐𝟕, 𝟔𝟒}

(ii) Since distinct elements in 𝐴 are mapped into distinct images in 𝐵, it is a one-one function.

2 ∈ 𝐵 is not the image of any element of 𝐴. So, it is Into function.

7. In each of the following cases state whether the function is bijective or not. Justify your answer.

(i) 𝒇: ℝ → ℝ defined by 𝒇(𝒙) = 𝟐𝒙 + 𝟏

𝑓(𝑥) = 2𝑥 + 1

𝑓(0) = 2(0) + 1 = 1

𝑓(1) = 2(1) + 1 = 3

𝑓(2) = 2(2) + 1 = 5

𝑓(−1) = 2(−1) + 1 = −1

𝑓(−2) = 2(−2) + 1 = −3

𝑓(−3) = 2(−3) + 1 = −5

Range of 𝑓 = {1, 3, 5, −1, −3, −5}

As distinct elements of 𝐴 have distinct

images in 𝐵 and every elements in 𝐵 has a pre-image in 𝐴. The function is bijective.

(ii) 𝒇: ℝ → ℝ defined by 𝒇(𝒙) = 𝟑 − 𝟒𝒙𝟐

𝑓(𝑥) = 3 − 4𝑥2

𝑓(0) = 3 − 4(0)2 = 3

𝑓(1) = 3 − 4(1)2 = −1

𝑓(2) = 3 − 4(2)2 = −13

𝑓(−1) = 3 − 4(−1)2 = −1

Thus two distinct elements 1 and −1 in 𝐴 have same image −1 in 𝐵. Hence 𝑓 is not a one-one function. But every elements in 𝐵 has a pre-image in 𝐴 . Hence 𝑓 is a onto function.

Therefore 𝑓 is not one-one but onto.

Hence 𝑓 is not bijective.

8. Let 𝑨 = {−𝟏, 𝟏} and 𝑩 = {𝟎, 𝟐}. If the function 𝒇: 𝑨 → 𝑩 defined by 𝒇(𝒙) = 𝒂𝒙 + 𝒃 is an onto

function? Find 𝒂 and 𝒃.

Given 𝐴 = {−1,1} and 𝐵 = {0,2} Then 𝑓: 𝐴 → 𝐵 defined by

𝑓(𝑥) = 𝑎𝑥 + 𝑏 is an onto function.

[Range of 𝑓 = co-domain]

That is, 𝑓(−1) = 0 ⇒ −𝑎 + 𝑏 = 0 𝑓(1) = 2 ⇒ 𝑎 + 𝑏 = 2 Now −𝑎 + 𝑏 = 0 …………..(1)

𝑎 + 𝑏 = 2 …………..(2) (1)+(2) 2𝑏 = 2

𝑏 = 1 Substitute 𝑏 = 1 in (2) 𝑎 + 1 = 2

𝑎 = 2 − 1 = 1 Thus, 𝒂 = 𝟏 and 𝒃 = 𝟏

Try Your Self…

5. Let 𝐴 = {1,2,3,4,5} and 𝐵 = 𝑁 and 𝑓: 𝐴 → 𝐵 be defined by

𝑓(𝑥) = 𝑥3 then i) find the range of 𝑓 ii) identify the type of

function.

Ans: i) Range of 𝑓 = {1, 8, 27, 64, 125}

ii) Since distinct elements are associated in to distinct elements, it

is a one-one function.

Try Your Self…

6. Let 𝐴 = {0,1} and

𝐵 = {1,3}. If the

function 𝑓: 𝐴 → 𝐵

defined by 𝑎𝑥 + 𝑏

is an onto function.

Find 𝑎 and 𝑏?

Ans: 𝑎 = 2 and 𝑏 = 1



9. If the function 𝒇 is defined by 𝒇(𝒙) = {𝒙 + 𝟐;

𝟐;𝒙 − 𝟏;

𝒙 > 𝟏

−𝟏 ≤ 𝒙 ≤ 𝟏−𝟑 < 𝒙 < −𝟏

find the values of

(i) 𝒇(𝟑) (ii) 𝒇(𝟎) (iii) 𝒇(−𝟏. 𝟓) (iv) 𝒇(𝟐) + 𝒇(−𝟐)

𝑓(𝑥) = {𝑥 + 2;

2;𝑥 − 1;

𝑥 > 1

−1 ≤ 𝑥 ≤ 1−3 < 𝑥 < −1

where 𝑥 = 2,3,4

where 𝑥 = 0where 𝑥 = −2

(i) 𝑓(3) When 𝑥 = 3,

𝑓(𝑥) = 𝑥 + 2 𝑓(3) = 3 + 2 = 𝟓

(ii) 𝑓(0) When 𝑥 = 0

𝑓(𝑥) = 2 𝑓(0) = 𝟐

(iii) 𝑓(−1.5)

When 𝑥 = −1.5

𝑓(𝑥) = 𝑥 − 1

𝑓(−1.5) = −1.5 − 1

= −𝟐. 𝟓

(iv) 𝑓(2) + 𝑓(−2) When 𝑥 = 2, 𝑓(𝑥) = 𝑥 + 2

𝑓(2) = 2 + 2 = 4 When 𝑥 = −2, 𝑓(𝑥) = 𝑥 − 1

𝑓(−2) = −2 − 1 = −3 𝑓(2) + 𝑓(−2) = 4 + (−3) = 𝟏

10. A function 𝒇: [−𝟓, 𝟗] → ℝ is defined as follows: 𝒇(𝒙) = {𝟔𝒙 + 𝟏;

𝟓𝒙𝟐 − 𝟏𝟑𝒙 − 𝟒;

; −𝟓 ≤ 𝒙 < 𝟐 𝟐 ≤ 𝒙 < 𝟔 𝟔 ≤ 𝒙 ≤ 𝟗

Find (i) 𝒇(−𝟑) + 𝒇(𝟐) (ii) 𝒇(𝟕) − 𝒇(𝟏) (iii) 𝟐𝒇(𝟒) + 𝒇(𝟖) (iv) 𝟐𝒇(−𝟐)−𝒇(𝟔)

𝒇(𝟒)+𝒇(−𝟐)

𝑓(𝑥) = {6𝑥 + 1;

5𝑥2 − 1;3𝑥 − 4;

−5 ≤ 𝑥 < 2 2 ≤ 𝑥 < 6 6 ≤ 𝑥 ≤ 9

; Where 𝑥 = −5, −4, −3, −2, −1,0,1

; Where 𝑥 = 2, 3,4,5 ; Where 𝑥 = 6,7,8,9

(i) 𝑓(−3) + 𝑓(2)

When 𝑥 = −3

𝑓(𝑥) = 6𝑥 + 1

𝑓(−3) = 6(−3) + 1 = −18 + 1 = −17

When 𝑥 = 2

𝑓(𝑥) = 5𝑥2 − 1

𝑓(2) = 5(2)2 − 1 = 20 − 1 = 19

𝑓(−3) + 𝑓(2) = −17 + 19 = 𝟐

(ii) 𝑓(7) − 𝑓(1)

When 𝑥 = 7

𝑓(𝑥) = 3𝑥 − 4

𝑓(7) = 3(7) − 4 = 21 − 4 = 17

When 𝑥 = 1

𝑓(𝑥) = 6𝑥 + 1

𝑓(1) = 6(1) + 1 = 6 + 1 = 7

∴ 𝑓(7) − 𝑓(1) = 17 − 7 = 𝟏𝟎

(iii) 2𝑓(4) + 𝑓(8)

When 𝑥 = 4,

𝑓(𝑥) = 5𝑥2 − 1

𝑓(4) = 5(4)2 − 1

= 80 − 1

= 79

When 𝑥 = 8, 𝑓(𝑥) = 3𝑥 − 4

𝑓(8) = 3(8) − 4

= 24 − 4 = 20

2𝑓(4) + 𝑓(8) = 2(79) + 20

= 158 + 20 = 𝟏𝟕𝟖

(iv) 2𝑓(−2)−𝑓(6)

𝑓(4)+𝑓(−2)

When 𝑥 = −2, 𝑓(𝑥) = 6𝑥 + 1

𝑓(−2) = 6(−2) + 1

= −12 + 1 = −11

When 𝑥 = 6, 𝑓(𝑥) = 3𝑥 − 4

𝑓(6) = 3(6) − 4

= 18 − 4 = 14

When 𝑥 = 4, 𝑓(𝑥) = 5𝑥2 − 1

𝑓(4) = 5(4)2 − 1 = 80 − 1 = 79

2𝑓(−2)−𝑓(6)

𝑓(4)+𝑓(−2)=

2(−11)−14

79+(−11) =

−22−14

79−11=

−36

68= −

𝟗

𝟏𝟕

PTA-4

PTA-4

Try Your Self…

7. A function 𝑓 is defined by

𝑓(𝑥) = {4𝑥2 − 1; −3 ≤ 𝑥 < 23𝑥 − 2; 2 ≤ 𝑥 ≤ 42𝑥 − 3; 4 < 𝑥 < 7

Find (i) 𝑓(5) + 𝑓(6)

(ii) 𝑓(1) − 𝑓(−3)

(iii) 𝑓(−2) − 𝑓(4)

Ans: (i) 𝑓(5) + 𝑓(6) = 16 (ii) 𝑓(1) − 𝑓(−3) = −32

(iii) 𝑓(−2) − 𝑓(4) = 5



27

11. The distance 𝑺 an object travels under the influence of gravity in the time 𝒕 seconds is given by

𝑺(𝒕) =𝟏

𝟐𝒈𝒕𝟐 + 𝒂𝒕 + 𝒃 where, ( 𝒈 is the acceleration due to gravity), 𝒂, 𝒃 are constants. Verify

whether the function 𝑺(𝒕) is one-one or not.

Given 𝑆(𝑡) =1

2𝑔𝑡2 + 𝑎𝑡 + 𝑏 (𝑎, 𝑏 constants)

Now take 𝑡 = 1,2,3, … seconds

𝑡 = 1, 𝑆(1) =1

2𝑔(1)2 + 𝑎(1) + 𝑏

=1

2𝑔 + 𝑎 + 𝑏 = 𝟎. 𝟓𝒈 + 𝒂 + 𝒃

𝑡 = 2, 𝑆(2) =1

2𝑔(2)2 + 𝑎(2) + 𝑏

= 𝟐𝒈 + 𝟐𝒂 + 𝒃

𝑡 = 3, 𝑆(3) =1

2𝑔(3)2 + 𝑎(3) + 𝑏

= 𝟒. 𝟓𝒈 + 𝟑𝒂 + 𝒃

Since distinct elements of 𝐴 have distinct image in 𝐵. Yes, it is an one-one function.

12. The function ′𝒕′ which maps temperature in Celsius (𝑪) into temperature in Fahrenheit (𝑭) is

defined by 𝒕(𝑪) = 𝑭 where 𝑭 =𝟗

𝟓𝑪 + 𝟑𝟐. Find

(i) 𝒕(𝟎) (ii) 𝒕(𝟐𝟖) (iii) 𝒕(−𝟏𝟎) (iv) the value of 𝑪 when 𝒕(𝑪) = 𝟐𝟏𝟐

(v) the temperature when the Celsius value is equal to the Fahrenheit value

The function 𝑡 is defined by, 𝑡(𝐶) = 𝐹, where 𝐹 =9

5𝐶 + 32

(i) 𝑡(0) =9

5(0) + 32 = 𝟑𝟐°𝑭

(ii) 𝑡(28) =9

5(28) + 32

= 9(5.6) + 32

= 50.4 + 32

= 𝟖𝟐. 𝟒°𝑭

(iii) 𝑡(−10) =9

5(−10) + 32

= −18 + 32

= 𝟏𝟒°𝑭

(iv) When 𝑡(𝐶) = 212

9

5𝐶 + 32 = 212

9

5𝐶 = 212 − 32 = 180

𝐶 =180×5

9= 𝟏𝟎𝟎°𝑪

(v) we know that

𝑡(𝐶) = 𝐹 where 𝐹 =9

5𝐶 + 32

𝑡(𝐹) = 𝐶 where 𝐶 =9

5𝐹 + 32

If the temperatures are same then two ′𝑡′s in the formula should represent the same temperature. So then we multiply

each side by (−5

4)

𝑡 =9

5𝑡 + 32°

𝑡 −9

5𝑡 = 32°

Multiply each side by (−5

4)

−5

4(𝑡 −

9

5𝑡) = 32° × (−

5

4)

−5

4𝑡 +

9

4𝑡 = −40°

−5𝑡+9𝑡

4= −40°

4𝑡

4= −40°

𝒕 = −𝟒𝟎°

PTA-1

PTA-3



1. 𝑹 = {(𝒙, −𝟐), (−𝟓, 𝒚)} represents the identity function, find the values of 𝒙 and 𝒚

𝑅 represents the identity function

∴ 𝑥 = −2, 𝑦 = −5

2. Let 𝑨 = {𝟏, 𝟐, 𝟑, 𝟒} , 𝑩 = ℕ. Let 𝒇: 𝑨 → 𝑩 be defined by 𝒇(𝒙) = 𝒙𝟐.

Find (i) the range of 𝒇 (ii) identify the type of function

Now 𝐴 = {1,2,3,4} , 𝐵 = {1,2,3, … } Given 𝑓: 𝐴 → 𝐵 and 𝑓(𝑥) = 𝑥3

𝑓(1) = 12 = 1, 𝑓(3) = 32 = 9 𝑓(2) = 22 = 4, 𝑓(4) = 42 = 16

(i) Range of 𝑓 = {1, 4, 9, 16} (ii) Since distinct elements in 𝐴 are mapped

into distinct images in 𝐵 , it is a one-one function. 2 ∈ 𝐵 is not the image of any element of 𝐴. So, it is Into function.

3. Using horizontal line test determine which of the following functions are one – one.

i) ii)

i)

The curve represent a one – one function as the

horizontal line meet the curve in only one point 𝑃.

ii)

The curve does not represent a one – one function,

since, the horizontal line meet the curve in two points 𝑃 and 𝑄.

4. Let 𝑨 = {𝟎, 𝟏, 𝟐, 𝟑} and 𝑩 = {𝟏, 𝟑, 𝟓, 𝟕, 𝟗} be two sets. Let 𝒇: 𝑨 → 𝑩 be a function. given by 𝒇(𝒙) = 𝟐𝒙 + 𝟏. Represent this function i) by arrow diagram ii) in a table form iii) as a set of ordered pairs iv) in a graphical form.

𝐴 = {0,1,2,3}

𝐵 = {1,3,5,7,9}

𝑓(𝑎) = 2𝑥 + 1

𝑓(0) = 2(0) + 1 = 0 + 1 = 1

𝑓(1) = 2(1) + 1 = 2 + 1 = 3

𝑓(2) = 2(2) + 1 = 4 + 1 = 5

𝑓(3) = 2(3) + 1 = 6 + 1 = 7

i) Arrow diagram: Let us represent the function 𝑓: 𝐴 → 𝐵 by an arrow diagram

ii) Table form: Let us represent 𝑓 using a table as shown below

𝑥 0 1 2 3 𝑓(𝑥) 1 3 5 7

iii) Set of ordered pairs: The given function 𝑓 can represented as a set of ordered pairs

𝑓 = {(0,1), (1,3), (2,5), (3,7)}

iv) Graphical form: In the adjacent 𝑥𝑦 – plane the points (0,1), (1,3), (2,5) , (3,7) are plotted.

Creative Questions

PTA-5

PTA-6



29

5. Let 𝑨 = {𝟏, 𝟐, 𝟑}, 𝑩 = {𝒂, 𝒃, 𝒄, 𝒅} and 𝒇 = {(𝟏, 𝒂) (𝟐, 𝒃) (𝟑, 𝒅)} be a function from 𝑨 to 𝑩. Show that 𝒇 is one – one but not onto function. 𝐴 = {1,2,3}, 𝐵 = {𝑎, 𝑏, 𝑐, 𝑑}, 𝑓 = {(1, 𝑎) (2, 𝑏) (3, 𝑑)}

Then 𝑓 is a function from 𝐴 → 𝐵 and for different elements in 𝐴 there are different images in 𝐵. Hence 𝑓 is one – one function. But the element 𝑐 is the Co-domain does not have any pre-image in the Domain. Hence 𝑓 is not onto. Therefore 𝑓 is one – one but not an onto function.

6. If 𝑨 = {𝟏, −𝟐, −𝟑, 𝟒. −𝟓} and 𝒇: 𝑨 → 𝑩 is onto function defined by 𝒇(𝒙) = 𝒙𝟐 + 𝒙 + 𝟏 then find 𝑩.

Given, 𝐴 = {1, −2, −3,4, −5} 𝑓(𝑥) = 𝑥2 + 𝑥 + 1 𝑓(1) = 12 + 1 + 1 = 3 𝑓(−2) = (−2)2 − 2 + 1 = 4 − 2 + 1 = 3 𝑓(−3) = (−3)2 − 3 + 1 = 9 − 3 + 1 = 7 𝑓(4) = (4)2 + 4 + 1 = 16 + 4 + 1 = 21 𝑓(−5) = (−5)2 − 5 + 1 = 25 − 5 + 1 = 21 Since, 𝑓 is an onto function, the range of 𝑓 = 𝐵 = co domain of 𝑓 𝐵 = {3,7,21}

7. Let 𝒇 be a function from 𝑹 to 𝑹 defined by

𝒇(𝒙) = 𝟑𝒙 − 𝟐. Find the value of 𝒂 and 𝒃 given that (𝒂, 𝟒) and (𝟏, 𝒃) belong to 𝒇. 𝑓(𝑥) = 3𝑥 − 2; 𝑓 = {(𝑥, 3𝑥 − 2)/ 𝑥 ∈ 𝑅} (𝑎, 4) means the image of 𝑎 is 4 That is, 𝑓(𝑎) = 4 3𝑎 − 2 = 4 3𝑎 = 4 + 2 = 6 𝑎 = 2. (1, 𝑏) means the image of 1 is 𝑏 That is, 𝑓(1) = 𝑏 3(1) − 2 = 𝑏 𝑏 = 1

8. Let 𝑨 = {𝟓, 𝟔, 𝟕, 𝟖}, 𝑩 = {−𝟏𝟏, 𝟒, 𝟕, −𝟏𝟎, −𝟗, −𝟕, −𝟏𝟑} and 𝒇 = {(𝒙, 𝒚)/𝒚 = 𝟑 − 𝟐𝒙, 𝒙 ∈ 𝑨, 𝒚 ∈ 𝑩}

i) write down the elements of 𝒇. ii) what is the Co-domain? iii) what is the range? iv) Identify the type of function? v) Find the pre-images of −𝟗 and −𝟏𝟑

The function 𝑓: 𝐴 → 𝐵 is defined by 𝑓(𝑥) = 3 − 2𝑥

i) if 𝑥 = 5, 𝑓(5) = 3 − 2(5) = −7 if 𝑥 = 6, 𝑓(6) = 3 − 2(6) = −9 if 𝑥 = 7, 𝑓(7) = 3 − 2(7) = −11 if 𝑥 = 8, 𝑓(8) = 3 − 2(8) = −13

element of 𝑓 = {(5, −7) (6, −9) (7, −11) (8, −13)} ii) Co domain = {−11,4,7, −10, −9, −7, −13} iii) Range of 𝑓 = {−7, −9, −11, −13} iv) Since different elements of 𝐴 have different

image in the Co – domains, the function 𝑓 is one – one function.

v) The pre-images of −9 and −13 are 6 and 8 respectively.

9. Forensic scientists can determine the height (in

inches) of a person based on the length of their forehand. They usually do so using the function 𝒉(𝒃) = 𝟎. 𝟗𝒃 + 𝟐𝟒. 𝟓 where 𝒃 is the length of the forehand. i) check if the function 𝒉 is one – one. ii) Also find the height of a person if the length of his forehand is 𝟓𝟎 𝒄𝒎𝒔. iii) Find the length of the forehand if the height of a person is 𝟕𝟓 inches. i) To check if ℎ is one – one. we assume that, ℎ(𝑏1) = ℎ(𝑏2) Then we get, 0.9𝑏1 + 24.5 = 0.9𝑏2 + 24.5 0.9𝑏1 = 0.9𝑏2 𝑏1 = 𝑏2 Thus ℎ(𝑏1) = ℎ(𝑏2) ⇒ 𝑏1 = 𝑏2 So the function ℎ is one – one. ii) If the length of his forehand 𝑏 = 50 𝑐𝑚, then

the height is ℎ(50) = (0.9 × 50) + 24.5 = 45 + 24.5 = 69.5 inches. iii) If the height of a person is 75 inches, then

ℎ(𝑏) = 75 and do the length of the forehand is given by, 0.9𝑏 + 24.5 = 75 0.9𝑏 = 75 − 24.5 = 50.5

𝑏 =50.5

0.9 = 56.11 𝑐𝑚𝑠



10. If the function 𝒇: 𝑹 → 𝑹 difined by 𝒇(𝒙) = {𝟏 + 𝒙, 𝒙 < −𝟐

𝟐𝒙 − 𝟏, − 𝟐 ≤ 𝒙 < 𝟒

𝟑𝒙𝟐 − 𝟏𝟎 𝒙 ≥ 𝟒

then find the value of,

i) 𝒇(𝟑) ii) 𝒇(−𝟒) iii) 𝒇(𝟑) + 𝟐𝒇(𝟏) iv) 𝟐𝒇(𝟑)−𝟑𝒇(−𝟐)

𝒇(𝟓)

The function 𝑓 is defined by three values in intervals I, II, III as shown below.

i) Now, see that 𝑥 = 3 lies in the second intervel

𝑓(𝑥) = 2𝑥 − 1

𝑓(3) = 2(3) − 1

= 6 − 1 = 5

ii) 𝑥 = −4 lies in the first interval

𝑓(𝑥) = 1 + 𝑥

= 1 + (−4) = −3.

iii) From (i) 𝑓(3) = 5

𝑥 = 1 lies in the second interval

𝑓(𝑥) = 2𝑥 − 1

𝑓(1) = 2(1) − 1 = 2 − 1 = 1

𝑓(𝑥) + 2𝑓(1) = 5 + 2(1) = 5 + 2 = 7

iv) we know 𝑓(3) = 5

To finding 𝑓(−2), 𝑥 = −2 lies in the second interval

𝑓(𝑥) = 2𝑥 − 1

𝑓(−2) = 2(−2) − 1

= −4 − 1

= −5

𝑥 = 5 lies in the third interval,

𝑓(𝑥) = 3𝑥2 − 10

= 3(5)2 − 10

= 75 − 10

= 65

Hence, 2𝑓(3)−3𝑓(−2)

𝑓(5)

=2(5)−3(−5)

𝑓(5)

=10+15

65

=25

65

=5

13

Definition: Let 𝑓: 𝐴 → 𝐵 and 𝑔: 𝐵 → 𝐶 be two functions. Then the composition of 𝑓 and 𝑔 denoted by 𝑔 ∘ 𝑓 is defined as the function 𝑔 ∘ 𝑓(𝑥) = 𝑔(𝑓(𝑥)) for all 𝑥 ∈ 𝐴. The composition 𝑔 ∘ 𝑓(𝑥) exists only when range of 𝑓 is a subset of 𝑔 𝑓 ∘ 𝑔 ≠ 𝑔 ∘ 𝑓 Composition of function is not commutative. Composition of three functions is always associative. That is 𝑓 ∘ (𝑔 ∘ ℎ) = (𝑓 ∘ 𝑔) ∘ ℎ. A function 𝑓: 𝑅 → 𝑅 defined by 𝑓(𝑥) = 𝑚𝑥 + 𝑐, 𝑚 ≠ 0 is called a linear function.

Some specific linear functions and their graphs are given below. No. Function Domain and Definition Graph 1 The identity function 𝑓: 𝑅 → 𝑅 defined by 𝑓(𝑥) = 𝑥

Concept corner




31

2 Additive inverse function 𝑓: 𝑅 → 𝑅 defined by 𝑓(𝑥) = −𝑥

A function 𝑓: 𝑅 → 𝑅 defined by 𝑓(𝑥) = 𝑎𝑥2 + 𝑏𝑥 + 𝑐 (𝑎 ≠ 0) is called a quadratic function. Function, Domain, Range and Definition Graph

𝑓: 𝑅 → 𝑅 defined by 𝑓(𝑥) = 𝑥2, 𝑥 ∈ 𝑅, 𝑓(𝑥) ∈ [0, ∞ )

𝑓: 𝑅 → 𝑅 defined by 𝑓(𝑥) = −𝑥2, 𝑥 ∈ 𝑅. 𝑓(𝑥) ∈ (−∞, 0]

A function 𝑓: 𝑅 → 𝑅 defined by 𝑓(𝑥) = 𝑎𝑥3 + 𝑏𝑥2 + 𝑐𝑥 + 𝑑 (𝑎 ≠ 0) is called a cubic function.

A function 𝑓: 𝑅 − {0} → 𝑅 defined by 𝑓(𝑥) =1

𝑥 is called a

reciprocal function.

A function 𝑓: 𝑅 → 𝑅 defined by 𝑓(𝑥) = 𝑐 for all 𝑥 ∈ 𝑅 is called a constant function.

Modulus or Absolute Valued Function: 𝑓: 𝑅 → [0, ∞) defined by

𝑓(𝑥) = |𝑥| = {𝑥 ; 𝑥 ≥ 0

−𝑥 ; 𝑥 < 0

Modulus function is not a linear function but it is composed of two linear functions 𝑥 and −𝑥



Pg.No 28 State your answer for the following questions by selectiong the correct option.

1. Composition of function is commutative (a) Always true (b) Never true (c) Sometimes true Ex: 𝑓(𝑥) = 3 + 𝑥 , 𝑔(𝑥) = 𝑥 − 4

𝑓 ∘ 𝑔 = 𝑓(𝑔(𝑥))

= 𝑓(𝑥 − 4) = 3 + 𝑥 − 4 = 𝑥 − 1

𝑔 ∘ 𝑓 = 𝑔(𝑓(𝑥))

= 𝑔(3 + 𝑥) = 3 + 𝑥 − 4 = 𝑥 − 1 Here , 𝑓 ∘ 𝑔 = 𝑔 ∘ 𝑓 2. Composition of function is assoctative (a) Always true (b) Never true (c) Sometimes true Composition of the function is always associative. That is 𝑓 ∘ (𝑔 ∘ ℎ) = (𝑓 ∘ 𝑔) ∘ ℎ

31 1. Is a constant function a linear function? Yes. If the constant function is constantly ‘o’, then it’s linear. 2. Is quadratic function a one – one function? No. 𝑓(𝑥) = 𝑥2 − 1 Here , -2 and 2 have same image 3. Is cubic function a one – one function? Yes, The cubic function is indeed a function as it passes the vertical line test. In

addition this function possesses the property that each 𝑥 value has one unique 𝑦 value. This characteristic is referred to as being a one – one function.

4. Is the reciprocal function a bijection? Yes, the reciprocal function is a bijection 5. If 𝑓: 𝐴 → 𝐵 is a constant function, then the range of f will have singleton (or) one

element.

Text Book Page Number: 26

If 𝑓(𝑥) = 𝑥𝑚 and 𝑔(𝑥) = 𝑥𝑛 does 𝑓 ∘ 𝑔 = 𝑔 ∘ 𝑓 ?

𝑓 ∘ 𝑔 = 𝑓(𝑔(𝑥)) = 𝑓(𝑥𝑛) = (𝑥𝑛)𝑚 = 𝑥𝑛𝑚

𝑔 ∘ 𝑓 = 𝑔(𝑓(𝑥)) = 𝑔(𝑥𝑚) = (𝑥𝑚)𝑛 = 𝑥𝑚𝑛

Hence, 𝑓 ∘ 𝑔 = 𝑔 ∘ 𝑓

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

𝑓 ∘ 𝑔 = 𝑓 ∘ 𝑔(𝑥)

= 𝑓(𝑔(𝑥))

= 𝑓(𝑥2)

= 𝑥2 − 6…………..(1)

𝑔 ∘ 𝑓(𝑥) = 𝑔(𝑓(𝑥))

= 𝑔(𝑥 − 6)

= (𝑥 − 6)2……………(2)

From (1) and (2) 𝒇 ∘ 𝒈 ≠ 𝒈 ∘ 𝒇

(ii) 𝒇(𝒙) =𝟐

𝒙 , 𝒈(𝒙) = 𝟐𝒙𝟐 − 𝟏

𝑓 ∘ 𝑔(𝑥) = 𝑓(𝑔(𝑥))

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