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Samacheer Kalvi 10th Maths Exercise 1.1 Solutions

10th Maths Exercise 1.1 solutions (Chapter 1 Relations and Functions) with step-by-step answers for all 12 questions, as per the 2026-27 Samacheer Kalvi textbook.

Answers at a glance4 answers · full steps below
  1. Q1 (ii)A×B=A×A=B×A={(p,p),(p,q),(q,p),(q,q)}A\times B=A\times A=B\times A=\{(p,p),(p,q),(q,p),(q,q)\}
  2. Q1 (iii)A×B=ϕA\times B=\phi; B×A=ϕB\times A=\phi; A×A={(m,m),(m,n),(n,m),(n,n)}A\times A=\{(m,m),(m,n),(n,m),(n,n)\}
  3. Q3A={3,4}A=\{3,4\} and B={−2,0,3}B=\{-2,0,3\}
  4. Q5Yes, (A∩C)×(B∩D)=(A×B)∩(C×D)={(3,3),(3,5)}(A\cap C)\times(B\cap D)=(A\times B)\cap(C\times D)=\{(3,3),(3,5)\}

Q1. Find A×BA \times B , A×AA \times A and B×AB \times A

Find A×BA \times B, A×AA \times A and B×AB \times A

(i)

A={2,−2,3}A=\{2,-2,3\} and B={1,−4}B=\{1,-4\}

Show steps (6)

  1. A×B={2,−2,3}×{1,−4}A\times B=\{2,-2,3\}\times\{1,-4\}

  2. A×B={(2,1),(2,−4),(−2,1),(−2,−4),(3,1),(3,−4)}A\times B=\{(2,1),(2,-4),(-2,1),(-2,-4),(3,1),(3,-4)\}

  3. A×A={2,−2,3}×{2,−2,3}A\times A=\{2,-2,3\}\times\{2,-2,3\}

  4. A×A={(2,2),(2,−2),(2,3),(−2,2),(−2,−2),(−2,3),(3,2),(3,−2),(3,3)}A\times A=\{(2,2),(2,-2),(2,3),(-2,2),(-2,-2),(-2,3),(3,2),(3,-2),(3,3)\}

  5. B×A={1,−4}×{2,−2,3}B\times A=\{1,-4\}\times\{2,-2,3\}

  6. B×A={(1,2),(1,−2),(1,3),(−4,2),(−4,−2),(−4,3)}B\times A=\{(1,2),(1,-2),(1,3),(-4,2),(-4,-2),(-4,3)\}

Answer

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

(ii)

A=B={p,q}A = B = \{p, q\}

  1. A=B={p,q}A=B=\{p,q\}

  2. A×B={p,q}×{p,q}={(p,p),(p,q),(q,p),(q,q)}A\times B=\{p,q\}\times\{p,q\}=\{(p,p),(p,q),(q,p),(q,q)\}

  3. Since A=BA=B, A×AA\times A and B×AB\times A are the same product.

  4. A×A={(p,p),(p,q),(q,p),(q,q)}A\times A=\{(p,p),(p,q),(q,p),(q,q)\}

  5. B×A={(p,p),(p,q),(q,p),(q,q)}B\times A=\{(p,p),(p,q),(q,p),(q,q)\}

Answer

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

(iii)

A={m,n}A = \{m, n\}; B=ϕB = \phi

  1. B=ϕB=\phi, so there is no second coordinate to form ordered pairs.

  2. A×B={m,n}×ϕ=ϕA\times B=\{m,n\}\times\phi=\phi

  3. B×A=ϕ×{m,n}=ϕB\times A=\phi\times\{m,n\}=\phi

  4. A×A={m,n}×{m,n}={(m,m),(m,n),(n,m),(n,n)}A\times A=\{m,n\}\times\{m,n\}=\{(m,m),(m,n),(n,m),(n,n)\}

Answer

A×B=ϕA\times B=\phi; B×A=ϕB\times A=\phi; A×A={(m,m),(m,n),(n,m),(n,n)}A\times A=\{(m,m),(m,n),(n,m),(n,n)\}

Q2. Let A={1,2,3}A = \{1,2,3\} and B={x∣xB = \{x \mid x is a prime number less than 10}\}. Find …

Let A={1,2,3}A = \{1,2,3\} and B={x∣xB = \{x \mid x is a prime number less than 10}\}. Find A×BA \times B and B×AB \times A.

  1. Prime numbers less than 10 are 2,3,5,72,3,5,7, so B={2,3,5,7}B=\{2,3,5,7\}.

  2. A×B={1,2,3}×{2,3,5,7}A\times B=\{1,2,3\}\times\{2,3,5,7\}

  3. 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)}A\times 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. B×A={2,3,5,7}×{1,2,3}B\times A=\{2,3,5,7\}\times\{1,2,3\}

  5. 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)}B\times 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)\}

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)}A\times 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)}B\times 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)\}

Q3. If B×A={(−2,3),(−2,4),(0,3),(0,4),(3,3),(3,4)}B \times A = \{(-2,3),(-2,4),(0,3),(0,4),(3,3),(3,4)\} find AA and BB .

  1. In B×AB\times A, the first coordinates come from BB and the second coordinates come from AA.

  2. First coordinates: −2,−2,0,0,3,3-2,-2,0,0,3,3, so B={−2,0,3}B=\{-2,0,3\}.

  3. Second coordinates: 3,4,3,4,3,43,4,3,4,3,4, so A={3,4}A=\{3,4\}.

Answer

A={3,4}A=\{3,4\} and B={−2,0,3}B=\{-2,0,3\}

Q4. If A={5,6}A = \{5,6\} , B={4,5,6}B = \{4,5,6\} , C={5,6,7}C = \{5,6,7\} , Show that A×A=(B×B)∩(C×C)A \times A = (B \times B) \cap (C \times C) .

Show steps (6)

  1. A×A={5,6}×{5,6}={(5,5),(5,6),(6,5),(6,6)}A\times A=\{5,6\}\times\{5,6\}=\{(5,5),(5,6),(6,5),(6,6)\} ...(1)

  2. B×B={(4,4),(4,5),(4,6),(5,4),(5,5),(5,6),(6,4),(6,5),(6,6)}B\times B=\{(4,4),(4,5),(4,6),(5,4),(5,5),(5,6),(6,4),(6,5),(6,6)\}

  3. C×C={(5,5),(5,6),(5,7),(6,5),(6,6),(6,7),(7,5),(7,6),(7,7)}C\times C=\{(5,5),(5,6),(5,7),(6,5),(6,6),(6,7),(7,5),(7,6),(7,7)\}

  4. Common pairs of B×BB\times B and C×CC\times C:

  5. (B×B)∩(C×C)={(5,5),(5,6),(6,5),(6,6)}(B\times B)\cap(C\times C)=\{(5,5),(5,6),(6,5),(6,6)\} ...(2)

  6. From (1) and (2) both sets are the same.

Answer

A×A=(B×B)∩(C×C)={(5,5),(5,6),(6,5),(6,6)}A\times A=(B\times B)\cap(C\times C)=\{(5,5),(5,6),(6,5),(6,6)\}, hence shown.

Q5. Given A={1,2,3}\{1,2,3\}, B={2,3,5}B = \{2,3,5\} , C={3,4}C = \{3,4\} and D={1,3,5}D = \{1,3,5\} , check if (A∩C)×(B∩D)=(A×B)∩(C×D)(A \cap C) \times (B \cap D) = (A \times B) \cap (C \times D) is …

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

Show steps (7)

  1. A∩C={1,2,3}∩{3,4}={3}A\cap C=\{1,2,3\}\cap\{3,4\}=\{3\}

  2. B∩D={2,3,5}∩{1,3,5}={3,5}B\cap D=\{2,3,5\}\cap\{1,3,5\}=\{3,5\}

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

  4. A×B={(1,2),(1,3),(1,5),(2,2),(2,3),(2,5),(3,2),(3,3),(3,5)}A\times B=\{(1,2),(1,3),(1,5),(2,2),(2,3),(2,5),(3,2),(3,3),(3,5)\}

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

  6. (A×B)∩(C×D)={(3,3),(3,5)}(A\times B)\cap(C\times D)=\{(3,3),(3,5)\} ...(2)

  7. (1) and (2) are equal.

Answer

Yes, (A∩C)×(B∩D)=(A×B)∩(C×D)={(3,3),(3,5)}(A\cap C)\times(B\cap D)=(A\times B)\cap(C\times D)=\{(3,3),(3,5)\}

Q6. Let A={x∈W∣x<2}A = \{x \in W \mid x < 2\} , B={x∈N∣1<x≤4}B = \{x \in N \mid 1 < x \le 4\} and C={3,5}C = \{3,5\} . Verify that

Let A={x∈W∣x<2}A = \{x \in W \mid x < 2\}, B={x∈N∣1<x≤4}B = \{x \in N \mid 1 < x \le 4\} and C={3,5}C = \{3,5\}. Verify that

(i)

A×(B∪C)=(A×B)∪(A×C)A \times (B \cup C) = (A \times B) \cup (A \times C)

Show steps (8)

  1. A={x∈W∣x<2}={0,1}A=\{x\in W\mid x<2\}=\{0,1\}, B={x∈N∣1<x≤4}={2,3,4}B=\{x\in N\mid 1<x\le4\}=\{2,3,4\}, C={3,5}C=\{3,5\}

  2. B∪C={2,3,4,5}B\cup C=\{2,3,4,5\}

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

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

  5. A×B={(0,2),(0,3),(0,4),(1,2),(1,3),(1,4)}A\times B=\{(0,2),(0,3),(0,4),(1,2),(1,3),(1,4)\}

  6. A×C={(0,3),(0,5),(1,3),(1,5)}A\times C=\{(0,3),(0,5),(1,3),(1,5)\}

  7. (A×B)∪(A×C)={(0,2),(0,3),(0,4),(0,5),(1,2),(1,3),(1,4),(1,5)}(A\times B)\cup(A\times C)=\{(0,2),(0,3),(0,4),(0,5),(1,2),(1,3),(1,4),(1,5)\} ...(2)

  8. (1) = (2).

Answer

A×(B∪C)=(A×B)∪(A×C)A\times(B\cup C)=(A\times B)\cup(A\times C) is verified.

(ii)

A×(B∩C)=(A×B)∩(A×C)A \times (B \cap C) = (A \times B) \cap (A \times C)

Show steps (7)

  1. A={0,1}A=\{0,1\}, B={2,3,4}B=\{2,3,4\}, C={3,5}C=\{3,5\}

  2. B∩C={3}B\cap C=\{3\}

  3. A×(B∩C)={0,1}×{3}={(0,3),(1,3)}A\times(B\cap C)=\{0,1\}\times\{3\}=\{(0,3),(1,3)\} ...(1)

  4. A×B={(0,2),(0,3),(0,4),(1,2),(1,3),(1,4)}A\times B=\{(0,2),(0,3),(0,4),(1,2),(1,3),(1,4)\}

  5. A×C={(0,3),(0,5),(1,3),(1,5)}A\times C=\{(0,3),(0,5),(1,3),(1,5)\}

  6. (A×B)∩(A×C)={(0,3),(1,3)}(A\times B)\cap(A\times C)=\{(0,3),(1,3)\} ...(2)

  7. (1) = (2).

Answer

A×(B∩C)=(A×B)∩(A×C)={(0,3),(1,3)}A\times(B\cap C)=(A\times B)\cap(A\times C)=\{(0,3),(1,3)\} is verified.

(iii)

(A∪B)×C=(A×C)∪(B×C)(A \cup B) \times C = (A \times C) \cup (B \times C)

Show steps (8)

  1. A={0,1}A=\{0,1\}, B={2,3,4}B=\{2,3,4\}, C={3,5}C=\{3,5\}

  2. A∪B={0,1,2,3,4}A\cup B=\{0,1,2,3,4\}

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

  4. ={(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)\} ...(1)

  5. A×C={(0,3),(0,5),(1,3),(1,5)}A\times C=\{(0,3),(0,5),(1,3),(1,5)\}

  6. B×C={(2,3),(2,5),(3,3),(3,5),(4,3),(4,5)}B\times C=\{(2,3),(2,5),(3,3),(3,5),(4,3),(4,5)\}

  7. (A×C)∪(B×C)={(0,3),(0,5),(1,3),(1,5),(2,3),(2,5),(3,3),(3,5),(4,3),(4,5)}(A\times C)\cup(B\times C)=\{(0,3),(0,5),(1,3),(1,5),(2,3),(2,5),(3,3),(3,5),(4,3),(4,5)\} ...(2)

  8. (1) = (2).

Answer

(A∪B)×C=(A×C)∪(B×C)(A\cup B)\times C=(A\times C)\cup(B\times C) is verified.

Q7. Let AA = The set of all natural numbers less than 8 …

Let AA = The set of all natural numbers less than 8, BB = The set of all prime numbers less than 8, CC = The set of even prime number. Verify that

(i)

(A∩B)×C=(A×C)∩(B×C)(A \cap B) \times C = (A \times C) \cap (B \times C)

Show steps (7)

  1. A={1,2,3,4,5,6,7}A=\{1,2,3,4,5,6,7\}, B={2,3,5,7}B=\{2,3,5,7\}, C={2}C=\{2\}

  2. A∩B={2,3,5,7}A\cap B=\{2,3,5,7\}

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

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

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

  6. (A×C)∩(B×C)={(2,2),(3,2),(5,2),(7,2)}(A\times C)\cap(B\times C)=\{(2,2),(3,2),(5,2),(7,2)\} ...(2)

  7. (1) = (2).

Answer

(A∩B)×C=(A×C)∩(B×C)={(2,2),(3,2),(5,2),(7,2)}(A\cap B)\times C=(A\times C)\cap(B\times C)=\{(2,2),(3,2),(5,2),(7,2)\} is verified.

(ii)

A×(B−C)=(A×B)−(A×C)A \times (B - C) = (A \times B) - (A \times C)

Show steps (9)

  1. A={1,2,3,4,5,6,7}A=\{1,2,3,4,5,6,7\}, B={2,3,5,7}B=\{2,3,5,7\}, C={2}C=\{2\}

  2. B−C={3,5,7}B-C=\{3,5,7\}

  3. A×(B−C)={1,2,3,4,5,6,7}×{3,5,7}A\times(B-C)=\{1,2,3,4,5,6,7\}\times\{3,5,7\}

  4. ={(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)}=\{(1,3),(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)

  5. 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\times 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)\}

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

  7. Removing the pairs of A×CA\times C from A×BA\times B:

  8. (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)}(A\times B)-(A\times 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)

  9. (1) = (2).

Answer

A×(B−C)=(A×B)−(A×C)A\times(B-C)=(A\times B)-(A\times C) is verified.

Quick check

  1. Let ff and gg be two functions given by f={(0,1),(2,0),(3,−4),(4,2),(5,7)}f = \{(0,1),(2,0),(3,-4),(4,2),(5,7)\}, g={(0,2),(1,0),(2,4),(−4,2),(7,0)}g = \{(0,2),(1,0),(2,4),(-4,2),(7,0)\} then the range of f∘gf \circ g is

    1. (1) {0,2,3,4,5}\{0,2,3,4,5\}
    2. (2) {−4,1,0,2,7}\{-4,1,0,2,7\}
    3. (3) {1,2,3,4,5}\{1,2,3,4,5\}
    4. (4) {0,1,2}\{0,1,2\}
    Show answer

    Answer: (4) {0,1,2}\{0,1,2\}

    1. (f∘g)(0)=f(2)=0(f\circ g)(0)=f(2)=0, (f∘g)(1)=f(0)=1(f\circ g)(1)=f(0)=1.

    2. (f∘g)(2)=f(4)=2(f\circ g)(2)=f(4)=2, (f∘g)(−4)=f(2)=0(f\circ g)(-4)=f(2)=0, (f∘g)(7)=f(0)=1(f\circ g)(7)=f(0)=1.

    3. Range ={0,1,2}=\{0,1,2\}.

    Question 1 of 5

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

    1. (1) 3
    2. (2) 2
    3. (3) 4
    4. (4) 8
    Show answer

    Answer: (2) 2

    1. Number of relations from AA to BB is 2n(A)⋅n(B)2^{n(A)\cdot n(B)}.

    2. 25 n(B)=1024=2102^{5\,n(B)}=1024=2^{10}.

    3. 5 n(B)=10⇒n(B)=25\,n(B)=10\Rightarrow n(B)=2.

    Question 2 of 5

  3. Let A={1,2,3,4}A = \{1,2,3,4\} and B={4,8,9,10}B = \{4,8,9,10\}. A function f:A→Bf : A \to B given by f={(1,4),(2,8),(3,9),(4,10)}f = \{(1,4),(2,8),(3,9),(4,10)\} is a

    1. (1) Many-one function
    2. (2) Identity function
    3. (3) One-to-one function
    4. (4) Into function
    Show answer

    Answer: (3) One-to-one function

    1. Images are 4,8,9,104,8,9,10 — all different.

    2. Different elements of AA have different images.

    3. So ff is one-to-one.

    Question 3 of 5

  4. If f(x)=2x2f(x) = 2x^2 and g(x)=13xg(x) = \dfrac{1}{3x}, then f∘gf \circ g is

    1. (1) 32x2\dfrac{3}{2x^2}
    2. (2) 23x2\dfrac{2}{3x^2}
    3. (3) 29x2\dfrac{2}{9x^2}
    4. (4) 16x2\dfrac{1}{6x^2}
    Show answer

    Answer: (3) 29x2\dfrac{2}{9x^2}

    1. (f∘g)(x)=f(g(x))=f(13x)(f\circ g)(x)=f\big(g(x)\big)=f\left(\dfrac{1}{3x}\right).

    2. =2(13x)2=2\left(\dfrac{1}{3x}\right)^{2}.

    3. =29x2=\dfrac{2}{9x^{2}}.

    Question 4 of 5

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

    1. (1) 8
    2. (2) 20
    3. (3) 12
    4. (4) 16
    Show answer

    Answer: (3) 12

    1. A∪C={a,b,p,q,r,s}A\cup C=\{a,b,p,q,r,s\}, so n(A∪C)=6n(A\cup C)=6.

    2. n(B)=2n(B)=2.

    3. n[(A∪C)×B]=6×2=12n[(A\cup C)\times B]=6\times 2=12.

    Question 5 of 5
Chapter 1 formulas
  • Cartesian product of A with B

    A×B={(a,b)∣for all a∈A, b∈B}A \times B = \{(a,b) \mid \text{for all } a \in A,\ b \in B\}

  • A relation R from A to B is a subset of A × B

    R⊆A×BR \subseteq A \times B

  • Condition for a relation R from X to Y to be a function

    for every x∈X there exists only one y∈Y\text{for every } x \in X \text{ there exists only one } y \in Y

  • Identity function

    f(x)=xf(x) = x

  • Reciprocal function

    f(x)=1xf(x) = \dfrac{1}{x}

  • Constant function

    f(x)=cf(x) = c

  • Linear function

    f(x)=ax+b, a≠0f(x) = ax + b,\ a \neq 0

  • Quadratic function

    f(x)=ax2+bx+c, a≠0f(x) = ax^{2} + bx + c,\ a \neq 0

  • Cubic function

    f(x)=ax3+bx2+cx+d, a≠0f(x) = ax^{3} + bx^{2} + cx + d,\ a \neq 0

  • Composition of two functions f : A → B and g : B → C

    g∘f:A→C,(g∘f)(x)=g(f(x))  for all x∈Ag \circ f : A \to C,\quad (g \circ f)(x) = g(f(x)) \ \text{ for all } x \in A

  • Composition is not commutative in general

    f∘g≠g∘ff \circ g \neq g \circ f

  • Composition is associative

    f∘(g∘h)=(f∘g)∘hf \circ (g \circ h) = (f \circ g) \circ h

Answers checked against the Samacheer Kalvi textbook · Updated

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