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Samacheer Kalvi 10th Maths Exercise 1.6 Solutions

10th Maths Exercise 1.6 solutions (Chapter 1 Relations and Functions) with step-by-step answers for all 15 questions, as per the 2026-27 Samacheer Kalvi textbook.

Answers at a glance15 answers · full steps below
  1. Q1(C) 3
  2. Q2(C) 12
  3. Q3(A) (A×C)⊂(B×D)(A \times C) \subset (B \times D)
  4. Q4(B) 2
  5. Q5(C) {4,9,25,49,121}\{4,9,25,49,121\}
  6. Q6(D) (3,−2)(3, -2)
  7. Q7(C) 2mn−12^{mn} - 1
  8. Q8(A) (8,6)(8,6)
  9. Q9(C) One-to-one function
  10. Q10(C) 29x2\dfrac{2}{9x^2}
  11. Q11(A) 7
  12. Q12(D) {0,1,2}\{0,1,2\}
  13. Q13(C) f(xy)≤f(x).f(y)f(xy) \leq f(x).f(y)
  14. Q14(B) (2,−1)(2, -1)
  15. Q15(D) quadratic

Q1. If n(A×B)=6n(A \times B) = 6 and A={1,3}A = \{1,3\} then n(B)n(B) is

  1. (A) 1
  2. (B) 2
  3. (C) 3
  4. (D) 6

  1. n(A×B)=n(A)×n(B)n(A\times B)=n(A)\times n(B).

  2. 6=2×n(B)6=2\times n(B).

  3. n(B)=3n(B)=3.

Answer

(C) 3

Q2. 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. (A) 8
  2. (B) 20
  3. (C) 12
  4. (D) 16

  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.

Answer

(C) 12

Q3. If A={1,2}A = \{1,2\} , B={1,2,3,4}B = \{1,2,3,4\} , C={5,6}C = \{5,6\} and D={5,6,7,8}D = \{5, 6, 7, 8\} then state which of …

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

  1. (A) (A×C)⊂(B×D)(A \times C) \subset (B \times D)
  2. (B) (B×D)⊂(A×C)(B \times D) \subset (A \times C)
  3. (C) (A×B)⊂(A×D)(A \times B) \subset (A \times D)
  4. (D) (D×A)⊂(B×A)(D \times A) \subset (B \times A)

  1. A={1,2}⊂B={1,2,3,4}A=\{1,2\}\subset B=\{1,2,3,4\} and C={5,6}⊂D={5,6,7,8}C=\{5,6\}\subset D=\{5,6,7,8\}.

  2. So every pair of A×CA\times C is also a pair of B×DB\times D.

  3. Hence (A×C)⊂(B×D)(A\times C)\subset (B\times D).

Answer

(A) (A×C)⊂(B×D)(A \times C) \subset (B \times D)

Q4. If there are 1024 relations from a set A={1,2,3,4,5}A = \{1, 2, 3, 4, 5\} to a set …

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. (A) 3
  2. (B) 2
  3. (C) 4
  4. (D) 8

  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.

Answer

(B) 2

Q5. The range of the relation R={(x,x2)∣xR = \{(x, x^2) \mid x is a prime number less than …

The range of the relation R={(x,x2)∣xR = \{(x, x^2) \mid x is a prime number less than 13}\} is

  1. (A) {2,3,5,7}\{2,3,5,7\}
  2. (B) {2,3,5,7,11}\{2,3,5,7,11\}
  3. (C) {4,9,25,49,121}\{4,9,25,49,121\}
  4. (D) {1,4,9,25,49,121}\{1,4,9,25,49,121\}

  1. Primes less than 13 are 2,3,5,7,112,3,5,7,11.

  2. Range consists of the squares x2x^{2}.

  3. Range ={4,9,25,49,121}=\{4,9,25,49,121\}.

Answer

(C) {4,9,25,49,121}\{4,9,25,49,121\}

Q6. If the ordered pairs (a+2,4)(a + 2, 4) and (5,2a+b)(5, 2a + b) are equal then (a,b)(a,b) is

  1. (A) (2,−2)(2, -2)
  2. (B) (5,1)(5,1)
  3. (C) (2,3)(2,3)
  4. (D) (3,−2)(3, -2)

  1. Equal ordered pairs: a+2=5a+2=5 and 2a+b=42a+b=4.

  2. a=3a=3.

  3. 2(3)+b=4⇒b=−22(3)+b=4\Rightarrow b=-2.

Answer

(D) (3,−2)(3, -2)

Q7. Let n(A)=mn(A) = m and n(B)=nn(B) = n then the total number of non-empty relations that …

Let n(A)=mn(A) = m and n(B)=nn(B) = n then the total number of non-empty relations that can be defined from AA to BB is

  1. (A) mnm^n
  2. (B) nmn^m
  3. (C) 2mn−12^{mn} - 1
  4. (D) 2mn2^{mn}

  1. Total number of relations from AA to BB is 2mn2^{mn}.

  2. Remove the empty relation.

  3. Non-empty relations =2mn−1=2^{mn}-1.

Answer

(C) 2mn−12^{mn} - 1

Q8. If {(a,8),(6,b)}\{(a,8),(6,b)\} represents an identity function, then the value of aa and …

If {(a,8),(6,b)}\{(a,8),(6,b)\} represents an identity function, then the value of aa and bb are respectively

  1. (A) (8,6)(8,6)
  2. (B) (8,8)(8,8)
  3. (C) (6,8)(6,8)
  4. (D) (6,6)(6,6)

  1. For an identity function f(x)=xf(x)=x, each pair is of the form (x,x)(x,x).

  2. (a,8)⇒a=8(a,8)\Rightarrow a=8.

  3. (6,b)⇒b=6(6,b)\Rightarrow b=6.

Answer

(A) (8,6)(8,6)

Q9. 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 …

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. (A) Many-one function
  2. (B) Identity function
  3. (C) One-to-one function
  4. (D) Into 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.

Answer

(C) One-to-one function

Q10. If f(x)=2x2f(x) = 2x^2 and g(x)=13xg(x) = \dfrac{1}{3x} , then f∘gf \circ g is

  1. (A) 32x2\dfrac{3}{2x^2}
  2. (B) 23x2\dfrac{2}{3x^2}
  3. (C) 29x2\dfrac{2}{9x^2}
  4. (D) 16x2\dfrac{1}{6x^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}}.

Answer

(C) 29x2\dfrac{2}{9x^2}

Q11. If f:A→Bf : A \to B is a bijective function and if n(B)=7n(B) = 7 , then n(A)n(A) …

If f:A→Bf : A \to B is a bijective function and if n(B)=7n(B) = 7, then n(A)n(A) is equal to

  1. (A) 7
  2. (B) 49
  3. (C) 1
  4. (D) 14

  1. A bijection pairs each element of AA with exactly one element of BB.

  2. So n(A)=n(B)n(A)=n(B).

  3. n(A)=7n(A)=7.

Answer

(A) 7

Q12. 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)\} …

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. (A) {0,2,3,4,5}\{0,2,3,4,5\}
  2. (B) {−4,1,0,2,7}\{-4,1,0,2,7\}
  3. (C) {1,2,3,4,5}\{1,2,3,4,5\}
  4. (D) {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\}.

Answer

(D) {0,1,2}\{0,1,2\}

Q13. Let f(x)=1+x2f(x) = \sqrt{1 + x^2} then

  1. (A) f(xy)=f(x).f(y)f(xy) = f(x).f(y)
  2. (B) f(xy)≥f(x).f(y)f(xy) \geq f(x).f(y)
  3. (C) f(xy)≤f(x).f(y)f(xy) \leq f(x).f(y)
  4. (D) None of these

  1. f(xy)=1+x2y2f(xy)=\sqrt{1+x^{2}y^{2}}.

  2. f(x)⋅f(y)=(1+x2)(1+y2)=1+x2+y2+x2y2f(x)\cdot f(y)=\sqrt{(1+x^{2})(1+y^{2})}=\sqrt{1+x^{2}+y^{2}+x^{2}y^{2}}.

  3. Since x2+y2≥0x^{2}+y^{2}\ge 0,   1+x2y2≤1+x2+y2+x2y2\;1+x^{2}y^{2}\le 1+x^{2}+y^{2}+x^{2}y^{2}.

  4. Hence f(xy)≤f(x)⋅f(y)f(xy)\le f(x)\cdot f(y).

Answer

(C) f(xy)≤f(x).f(y)f(xy) \leq f(x).f(y)

Q14. If g={(1,1),(2,3),(3,5),(4,7)}g = \{(1,1),(2,3),(3,5),(4,7)\} is a function given by g(x)=αx+βg(x) = \alpha x + \beta then the values of …

If g={(1,1),(2,3),(3,5),(4,7)}g = \{(1,1),(2,3),(3,5),(4,7)\} is a function given by g(x)=αx+βg(x) = \alpha x + \beta then the values of α\alpha and β\beta are

  1. (A) (−1,2)(-1,2)
  2. (B) (2,−1)(2, -1)
  3. (C) (−1,−2)(-1, -2)
  4. (D) (1,2)(1,2)

  1. g(1)=1⇒α+β=1g(1)=1\Rightarrow \alpha+\beta=1.

  2. g(2)=3⇒2α+β=3g(2)=3\Rightarrow 2\alpha+\beta=3.

  3. Subtracting, α=2\alpha=2, then β=−1\beta=-1.

Answer

(B) (2,−1)(2, -1)

Q15. f(x)=(x+1)3−(x−1)3f(x) = (x + 1)^3 - (x - 1)^3 represents a function which is

  1. (A) linear
  2. (B) cubic
  3. (C) reciprocal
  4. (D) quadratic

  1. (x+1)3=x3+3x2+3x+1(x+1)^{3}=x^{3}+3x^{2}+3x+1, (x−1)3=x3−3x2+3x−1(x-1)^{3}=x^{3}-3x^{2}+3x-1.

  2. Subtracting, f(x)=6x2+2f(x)=6x^{2}+2.

  3. Highest power is 22, so it is quadratic.

Answer

(D) quadratic

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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