# 10th Maths Chapter 1 Exercise 1.6 Guide

10th Standard Maths Chapter 1 Exercise 1.6 Relations and Functions Guide Book Back Answers Solutions. TN 10th SSLC Samacheer Kalvi Guide. 10th All Subject Guide – Click Here. Class 1 to 12 All Subject Book Back Answers – Click Here.

## 10th Maths – Chapter 1 Relations and Functions – Exercise 1.6 Guide

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

(1) 1

(2) 2

(3) 3

(4) 6

**Solu.:**

(3) 3

Hint:

If n(A × B) = 6

A = {1, 1}, n(A) = 2

n(B) = 3

### 2.A = {a, b,p}, B = {2, 3}, C = {p, q, r, s)

then n[(A ∪ C) × B] is ………….

(1) 8

(2) 20

(3) 12

(4) 16

**Solu.:**

(3) 12

Hint: A ∪ C = [a, b, p] ∪ [p, q, r, s]

= [a, b, p, q, r, s]

n (A ∪ C) = 6

n(B) = 2

∴ n [(A ∪ C)] × 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.

(1) (A × C) ⊂ (B × D)

(2) (B × D) ⊂ (A × C)

(3) (A × B) ⊂ (A × D)

(4) (D × A) ⊂ (B × A)

**Solu.:**

(1) (A × C) ⊂ (B × D)]

Hint:

A = {1, 2}, B = {1, 2, 3, 4},

C = {5, 6}, D ={5, 6, 7, 8}

A × C ={(1,5), (1,6), (2, 5), (2, 6)}

B × D = {(1, 5),(1, 6),(1, 7),(1, 8),(2, 5),(2, 6), (2, 7), (2, 8), (3, 5), (3, 6), (3, 7), (3, 8)}

∴ (A × C) ⊂ B × D it is true

### 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 ………………….

(1) 3

(2) 2

(3) 4

(4) 8

**Solu.:**

(2) 2

Hint: n(A) = 5

n(A × B) = 10

(consider 1024 as 10)

n(A) × n(B) = 10

5 × n(B) = 10

n(B) = 105 = 2

n(B) = 2

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

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

(2) {2, 3, 5, 7, 11}

(3) {4, 9, 25, 49, 121}

(4) {1, 4, 9, 25, 49, 121}

**Solu.:**

(3) {4, 9, 25, 49, 121}]

Hint:

R = {(x, x2)/x is a prime number < 13}

The squares of 2, 3, 5, 7, 11 are

{4, 9, 25, 49, 121}

### 6. If the ordered pairs (a + 2,4) and (5, 2a + 6) are equal then (a, b) is ………

(1) (2, -2)

(2) (5, 1)

(3) (2, 3)

(4) (3, -2)

**Solu.:**

(4) (3, -2)

Hint:

The value of a = 3 and b = -2

### 7. Let n(A) = m and n(B) = n then the total number of non-empty relations that can be defined from A to B is

(1) mn

(2) nm

(3) 2mn – 1

(4) 2mn

**Solu.:**

(4) 2mn

Hint:

n(A) = m, n(B) = n

n(A × B) = 2mn

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

(1) (8,6)

(2) (8,8)

(3) (6,8)

(4) (6,6)

**Solu.:**

(1) (8,6)

Hint: f = {{a, 8) (6, 6)}. In an identity function each one is the image of it self.

∴ a = 8, b = 6

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

(1) Many-one function

(2) Identity function

(3) One-to-one function

(4) Into function

**Solu.:**

(3) One-to one function

Hint:

A = {1, 2, 3, 4), B = {4, 8, 9,10}

### 10. If f(x) = 2×2 and g (x) = 13x, Then fog is

**Solu.:**

(3) 29×2

Hint:

f(x) = 2×2

g(x) = 13x

fog = f(g(x)) = f(13x)=2(13x)2

= 2 × 19×2=29×2

### 11. If f: A → B is a bijective function and if n(B) = 7, then n(A) is equal to …………..

(1) 7

(2) 49

(3) 1

(4) 14

**Solu.:**

(1) 7

Hint:

n(B) = 7

Since it is a bijective function, the function is one – one and also it is onto.

n(A) = n(B)

∴ n(A) = 7

### 12. Let f and g be two functions given by f = {(0, 1), (2, 0), (3, -4), (4, 2), (5, 7)} g = {(0, 2), (1, 0), (2, 4), (-4, 2), (7, 0)} then the range of fog is

(1) {0, 2, 3, 4, 5}

(2) {-4, 1, 0, 2, 7}

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

(4) {0, 1, 2}

**Solu.:**

(4) {0, 1, 2}

Hint:

gof = g(f(x))

fog = f(g(x))

= {(0, 2),(1, 0),(2, 4),(-4, 2),(7, 0)}

Range of fog = {0, 1, 2}

### 13. Let f (x) = 1+x2−−−−−√ then ………………..

(1) f(xy) = f(x) f(y)

(2) f(xy) __>__ f(x).f(y)

(3) f(xy) __<__ f(x). f(y)

(4) None of these

**Solu.:**

(3) f(xy) __<__ f(x) . f(y)

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

(1) (-1, 2)

(2) (2, -1)

(3) (-1, -2)

(4) (1, 2)

**Solu.:**

(2) (2,-1)

Hint:

g(x) = αx + β

α = 2

β = -1

g(x) = 2x – 1

g(1) = 2(1) – 1 = 1

g(2) = 2(2) – 1 = 3

g(3) = 2(3) – 1 = 5

g(4) = 2(4) – 1 = 7

### 15. f(x) = (x + 1)3 – (x – 1)3 represents a function which is …………….

(1) linear

(2) cubic

(3) reciprocal

(4) quadratic

**Solu.:**

(4) quadratic

Hint: f(x) = (x + 1)3 – (x – 1)3

[using a3 – b3 = (a – b)3 + 3 ab (a – b)]

= (x + 1 – x + 1)3 + 3(x + 1) (x – 1)

(x + 1 – x + 1)

= 8 + 3 (x2 – 1)2

= 8 + 6 (x2 – 1)

= 8 + 6×2 – 6

= 6×2 + 2

It is quadratic polynomial