Question

Let f be a functions f : N to N be defined by f(x) = 3x +2,x belongs to N .
(i) Find the images of 1, 2, 3 (ii) Find the pre-images of 29, 53
(ii) Identify the type of function​

Answer 1

Step-by-step explanation:

(i) f(1)=3*1+2=5

f(2)=8

f(3)=11

(ii) 29=3x+2

27=3x

x=27/3

x=9

53=3x+2

51=3x

x=51/3

x=17

one to one function

Answer 2

Given:

f : N→N

f(x) = 3x + 2, xξN

Find:

(i) The images of 1, 2 and 3.

(ii) The pre-images of 29 and 53.

(iii) Type of function.

Solution:

f(x) = 3x + 2

(i) For finding the images we will put x = 1, 2 and 3

Putting x = 1, we have

f(1) = 3(1) + 2

     = 3 + 2 = 5

∴ Image of 1 is 5

Putting x = 2, we have

f(1) = 3(2) + 2

     = 6 + 2 = 8

∴ Image of 2 is 8

Putting x = 3, we have

f(1) = 3(3) + 2

     = 9 + 2 = 11

∴ Image of 3 is 11

Hence, the images of 1, 2 and 3 are 5, 8 and 11 respectively.

(ii) For finding the pre-images of 29 and 53, we will put f(x) = 29 and 53.

Putting f(x) = 29, we have

29 = 3x + 2

29 - 2 = 3x

27 = 3x

27/3 = x

x = 9

∴ The pre-image of 29 is 9.

Putting f(x) = 53, we have

53 = 3x + 2

53 - 2 = 3x

51 = 3x

51/3 = x

x = 17

∴ The pre-image of 53 is 17.

Hence, the pre-images of 29 and 53 are 9 and 17 respectively.

(iii) Let a and b belongs to N for which

f(a) = f(b)

3a+2 = 3b+2

3a = 3b

a = b

∴ It is a one-one function.

Now, let f(x) = y = 3x+2

3x = y - 2

x = (y-2)/3

By putting y = 2, we get

x = (2-2)/3 = 0

x = 0 but this does not belong to N that means the set Co-domain is not fully used up.

∴ It is also an into function.

Hence, f(x) is one-one into function.

#SPJ3