HOW TO DETERMINE WHICH TYPE OF FUNCTION IS GIVEN

Question 1 :

Show that the function f : N -> N defined by f (x) = 2x – 1 is one-one but not onto

Solution :

If for all a₁, a₂ ∈ A, f(a₁) = f(a₂) implies a₁ = a₂ then f is called one – one function.

Let x, y ∈ N, f(x)  =  f(y)

f(x)  =  2x - 1  -----(1)

f(y)  =  2y - 1  -----(2)

(1)  =  (2)

2x - 1  =  2y - 1 

2x  =  2y

x  =  y

Hence the function is one to one.

It is not onto :

If co-domain of the function = range of function, then the function is said to be onto.

Even numbers in the co-domain are not associated with the elements of domain. Hence it is not onto.

Question 2 :

Show that the function f : N -> N defined by f (m) = m² + m + 3 is one-one function.

Solution :

Let x, y ∈ N, f(x)  =  f(y)

f (m) = m² + m + 3

f(x)  =  x² + x + 3  -----(1)

f(y)  =  y² + y + 3  -----(2)

(1)  =  (2)

x² + x + 3  =  y² + y + 3

x² + x  =  y² + y

x² - y² + x - y = 0

(x + y) (x - y) + (x - y)  =  0

(x - y) (x + y + 1)  =  0

x - y  =  0 

x  =  y

Hence it is one to one function.

Question 3 :

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

Solution :

Given that :

f (x) = x³ 

Range of f  =  {1, 8, 27, 64}

Every element in A has associated with different elements of B. Hence it is one to one.

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