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 a_{1}, a_{2} ∈ A, f(a_{1}) = f(a_{2}) implies a_{1 }= a_{2 }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) = m^{2} + m + 3 is one-one function.
Solution :
Let x, y ∈ N, f(x) = f(y)
f (m) = m^{2} + m + 3
f(x) = x^{2} + x + 3 -----(1)
f(y) = y^{2} + y + 3 -----(2)
(1) = (2)
x^{2} + x + 3 = y^{2} + y + 3
x^{2} + x = y^{2} + y
x^{2} - y^{2} + 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) = x^{3} then, (i) find the range of f (ii) identify the type of function
Solution :
Given that :
f (x) = x^{3}
f (x) = x^{3} x = 1 f (1) = 1^{3} = 1 |
f (x) = x^{3} x = 2 f (2) = 2^{3} = 8 |
f (x) = x^{3} x = 3 f (3) = 3^{3} = 27 |
f (x) = x^{3} x = 4 f (4) = 4^{3} = 64 |
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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