Here we are going to see, how to check if function is bijective.
If a function f : A -> B is both one–one and onto, then f is called a bijection from A to B.
A function f : A -> B is called one – one function if distinct elements of A have distinct images in B.
A function f : A -> B is said to be onto function if the range of f is equal to the co-domain of f.
Question 1 :
In each of the following cases state whether the function is bijective or not. Justify your answer.
(i) f : R -> R defined by f (x) = 2x +1
Solution :
Testing whether it is one to one :
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 ∈ R, f(x) = f(y)
f(x) = 2x + 1 ------(1)
f(y) = 2y + 1 ------(2)
(1) = (2)
2x + 1 = 2y + 1
2x = 2y
x = y
So, it is one to one.
Testing whether it is onto :
Range of f = co-domain
If f : A -> B is an onto function then, the range of f = B . That is, f(A) = B.
Let x ∈ A, y ∈ B and x, y ∈ R. Then, x is pre-image and y is image.
Then, we have
y = 2x + 1
Solve for x.
x = (y - 1) /2
Here, y is a real number. When we subtract 1 from a real number and the result is divided by 2, again it is a real number.
For every real number of y, there is a real number x.
So, range of f(x) is equal to co-domain.
It is onto function.
Hence it is bijective function.
(ii) f : R -> R defined by f (x) = 3 – 4x^{2}
Solution :
Testing whether it is one to one :
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 ∈ R, f(x) = f(y)
f(x) = 3 – 4x^{2} ------(1)
f(y) = 3 – 4y^{2} ------(2)
(1) = (2)
3 – 4x^{2} = 3 – 4y^{2}
-4x^{2} = –4y^{2}
x^{2} - y^{2 }= 0
(x - y)(x + y) = 0
x - y = 0 (or) x + y = 0
x = y (or) x = -y
It is not one to one.Hence it is not bijective function.
Question 2 :
Let A = {−1, 1}and B = {0, 2} . If the function f : A -> B defined by f(x) = ax + b is an onto function? Find a and b.
Solution :
f(x) = ax + b
f(-1) = a(-1) + b = 0
-a + b = 0 ------(1)
f(1) = a(1) + b = 2
a + b = 2 ------(2)
(1) + (2)
2b = 2
b = 1
By applying the value of b in (1), we get
-a + 1 = 0
-a = -1
a = 1
Hence the values of a and b are 1 and 1 respectively.
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