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High School Mathematics - 2
4.4 Many-one, One-One Functions

Many - One Function: If the function f: A->B is such that 2 or more elements a1, a2.... of A have the same f-image in B, then the mapping is called many-one mapping or many-one function.
Note that f:A->B is not one-one then it is many-one.

One - One Onto Function
If the function f is both one-one and onto then f is called one-one onto function. It is also called a bijection or a bijective function.

In other words, a function f: A ->B is bijective if
  1. if it is one-one, i.e f(x) = f(y) which implies x = y for all x,y A
  2. it is onto,i.e, for all y B, there exists x A such that f(x) = y.

Example: If R->R be a function defined by f(x) = 5x3-8, show that the function is a bijective function.
Solution: f(x) = 5x3-8 , x R be any two elements of R (domain)
f is one-one (injective)
Let x,y R
then f(x) = f(y) which implies 5x3 - 8 = 5y3 - 8
From the above we get x3 = y3
Hence x3 = y3 = 0
(x-y) = 0, because the second factor is not equal to zero.
x = y
Hence is one-one.
f is onto, i.e surjective.
Let k be any real number in the codomain.
Then f(x) = k = 5x3-8
k => x = {(k+8)/5}1/3
Now, f[(k+8)/5]1/3


Directions: Answer the following.
Q 1: {(1,2), (2,2), (3,2)}. find the range
{1,2,3}
{2}
not a function

Q 2: f(x) = x+1, A function R->R, is onto or not.
yes
no

Q 3: f(x) = x3, A function R->R, is onto or not.
yes
no

Q 4: Let f: A->B be one-to-one function such that range of f is {b}. Determine the number of elements in A.
such a function cannot exist
0
1

Q 5: {(x, - |x|): x R}. Find the domain
R - {2}
not a function
R - {1}

Q 6: Find the range of the function [(x, (x2-1)/(x-1): x R, x not equal to 1}
R - {2}
R - {1}
not a function

Q 7: If A = {1,2,3}, B = {4,5) and f = {(1,4), (1,5),(2,4), (3,5). Is f a function from A into B?
yes
no

Q 8: For a function f: A->B an onto function, in this case n(B) ___ n(A)
less than
less than or equal to
more than

Question 9: This question is available to subscribers only!

Question 10: This question is available to subscribers only!


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