LEARN MATH STEP BY STEP THROUGH VERY EASY PROCESS

RELATION & FUNCTION - TYPES OF FUNCTION - MANY-ONE FUNCTION

**Types Of Function- Many-One Function –**

**If the function f : A → B is such that two or
more elements a₁, a₂,……………. of ‘A’ have the same f-image in ‘B’, then function
is called many-one function.**

**Note that if
the function f : A → B is not one-one, then f is many one.**

**Illustrations
-**

**1) Let f : A → B be a function represented by the diagram shown
in below figure.**

**Clearly, f(a)
= 1, f(b) = 1, f(c) = 1, f(d) = 2, f(e) = 2 and a ≠ b ≠ c, d ≠ e. So, f is a many-one function.**

**2) f : R → R,
defined by f(x) = x²+ 5 is a many-one function, since f(- 1) = (- 1)² + 5 = 6**

**And f(1) =
(1)² + 5 = 6, that is, two distinct elements – 1 and 1 have the same image is
6.**

**Example.1) Prove that the greatest
integer function f : R → R, given by
f(x) = [x] is a many-one function.**

**Ans.) The
diagram shows that even though 1.3 ≠ 1.4 yet f(1.3) = f(1.4) = 1**

**Thus, two
elements 1.3 and 1.4 in set A have the same image 1 in set B. Hence, f : A → B
is not a one-one function. It is a many-one function.**

**Method to check whether the
given function is Many-One -**

__Step.1)__ Consider any two
arbitrary elements a₁, a₂ ∈ A

__Step.2)__ Put f(a₁) = f(a₂)
and solve the equation

__Step.3)__ If we get a₁ ≠ a₂,
then f is many-one.

__Note.__ Recall
that if we get a₁ = a₂, then f is a one-one function.

**Onto function
(Surjective Function) –**

**The mapping f
: A → B is called an onto function if the set B is entirely used up, i.e., if
every element of B is the image of at least one element of A.**

**=> for
every b ∈ B there exists at least one element a ∈ A such that f(a) = b**

**=> Range
of f is the co-domain of f (Range = Co-domain)**

**If function f
: A → B is not onto, that is, some of the elements of B remain unmated, then
‘f’ is called an into function.**

__Remarks.1)__ An
onto function is also called surjective function or a surjection.

__Remarks.2)__ A
one-one many-one function may be both onto and into.

**Illustration
–**

**1) Let A = {-
2, 2, - 3, 3}, B = {4, 9} and f : A → B be a functioned defined by f(x) = x²,
then f is onto, because f(- 2) = 4, f(2) = 4, f(- 3) = 9, f(3) = 9, i.e., f(A)
= {4, 9} = B**

**Here, range =
{4, 9} = B = co-domain.**

**2) A function
f : N → N denoted by f(x) = 7x is not an onto function, because f(N) = {7, 14,
21,………..} ≠ N.**

**=> Range ≠
co-domain (N)**

**3) The
function f : R → R defined by f(x) = ǀ x ǀ, x ∈ R is not onto since ǀ x ǀ is
never negative and therefore no negative real number can be the image of any
real number under ‘f’.**

**Here, also
range = set of non-negative real numbers ≠ co-domain (R)**

**4) Consider
the function f : R → R₁ defined by f(x) = √x², x ∈ R.**

**Let, x ∈ R₁,
then x is non-negative real number,**

**So, √x² = x**

**=> f(x)
= x,**

**Also, √(-x)²
= √x² = x**

**=> f(-x)
= x**

**Thus, all
numbers in R (domain) have an image in R¹ (Co-domain), i.e., range (all numbers
in R) = co-domain (R¹)**

**So, f is
onto.**

**5) Let the
function f : R → R be defined by the formula f(x) = x². The f is not an onto
function since the negative numbers do not appear in the range of f, that is, no
negative number is the square of a real number. Here, range of f ≠ R
(co-domain). It is subset of R.**

**Example.1) Find whether the
following are onto functions (surjections) or not **

**(i) f : R → R defined by f(x) = x³ +
5 for all x ∈ R.**

**(ii) f : R → R defined by f(x) = x²
+ 3 for all x ∈ R.**

**(iii) f : Z → Z defined by f(x) = 5x
– 9 for all x ∈ Z.**

**Method –**

**Suppose f : A → B is a given
function **

__Step.1)__ Choose any arbitrary
element y in B.

__Step.2)__ Put f(x) = y

__Step.3)__ Solve the equation
f(x) = y for x and obtained
x in terms of ‘y’.

**Let, x = g(y)**

* Step.4) If for all values of
y ∈ B, the values of x obtained from* x =

**Ans.) Let there be an arbitrary element y in R.
Then f(x) = y = x³ + 5**

**
=> x³ = y – 5**

** => x = (y – 5)⅓ **

**Now, for all
y ∈ R, (y – 5)⅓ is a real number. So, for all y ∈ R (co-domain), there exists **

**x = (y – 5)⅓. In R (domain) such that f(x) =
x³ + 5**

**Here, f : R →
R is an onto function.**

**(ii) f(x) = x² + 3 ≥ 3 for all x
∈ R. If we
take f(x) = a real number < 3, say = 2,**

**then x² + 3 =
2**

** => x² = 3 – 2 = - 1 which is not true for any x ∈ R (domain).**

**So, 2 is not
the image of any element in the domain.**

**So, ‘f ’ is
not onto.**

**(iii) Let y
be an arbitrary element of Z
(co-domain). Then**

**f(x) = y = 3x
+ 2**

** (y – 2)**

** =>
x = ---------- **

** 3 **

** - 2**

**Now, if y =
0, then x = ------- ∉ Z. Thus, y = 0 **

**3 **

**in Z (co-domain) does not have its pre-image**

** in Z (domain).**

**Hence, f is
not an onto function.**