Mathematics Functions For CBSE-NCERT

### Topics Covered

star function
star One -One (or injective ) and Many- One
star Onto (or surjective )
star Bijective Function

### function

\color{purple} "★ DEFINITION ALERT"

\color{fuchsia} \mathbf\ul"FUNCTION"

\color{green} ✍️ A \color{blue}ul"function" is a relation between a set of inputs and a set of permissible outputs with the property that \color{green}ul"each input is related to exactly one output."

\color{green} ✍️ The output of a function f corresponding to an input x is denoted by f(x) "(read f of x )."

A function is a special relationship where each input has a single output.

It is often written as f(x) where x is the input value.

\color{green} "Example:" f(x) = x/2 is a function, because each input "x" has a single output "x/2":
• f(2) = 1
• f(16) = 8
• f(−10) = −5

\color{fuchsia} \mathbf\ul"NOTATION"

A function f is commonly declared by stating its domain X and codomain Y using the expression

\color{green}( f\ : X\rightarrow Y)
or

\color{green} (Xstackrel{f}{\rightarrow }Y)

### One -One (or injective ) and Many- One

\color{purple} "★ DEFINITION ALERT"

\color{fuchsia} \mathbf\ul"One- One and Many- One"

\color{fuchsia} ✍️ A function f : X → Y is defined to be \color{green}ul"one-one (or injective)," if the images of distinct elements of X under f are distinct, i.e., for every x_1, x_2 ∈ X, \ \ f (x_1) = f (x_2) implies x_1 = x_2. Otherwise, f is called \color{green}ul"many-one".

\color{fuchsia} ✍️ A relation is a function if for every x in the domain there is \color{green}ul"exactly one" y in the codomain.

\color{fuchsia} ✍️ A \color{green}ul"function" for which \color{green}ul"every element" of the \color{green}ul"range" of the function corresponds to \color{green}ul"exactly one element" of the domain.

\color{green} ✍️ One-to-one is often written 1-1.

\color{green} ✍️ The function f_1 and f_4 in Fig (i) and (ii) are one-one and the function f_2 and f_3 in Fig (iii) and (iv) are many-one.

\color { maroon} ® \color{maroon} ul (" REMEMBER")

\color { blue} (✍️ "A function is one to one if it is either strictly increasing or strictly decreasing.")

### Onto (or surjective )

\color{purple} "★ DEFINITION ALERT"

\color{fuchsia} \mathbf\ul"Onto Function"

\color{fuchsia} ✍️ A function f : X → Y is said to be \color{green}ul"onto (or surjective)", if every element of Y is the image of some element of X under f, i.e., for every y ∈ Y, there exists an element x in X such that f (x) = y.

\color{fuchsia} ✍️ A function f from a set X to a set Y is surjective (or onto), or a surjection, if for every element y in the codomain Y of f there is at least one element x in the domain X of f such that f(x) = y. It is not required that x is unique .

The function f_3 and f_4 in Fig (i), (ii) are onto and the function f_1 in Fig (iii) is not onto as elements e, f in X_2 are not the image of any element in X_1 under f_1.

\color { maroon} ® \color{maroon} ul (" REMEMBER") X → Y is onto if and only if Range of f = Y.

\color { blue}(✍️ "For onto function, range and co-domain are equal.")
Q 3104056858

Show that the function f : N→ N, given by f (x) = 2x, is one-one but not
onto.

Solution:

The function f is one-one, for f (x_1) = f (x_2) ⇒ 2x_1 = 2x_2 ⇒ x_1 = x_2. Further,
f is not onto, as for 1 ∈ N, there does not exist any x in N such that f (x) = 2x = 1 .
Q 3114156950

Prove that the function f : R→R, given by f (x) = 2x, is one-one and onto.

Solution:

f is one-one, as f (x_1) = f (x_2) ⇒ 2x_1 = 2x_2 ⇒ x_1 = x_2. Also, given any real

number y in R, there exists y/2 in R such that f (y/2) = 2 * (y/2) = y . Hence, f is onto.
Q 3124156951

Show that the function f : N→N, given by f (1) = f (2) = 1 and f (x) = x – 1,
for every x > 2, is onto but not one-one.

Solution:

f is not one-one, as f (1) = f (2) = 1. But f is onto, as given any y ∈ N, y ≠ 1,
we can choose x as y + 1 such that f (y + 1) = y + 1 – 1 = y. Also for 1 ∈ N, we
have f (1) = 1.
Q 3184156957

Show that the function f : R →R,
defined as f (x) = x^2 , is neither one-one nor onto.

Solution:

Since f (– 1) = 1 = f (1), f is not one one.
Also, the element – 2 in the co-domain R is
not image of any element x in the domain R
(Why?). Therefore f is not onto.
Q 3134267152

Show that an onto function f : {1, 2, 3} → {1, 2, 3} is always one-one.

Solution:

Suppose f is not one-one. Then there exists two elements, say 1 and 2 in the
domain whose image in the co-domain is same. Also, the image of 3 under f can be
only one element. Therefore, the range set can have at the most two elements of the
co-domain {1, 2, 3}, showing that f is not onto, a contradiction. Hence, f must be one-one.
Q 3154267154

Show that a one-one function f : {1, 2, 3} → {1, 2, 3} must be onto.

Solution:

Since f is one-one, three elements of {1, 2, 3} must be taken to 3 different
elements of the co-domain {1, 2, 3} under f. Hence, f has to be onto.

### Bijective/Invertible Function

\color{purple} "★ DEFINITION ALERT"

\color{fuchsia} \mathbf\ul"Bijective Function"

\color{fuchsia} ✍️ A function f : X → Y is said to be one-one and onto (or bijective), if f is \color{green}ul"both one-one and onto."

\color{green} ✍️ The function f_4 in Fig is one-one and onto.

\color { maroon} ® \color{maroon} ul (" REMEMBER")

\color{blue} (✍️ "A function f is not bijection, inverse function of f cannot be defined.")
Q 3114167059

Show that f : N→ N, given by

f(x) = {tt ( (x+1 , text (if) x text (is odd) ), ( x-1 , text (if) x text (is even) ) )

is both one-one and onto.

Solution:

Suppose f (x_1) = f (x_2). Note that if x_1 is odd and x_2 is even, then we will have
x_1 + 1 = x_2 – 1, i.e., x_2 – x_1 = 2 which is impossible. Similarly, the possibility of x_1 being
even and x_2 being odd can also be ruled out, using the similar argument. Therefore,
both x_1 and x_2 must be either odd or even. Suppose both x_1 and x_2 are odd. Then
f (x_1) = f (x_2) ⇒x_1 + 1 = x_2 + 1 ⇒x_1 = x_2. Similarly, if both x_1 and x_2 are even, then also
f (x_1) = f (x_2) ⇒ x_1 – 1 = x_2 – 1 ⇒ x_1 = x_2. Thus, f is one-one. Also, any odd number
2r + 1 in the co-domain N is the image of 2r+ 2 in the domain N and any even number
2r in the co-domain N is the image of 2r – 1 in the domain N. Thus, f is onto.