Mathematics Functions For CBSE-NCERT
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Class 12 Chapter 1 - RELATIONS AND FUNCTIONS

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.
 
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