1.Types of Functions in Class 12

Types of Functions in Class 12

In this article Types of Functions in Class 12, we will read specific types functions viz one-one function,onto function and one-one onto function examples with Solution.
“Before diving into these solutions, make sure you understand the foundational concept of “Relations Class 12 Examples with Solution” and have reviewed the essential “Relations in Class 12 Examples with Solutions“. This will make following today’s step-by-step solutions much easier.”

Also Read This Article:-  Function (mathematics)

2.Illustrations Based on Types of Functions in Class 12

Illustration 1: Show that the function f : R_* \to R_* defined by f(x) = \frac{1}{x} is one-one and onto, where R_* is the set of all non-zero numbers. Is the result true, if the domain R_* is replaced by N with co-domain being same as R_*?
Solution: Given Function
f(x) = \frac{1}{x}, f : R_* \to R_*
Let x_1, x_2 \in R_* (Domain)
then f(x_1) = f(x_2)
\Rightarrow \frac{1}{x_1} = \frac{1}{x_2} \\ \Rightarrow x_1 = x_2
So function is one-one.
Again if x is arbitrary element of co-domain.
If f(x) = y, then
y = \frac{1}{x} \Rightarrow x = \frac{1}{y}, y \neq 0
which is element of domain (\forall y \in R_*, x \in R_*)
So f(\frac{1}{y}) = \frac{1}{\frac{1}{y}} = y \in R_* (codomain)
i.e., \forall y \in R_* (codomain),
f(\frac{1}{y}) = y \quad \therefore f(R_*) = R_*
But each element of co-domain is only image of one element.
\therefore f is onto
so given function is one-one and onto.
(ii) If the domain R_* is replaced by N with co-domain being same as R_* then
f : N \to R_*
and f(x) = \frac{1}{x}, where x \in N
If x_1, x_2 \in N then
f(x_1) = f(x_2) \Rightarrow \frac{1}{x_1} = \frac{1}{x_2} \\ \Rightarrow x_1 = x_2, \quad x_1, x_2 \in N
\therefore function is one-one.
Domain is set of natural numbers and codomain is set of non-negative real numbers.
f : N \to R_+ and f(x) = \frac{1}{x}
Let y \in R_* (Codomain)
and f(x) = y
\Rightarrow y = \frac{1}{x} \Rightarrow x = \frac{1}{y}
when y = 1, then x = \frac{1}{1} \in N
but y = \frac{5}{4} than x = \frac{1}{\frac{5}{4}} = \frac{4}{5} \notin N
i.e. \frac{4}{5} is not in domain
\therefore Range of f \subset R_*
\therefore f is not onto
Thus f is one-one but not onto. So result is not same.
Illustration 2. Check the injectivity and surjectivity of the following functions:
Illustration 2(i). f : N \to N given by f(x) = x^2
Solution: f(x) = x^2, f : N \to N given by f(x) = x^2
Let x_1, x_2 \in N then
f(x_1) = f(x_2) \Rightarrow x_1^2 = x_2^2\\ \Rightarrow x_1 = x_2  \\ \therefore function is injective.
[\because x_1 = \pm x_2 \text{ but we cannot take } x_2 \text{ as negative because } x_2 \in N]
Again f(x) = x^2, put x = 1, 2, 3, \cdots \\ f(1) = 1^2 = 1, f(2) = 2^2 = 4, f(3) = 3^2 = 9
i.e. codomain has such element which has is not image of any element of domain. example 7
\therefore f is not onto-surjective
\therefore f is injective but not surjective.
Illustration 2(ii). f : \mathbb{Z} \to \mathbb{Z} given by f(x) = x^2
Solution: f : \mathbb{Z} \to \mathbb{Z}, where f(x) = x^2
Let x_1, x_2 \in \mathbb{Z} (domain)
then f(x_1) = f(x_2) \Rightarrow x_1^2 = x_2^2
\Rightarrow x_1^2 - x_2^2 = 0 \Rightarrow (x_1 - x_2)(x_1 + x_2) = 0 \\ \Rightarrow x_1 = x_2 \text{ or } -x_2 \\ \Rightarrow f(x_1) = f(x_2) \Rightarrow x_1 \neq x_2
(\because x_1 = -x_2)
\therefore f is not one-one injective
Again Let y \in \mathbb{Z} codomain is any element.
Range of f \subset \mathbb{Z} (Codomain)
\therefore f is not surjective
so f is neither one-one nor surjective (injective).
Illustration 2(iii). f : R \to R given by f(x) = x^2
Solution: f : R \to R given by f(x) = x^2
Let x_1, x_2 \in R then
f(x_1) = f(x_2) \Rightarrow x_1^2 = x_2^2
\Rightarrow \pm x_1 = x_2
i.e. different element of domain have same image in codomain
\therefore f is not one-one injective
we see that some element of codomain has not pre-image in domain.
\therefore f is not onto-surjective
\therefore f is neither one-one nor onto-surjective (injective).
Illustration 2(iv). f : N \to N given by f(x) = x^3
Solution: f : N \to N where f(x) = x^3
If x_1, x_2 \in N (domain) then
f(x_1) = f(x_2) \Rightarrow x_1^3 = x_2^3 \\ \Rightarrow x_1 = x_2
\therefore f is injective
Again codomain has many element which is not image of any element of domain. ex. 2, 3, 4, 5, \dots etc.
Range of f \subset N
i.e. f(N) \subset N
\therefore function f is not surjective
\therefore f is neither injective nor surjective.
Illustration 2(v). f : \mathbb{Z} \to \mathbb{Z} given by f(x) = x^3
Solution: f : \mathbb{Z} \to \mathbb{Z} given by f(x) = x^3
If x_1, x_2 \in \mathbb{Z} (domain) then
f(x_1) = f(x_2) \Rightarrow x_1^3 = x_2^3 \\ \Rightarrow x_1 = x_2
\therefore f is injective
Again f(1) = 1^3 = 1, f(-1) = (-1)^3 = -1 \\ f(2) = 2^3 = 8, f(-2) = (-2)^3 = -8 \\ f(3) = 3^3 = 27, f(-3) = (-3)^3 = -27
Range of f \subset \mathbb{Z}
i.e. f(\mathbb{Z}) \subset \mathbb{Z} (codomain)
\therefore f is not surjective
Illustration 3. Prove that the Greatest Integer Function f : R \to R, given by f(x) = [x], is neither one-one nor onto, where [x] denotes the greatest integer less than or equal to x.
Solution: Given f : R \to R and f(x) = [x]
then f(1.8) = 1 and f(1.9) = 1
Here 1.8 and 1.9 both has image 1.
\therefore f is not one-one.
Codomain of f is set of real number R and it has not only integers. But image of x \in R (domain) is integers.
\therefore f That element of codomain, which is not integer, is not image.
element of any domain
i.e. f(R) \subset R (codomain)
\therefore f is not onto
so f is neither one-one nor onto
Illustration 4. Show that the modulus function f : R \to R given by f(x) = |x|, is neither one-one nor onto, where |x| is x, if x is positive or 0 and x is -x, if x is negative.
Solution: Here f : R \to R and f(x) = |x| \\ f(1) = |1| = 1 and f(-1) = |-1| = 1
Here 1 \neq -1 \Rightarrow f(1) = f(-1) = 1
1 and -1 have same image.
\therefore f is not one-one
codomain of f has negative numbers but any negative number f has not pre-image in domain.
\therefore f is not onto.
\therefore f is neither one-one nor onto.
Illustration 5. Show that the signum function f : R \to R, given by
f(x) = \begin{cases} 1, & \text{if } x > 0 \\ 0, & \text{if } x = 0 \\ -1, & \text{if } x < 0 \end{cases}
is neither one-one nor onto.
Solution: Here f : R \to R and
f(x) = \begin{cases} 1, & \text{if } x > 0 \\ 0, & \text{if } x = 0 \\ 1, & \text{if } x < 0 \end{cases}
Here f(1) = 1, f(2) = 1
because 1 > 0, 2 > 0
\therefore 1 \neq 2 \Rightarrow f(1) = f(2)
i.e. 1 and 2 have same image 
Hence f is not one-one
Codomain of f is set of real numbers and domain images of element of domain is only -1, 0, 1, then 
f(R) = \{-1, 0, 1\} \subset R (codomain) 
or Range of f = \{-1, 0, 1\} \subset R (domain) 
\therefore f is not onto 
Hence f is neither one-one nor onto.

Types of Functions in Class 12

Illustration 6. Let A = \{1, 2, 3\}, B = \{4, 5, 6, 7\} and let f = \{(1, 4), (2, 5), (3, 6)\} be a function from A to B. Show that f is one-one.
Solution: A = \{1, 2, 3\}, B = \{4, 5, 6, 7\} \\ f : A \to B like that
f = \{(1, 4), (2, 5), (3, 6)\}
Each element of A has different image in B.
\therefore f is one-one
Illustration 7. In each of the following cases, state whether the function is one-one, onto or bijective. Justify your answer.
Illustration 7(i). f : R \to R defined
by f(x) = 3 - 4x
Solution: Here f : R \to R and f(x) = 3 - 4x
If x_1, x_2 \in R (domain)
\Rightarrow f(x_1) = f(x_2) \Rightarrow 3 - 4x_1 = 3 - 4x_2 \\ \Rightarrow -4x_1 = -4x_2 \\ \Rightarrow x_1 = x_2
\therefore f is one-one
Again if any arbitrary element of codomain like that
f(x) = y \text{ then } y = 3 - 4x \\ \Rightarrow y - 3 = -4x \\ \Rightarrow 3 - y = 4x \\ \Rightarrow x = \frac{3 - y}{4}
Now f\left(\frac{3 - y}{4}\right) = 3 - 4\left(\frac{3 - y}{4}\right) \\ = 3 - (3 - y) = y
i.e. \forall y \in R (codomain)
\exists \left(\frac{3 - y}{4}\right) \in R : f\left(\frac{3 - y}{4}\right) = y \\ \therefore f(R) = R
\therefore f Each element of codomain has preimage in domain.
\therefore f is onto
Hence f is one-one and onto.
Illustration 7(ii). f : R \to R defined by f(x) = 1 + x^2
Solution: f : R \to R and f(x) = 1 + x^2
Let x_1, x_2 \in R (domain) then
f(x_1) = f(x_2) \Rightarrow 1 + x_1^2 = 1 + x_2^2
\Rightarrow x_1^2 = x_2^2
\Rightarrow x_1^2 - x_2^2 = 0 \Rightarrow (x_1 - x_2)(x_1 + x_2) = 0
\Rightarrow x_1 = x_2, -x_2
i.e. image of elements are equal but elements are not equal.
\therefore f is not one-one
Negative numbers of codomain of f, which has not any preimage in domain.
\therefore f is not onto.
Let y \in R (codomain) then
f(x) = y = 1 + x^2
\Rightarrow x^2 = y - 1
then x = \pm\sqrt{y - 1}
If y = 0, then x = \pm\sqrt{-1} \notin R
\therefore f is not onto
Hence f is neither one-one nor onto.
Illustration 8. Let A and B be sets. Show that f : A \times B \to B \times A such that f(a, b) = (b, a) is bijective function.
Solution: f : A \times B \to B \times A iff f(a, b) = (b, a)
Let (a_1, b_1) and (a_2, b_2) \in A \times B, then
f(a_1, b_1) = f(a_2, b_2) \\ \Rightarrow (b_1, a_1) = (b_2, a_2) \\ \Rightarrow b_1 = b_2, a_1 = a_2 \\ \Rightarrow (a_1, b_1) = (a_2, b_2)
\therefore f(a_1, b_1) = f(a_2, b_2)
where (a_1, b_1), (a_2, b_2) \in A \times B
\therefore f is one-one
Again let (b, a) is any arbitrary element of set B \times A
\Rightarrow b \in B \text{ and } a \in A
\Rightarrow a \in A \text{ and } b \in B
\Rightarrow (a, b) \in A \times B
\therefore \forall (b, a) \in B \times A \exists (a, b) \in A \times B like that f(a, b) = (a, b)
\therefore f : A \times B \to B \times A is onto
Hence given function is one-one and bijective.
Illustration 9. Let f : N \to N be defined by
f(n) = \begin{cases} \frac{n+1}{2}, & \text{if } n \text{ is odd} \\ \frac{n}{2}, & \text{if } n \text{ is even} \end{cases} for all n \in N
State whether the function f is bijective. Justify your answer.
Solution: f(n) = \begin{cases} \frac{n+1}{2}, & \text{if } n \text{ is odd} \\ \frac{n}{2}, & \text{if } n \text{ is even} \end{cases}
Here n = 1, when n is odd
f(1) = \frac{1+1}{2} = \frac{2}{2} = 1
and n = 2, when n is even
f(2) = \frac{2}{2} = 1 \\ \therefore 1 \neq 2 \Rightarrow f(1) = f(2) = 1
Two different elements have same image.
\therefore f is not one-one i.e. it is many-one.
Let any arbitrary element n of set N i.e. n \in N
If n is odd then 2n-1 is also odd, then
f(2n-1) = \frac{(2n-1)+1}{2} = \frac{2n}{2} = n \\ [\because f(n) = \frac{n+1}{2}, \text{ when } n \text{ is odd}]
and when n is even, then 2n will also even
\therefore f(2n) = \frac{2n}{2} = n \\ [\because f(n) = \frac{n}{2}, \text{ when } n \text{ is even}]
we see that n is either even or odd f(N) = N
\therefore f is onto 
\therefore f Given function is bijective.

Types of Functions in Class 12

Illustration 10. Let A = R - \{3\} and B = R - \{1\}. Consider the
function f : A \to B defined by f(x) = \left(\frac{x-2}{x-3}\right). Is f one-one and onto? Justify your answer.
Solution: A = R - \{3\} and B = R - \{1\}
and f : A \to B, then
f(x) = \frac{x-2}{x-3}
Let x_1, x_2 \in A, then
Let f(x_1) = f(x_2) \Rightarrow \frac{x_1-2}{x_1-3} = \frac{x_2-2}{x_2-3}
\Rightarrow (x_1-2)(x_2-3) = (x_2-2)(x_1-3) \\ \Rightarrow x_1x_2 - 3x_1 - 2x_2 + 6 = x_1x_2 - 3x_2 - 2x_1 + 6 \\ \Rightarrow 3x_2 - 2x_2 = 3x_1 - 2x_1 \\ \Rightarrow x_2 = x_1 \Rightarrow x_1 = x_2
\therefore f function f is one-one.
Again let y \in B any arbitrary element like that
y = f(x) \Rightarrow y = \frac{x-2}{x-3} \quad (x \neq 3) \\ \Rightarrow (x-3)y = x-2 \\ \Rightarrow xy - x = 3y - 2 \Rightarrow x(y-1) = 3y-2 \\ \therefore x = \frac{3y-2}{y-1}
clearly x is not defined for y \neq 1
\therefore x \neq 3
then f(x) = f\left(\frac{3y-2}{y-1}\right) \\ = \frac{\frac{3y-2}{y-1} - 2}{\frac{3y-2}{y-1} - 3} \\ = \frac{(3y-2) - 2(y-1)}{(3y-2) - 3(y-1)} \\ = \frac{3y-2 - 2y+2}{3y-2 - 3y+3} \\ = \frac{y}{1} = y \\ \forall y \in B, \exists x \in A \quad f(x) = y
\therefore f function f is onto
Hence given function f is one-one onto.
Illustration 11. Let f : R \to R be defined as f(x) = x^4
choose the correct answer.
(A) f is one-one onto (B) f is many-one onto
(C) f is one-one but not onto (D) f is neither one-one nor onto
Solution: f : R \to R, f(x) = x^4
Let x_1, x_2 \in R (domain)
\because f(x_1) = f(x_2) \Rightarrow x_1^4 = x_2^4 \\ \Rightarrow x_1^4 - x_2^4 = 0 \\ \Rightarrow (x_1^2 - x_2^2)(x_1^2 + x_2^2) = 0 \\ \Rightarrow (x_1^2 - x_2^2) = 0
but x_1^2 + x_2^2 \neq 0, because sum of square of two positive real number is not equal to zero.
\Rightarrow x_1^2 - x_2^2 = 0 \\ \Rightarrow (x_1 - x_2)(x_1 + x_2) = 0 \\ \Rightarrow x_1 - x_2 = 0 \text{ or } x_1 + x_2 = 0 \\ \Rightarrow x_1 = x_2 \text{ or } -x_2
\therefore f is one-one \left[ 1^4 = 1, (-1)^4 = 1 \right]  
Let y \in R (codomain) any arbitrary element and f(x) = y
then y = x^4 \Rightarrow x = (y)^{\frac{1}{4}}, y \geq 0
Negative number in codomain has not preimage in domain.
i.e. Range of f \subset R (domain)
\Rightarrow f(R) \subset R (domain)
\therefore f is not onto function.
Hence f is neither one-one nor onto. Alternate (D) is true.
Illustration 12. Let f : R \to R be defined as f(x) = 3x. Choose the correct answer.
(A) f is one-one onto (B) f is many-one onto
(C) f is one-one but not onto (D) f is neither one-one nor onto.
Solution: f : R \to R, \text{ if } f(x) = 3x
Let x_1, x_2 \in R (domain)
then f(x_1) = f(x_2) \Rightarrow 3x_1 = 3x_2 \\ \Rightarrow x_1 = x_2
\therefore f is one-one
Let y \in R (codomain) any arbitrary element like that y = f(x), then
y = f(x) \Rightarrow y = 3x \Rightarrow x = \frac{y}{3} \in R (domain)
then f\left(\frac{y}{3}\right) = 3 \cdot \left(\frac{y}{3}\right) \quad [f(x) = 3x]
= y \in R (codomain)
\therefore y element of codomain is image of \frac{y}{3} element of domain.
\therefore f function f is onto.
Hence given function is one-one onto. Alternate (A) is true.
With the above illustrations,one can understand the Types of Functions in Class 12.

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3.Frequently Asked Questions Related to Types of Functions in Class 12

**छात्र-छात्राओं से आज का प्रश्न**

*"In a sign language \bigcirc \triangle \square ,stands for 'children watch television',stands for 'television and cartoons' and \square \times + means 'children play games', then which of the following symbols will appear for 'sports' in the same sign language?"*
(A)×  (B)+  (C) Can't be fixed  (E)None of these
*Previous Quiz Solution*
The cupboard fills up in 24 days=1 (full)
The cupboard is filled in 23 days=\frac{1}{2}
The cupboard is filled in 22 days=\frac{1}{4}

*"This article has been prepared by **Satyam Coaching Centre** on the **Satyam Mathematics** blog."*