**Question:**

Define each of the following:

(i) injective function

(ii) surjective function

(iii) bijective function

(iv) many - one function

(v) into function

Give an example of each type of functions.

**Solution:**

1)injective function

Definition: $A$ function $f: A \rightarrow B$ is said to be a one - one function or injective mapping if different elements of $A$ have different $f$ images in $B$.

A function $f$ is injective if and only if whenever $f(x)=f(y), x=y$.

Example: $f(x)=x+9$ from the set of real number $R$ to $R$ is an injective function. When $x=3$, then : $f(x)=$ 12, when $f(y)=8$, the value of $y$ can only be 3, so $x=y$.

(ii) surjective function

Definition: If the function $\mathrm{f}: \mathrm{A} \rightarrow \mathrm{B}$ is such that each element in $\mathrm{B}$ (co - domain) is the ' $\mathrm{f}$ ' image of atleast one element in $A$, then we say that $f$ is a function of $A$ 'onto' $B$. Thus $f$ : $A \rightarrow B$ is surjective if, for all b $\in B$, there are some $a \in A$ such that $f(a)=b$.

Example: The function $f(x)=2 x$ from the set of natural numbers $N$ to the set of non negative even numbers is a surjective function.

(iii) bijective function

Definition: A function $f$ (from set $A$ to $B$ ) is bijective if, for every $y$ in $B$, there is exactly one $x$ in $A$ such that $f(x)=y .$ Alternatively, $f$ is bijective if it is a one - to - one correspondence between those sets, in other words, both injective and surjective.

Example: If $f(x)=x^{2}$, from the set of positive real numbers to positive real numbers is both injective and surjective.Thus it is a bijective function.

(iv)many - one function

Defintion : A function $\mathrm{f}: \mathrm{A} \rightarrow \mathrm{B}$ is said to be a many one functions if two or more elements of $\mathrm{A}$ have the same $\mathrm{f}$ image in $\mathrm{B}$.

trigonometric functions such as $\sin x$ are many - to - one since $\sin x=\sin (2 \pi+x)=\sin (4 \pi+x)$ and so one...

(v) into function

Definition: If $f: A \rightarrow B$ is such that there exists atleast one element in co - domain, which is not the image of any element in the domain, then $f(x)$ is into.

Let $f(x)=y=x-1000$

$\Rightarrow x=y+1000=g(y)$ (say)

Here $g(y)$ is defined for each $y \in I$, but $g(y) \notin N$ for $y \leq-1000 .$ Hence, $f$ is into.

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