- What is Function?
- Type Of Function
- One to One Function or Injective Function
- Onto or Surjective Relation
- Bijective (One on One and onto) Function
- Solved Examples

- A function is a "well-behaved" relation
- A function \(f\) is a relation from a non-empty set \(A\) to a non-empty set \(B\) such that the domain of \(f\) is \(A\) and no two distinct ordered pairs in \(f\) have the same first element.
- For a relation to be a function, there must be only and exactly one \(y\) that corresponds to a given \(x\)
- If \(f\) is a function from \(A\) to \(B\) and \(\left( {a,{\rm{ }}b} \right) \in f\), then \(f\left( a \right) = b\), where \(b\) is called the image of \(a\) under \(f\) and \(a\) is called the preimage of \(b\) under\(f\).

For functions \(f:{\rm{ }}X - > {\bf{R}}\) and \(g:{\rm{ }}X - > {\bf{R}}\), we have

*Addition*

\(\left( {f + g} \right)\left( x \right) = f\left( x \right) + g\left( x \right),x \in X\)*Substraction*

\(\left( {f - g} \right)\left( x \right) = f\left( x \right)-g\left( x \right),x \in X\)*Multiplication*

\(\left( {f.g} \right)\left( x \right) = f\left( x \right).g\left( x \right),x \in X\)*Multiplication by real number*\(\left( {kf} \right)\left( x \right) = k{\rm{ }}f\left( x \right),x \in X\), where \(k\) is a real number.*Division*

\(\frac{f}{g}\left( x \right) = \frac{{f(x)}}{{g(x)}}\)

\(x \in X\) and \(g\left( x \right) \ne 0\)

Let f: A -> B, a function from a set A to a set B, f is called a one-to-one function or injection, if, and only if, for all elements a

if f(a

Equivalently,

if a

A function is not one on one if this condition is met, then it is called

f(a

if f(a

(2) Let us take example as

f : R-> R

f(x) =3x-2 for x ∈ R

Then f(a

f(a

Now f(a

3a

or a

So our first point is proved,so this is one to one function

if f(a

(2) We can prove this by giving counter example

f : R-> R

f(x) =x

Now for x=2 f(2)=4

for x=-2 ,f(-2)=4

Now f(2)=f(-2) Hence Many one

for every y ∈ B, there exists a element x in A where f(x)=y

(2) Let us take example as

f : R-> R

f(x) =3x-2 for x ∈ R

(3) Let y ∈ R, then we first need to prove that we can find x ∈ R such that f(x)=y

If such real number x exists then y=3x-2

or x=(y+2)/3

Now sum and division of real number is a real number,So x ∈ R

it follows

f((y+2)/3 )= 3(y+2)/3 -2=y

hence f is onto function

(4) if we have to prove that function is not onto,then easiast way would be to take counter examples

"Injective" means no two elements in the domain of the function gets mapped to the same image.The method has been already described above.

"Surjective" means that any element in the range of the function is hit by the function.The method has been already described above.

A function f : N -> N, given by f(x) = 2x, Show that it is one-one but not onto

for f(x

Or x

or x

So it is one-one

f is not onto, as for 1 ∈ N, there does not exist any x in N such that f(x) =2 x = 1

A function f : R -> R, given by f(x) = x

f(1) =1

So it is not one-one

f is not onto, as for -1 ∈ R, there does not exist any x in R such that f(x) = x

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

Lets take this by contradiction method

So we assume that 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.

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

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

**Notes****NCERT Solutions & Worksheets**

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