Question

1. Give an example of a function
(i) which is one-one but not onto.
(ii) which is not one-one but onto.
(iii) which is neither one-one nor onto.​

Answer 1

Step-by-step explanation:

(i) Let function f:N→N, given by f(x)=2x

Calculate f(x

1

):

⇒ f(x

1

)=2x

1

Now, calculate f(x

2

):

⇒ f(x

2

)=2x

2

Now, f(x

1

)=f(x

2

)

⇒ 2x

1

=2x

2

⇒ x

1

=x

2

Hence, if f(x

1

)=f(x

2

), x

1

=x

2

the function f is one−one.

Now, f(x)=2x

Let f(x)=y, such that y∈N

⇒ 2x=y

⇒ x=

2

y

If y=1

x=

2

1

=0.5, which is not possible as x∈N

Hence, f is not onto.

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

(ii) Let the function f:N→N, given by f(1)=f(2)=1

Here, f(x)=f(1)=1 and

⇒ f(x)=f(2)=1

Since, different elements 1,2 have same image 1,

∴ f is not one-one.

Let f(x)=y, such that y∈N

Here, y is a natural number and for every y, there is a value of x which is natural number.

Hence f is onto.

So, the function f:N→N, given by f(1)=f(2)=1 is not one-one but onto.

(iii) Let function f:R→R, given by f(x)=x

2

Calculate f(x

1

):

⇒ f(x

1

)=(x

1

)

2

Calculate f(x

2

):

⇒ f(x

2

)=(x

2

)

2

Now, f(x

1

)=f(x

2

)

⇒ (x

1

)

2

=(x

2

)

2

⇒ x

1

=x

2

or x

1

=−x

2

Since, x

1

does not have unique image, it is not one-one.

Now, f(x)=x

2

Let f(x)=y, such that y∈R

⇒ x

2

=y

⇒ x−±

y

Since, y is real number, then it can be negative also.

Putting y=−5

x=±

−5

Which is not possible as the root of a negative number is not real.

Hence, x is not real, so f is not onto.

∴ Function f:R→R, given by f(x)=x

2

is neither one-one nor onto.

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