Relations & Functions(Ex. 1.2)

 Exercise 1.2

1. Show that the function f: R*→R* defined by f(x)=1/x is one-one and onto, where R* is the set of all non-zero real numbers. Is the result true, if the domain R* is replaced by N with co-domain being same as R*?

Solun:- Given set is R* set of all non-zero real numbers.

Given function is f(x)=1/x

We know that for one-one function:- if f(x1)=f(x2) then x1=x2

⇒ f(x1)=f(x2)

⇒ 1/x1=1/x2

⇒ x1=x2 

Hence given function is a one-one function.

We know that for onto function:- codomain of f=Range of f

⇒ f(x)=1/x

⇒ y=1/x

⇒ x=1/y

Hence given function is onto function because there is a unique preimage of y.

Now, the domain R* is replaced by N with the co-domain being the same as R*.

For one-one:-

⇒ f(x1)=f(x2)

⇒ 1/x1=1/x2

⇒ x1=x2 ∈ N

Hence given function is a one-one function in domain N.

for onto function:- codomain of f=Range of f

⇒ x=1/y but given domain is N

But some elements of Range in not included in the co-domain.

Hence given function is not an onto function.

2. Check the injective and surjective of the following functions:

(i) f:N→N given by f(x)=x2

Solun:- Given set is N set of natural numbers.

Given function is f(x)=x2

We know that for one-one:- if f(x1)=f(x2) then x1=x2

⇒ f(x1)=f(x2)

⇒ x12 = x22

⇒ x1=x2 

Hence given function is a one-one function.

We know that for onto function:- codomain of f=Range of f

⇒ f(x)=x2

⇒ y=x2

⇒ x=√y

For example:- x=√3 is not a natural number. 

But given domain is a set of natural numbers and some elements of Range in not included in the co-domain.

Hence given function is not onto function.

So, f is injective but not a surjective function.

(ii) f: Z→Z given by f(x)=x2

Solun:- Given set is Z set of integers.

Given function is f(x)=x2

We know that for one-one:- if f(x1)=f(x2) then x1=x2

⇒ f(x1)=f(x2)

Put x1=1 and x1 = - 1

⇒ x12 = x22=1

Hence given function is not a one-one function because according to the definition, the image of the distinct elements is also distinct but here the image of 1 and -1 is same.

We know that for onto function:- codomain of f=Range of f

⇒ f(x)=x2

⇒ y=x2

⇒ x=√y

For example:- x=√3 is not an integer.

But given domain is set of integers and some elements of Range in not included in the co-domain.

Hence given function is not onto function.

So, f is neither injective nor surjective.

(iii) f: R→R given by f(x)=x2

Solun:- Given set is R set of real numbers.

Given function is f(x)=x2

We know that for one-one:- if f(x1)=f(x2) then x1=x2

⇒ f(x1)=f(x2)

Put x1=1 and x1 = - 1

⇒ x12 = x22=1

Hence given function is not a one-one function because according to the definition, the image of the distinct elements is also distinct but here the image of 1 and -1 is same.

We know that for onto function:- codomain of f=Range of f

⇒ f(x)=x2

⇒ y=x2

⇒ x=√y

And also some elements of Range in not included in the co-domain.

For example:- x=√-2 is not a real number.

Hence given function is not onto function.

So, f is neither injective nor surjective.

(iv) f: N→N given by f(x)=x3

Solun:- Given set is N set of natural numbers.

Given function is f(x)=x3

We know that for one-one:- if f(x1)=f(x2) then x1=x2

⇒ f(x1)=f(x2)

⇒ x13= x23

⇒ x1=x2 

Hence given function is a one-one function.

We know that for onto function:- codomain of f=Range of f

⇒ f(x)=x3

⇒ y=x3

⇒ x=3√y

For example, x=3√3 is not a natural number.

But given domain is a set of natural numbers and some elements of Range in not included in the co-domain.

Hence given function is not onto function.

So, f is injective but not surjective.

(v) f: Z→Z given by f(x)=x3

Solun:- Given set is N set of natural numbers.

Given function is f(x)=x3

We know that for one-one:- if f(x1)=f(x2) then x1=x2

⇒ f(x1)=f(x2)

⇒ x13= x23

⇒ x1=x2 

Hence given function is a one-one function.

We know that for onto function:- codomain of f=Range of f

⇒ f(x)=x3

⇒ y=x3

⇒ x=3√y

For example:- x=3√3 is not an integer.

But given domain is set of integers and some elements of Range in not included in the co-domain.

Hence given function is not onto function.

So, f is injective but not surjective.

3. Prove that the Greatest Integer Function f: R→R, given by f(x) = [x], is neither one-one nor onto, where [x] denotes the greatest integer less than or equal to x.

Solun:- Given set is R set of real numbers.

Given function is f(x)=[x]

We know that for one-one:- if f(x1)=f(x2) then x1=x2

⇒ f(x1)=f(x2)

For example:- x1=1.2 and x2=1.5

⇒ [x1] = [x2]

⇒ [1.2] = [1.5]=1

Hence given function is not a one-one function because according to the definition, the image of the distinct elements is also distinct but here the image of 1.2 and 1.5 is same.

We know that for onto function:- codomain of f=Range of f

⇒ f(x)=[x] is integers

But given domain is set of real numbers and also some elements of Range in not included in the co-domain.

Hence given function is not onto function.

So, f is neither one-one nor onto.

4. Show that the Modulus Function f: R→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.

Solun:- Given set is R set of real numbers.

Given function is f(x)=|x|

We know that for one-one:- if f(x1)=f(x2) then x1=x2

⇒ f(x1)=f(x2)

For example:- x1=1 and x2 = -1

⇒ |x1| = |x2|

⇒ |1| = |- 1|=1

Hence given function is not a one-one function because according to the definition, the image of the distinct elements is also distinct but here the image of 1 and -1 is same.

We know that for onto function:- codomain of f=Range of f

⇒ f(x)=|x| is integers

But given domain is set of real numbers so some elements of Range in not included in the co-domain.

Hence given function is not onto function.

So, f is neither one-one nor onto.

5. Show that the Signum Function f: R→R, given by 

is neither one-one nor onto.

Solun:- Given set is R set of real numbers.

Given function is signum function

We know that for one-one:- if f(x1)=f(x2) then x1=x2

⇒ f(x1)=f(x2)

For example:- x1=1 and x2 = 2

⇒ |x1| = |x2|

⇒ |1| = |2|=1

Hence given function is not a one-one function because according to the definition, the image of the distinct elements is also distinct but here the image of 1 and 2 is same.

We know that for onto function:- codomain of f=Range of f

We know that range of signum function is an integer.

But given domain is set of real numbers so some elements of Range in not included in the co-domain.

Hence given function is not onto function.

So, f is neither one-one nor onto.

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.

Solun:- Given function is f={(1, 4), (2, 5), (3, 6)} from A to B

Given function is a one-one function because every element has its distinct image.

7. In each of the following cases, state whether the function is one-one, onto or bijective. Justify your answer. 

(i) f: R→R defined by f(x) = 3-4x

(i) f: R→R defined by f(x) = 1+x2

Solun:- Given set is R set of real numbers.

(i) Given function is f(x)=3-4x

We know that for one-one:- if f(x1)=f(x2) then x1=x2

⇒ f(x1)=f(x2)

⇒ 3-4x1=3-4x2

⇒ x1=x2

Hence given function is a one-one function because according to the definition, the image of the distinct elements is also distinct.

We know that for onto function:- codomain of f=Range of f

⇒ f(x)=3-4x

⇒ y=3-4x

⇒ 4x=3-y

⇒ x=(3-y)/4 ∈ R

Hence given function is not onto function.

So, the given function is one-one and onto function and also a bijective function.

(ii) Given function is f(x)=1+x2

We know that for one-one:- if f(x1)=f(x2) then x1=x2

⇒ f(x1)=f(x2)

For example:- x1=1 and x2 = -1

⇒ 1+x21=1+x22 = 2

Hence given function is not a one-one function because according to the definition, the image of the distinct elements is also distinct but here the image of 1 and -1 is same.

We know that for onto function:- codomain of f=Range of f

⇒ f(x)=1+x2

⇒ y=1+x2

⇒ x2=y-1

⇒ x=√(y-1)

Put y=-1

⇒ x=√-2 is not a real number but the given domain is R.

Hence given function is not onto function.

So, the given function is neither one-one nor onto function.

8. Let A and B be sets. Show that f: AXB→BXA such that f(a, b)=(b, a) is bijective function.

Solun:- Given function is f(a, b)=(b, a)

We know that for one-one:- if f(x1)=f(x2) then x1=x2

⇒ f(a1, b1)=f(a2, b2)

⇒ (b1, a1) = (b2, a2)

then b1=b2 and a1=a2

Hence given function is a one-one function because according to the definition, the image of the distinct elements is also distinct.

We know that for onto function:- codomain of f=Range of f

Hence given function is onto function because there is a preimage of every y.

So, the given function is one-one and onto function and also a bijective function.

9. Let f: N→N be defined by 

for all n ∈ N. State whether the function f is bijective. Justify your answer.

Solun:- We know that for one-one:- if f(x1)=f(x2) then x1=x2

⇒ f(x1)=f(x2)

For example:- x1=1 and x2 = 2

⇒ f(x1)=f(x2)=1

Hence given function is not a one-one function because according to the definition, the image of the distinct elements is also distinct but here the image of 1 and 2 is same.

We know that for onto function:- codomain of f=Range of f

Hence given function is onto function because there is a preimage of every y.

So, the given function is an onto function but not a one-one function.

Hence, the given function is not a bijective function.

10. Let A=R-{3} and B=R-{1}. Consider the function f: A→B be defined by 

. Is f one-one and onto? Justify your answer.

Solun:- We know that for one-one:- if f(x1)=f(x2) then x1=x2

 ⇒ f(x1)=f(x2)

 ⇒ (x1-2)(x2-3)=(x1-3)(x2-2)

 ⇒ x1x2-3x1-2x2+6=x1x2-2x1-3x2+6

 ⇒ -3x1-2x2=-2x1-3x2

 ⇒  x1=x2

Hence given function is a one-one function because according to the definition, the image of the distinct elements is also distinct.

We know that for onto function:- codomain of f=Range of f

⇒ y=(x-2)/(x-3)

⇒ y(x-3)=(x-2)

⇒ xy-3y=x-2

⇒ xy-x=3y-2

⇒ x(y-1)=3y-2

⇒ x=(3y-2)/y-1 

Hence given function is onto function because there is a preimage of every y.

So, the given function is one-one and onto function.

Hence, the given function is a bijective function.

11. Let f: R→R be defined as f(x)=x4. 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

Solun:- Given function is f(x)=x4

We know that for one-one:- if f(x1)=f(x2) then x1=x2

⇒ f(x1)=f(x2)

For example:- x1=1 and x2= - 1

⇒ x41=x42=1

Hence given function is not a one-one function because according to the definition, the image of the distinct elements is also distinct but here the image of 1 and -1 is same.

We know that for onto function:- codomain of f=Range of f

⇒ f(x)=x4

⇒ y=x4

⇒ x=4√y

Put y=-3

⇒ x=4√-3 is not a real number but the given domain is R.

Hence given function is not an onto function.

F is neither one-one nor onto.

The correct answer is (D).

12. Let f: R→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

Solun:- Given function is f(x)=3x

We know that for one-one:- if f(x1)=f(x2) then x1=x2

⇒ f(x1)=f(x2)

⇒ 3x1=3x2

⇒ x1=x2

Hence given function is a one-one function because according to the definition, the image of the distinct elements is also distinct.

We know that for onto function:- codomain of f=Range of f

⇒ f(x)=3x

⇒ y=3x

⇒ x=y/3

Hence given function is onto function because there is a preimage of every y.

Hence f is one-one and onto function.

The correct answer is (A).

Download PDF of Exercise 1.2

SEE ALSO:-

Notes of Relations & Functions

Exercise 1.1

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