# Exercise 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∗?

Solution: 2. Check the injectivity and surjectivity of the following functions:
(i) f : N → N given by f(x) = x2
(ii) f : Z → Z given by f(x) = x2
(iii) f : R → R given by f(x) = x2
(iv) f : N → N given by f(x) = x3
(v) f : Z → Z given by f(x) = x3

Solution:

(i) f : N → N given by f(x) = x2
For x, y ∈ N => f(x) = f(y) which implies x2 = y2
x = y
Therefore f is injective.
There are such numbers of co-domain which have no image in domain N.
Say, 3 ∈ N, but there is no pre-image in domain of f. such that f(x) = x2 = 3.
f is not surjective.
Therefore, f is injective but not surjective.

(ii) Given, f : Z → Z given by f(x) = x2
Here, Z = {0, ±1, ±2, ±3, ±4, …..}
f(-1) = f(1) = 1
But -1 not equal to 1.
f is not injective.
There are many numbers of co-domain which have no image in domain Z.
For example, -3 ∈ co-domain Z, but -3 ∉ domain Z
f is not surjective.
Therefore, f is neither injective nor surjective.

(iii) f : R → R given by f(x) = x2
f(-1) = f(1) = 1
But -1 not equal to 1.
f is not injective.
There are many numbers of co-domain which have no image in domain R.
For example, -3 ∈ co-domain R, but there does not exist any x in domain R where x
2
= -3
f is not surjective.
Therefore, f is neither injective nor surjective.
(iv) f : N → N given by f(x) = x3
For x, y ∈ N => f(x) = f(y) which implies x
3 = y3
x = y
Therefore f is injective.
There are many numbers of co-domain which have no image in domain N.
For example, 4 ∈ co-domain N, but there does not exist any x in domain N where x3= 4.
f is not surjective.
Therefore, f is injective but not surjective.
(v) f : Z → Z given by f(x) = x3
For x, y ∈ Z => f(x) = f(y) which implies x
3 = y3
x = y
Therefore f is injective.
There are many numbers of co-domain which have no image in domain Z.
For example, 4 ∈ co-domain N, but there does not exist any x in domain Z where x3= 4.
f is not surjective.
Therefore, f is injective but not surjective.

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

Solution:

Function f : R → R, given by f(x) = [x]
f(x) = 1, because 1 ≤ x ≤ 2
f(1.2) = [1.2] = 1
f(1.9) = [1.9] = 1
But 1.2 ≠ 1.9
f is not one-one.
There is no fraction proper or improper belonging to co-domain of f has any pre-image in its
domain.
For example, f(x) = [x] is always an integer
for 0.7 belongs to R there does not exist any x in domain R where f(x) = 0.7
f is not onto.
Hence proved, the Greatest Integer Function 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.

Solution:

f : R → R, given by f(x) = | x |, defined as  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} and
f = {(1, 4), (2, 5), (3, 6)}
f(1) = 4, f(2) = 5 and f(3) = 6
Here, also distinct elements of A have distinct images in B.
Therefore, f is one-one.

7. In each of the following cases, state whether the function is one-one, onto or
(i) f : R → R defined by f(x) = 3 – 4x
(ii) f : R → R defined by f(x) = 1 + x2

Solution:

(i) f : R → R defined by f(x) = 3 – 4x
If x1, x2 ∈ R then
f(x1) = 3 – 4x1 and
f(x2) = 3 – 4x2
If f(x1) = f(x2) then x1 = x2
Therefore, f is one-one.
Again,
f(x) = 3 – 4x
or y = 3 – 4x
or x = (3-y)/4 in R
f((3-y)/4) = 3 – 4((3-y)/4) = y
f is onto.
Hence f is onto or bijective.

(ii) f : R → R defined by f(x) = 1 + x2
If x1, x2 ∈ R then
f(x1) = 1 + 𝑥12 and
f(x2) = 1 + 𝑥2
2
If f(x1) = f(x2) then 𝑥12 = 𝑥22
This implies x1 ≠ x2
Therefore, f is not one-one
Again, if every element of co-domain is image of some element of Domain under f, such that
f(x) = y
f(x) = 1 + x2

y = f(x) = 1 + x2
or x = ±√1 − 𝑦
Therefore, f(√1 − 𝑦 ) = 2 – y ≠ y
Therefore, f is not onto or bijective.

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

Solution: Step 2: Check for Surjectivity:
Let (b, a) be any element of B x A. Then a ∈ A and b ∈ B
This implies (a, b) ∈ A x B
For all (b, a) ∈ B x A, their exists (a, b) ∈ A x B
Therefore, f: A x B -> B x A is bijective function.

9. Let f : N → N be defined by Solution: For n = 1, 2
f(1) = (n+1)/2 = (1+1)/2 = 1 and
f(2) = (n)/2 = (2)/2 = 1
f(1) = f(2), but 1 ≠ 2
f is not one-one.
For a natural number, “a” in co-domain N
If n is odd
n = 2k + 1 for k ∈ N , then 4k + 1 ∈ N such that
f(4k+1) = (4k+1+1)/2 = 2k + 1
If n is even
n= 2k for some k ∈ N such that
f(4k) = 4k/2 = 2k
f is onto
Therefore, f is onto but not bijective function.

10. Let A = R – {3} and B = R – {1}. Consider the function f : A → B defined by
f(x) = (x-2)/(x-3)  11. Let f : R → R be defined as f(x) = x4
(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 → R be defined as f(x) = x4
let x and y belongs to R such that, f(x) = f(y)
x
4 = y4 or x = ± y
f is not one-one function.
Now, y = f(x) = x
4
Or x = ± y1/4
f(y1/4 ) = y and f(-y1/4 ) = -y
Therefore, f is not onto function.
Option D is correct.

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.
Solution: f : R → R be defined as f(x) = 3x
let x and y belongs to R such that f(x) = f(y)
3x = 3y or x = y
f is one-one function.
Now, y = f(x) = 3x
Or x = y/3
f(x) = f(y/3) = y
Therefore, f is onto function.
Option (A) is correct.