Raji visited the Exhibition along with her family. The Exhibition had a huge swing, which attracted many children. Raji found that the swing traced the path of a Parabola as given by π¦ = π₯2 .

Answer the following questions using the above information.

Question 1

Let π:
π
→
π
be defined by π(π₯) = π₯
^{
2
}
is_________
(a) Neither Surjective nor Injective (b) Surjective
(c) Injective (d) Bijective

Question 2

Let π:
π
→
π
be defined by π(π₯) = π₯
^{
2
}
is ________
(a) Surjective but not Injective (b) Surjective
(c) Injective (d) Bijective

Question 3

Let f: {1, 2, 3,….} → {1, 4, 9,….} be defined by π(π₯) = π₯
^{
2
}
is _________
(a) Bijective (b) Surjective but not Injective
(c) Injective but Surjective (d) Neither Surjective nor Injective

Question 4

Let : π → π be defined by π(π₯) = π₯
^{
2
}
. Range of the function among the following is _________
(a) {1, 4, 9, 16,…}
(b) {1, 4, 8, 9, 10,…}
(c) {1, 4, 9, 15, 16,…}
(d) {1, 4, 8, 16,…}

Question 5

The function f:
Z
→
Z
defined by π(π₯) = π₯
^{
2
}
is__________
(a) Neither Injective nor Surjective (b) Injective
(c) Surjective (d) Bijective

Question Raji visited the Exhibition along with her family. The Exhibition had a huge swing, which attracted many children. Raji found that the swing traced the path of a Parabola as given by π¦ = π₯2 . Answer the following questions using the above information.
Question 1 Let π: π β π be defined by π(π₯) = π₯2 is_________ (a) Neither Surjective nor Injective (b) Surjective (c) Injective (d) Bijective
f(x) = x2, where π: π β π
Checking one-one
f (x1) = (x1)2
f (x2) = (x2)2
Putting f (x1) = f (x2)
(x1)2 = (x2)2
x1 = x2 or x1 = βx2
Since x1 does not have unique image,
It is not one-one
Example:
f(β1) = (β1)2 = 1
f(1) = (1)2 = 1
Here, f(β1) = f(1) , but β1 β 1
Hence, it is not one-one
Check onto
f(x) = x2
Let f(x) = y , such that y β R
x2 = y
x = Β±βπ¦
Note that y is a real number, so it can be negative also
Putting y = β3
x = Β±β((β3))
which is not possible as root of negative number is not real
Hence, x is not real
So, f is not onto
Thus, f(x) is neither Surjective nor Injective
So, the correct answer is (a)
Question 2 Let π: π β π be defined by π(π₯) = π₯2 is ________ (a) Surjective but not Injective (b) Surjective (c) Injective (d) Bijective
f(x) = x2, where π: N β N
Checking one-one
f (x1) = (x1)2
f (x2) = (x2)2
Putting f (x1) = f (x2)
(x1)2 = (x2)2
x1 = x2 or x1 = βx2
But x1 cannot be negative as x1 and x2 are natural numbers
So, only option is
When f(x1) = f(x2), then x1 = x2
Hence, f is one-one
Check onto
f(x) = x2
Let f(x) = y , such that y β N
x2 = y
x = Β±βπ¦
As x is natural number, it will be positive
β΄ x = βπ
Also, since y is a natural number, letβs put y = 2
Putting y = 2
x = β2
which is not possible as x is a natural number
So, f is not onto
Thus, f(x) is only injective
So, the correct answer is (c)
Question 3 Let f: {1, 2, 3,β¦.} β {1, 4, 9,β¦.} be defined by π(π₯) = π₯2 is _________ (a) Bijective (b) Surjective but not Injective (c) Injective but Surjective (d) Neither Surjective nor Injective
f(x) = x2, where π: {1, 2, 3,β¦.} β {1, 4, 9,β¦.}
Checking one-one
f (x1) = (x1)2
f (x2) = (x2)2
Putting f (x1) = f (x2)
(x1)2 = (x2)2
x1 = x2 or x1 = βx2
Since x β {1, 2, 3,β¦.}
β΄ x1 cannot be negative
So, only option is
When f(x1) = f(x2), then x1 = x2
Hence, f is one-one
Check onto
f(x) = x2
Let f(x) = y , such that y β {1, 4, 9,β¦.}
x2 = y
x = Β±βπ¦
As x β {1, 2, 3,β¦.} , it will be positive
β΄ x = βπ
Also, since y β {1, 4, 9,β¦.}
For all values of y, we will get a value of x,
Where x β {1, 2, 3,β¦.}
So, f is onto
Thus, f(x) is one-one and onto
β΄ f(x) is Bijective
So, the correct answer is (a)
Question 4 Let : π β π be defined by π(π₯) = π₯2 . Range of the function among the following is _________ (a) {1, 4, 9, 16,β¦} (b) {1, 4, 8, 9, 10,β¦} (c) {1, 4, 9, 15, 16,β¦} (d) {1, 4, 8, 16,β¦}
For f(x) = x2, where π: N β R
Range will be = {12, 22, 32, 42, 52, 62, β¦}
= {1, 4, 9, 16, 25, 36, β¦}
So, the correct answer is (a)
Question 5 The function f: Z β Z defined by π(π₯) = π₯2 is__________ (a) Neither Injective nor Surjective (b) Injective (c) Surjective (d) Bijective
f(x) = x2, where π: Z β Z
Checking one-one
f (x1) = (x1)2
f (x2) = (x2)2
Putting f (x1) = f (x2)
(x1)2 = (x2)2
x1 = x2 or x1 = βx2
Since x1 does not have unique image,
It is not one-one
Example:
f(β1) = (β1)2 = 1
f(1) = (1)2 = 1
Here, f(β1) = f(1) , but β1 β 1
Hence, it is not one-one
Check onto
f(x) = x2
Let f(x) = y , such that y β Z
x2 = y
x = Β±βπ¦
Note that y is an integer, letβs put y = 5
Putting y = 5
x = Β±β5
But this is not possible as Β±β5 is not an integer
So, f is not onto
Thus, f(x) is neither Surjective nor Injective
So, the correct answer is (a)

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.