Question 5 - Case Based Questions (MCQ) - Chapter 1 Class 12 Relation and Functions

Last updated at April 16, 2024 by Teachoo

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 𝑦 = 𝑥
^{
2
}
.

Answer the following questions using the above information.

Question 1

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

Question 2

Let 𝑓:
𝑁
→
𝑁
be defined by 𝑓(𝑥) = 𝑥
^{
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 𝑓(𝑥) = 𝑥
^{
2
}
is _________
(a) Bijective (b) Surjective but not Injective
(c) Injective but Surjective (d) Neither Surjective nor Injective

Question 4

Let : 𝑁 → 𝑅 be defined by 𝑓(𝑥) = 𝑥
^{
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 𝑓(𝑥) = 𝑥
^{
2
}
is__________
(a) Neither Injective nor Surjective (b) Injective
(c) Surjective (d) Bijective

Transcript

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 𝑦 = 𝑥2 . Answer the following questions using the above information.
Question 1 Let 𝑓: 𝑅 → 𝑅 be defined by 𝑓(𝑥) = 𝑥2 is_________ (a) Neither Surjective nor Injective (b) Surjective (c) Injective (d) Bijective
f(x) = x2, where 𝑓: 𝑅 → 𝑅
Checking one-one
f (x1) = (x1)2
f (x2) = (x2)2
Putting f (x1) = f (x2)
(x1)2 = (x2)2
x1 = x2 or x1 = –x2
Since x1 does not have unique image,
It is not one-one
Example:
f(–1) = (–1)2 = 1
f(1) = (1)2 = 1
Here, f(–1) = f(1) , but –1 ≠ 1
Hence, it is not one-one
Check onto
f(x) = x2
Let f(x) = y , such that y ∈ R
x2 = y
x = ±√𝑦
Note that y is a real number, so it can be negative also
Putting y = −3
x = ±√((−3))
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 𝑓: 𝑁 → 𝑁 be defined by 𝑓(𝑥) = 𝑥2 is ________ (a) Surjective but not Injective (b) Surjective (c) Injective (d) Bijective
f(x) = x2, where 𝑓: N → 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 = –x2
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 ∈ N
x2 = y
x = ±√𝑦
As x is natural number, it will be positive
∴ x = √𝒚
Also, since y is a natural number, let’s put y = 2
Putting y = 2
x = √2
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,….} → {1, 4, 9,….} be defined by 𝑓(𝑥) = 𝑥2 is _________ (a) Bijective (b) Surjective but not Injective (c) Injective but Surjective (d) Neither Surjective nor Injective
f(x) = x2, where 𝑓: {1, 2, 3,….} → {1, 4, 9,….}
Checking one-one
f (x1) = (x1)2
f (x2) = (x2)2
Putting f (x1) = f (x2)
(x1)2 = (x2)2
x1 = x2 or x1 = –x2
Since x ∈ {1, 2, 3,….}
∴ 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 ∈ {1, 4, 9,….}
x2 = y
x = ±√𝑦
As x ∈ {1, 2, 3,….} , it will be positive
∴ x = √𝒚
Also, since y ∈ {1, 4, 9,….}
For all values of y, we will get a value of x,
Where x ∈ {1, 2, 3,….}
So, f is onto
Thus, f(x) is one-one and onto
∴ f(x) is Bijective
So, the correct answer is (a)
Question 4 Let : 𝑁 → 𝑅 be defined by 𝑓(𝑥) = 𝑥2 . 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,…}
For f(x) = x2, where 𝑓: N → R
Range will be = {12, 22, 32, 42, 52, 62, …}
= {1, 4, 9, 16, 25, 36, …}
So, the correct answer is (a)
Question 5 The function f: Z → Z defined by 𝑓(𝑥) = 𝑥2 is__________ (a) Neither Injective nor Surjective (b) Injective (c) Surjective (d) Bijective
f(x) = x2, where 𝑓: Z → 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 = –x2
Since x1 does not have unique image,
It is not one-one
Example:
f(–1) = (–1)2 = 1
f(1) = (1)2 = 1
Here, f(–1) = f(1) , but –1 ≠ 1
Hence, it is not one-one
Check onto
f(x) = x2
Let f(x) = y , such that y ∈ Z
x2 = y
x = ±√𝑦
Note that y is an integer, let’s put y = 5
Putting y = 5
x = ±√5
But this is not possible as ±√5 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 has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.

Hi, it looks like you're using AdBlock :(

Displaying ads are our only source of revenue. To help Teachoo create more content, and view the ad-free version of Teachooo... please purchase Teachoo Black subscription.

Please login to view more pages. It's free :)

Teachoo gives you a better experience when you're logged in. Please login :)

Solve all your doubts with Teachoo Black!

Teachoo answers all your questions if you are a Black user!