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 = π‘₯2 .

Raji Visited the Exhibition - Teachoo.jpg

Answer the following questions using the above information.

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Question 1

Let 𝑓: 𝑅 𝑅 be defined by 𝑓(π‘₯) = π‘₯ 2 is_________
(a) Neither Surjective nor Injective   (b) Surjective
(c) Injective   (d) Bijective

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Question 2

Let 𝑓: 𝑁 𝑁 be defined by 𝑓(π‘₯) = π‘₯ 2 is ________
(a) Surjective but not Injective   (b) Surjective
(c) Injective   (d) Bijective

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Question 3

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

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Question 4

Let : 𝑁 → 𝑅 be defined by 𝑓(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,…}

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Question 5

The function f: Z Z defined by 𝑓(π‘₯) = π‘₯ 2 is__________
(a) Neither Injective nor Surjective    (b) Injective
(c) Surjective   (d) Bijective

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  1. Chapter 1 Class 12 Relation and Functions
  2. Serial order wise

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 = π‘₯2 . Answer the following questions using the above information. Question 1 Let 𝑓: 𝑅 β†’ 𝑅 be defined by 𝑓(π‘₯) = π‘₯2 is_________ (a) Neither Surjective nor Injective (b) Surjective (c) Injective (d) Bijective f(x) = x2, where 𝑓: 𝑅 β†’ 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 β‰  1 Hence, it is not one-one Check onto f(x) = x2 Let f(x) = y , such that y ∈ R x2 = y x = Β±βˆšπ‘¦ Note that y is a real number, so it can be negative also Putting y = βˆ’3 x = ±√((βˆ’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 β†’ 𝑁 be defined by 𝑓(π‘₯) = π‘₯2 is ________ (a) Surjective but not Injective (b) Surjective (c) Injective (d) Bijective f(x) = x2, where 𝑓: 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 = –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 = Β±βˆšπ‘¦ As x is natural number, it will be positive ∴ x = βˆšπ’š 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,….} β†’ {1, 4, 9,….} be defined by 𝑓(π‘₯) = π‘₯2 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,….} β†’ {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 = Β±βˆšπ‘¦ As x ∈ {1, 2, 3,….} , it will be positive ∴ x = βˆšπ’š 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 β†’ 𝑅 be defined by 𝑓(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,…} For f(x) = x2, where 𝑓: N β†’ 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 β†’ Z defined by 𝑓(π‘₯) = π‘₯2 is__________ (a) Neither Injective nor Surjective (b) Injective (c) Surjective (d) Bijective f(x) = x2, where 𝑓: 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 = –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 β‰  1 Hence, it is not one-one Check onto f(x) = x2 Let f(x) = y , such that y ∈ Z x2 = y x = Β±βˆšπ‘¦ 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)

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Davneet Singh
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.