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A function f : A → B, where A = {x : −1 ≤ x ≤ 1} and B = {y : 1 ≤ y ≤ 2}, is defined by the rule y = f(x) = 1 + x2. Which of the following statement is true?
- a)f is injective but not surjective
- b)f is surjective but not injective
- c)f is both injective and surjective
- d)f is neither injective nor surjective
Correct answer is option 'B'. Can you explain this answer?
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A function f : A → B, where A = {x : −1 ≤ x ≤ 1}and B...
Surjectivity and Injectivity of the Function f(x) = 1 + x^2
Injectivity:
- To check injectivity, we need to see if each element in set A maps to a unique element in set B.
- Let's assume two different values, a and b, in set A such that f(a) = f(b).
- We have: 1 + a^2 = 1 + b^2.
- This implies a^2 = b^2, which further implies a = ±b.
- Since a and b are within the range of -1 to 1, a = b is the only valid solution.
- Thus, f(x) = 1 + x^2 is injective.
Surjectivity:
- To check surjectivity, we need to see if every element in set B has a pre-image in set A.
- In this case, the range of f(x) is [1, 2], which means all y values between 1 and 2 are covered.
- Since any real number squared is non-negative, the minimum value of f(x) is 1.
- However, the maximum value of f(x) is 2, which is achieved when x = 1.
- Therefore, there is no x in set A that maps to the value 2 in set B.
- Hence, f(x) = 1 + x^2 is not surjective.
Therefore, the correct statement is option 'B': f is surjective but not injective.
Injectivity:
- To check injectivity, we need to see if each element in set A maps to a unique element in set B.
- Let's assume two different values, a and b, in set A such that f(a) = f(b).
- We have: 1 + a^2 = 1 + b^2.
- This implies a^2 = b^2, which further implies a = ±b.
- Since a and b are within the range of -1 to 1, a = b is the only valid solution.
- Thus, f(x) = 1 + x^2 is injective.
Surjectivity:
- To check surjectivity, we need to see if every element in set B has a pre-image in set A.
- In this case, the range of f(x) is [1, 2], which means all y values between 1 and 2 are covered.
- Since any real number squared is non-negative, the minimum value of f(x) is 1.
- However, the maximum value of f(x) is 2, which is achieved when x = 1.
- Therefore, there is no x in set A that maps to the value 2 in set B.
- Hence, f(x) = 1 + x^2 is not surjective.
Therefore, the correct statement is option 'B': f is surjective but not injective.
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Community Answer
A function f : A → B, where A = {x : −1 ≤ x ≤ 1}and B...
Since, A = {x : −1 ≤ x ≤ 1}, B = {y : 1 ≤ y ≤ 2} and
y = f(x) = 1 + x2
For x = −1, y = 1 + (−1)2 = 2
And for x = 1, y = 1 + 12 = 2
∴ f is not injective. (one-one)
Here,∀y ∈ B, there is a preimage.
Hence, f is surjective.
y = f(x) = 1 + x2
For x = −1, y = 1 + (−1)2 = 2
And for x = 1, y = 1 + 12 = 2
∴ f is not injective. (one-one)
Here,∀y ∈ B, there is a preimage.
Hence, f is surjective.
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A function f : A → B, where A = {x : −1 ≤ x ≤ 1}and B = {y : 1 ≤ y ≤ 2}, is defined by the rule y = f(x) = 1 + x2. Which of the following statement is true?a)fis injective but not surjectiveb)fis surjective but not injectivec)fis both injective and surjectived)fis neither injective nor surjectiveCorrect answer is option 'B'. Can you explain this answer?
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A function f : A → B, where A = {x : −1 ≤ x ≤ 1}and B = {y : 1 ≤ y ≤ 2}, is defined by the rule y = f(x) = 1 + x2. Which of the following statement is true?a)fis injective but not surjectiveb)fis surjective but not injectivec)fis both injective and surjectived)fis neither injective nor surjectiveCorrect answer is option 'B'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about A function f : A → B, where A = {x : −1 ≤ x ≤ 1}and B = {y : 1 ≤ y ≤ 2}, is defined by the rule y = f(x) = 1 + x2. Which of the following statement is true?a)fis injective but not surjectiveb)fis surjective but not injectivec)fis both injective and surjectived)fis neither injective nor surjectiveCorrect answer is option 'B'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A function f : A → B, where A = {x : −1 ≤ x ≤ 1}and B = {y : 1 ≤ y ≤ 2}, is defined by the rule y = f(x) = 1 + x2. Which of the following statement is true?a)fis injective but not surjectiveb)fis surjective but not injectivec)fis both injective and surjectived)fis neither injective nor surjectiveCorrect answer is option 'B'. Can you explain this answer?.
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