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Let R be the set of all real numbers. Consider the sets P = {x ∈ R : (x – 1)(x2 + 1) = 0}, Q = {x ∈ R : x2 – 9x + 2) = 0} and S = {x ∈ R : x = 5y for some y ∈ R} Then the set (P∩ S) ∪ Q contains
- a)exactly two elements
- b)exactly three elements
- c)exactly four elements
- d)infinitely many elements
Correct answer is option 'B'. Can you explain this answer?
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Most Upvoted Answer
Let R be the set of all real numbers. Consider the sets P = {x R : (x...
**Solution:**
To find the set (PS) Q, we need to find the elements that are common to both sets PS and Q.
**Set P:**
The equation (x - 1)(x^2 - 1) = 0 can be factorized as (x - 1)(x - 1)(x + 1) = 0. So, the solutions to this equation are x = 1 and x = -1.
Therefore, set P = {1, -1}.
**Set Q:**
The equation x^2 - 9x - 2 = 0 can be solved using the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / (2a)
Plugging in the values a = 1, b = -9, and c = -2, we get:
x = (9 ± √(81 + 8)) / 2
Simplifying further, we get:
x = (9 ± √89) / 2
Therefore, the solutions to this equation are x = (9 + √89) / 2 and x = (9 - √89) / 2.
Therefore, set Q = {(9 + √89) / 2, (9 - √89) / 2}.
**Set PS:**
To find the set PS, we need to find the elements that satisfy both sets P and S.
Set S is defined as {x ∈ R : x = 5y for some y ∈ R}.
So, for an element x to be in set PS, it must satisfy both the conditions x ∈ P and x ∈ S.
For x = 1, it satisfies the condition x ∈ P but not the condition x ∈ S.
For x = -1, it satisfies the condition x ∈ P but not the condition x ∈ S.
Therefore, the set PS = ∅ (empty set).
**Set (PS) Q:**
Now, we need to find the elements that are common to both sets PS and Q.
Since the set PS is empty, there are no elements common to both PS and Q.
Therefore, the set (PS) Q = ∅ (empty set).
Hence, the correct option is (B) exactly three elements.
To find the set (PS) Q, we need to find the elements that are common to both sets PS and Q.
**Set P:**
The equation (x - 1)(x^2 - 1) = 0 can be factorized as (x - 1)(x - 1)(x + 1) = 0. So, the solutions to this equation are x = 1 and x = -1.
Therefore, set P = {1, -1}.
**Set Q:**
The equation x^2 - 9x - 2 = 0 can be solved using the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / (2a)
Plugging in the values a = 1, b = -9, and c = -2, we get:
x = (9 ± √(81 + 8)) / 2
Simplifying further, we get:
x = (9 ± √89) / 2
Therefore, the solutions to this equation are x = (9 + √89) / 2 and x = (9 - √89) / 2.
Therefore, set Q = {(9 + √89) / 2, (9 - √89) / 2}.
**Set PS:**
To find the set PS, we need to find the elements that satisfy both sets P and S.
Set S is defined as {x ∈ R : x = 5y for some y ∈ R}.
So, for an element x to be in set PS, it must satisfy both the conditions x ∈ P and x ∈ S.
For x = 1, it satisfies the condition x ∈ P but not the condition x ∈ S.
For x = -1, it satisfies the condition x ∈ P but not the condition x ∈ S.
Therefore, the set PS = ∅ (empty set).
**Set (PS) Q:**
Now, we need to find the elements that are common to both sets PS and Q.
Since the set PS is empty, there are no elements common to both PS and Q.
Therefore, the set (PS) Q = ∅ (empty set).
Hence, the correct option is (B) exactly three elements.
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Let R be the set of all real numbers. Consider the sets P = {x R : (x 1)(x2 + 1) = 0}, Q = {x R : x2 9x + 2) = 0} and S = {x R : x = 5y for some y R} Then the set (PS) Q containsa)exactly two elementsb)exactly three elementsc)exactly four elementsd)infinitely many elementsCorrect answer is option 'B'. Can you explain this answer?
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Let R be the set of all real numbers. Consider the sets P = {x R : (x 1)(x2 + 1) = 0}, Q = {x R : x2 9x + 2) = 0} and S = {x R : x = 5y for some y R} Then the set (PS) Q containsa)exactly two elementsb)exactly three elementsc)exactly four elementsd)infinitely many elementsCorrect answer is option 'B'. Can you explain this answer? for IIT JAM 2024 is part of IIT JAM preparation. The Question and answers have been prepared according to the IIT JAM exam syllabus. Information about Let R be the set of all real numbers. Consider the sets P = {x R : (x 1)(x2 + 1) = 0}, Q = {x R : x2 9x + 2) = 0} and S = {x R : x = 5y for some y R} Then the set (PS) Q containsa)exactly two elementsb)exactly three elementsc)exactly four elementsd)infinitely many elementsCorrect answer is option 'B'. Can you explain this answer? covers all topics & solutions for IIT JAM 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Let R be the set of all real numbers. Consider the sets P = {x R : (x 1)(x2 + 1) = 0}, Q = {x R : x2 9x + 2) = 0} and S = {x R : x = 5y for some y R} Then the set (PS) Q containsa)exactly two elementsb)exactly three elementsc)exactly four elementsd)infinitely many elementsCorrect answer is option 'B'. Can you explain this answer?.
Let R be the set of all real numbers. Consider the sets P = {x R : (x 1)(x2 + 1) = 0}, Q = {x R : x2 9x + 2) = 0} and S = {x R : x = 5y for some y R} Then the set (PS) Q containsa)exactly two elementsb)exactly three elementsc)exactly four elementsd)infinitely many elementsCorrect answer is option 'B'. Can you explain this answer? for IIT JAM 2024 is part of IIT JAM preparation. The Question and answers have been prepared according to the IIT JAM exam syllabus. Information about Let R be the set of all real numbers. Consider the sets P = {x R : (x 1)(x2 + 1) = 0}, Q = {x R : x2 9x + 2) = 0} and S = {x R : x = 5y for some y R} Then the set (PS) Q containsa)exactly two elementsb)exactly three elementsc)exactly four elementsd)infinitely many elementsCorrect answer is option 'B'. Can you explain this answer? covers all topics & solutions for IIT JAM 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Let R be the set of all real numbers. Consider the sets P = {x R : (x 1)(x2 + 1) = 0}, Q = {x R : x2 9x + 2) = 0} and S = {x R : x = 5y for some y R} Then the set (PS) Q containsa)exactly two elementsb)exactly three elementsc)exactly four elementsd)infinitely many elementsCorrect answer is option 'B'. Can you explain this answer?.
Solutions for Let R be the set of all real numbers. Consider the sets P = {x R : (x 1)(x2 + 1) = 0}, Q = {x R : x2 9x + 2) = 0} and S = {x R : x = 5y for some y R} Then the set (PS) Q containsa)exactly two elementsb)exactly three elementsc)exactly four elementsd)infinitely many elementsCorrect answer is option 'B'. Can you explain this answer? in English & in Hindi are available as part of our courses for IIT JAM.
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Let R be the set of all real numbers. Consider the sets P = {x R : (x 1)(x2 + 1) = 0}, Q = {x R : x2 9x + 2) = 0} and S = {x R : x = 5y for some y R} Then the set (PS) Q containsa)exactly two elementsb)exactly three elementsc)exactly four elementsd)infinitely many elementsCorrect answer is option 'B'. Can you explain this answer?, a detailed solution for Let R be the set of all real numbers. Consider the sets P = {x R : (x 1)(x2 + 1) = 0}, Q = {x R : x2 9x + 2) = 0} and S = {x R : x = 5y for some y R} Then the set (PS) Q containsa)exactly two elementsb)exactly three elementsc)exactly four elementsd)infinitely many elementsCorrect answer is option 'B'. Can you explain this answer? has been provided alongside types of Let R be the set of all real numbers. Consider the sets P = {x R : (x 1)(x2 + 1) = 0}, Q = {x R : x2 9x + 2) = 0} and S = {x R : x = 5y for some y R} Then the set (PS) Q containsa)exactly two elementsb)exactly three elementsc)exactly four elementsd)infinitely many elementsCorrect answer is option 'B'. Can you explain this answer? theory, EduRev gives you an
ample number of questions to practice Let R be the set of all real numbers. Consider the sets P = {x R : (x 1)(x2 + 1) = 0}, Q = {x R : x2 9x + 2) = 0} and S = {x R : x = 5y for some y R} Then the set (PS) Q containsa)exactly two elementsb)exactly three elementsc)exactly four elementsd)infinitely many elementsCorrect answer is option 'B'. Can you explain this answer? tests, examples and also practice IIT JAM tests.
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