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For any two real numbers a and b, we define a R b if and only if sin2 a + cos2b = 1. The relation R is
- a)Reflexive but not Symmetric
- b)Symmetric but not transitive
- c)Transitive but not Reflexive
- d)an Equivalence relation
Correct answer is option 'D'. Can you explain this answer?
Verified Answer
For any two real numbers a and b, we define a R b if and only if sin2 ...
sin2a + cos2b = 1
Reflexive : sin2a + cos2a = 1
⇒ aRa
sin2a + cos2b = 1,
1 – cos2a + 1 – sin2b
= 1
sin2b + cos2a = 1
⇒ bRa
Hence symmetric Let aRb, bRc
sin2a + cos2b = 1 ............. (1)
sin2b + cos2c = 1 ............... (2)
(1) + (2)
sin2a + cos2c = 1
Hence transitive therefore equivalence relation.
Reflexive : sin2a + cos2a = 1
⇒ aRa
sin2a + cos2b = 1,
1 – cos2a + 1 – sin2b
= 1
sin2b + cos2a = 1
⇒ bRa
Hence symmetric Let aRb, bRc
sin2a + cos2b = 1 ............. (1)
sin2b + cos2c = 1 ............... (2)
(1) + (2)
sin2a + cos2c = 1
Hence transitive therefore equivalence relation.
Most Upvoted Answer
For any two real numbers a and b, we define a R b if and only if sin2 ...
Explanation:
To determine the properties of the given relation R, we need to check if it is reflexive, symmetric, and transitive.
Reflexive Property:
A relation R is reflexive if every element of the set is related to itself. In this case, we need to check if a R a holds for any real number a.
Let's substitute a = a in the given equation: sin^2(a) cos^2(a) = 1.
Using the identity sin^2(a) + cos^2(a) = 1, we can rewrite the equation as: sin^2(a) (1 - sin^2(a)) = 1.
Expanding the equation further: sin^2(a) - sin^4(a) = 1.
Rearranging the terms: sin^4(a) - sin^2(a) + 1 = 0.
This is a quadratic equation in sin^2(a). For any real number a, the quadratic equation does not have any real solutions. Therefore, a R a does not hold.
Since the relation is not reflexive, option A can be eliminated.
Symmetric Property:
A relation R is symmetric if whenever a R b, then b R a must also hold for any real numbers a and b.
Let's assume a R b is true, which means sin^2(a) cos^2(b) = 1. Now, we need to check if b R a holds, i.e., sin^2(b) cos^2(a) = 1.
Using the commutative property of multiplication, we can rearrange the equation as cos^2(a) sin^2(b) = 1.
Comparing this equation with the original equation, it is clear that they are the same. Therefore, b R a holds whenever a R b holds.
Since the relation is symmetric, option B can be eliminated.
Transitive Property:
A relation R is transitive if whenever a R b and b R c, then a R c must also hold for any real numbers a, b, and c.
Let's assume a R b and b R c are true, which means sin^2(a) cos^2(b) = 1 and sin^2(b) cos^2(c) = 1. Now, we need to check if a R c holds, i.e., sin^2(a) cos^2(c) = 1.
We can multiply the two equations to get sin^2(a) cos^2(b) sin^2(b) cos^2(c) = 1.
Using the identity sin^2(x) + cos^2(x) = 1, we can simplify the equation as sin^2(a) (1 - sin^2(a)) sin^2(c) (1 - sin^2(c)) = 1.
Expanding and simplifying further: sin^4(a) sin^2(c) - sin^6(a) sin^2(c) + sin^2(a) sin^4(c) - sin^6(a) sin^4(c) = 1.
This equation does not simplify to 1, and hence, a R c does not hold for all real numbers a, b, and c.
Since the relation is not transitive, option C
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For any two real numbers a and b, we define a R b if and only if sin2 a + cos2b = 1. The relation R isa)Reflexive but not Symmetricb)Symmetric but not transitive c)Transitive but not Reflexived)an Equivalence relationCorrect answer is option 'D'. Can you explain this answer? for JEE 2025 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about For any two real numbers a and b, we define a R b if and only if sin2 a + cos2b = 1. The relation R isa)Reflexive but not Symmetricb)Symmetric but not transitive c)Transitive but not Reflexived)an Equivalence relationCorrect answer is option 'D'. Can you explain this answer? covers all topics & solutions for JEE 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for For any two real numbers a and b, we define a R b if and only if sin2 a + cos2b = 1. The relation R isa)Reflexive but not Symmetricb)Symmetric but not transitive c)Transitive but not Reflexived)an Equivalence relationCorrect answer is option 'D'. Can you explain this answer?.
For any two real numbers a and b, we define a R b if and only if sin2 a + cos2b = 1. The relation R isa)Reflexive but not Symmetricb)Symmetric but not transitive c)Transitive but not Reflexived)an Equivalence relationCorrect answer is option 'D'. Can you explain this answer? for JEE 2025 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about For any two real numbers a and b, we define a R b if and only if sin2 a + cos2b = 1. The relation R isa)Reflexive but not Symmetricb)Symmetric but not transitive c)Transitive but not Reflexived)an Equivalence relationCorrect answer is option 'D'. Can you explain this answer? covers all topics & solutions for JEE 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for For any two real numbers a and b, we define a R b if and only if sin2 a + cos2b = 1. The relation R isa)Reflexive but not Symmetricb)Symmetric but not transitive c)Transitive but not Reflexived)an Equivalence relationCorrect answer is option 'D'. Can you explain this answer?.
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