Set Theory and types of Sets MCQ Quiz in मराठी - Objective Question with Answer for Set Theory and types of Sets - मोफत PDF डाउनलोड करा

Last updated on Mar 23, 2025

पाईये Set Theory and types of Sets उत्तरे आणि तपशीलवार उपायांसह एकाधिक निवड प्रश्न (MCQ क्विझ). हे मोफत डाउनलोड करा Set Theory and types of Sets एमसीक्यू क्विझ पीडीएफ आणि बँकिंग, एसएससी, रेल्वे, यूपीएससी, स्टेट पीएससी यासारख्या तुमच्या आगामी परीक्षांची तयारी करा.

Latest Set Theory and types of Sets MCQ Objective Questions

Set Theory and types of Sets Question 1:

समजा, \(P = \left \{\theta : \sin \theta - \cos \theta = \sqrt {2} \cos \theta \right \}\) आणि \(Q = \left \{\theta : \sin \theta + \cos \theta = \sqrt {2}\sin \theta \right \}\) असे दोन संच आहेत. तर:

  1. \(P\subset Q\) आणि \(Q - P \neq \phi\)
  2. \(Q\not{\subset} P\)
  3. \(P = Q\)
  4. \(P\not{\subset} Q\)

Answer (Detailed Solution Below)

Option 3 : \(P = Q\)

Set Theory and types of Sets Question 1 Detailed Solution

\(\sin \theta - \cos \theta = \sqrt {2} \cos \theta\)

\(\Rightarrow \sin \theta = \cos \theta + \sqrt {2} \cos \theta\)

\(= (\sqrt {2} + 1)\cos \theta = \left (\dfrac {2 - 1}{\sqrt {2} - 1}\right ) \cos \theta\)

\(\Rightarrow (\sqrt {2} - 1)\sin\theta = \cos \theta \)

\(\Rightarrow \sin \theta + \cos \theta = \sqrt {2}\sin \theta\)

\(\therefore P = Q\)

Top Set Theory and types of Sets MCQ Objective Questions

Set Theory and types of Sets Question 2:

समजा, \(P = \left \{\theta : \sin \theta - \cos \theta = \sqrt {2} \cos \theta \right \}\) आणि \(Q = \left \{\theta : \sin \theta + \cos \theta = \sqrt {2}\sin \theta \right \}\) असे दोन संच आहेत. तर:

  1. \(P\subset Q\) आणि \(Q - P \neq \phi\)
  2. \(Q\not{\subset} P\)
  3. \(P = Q\)
  4. \(P\not{\subset} Q\)

Answer (Detailed Solution Below)

Option 3 : \(P = Q\)

Set Theory and types of Sets Question 2 Detailed Solution

\(\sin \theta - \cos \theta = \sqrt {2} \cos \theta\)

\(\Rightarrow \sin \theta = \cos \theta + \sqrt {2} \cos \theta\)

\(= (\sqrt {2} + 1)\cos \theta = \left (\dfrac {2 - 1}{\sqrt {2} - 1}\right ) \cos \theta\)

\(\Rightarrow (\sqrt {2} - 1)\sin\theta = \cos \theta \)

\(\Rightarrow \sin \theta + \cos \theta = \sqrt {2}\sin \theta\)

\(\therefore P = Q\)

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