Sets, Relations & Functions
Sets
Grade 11
Question:
<p>Let \(P = \{\theta : \sin\theta - \cos\theta = \sqrt{2}\cos\theta\}\) and \(Q = \{\theta : \sin\theta + \cos\theta = \sqrt{2}\sin\theta\}\) be two sets. Then</p>
<p>\(P \subset Q\) and \(Q - P \neq \phi\)</p>
<p>\(Q \not\subset P\)</p>
<p>\(P = Q\)</p>
<p>\(P \not\subset Q\)</p>
Step-by-Step Solution
Key Concept: Isolate the trigonometric functions and square both sides strategically to find common solutions. Recognize that both sets can be reduced to finding when specific trigonometric relationships hold, then find their intersection or relationship.
<p><strong>Step 1:</strong> Simplify set P: sin θ - cos θ = √2 cos θ</p><p>⟹ sin θ = cos θ + √2 cos θ = (1 + √2) cos θ</p><p>⟹ tan θ = 1 + √2</p><p><strong>Step 2:</strong> Simplify set Q: sin θ + cos θ = √2 sin θ</p><p>⟹ cos θ = √2 sin θ - sin θ = (√2 - 1) sin θ</p><p>⟹ cot θ = √2 - 1</p><p>⟹ tan θ = 1/(√2 - 1) = (√2 + 1)/(2 - 1) = √2 + 1 = 1 + √2</p><p><strong>Step 3:</strong> Both sets P and Q give tan θ = 1 + √2</p><p>Therefore P = Q, and P ∩ Q = P = Q</p><p><strong>Answer: C (P = Q)</strong></p>
Correct Answer: C