Let A = \{ \theta∊R:(1/3\sin (\theta )+2/3cos (\theta ) )^2 =1/3\sin^2(\theta ) +2 / 3 \cos^2(\theta )\} . Then

Correct Answer is : A∩[0,π] has exactly one point

A \cap [0, \pi] has exactly one point

A \cap [0, \pi] has exactly two points

A \cap [0, \pi] has more than two points

KVPY 2019 SB SX Exam Previous Papers

Question : 2 of 120

Marks: +1, -0

Let

A = \{ θ∊R:(1/3\sin (θ )+2/3cos (θ ) )^2

=1/3\sin^2(θ ) +2 / 3 \cos^2(θ )\}

. Then

Solution:

{\sin^2θ+ 4\cos^2θ+ 4\sinθ \cosθ}/9

={\sin^2θ+2\cos^2θ}/3

⇒

2 \sin^2θ+ 2 \cos^2θ-4 \sinθ\cosθ= 0

⇒

\sin2θ= 1

⇒ 2θ = 2nπ+π/2

⇒ θ = nπ+π/4

⇒

A ∩ [0, π]=\{π/4\}

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