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# For 0≤x≤2π, then 2cosec2x √12y2−y+1≤√2

A
is satisfied by exactly one value of y
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B
is satisfied by exactly two values of x
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C
is satisfied by x for which cosx=12
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D
is satisfied by x for which sinx=0
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Solution

## The correct options are A is satisfied by exactly one value of y B is satisfied by exactly two values of x 2cosec2x √12y2−y+1≤√2 ⇒2cosec2x √y2−2y+22≤√2 ⇒2cosec2x √(y−1)2+1≤2 ⋯(1) Since, cosec2x≥1, ∀ x∈R, we have 2cosec2x≥2 Also, √(y−1)2+1≥1 ⇒2cosec2x √(y−1)2+1≥2 ⋯(2) From (1) and (2), equality holds only when 2cosec2x=2 and √(y−1)2+1=1 Thus, cosec2x=1 and (y−1)2+1=1 ⇒sinx=±1 and y=1 ⇒x=π2,3π2 and y=1  Suggest Corrections  0      Similar questions  Related Videos   General Solutions
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