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Question

If the equation cos4θ+sin4θ+λ=0 has real solutions for θ, then λ lies in the interval:

A
[1,12]
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B
(54,1)
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C
(12,14]
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D
[32,54]
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Solution

The correct option is A [1,12]
cos4θ+sin4θ+λ=0
λ={112sin2 2θ}
2(λ+1)=sin22θ
02 (λ+1)1
0λ+112
1λ12

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