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Question

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


A

(-12,-14]

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B

-1,-12

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C

-32,-54

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D

-54,-1

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Solution

The correct option is B

-1,-12


Explanation for correct option

Given equation is cos4θ+sin4θ+λ=0

λ=-cos4θ+sin4θλ=-cos2θ+sin2θ2-2cos2θsin2θa2+b2=a+b2-2abλ=-12+122cosθsinθ2cos2θ+sin2θ=1λ=12sin22θ-12cosθsinθ=sin2θ

We know that -1sinx1

Therefore 0sin2x1

So 12sin22θ0,12

for 12sin22θ=0λ=0-1=-1

for 12sin22θ=12λ=12-1=-12

λ-1,-12

Hence, Option(B) i.e. -1,-12 is correct


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