If the equation cos4θ+sin4θ+λ=0 has real solutions for θ, then λ lies in the interval
(-12,-14]
-1,-12
-32,-54
-54,-1
Explanation for correct option
Given equation is cos4θ+sin4θ+λ=0
⇒λ=-cos4θ+sin4θ⇒λ=-cos2θ+sin2θ2-2cos2θsin2θ∵a2+b2=a+b2-2ab⇒λ=-12+122cosθsinθ2∵cos2θ+sin2θ=1⇒λ=12sin22θ-1∵2cosθsinθ=sin2θ
We know that -1≤sinx≤1
Therefore 0≤sin2x≤1
So 12sin22θ∈0,12
for 12sin22θ=0⇒λ=0-1=-1
for 12sin22θ=12⇒λ=12-1=-12
∴λ∈-1,-12
Hence, Option(B) i.e. -1,-12 is correct