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Q.

If the equation $\cos^{4} θ + \sin^{4} θ + λ = 0$ has a real solution for $θ$ , then $λ$ lies in the interval

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a

(-1/2, -1/4]

b

(-5/4, -1]

c

[-1, -1/2]

d

[-3/2, -5/4]

answer is B.

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Detailed Solution

$\begin{array}{l} \sin^{4} θ + \cos^{4} θ + λ = 0 \Rightarrow - λ = \sin^{4} θ + \cos^{4} θ = {(\sin^{2} θ + \cos^{2} θ)}^{2} - 2 \sin^{2} {θcos}^{2} θ = 1 - \frac{\sin^{2} 2 θ}{2} \\ \Rightarrow λ = - 1 + \frac{1}{2} \sin^{2} 2 θ \in [- 1, - \frac{1}{2}] \end{array}$

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