If the equation sin4θ+cos4θ=a has a real solution then

A
$a \leq \frac{1}{2}$
B
$a \geq \frac{1}{2}$
C
$\frac{1}{2} \leq a \leq 1$
D
$a \geq 0$

Solution:

We have,

$a = \sin^{4} θ + \cos^{4} θ \leq \sin^{2} θ + \cos^{2} θ \leq 1$

Also,

$\begin{array}{l} a = \sin^{4} θ + \cos^{4} θ = {(\sin^{2} θ + \cos^{2} θ)}^{2} - 2 \sin^{2} θ \cos^{2} θ \\ \Rightarrow a = \sin^{4} θ + \cos^{4} 0 = 1 - \frac{1}{2} \sin^{2} 2 θ \\ \Rightarrow \sin^{2} 20 = 2 (1 - a) \Rightarrow 2 (1 - a) \leq 1 \Rightarrow a \geq \frac{1}{2} \end{array}$

Hence, $\frac{1}{2} \leq a \leq 1$

If the equation $\sin^{4} θ + \cos^{4} θ = a$ has a real solution then

Solution:

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