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The range of $\cos^{2} x + \sin^{4} x$ is

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[\frac{3}{4}, 1]

[\frac{3}{2}, 2]

[\frac{1}{2}, 1]

[1, \frac{3}{2}]

answer is D.

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Detailed Solution

\sin^{4} x + 1 - \sin^{2} x = {(\sin^{2} x - \frac{1}{2})}^{2} + \frac{3}{4}

$0 \leq \sin^{2} x \leq 1$

$- \frac{1}{2} \leq \sin^{2} x - \frac{1}{2} \leq \frac{1}{2}$

$0 \leq {(\sin^{2} x - \frac{1}{2})}^{2} \leq \frac{1}{4}$

$\Rightarrow \frac{3}{4} \leq {(\sin^{2} x - \frac{1}{2})}^{2} + \frac{3}{4} \leq 1$

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