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If $A = \sin^{2} θ + \cos^{4} θ$ ,then for all real values of A

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1 \leq A \leq 2

\frac{3}{4} \leq A \leq 1

\frac{13}{16} \leq A \leq 1

\frac{3}{4} \leq A \leq \frac{13}{16}

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detailed solution

Correct option is B

We have, $A = \sin^{2} θ + \cos^{4} θ$

$= \sin^{2} θ + \cos^{2} {θcos}^{2} θ$

$\leq \sin^{2} θ + \cos^{2} θ (since, \cos^{2} θ \leq 1)$

$\Rightarrow \sin^{2} θ + \cos^{4} θ \leq 1 \Rightarrow A \leq 1$

Again, $\begin{aligned} \sin^{2} θ + \cos^{4} θ = 1 - \cos^{2} θ + \cos^{4} θ \\ = \cos^{4} θ - \cos^{2} θ + 1 \\ = {(\cos^{2} θ - \frac{1}{2})}^{2} + \frac{3}{4} \geq \frac{3}{4} \end{aligned}$

Hence, $\frac{3}{4} \leq A \leq 1$

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If A=sin2⁡θ+cos4⁡θ,then for all real values of A