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If $\mathrm{A}=\sin ^{2} \theta+\cos ^{4} \theta$, the for all real $\theta$, which one of the $\begin{array}{ll}\text { following is correct? }
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The correct answer is:
$\frac{3}{4} \leq \mathrm{A} \leq 1$
$\mathrm{A}=\sin ^{2} \theta+\cos ^{4} \theta$
$=\sin ^{2} \theta+\left(1-\sin ^{2} \theta\right)^{2}$
$=1+\sin ^{4} \theta-\sin ^{2} \theta$
$=1-\sin ^{2} \theta\left(1-\sin ^{2} \theta\right)$
$=1-\sin ^{2} \theta \cdot \cos ^{2} \theta$
$\quad=\frac{4-4 \sin ^{2} \theta \cdot \cos ^{2} \theta}{4}=\frac{4-\sin ^{2}(2 \theta)}{4}$
As, we know, $0 \leq \sin ^{2} 2 \theta \leq 1$
$\therefore \quad A=\frac{4-0}{4}$ or $\frac{4-1}{4} \Rightarrow \frac{3}{4} \leq A \leq 1$
$=\sin ^{2} \theta+\left(1-\sin ^{2} \theta\right)^{2}$
$=1+\sin ^{4} \theta-\sin ^{2} \theta$
$=1-\sin ^{2} \theta\left(1-\sin ^{2} \theta\right)$
$=1-\sin ^{2} \theta \cdot \cos ^{2} \theta$
$\quad=\frac{4-4 \sin ^{2} \theta \cdot \cos ^{2} \theta}{4}=\frac{4-\sin ^{2}(2 \theta)}{4}$
As, we know, $0 \leq \sin ^{2} 2 \theta \leq 1$
$\therefore \quad A=\frac{4-0}{4}$ or $\frac{4-1}{4} \Rightarrow \frac{3}{4} \leq A \leq 1$
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