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If $\mathrm{P}=\frac{1}{2} \sin ^2 \theta+\frac{1}{3} \cos ^2 \theta$ then
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Verified Answer
The correct answer is:
$\frac{1}{3} \leq \mathrm{P} \leq \frac{1}{2}$
Hints: $P=\frac{1}{2} \sin ^2 \theta+\frac{1}{3} \cos ^2 \theta=\frac{1}{2} \sin ^2 \theta+\frac{1}{3}\left(1-\sin ^2 \theta\right)=\frac{1}{3}+\frac{1}{6} \sin ^2 \theta$
$$
\begin{aligned}
& 0 \leq \sin ^2 \theta \leq 1 \Rightarrow \frac{1}{3} \leq \frac{1}{3}+\frac{1}{6} \sin ^2 \theta \leq \frac{1}{3}+\frac{1}{6} \\
& \Rightarrow \frac{1}{3} \leq \mathrm{P} \leq \frac{1}{2}
\end{aligned}
$$
$$
\begin{aligned}
& 0 \leq \sin ^2 \theta \leq 1 \Rightarrow \frac{1}{3} \leq \frac{1}{3}+\frac{1}{6} \sin ^2 \theta \leq \frac{1}{3}+\frac{1}{6} \\
& \Rightarrow \frac{1}{3} \leq \mathrm{P} \leq \frac{1}{2}
\end{aligned}
$$
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