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The minimum value of $f(x)=a^{2} \cos ^{2} x+b^{2} \sin ^{2} x$ if $a^{2}>b^{2}$, is
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Verified Answer
The correct answer is:
$a^{2}$
Given $f(x)$
$=a^{2} \cos ^{2} x+b^{2} \sin ^{2} x$
$=a^{2}\left(\frac{1+\cos 2 x}{2}\right)+b^{2}\left(\frac{1-\cos 2 x}{2}\right)$
$=\left(\frac{a^{2}+b^{2}}{2}\right)+\left(\frac{a^{2}-b^{2}}{2}\right) \cos 2 x$
$f(x)$ will be maximum when $\cos 2 x=1$
$\mathrm{f}(\mathrm{x})$ will be minimum when $\cos 2 \mathrm{x}=-1$
Hence minimum value of $f(x)$ is
$$
\begin{aligned}
f(x) &=\frac{a^{2}+b^{2}}{2}+\left(\frac{a^{2}-b^{2}}{2}\right)(-1) \\
&=\frac{a^{2}+b^{2}}{2}-\frac{a^{2}-b^{2}}{2}=b^{2}
\end{aligned}
$$
$=a^{2} \cos ^{2} x+b^{2} \sin ^{2} x$
$=a^{2}\left(\frac{1+\cos 2 x}{2}\right)+b^{2}\left(\frac{1-\cos 2 x}{2}\right)$
$=\left(\frac{a^{2}+b^{2}}{2}\right)+\left(\frac{a^{2}-b^{2}}{2}\right) \cos 2 x$
$f(x)$ will be maximum when $\cos 2 x=1$
$\mathrm{f}(\mathrm{x})$ will be minimum when $\cos 2 \mathrm{x}=-1$
Hence minimum value of $f(x)$ is
$$
\begin{aligned}
f(x) &=\frac{a^{2}+b^{2}}{2}+\left(\frac{a^{2}-b^{2}}{2}\right)(-1) \\
&=\frac{a^{2}+b^{2}}{2}-\frac{a^{2}-b^{2}}{2}=b^{2}
\end{aligned}
$$
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