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If $u=\sqrt{a^2 \cos ^2 \theta+b^2 \sin ^2 \theta}+\sqrt{a^2 \sin ^2 \theta+b^2 \cos ^2 \theta}$, then the difference between the maximum and minimum values of $u^2$ is given by
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Verified Answer
The correct answer is:
$(a-b)^2$
$(a-b)^2$
$u=\sqrt{a^2 \cos ^2 \theta+b^2 \sin ^2 \theta}+\sqrt{a^2 \sin ^2 \theta+b^2 \cos ^2 \theta}$
$=\sqrt{\frac{a^2+b^2}{2}+\frac{a^2-b^2}{2} \cos 2 \theta}+\sqrt{\frac{a^2+b^2}{2}+\frac{b^2-a^2}{2} \cos 2 \theta}$
$\Rightarrow u^2=a^2+b^2+2 \sqrt{\left(\frac{a^2+b^2}{2}\right)^2-\left(\frac{a^2-b^2}{2}\right)^2 \cos ^2 2 \theta}$
min value of $u^2=a^2+b^2+2 a b$
max value of $u^2=2\left(a^2+b^2\right)$
$\Rightarrow u_{\max }^2-u_{\min }^2=(a-b)^2$.
$=\sqrt{\frac{a^2+b^2}{2}+\frac{a^2-b^2}{2} \cos 2 \theta}+\sqrt{\frac{a^2+b^2}{2}+\frac{b^2-a^2}{2} \cos 2 \theta}$
$\Rightarrow u^2=a^2+b^2+2 \sqrt{\left(\frac{a^2+b^2}{2}\right)^2-\left(\frac{a^2-b^2}{2}\right)^2 \cos ^2 2 \theta}$
min value of $u^2=a^2+b^2+2 a b$
max value of $u^2=2\left(a^2+b^2\right)$
$\Rightarrow u_{\max }^2-u_{\min }^2=(a-b)^2$.
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