Let \(M\) and \(m\) respectively denote the maximum and the minimum values of \([f(\theta)]^2\), where…
Let \(M\) and \(m\) respectively denote the maximum and the minimum values of \([f(\theta)]^2\), where \(f(\theta)=\sqrt{a^2 \cos ^2 \theta+b^2 \sin ^2 \theta}\) \(+\sqrt{a^2 \sin ^2 \theta+b^2 \cos ^2 \theta}\). Then \(M-m=\)
\(a^2+b^2\)
\((a-b)^2\)
\(a^2 b^2\)
\((a+b)^2\)
Solution
If
\(\begin{aligned}
& f(\theta)=\sqrt{a^2 \cos ^2 \theta+b^2 \sin ^2 \theta}+\sqrt{a^2 \sin ^2 \theta+b^2 \cos ^2 \theta} \\
& \therefore[f(\theta)]^2=a^2 \cos ^2 \theta+b^2 \sin ^2 \theta+a^2 \sin ^2 \theta+b^2 \cos ^2 \theta \\
& +2 \sqrt{\left(a^2 \cos ^2 \theta+b^2 \sin ^2 \theta\right)\left(a^2 \sin ^2 \theta+b^2 \cos ^2 \theta\right)}
\end{aligned}\)
\(\because[f(\theta)]^2\) will be maximum, if \(\sin ^2 \theta=\cos ^2 \theta=\frac{1}{2}\) and will be minimum, if either \(\sin ^2 \theta=0\) or \(\cos ^2 \theta=0\).
\(\begin{aligned}
& \therefore M=\left(a^2+b^2\right)+2 \sqrt{\left(\frac{a^2}{2}+\frac{b^2}{2}\right)^2}=2\left(a^2+b^2\right) \\
& \text { and } m=\left(a^2+b^2\right)+2 \sqrt{a^2 b^2}=a^2+b^2+2 a b \\
& \therefore M-m=2\left(a^2+b^2\right)-\left[a^2+b^2+2 a b\right] \\
& =a^2+b^2-2 a b=(a-b)^2
\end{aligned}\)
Hence, option (2) is correct.