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(B) Given $f(	heta) = \sqrt{a^2 \cos^2 	heta + b^2 \sin^2 	heta} + \sqrt{a^2 \sin^2 	heta + b^2 \cos^2 	heta}$.Squaring both sides,we get $[f(	h

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$(a-b)^2$

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(B) Given $f(	heta) = \sqrt{a^2 \cos^2 	heta + b^2 \sin^2 	heta} + \sqrt{a^2 \sin^2 	heta + b^2 \cos^2 	heta}$.Squaring both sides,we get $[f(	heta)]^2 = (a^2 \cos^2 	heta + b^2 \sin^2 	heta) + (a^2 \sin^2 	heta + b^2 \cos^2 	heta) + 2 \sqrt{(a^2 \cos^2 	heta + b^2 \sin^2 	heta)(a^2 \sin^2 	heta + b^2 \cos^2 	heta)}$.Simplifying the terms,$[f(	heta)]^2 = a^2 + b^2 + 2 \sqrt{a^4 \sin^2 	heta \cos^2 	heta + a^2 b^2 \cos^4 	heta + a^2 b^2 \sin^4 	heta + b^4 \sin^2 	heta \cos^2 	heta}$.$[f(	heta)]^2 = a^2 + b^2 + 2 \sqrt{a^2 b^2 (\sin^4 	heta + \cos^4 	heta) + (a^4 + b^4) \sin^2 	heta \cos^2 	heta}$.Using $\sin^4 	heta + \cos^4 	heta = 1 - 2 \sin^2 	heta \cos^2 	heta$,we get $[f(	heta)]^2 = a^2 + b^2 + 2 \sqrt{a^2 b^2 (1 - 2 \sin^2 	heta \cos^2 	heta) + (a^4 + b^4) \sin^2 	heta \cos^2 	heta}$.$[f(	heta)]^2 = a^2 + b^2 + 2 \sqrt{a^2 b^2 + (a^4 + b^4 - 2 a^2 b^2) \sin^2 	heta \cos^2 	heta} = a^2 + b^2 + 2 \sqrt{a^2 b^2 + (a^2 - b^2)^2 \sin^2 	heta \cos^2 	heta}$.Let $X = \sin^2 	heta \cos^2 	heta = \frac{\sin^2(2	heta)}{4}$. The range of $X$ is $[0, 1/4]$.For maximum value $M$,$X = 1/4$: $M = a^2 + b^2 + 2 \sqrt{a^2 b^2 + \frac{(a^2 - b^2)^2}{4}} = a^2 + b^2 + 2 \sqrt{\frac{4a^2 b^2 + a^4 + b^4 - 2a^2 b^2}{4}} = a^2 + b^2 + 2 \sqrt{\frac{(a^2 + b^2)^2}{4}} = a^2 + b^2 + (a^2 + b^2) = 2(a^2 + b^2)$.For minimum value $m$,$X = 0$: $m = a^2 + b^2 + 2 \sqrt{a^2 b^2} = a^2 + b^2 + 2ab = (a+b)^2$.Thus,$M - m = 2a^2 + 2b^2 - (a^2 + b^2 + 2ab) = a^2 + b^2 - 2ab = (a-b)^2$.

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 $M-m=$

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