A) \[2({{a}^{2}}+{{b}^{2}})\]
B) \[2\sqrt{{{a}^{2}}+{{b}^{2}}}\]
C) \[{{(a+b)}^{2}}\]
D) \[{{(a-b)}^{2}}\]
Correct Answer: D
Solution :
\[u=\sqrt{{{a}^{2}}{{\cos }^{2}}\theta +{{b}^{2}}{{\sin }^{2}}\theta }\] \[+\sqrt{{{a}^{2}}{{\sin }^{2}}\theta +{{b}^{2}}{{\cos }^{2}}\theta }\] \[{{u}^{2}}={{a}^{2}}{{\cos }^{2}}\theta +{{b}^{2}}{{\sin }^{2}}\theta +{{a}^{2}}{{\sin }^{2}}\theta +{{b}^{2}}{{\cos }^{2}}\theta \] \[+2\sqrt{({{a}^{2}}+{{\cos }^{2}}\theta +{{b}^{2}}{{\sin }^{2}}\theta )}\sqrt{({{a}^{2}}{{\sin }^{2}}\theta +{{b}^{2}}{{\cos }^{2}}\theta )}\] \[{{u}^{2}}={{a}^{2}}+{{b}^{2}}+2\sqrt{x({{a}^{2}}+{{b}^{2}}-x)}\] [where\[x={{a}^{2}}{{\cos }^{2}}\theta +{{b}^{2}}{{\sin }^{2}}\theta \]] \[{{u}^{2}}=({{a}^{2}}+{{b}^{2}})+2\sqrt{({{a}^{2}}+{{b}^{2}})x-{{x}^{2}}}\] \[\frac{d{{u}^{2}}}{d\theta }=\frac{1}{\sqrt{({{a}^{2}}+{{b}^{2}})x-{{x}^{2}}}}({{a}^{2}}+{{b}^{2}}-2x)\times \frac{dx}{d\theta }\] and\[\frac{dx}{d\theta }=({{b}^{2}}-{{a}^{2}})\sin 2\theta \] \[\frac{d{{u}^{2}}}{d\theta }=\frac{({{a}^{2}}+{{b}^{2}}-2x)}{\sqrt{({{a}^{2}}+{{b}^{2}})x-{{x}^{2}}}}\times ({{b}^{2}}-{{a}^{2}})\sin 2\theta \] For maxima and minima put \[\frac{{{d}^{2}}u}{d\theta }=0\] \[{{a}^{2}}+{{b}^{2}}=2[{{a}^{2}}co{{s}^{2}}\theta +{{b}^{2}}si{{n}^{2}}\theta ]\] and \[sin\,2\theta =0\] \[\Rightarrow \] \[\cos 2\theta ({{b}^{2}}-{{a}^{2}})=0\] \[\Rightarrow \] \[\theta =0,\cos 2\theta =0\] \[\Rightarrow \] \[2\theta =\frac{\pi }{2}\Rightarrow \theta =\frac{\pi }{4}\] \[{{u}^{2}}\]will be minimum at\[\theta =0\]and will be maximum at\[\theta =\frac{\pi }{4}\] \[\therefore \]\[u_{\min }^{2}={{(a+b)}^{2}}\]and\[u_{\max }^{2}=2({{a}^{2}}+{{b}^{2}})\] Hence, \[u_{\max }^{2}-u_{\min }^{2}=2({{a}^{2}}+{{b}^{2}})-{{(a+b)}^{2}}\] \[={{(a-b)}^{2}}\]
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