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 :
[d] \[u=\sqrt{{{a}^{2}}{{\cos }^{2}}\theta +{{b}^{2}}{{\sin }^{2}}\theta }+\sqrt{{{a}^{2}}{{\sin }^{2}}\theta +{{b}^{2}}{{\cos }^{2}}\theta }\] \[\therefore {{u}^{2}}={{a}^{2}}{{\cos }^{2}}\theta +{{b}^{2}}{{\sin }^{2}}\theta +{{a}^{2}}{{\sin }^{2}}\theta +{{b}^{2}}{{\cos }^{2}}\theta \] \[+2\sqrt{({{a}^{2}}co{{s}^{2}}\theta +{{b}^{2}}si{{n}^{2}}\theta )}\sqrt{({{a}^{2}}si{{n}^{2}}\theta +{{b}^{2}}co{{s}^{2}}\theta )}\] \[={{a}^{2}}+{{b}^{2}}+2\sqrt{x({{a}^{2}}+{{b}^{2}}-x)}\] [where \[x={{a}^{2}}{{\cos }^{2}}\theta +{{b}^{2}}{{\sin }^{2}}\theta \]] \[=({{a}^{2}}+{{b}^{2}})+2\sqrt{({{a}^{2}}+{{b}^{2}})x-{{x}^{2}}}\] \[\Rightarrow \frac{d{{u}^{2}}}{dx}=\frac{1}{\sqrt{({{a}^{2}}+{{b}^{2}})x-{{x}^{2}}}}({{a}^{2}}+{{b}^{2}}-2x)\] And \[\frac{dx}{d\theta }=({{b}^{2}}-{{a}^{2}})sin2\theta \] \[\therefore \frac{d{{u}^{2}}}{d\theta }=\frac{({{a}^{2}}+{{b}^{2}}-2x)}{\sqrt{({{a}^{2}}+{{b}^{2}})x-{{x}^{2}}}}\times ({{b}^{2}}-{{a}^{2}}).sin2\theta \] Putting \[d{{u}^{2}}/d\theta =0\] for maximum and minima and \[{{a}^{2}}+{{b}^{2}}=2[{{a}^{2}}co{{s}^{2}}\theta +{{b}^{2}}si{{n}^{2}}\theta ]\], we get \[\sin 2\theta =0,\cos 2\theta ({{b}^{2}}-{{a}^{2}})=0\] \[\Rightarrow \theta =0,\cos 2\theta =0\] \[\Rightarrow \theta =0\,or\,\theta =\pi /4\] \[{{u}^{2}}\] Will be minimum at \[\theta =0\] and will be maximum at \[\theta =\pi /4\] \[\therefore u_{\min }^{2}={{(a+b)}^{2}}and\,\,u_{\max }^{2}=2({{a}^{2}}+{{b}^{2}})\] \[Hence,\text{ }u_{\max }^{2}-u_{\min }^{2}=2({{a}^{2}}+{{b}^{2}})-{{(a+b)}^{2}}={{(a-b)}^{2}}\]
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