Trigonometric Identities Question 217
Question: 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
Options:
A) $ 2(a^{2}+b^{2}) $
B) $ 2\sqrt{a^{2}+b^{2}} $
C) $ {{(a+b)}^{2}} $
D) $ {{(a-b)}^{2}} $
Show Answer
Answer:
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}cos^{2}\theta +b^{2}sin^{2}\theta )}\sqrt{(a^{2}sin^{2}\theta +b^{2}cos^{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{du^{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{du^{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 $ du^{2}/d\theta =0 $ for maximum and minima and $ a^{2}+b^{2}=2[a^{2}cos^{2}\theta +b^{2}sin^{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}}andu_{\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}} $
BETA
