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If `u=sqrt(a^2cos^2theta+b^2sin^2theta)+sqrt(a^2sin^2theta+b^2cos^2theta),` then the difference between maximum and minimum values of `u^2` is
A. `2(a^(2)+b^(2))`
B. `2sqrt(a^(2)+b^(2))`
C. `(a+b)^(2)`
D. `(a-b)^(2)`

1 Answer

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Best answer
Correct Answer - D
`u^(2)=a^(2)+b^(2)+2sqrt(a^(2)cos^(2)theta+b^(2)sin theta)xxsqrt(a^(2)sin^(2)theta+b^(2)cos^(2)theta)`
`=a^(2)+b^(2)+2sqrt(sin^(2)thetacos^(2)theta(a^(4)+b^(4))+a^(2)b^(2)(sin^(4)theta+cos^(4)theta))`
`=a^(2)+b^(2)+2sqrt(a^(2)b^(2)(1-2sin^(2)thetacos^(2)theta)+(a^(4)+b^(4))sin^(2)thetacos^(2)theta)`
`=(a^(2)+b^(2))+2sqrt(a^(2)b^(2)+(a^(2)-b^(2))^(2)sin^(2)thetacos^(2)theta)`
`=(a^(2)+b^(2))+2sqrt(a^(2)b^(2)+((a^(2)-b^(2))^(2))/(4))`
Min. `u^(2)=(a^(2)+b^(2))+2ab`.
`rArr` difference `=2sqrt(a^(2)b^(2)+((a^(2)-b^(2))^(2))/(4))-2ab`
`=sqrt(4a^(2)b^(2)+a^(4)+b^(4)-2a^(2)b^(2))-2ab`
`=sqrt((a^(2)+b^(2))^(2))-2ab`.
`=a^(2)+b^(2)-2ab`
`=(a-b)^(2)`.

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