Question

If $u=\sqrt{a^{2}co{s}^2\theta +b^{2}si{n}^2\theta }+\sqrt{a^{2}si{n}^2\theta +b^{2}co{s}^2\theta }$, then the difference between the maximum and minimum values of $u^2$ is given by

• A.
• B.
• C.
• D.

Answer 1

The correct option is D

$u^2=a^2+b^2+2\sqrt{(a^{2}co{s}^2\theta +b^{2}si{n}^2\theta )(a^{2}si{n}^2\theta +b^{2}co{s}^2\theta )}=a^2+b^2+2\sqrt{a^{4}co{s}^2\theta si{n}^2\theta +b^{4}si{n}^2\theta co{s}^2\theta +a^2{b}^2(si{n}^4\theta +co{s}^4\theta )}=a^2+b^2+2\sqrt{(a^4+b^4)si{n}^2\theta co{s}^2\theta +a^2{b}^2(1-2si{n}^2\theta +co{s}^2\theta )}=a^2+b^2+2\sqrt{si{n}^2\theta co{s}^2\theta (a^2-b^2)+a^2{b}^2}=a^2+b^2+2\sqrt{a^2{b}^2+(a^2-b^2)\frac{si{n}^{2}2\theta }{4}}=(u^2{)}_{max}=a^2+b^2+2\sqrt{a^2{b}^2\frac{(a^2-b^2{)}^2}{4}}=2(a^2+b^2)=(u^2{)}_{min}=a^2+b^2+2\sqrt{a^2{b}^2}=(a+b{)}^2=(u^2{)}_{max}-(u^2{)}_{min}=2(a^2+b^2)-(a+b{)}^2=(a-b{)}^2$