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. $2(a^2+b^2)$
• B. $2\sqrt{a^2+b^2}$
• C. $(a+b{)}^2$
• D. $(a-b{)}^2$

Answer 1

Answer: (D)

$u=\sqrt{a^{2}co{s}^2\theta +b^{2}si{n}^2\theta }+\sqrt{a^{2}si{n}^2\theta +b^{2}co{s}^2\theta },$
$=\sqrt{\frac{a^2+b^2}{2}+\frac{a^2-b^2}{2}cos2\theta }+\sqrt{\frac{a^2+b^2}{2}+\frac{b^2-a^2}{2}cos2\theta }$
$\Rightarrow u^2=a^2+b^2+2\sqrt{{\left(\frac{a^2+b^2}{2}\right)}^2-{\left(\frac{a^2-b^2}{2}\right)}^{2}co{s}^{2}2\theta }$
min value of $u^2=a^2+b^2+2ab$
max value of $u^2=2\left(a^2+b^2\right)$
$\Rightarrow u_{max}^2-u_{min}^2={\left(a-b\right)}^2.$