Question

If $u = \sqrt{a^{2} cos^{2} θ + b^{2} sin^{2} θ} + \sqrt{a^{2} sin^{2} θ + b^{2} cos^{2} θ}, $ 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} cos^{2} θ + b^{2} sin^{2} θ} + \sqrt{a^{2} sin^{2} θ + b^{2} cos^{2} θ}, $
$= \sqrt{\frac{a^{2}+b^{2}}{2}+\frac{a^{2}-b^{2}}{2}cos \,2\theta}+\sqrt{\frac{a^{2}+b^{2}}{2}+\frac{b^{2}-a^{2}}{2}cos \,2\theta }$
$\Rightarrow u^{2} = a^{2}+b^{2}+2\sqrt{\left(\frac{a^{2}+b^{2}}{2}\right)^{2}-\left(\frac{a^{2}-b^{2}}{2}\right)^{2}cos^{2} \,2\theta }$
min value of $u^{2} = a^{2} + b^{2} + 2ab$
max value of $u^{2} = 2\left(a^{2}+b^{2} \right)$
$⇒ u^{2}_{max} − u^{2}_{min} = \left(a − b\right)^{2}.$