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

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

Answer 1

The correct option is D $(a-b{)}^2$

Given $u=\sqrt{a^2{\cos }^2\theta +b^2{\sin }^2\theta }+\sqrt{a^2{\sin }^2\theta +b^2{\cos }^2\theta }$

$\therefore u^2=(a^2+b^2){\cos }^2\theta +(a^2+b^2){\sin }^2\theta +2\sqrt{a^2{\cos }^2\theta +b^2{\sin }^2\theta }\sqrt{a^2{\sin }^2\theta +b^2{\cos }^2\theta }$

$u^2=a^2+b^2+2\sqrt{a^4{\cos }^2\theta {\sin }^2\theta +b^4{\cos }^2\theta {\sin }^2\theta +a^2{b}^2({\cos }^4\theta +{\sin }^4\theta )}$

$=a^2+b^2+2\sqrt{(a^4+b^4){\cos }^2\theta {\sin }^2\theta +a^2{b}^2(1-2{\cos }^2\theta {\sin }^2\theta )}$

$u^2=a^2+b^2+2\sqrt{\frac{a^4+b^4}{4}(\sin 2\theta {)}^2+a^2{b}^2(1-\frac{{\sin }^{2}2\theta }{2})}$

$u^2=a^2+b^2+2\sqrt{\frac{(a^2-b^2{)}^2}{4}(\sin 2\theta {)}^2+a^2{b}^2}$

For max. value of $u^2,$ ${\sin }^{2}2\theta =1$

$u^2=a^2+b^2+2\frac{\sqrt{(a^2+b^2{)}^2}}{2}=2(a^2+b^2)$

And for min. $u^2,\sin 2\theta =0$

$u^2=a^2+b^2+2ab=(a+b{)}^2$

$\therefore$ Difference between max. and min. value of $u^2$ is $2(a^2+b^2)-(a+b{)}^2=(a-b{)}^2$.