Question

If $\mu =\sqrt{a^2{\cos }^2\theta +b^2{\sin }^2\theta }+\sqrt{a^2{\sin }^2\theta +b^2{\cos }^2\theta }$ then the difference between maximum and minimum values of ${\mu }^2$ is:

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

Answer 1

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

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

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

$Case-I$ : at $\theta =0$ or $\theta =\frac{\pi }{2}$

${\mu }^2\left(min\right)=a^2+b^2+2ab$

$Case-II$ : at $\theta =\frac{\pi }{4}$

${\mu }^2\left(max\right)=a^2+b^2+(a^2+b^2)=2a^2+2b^2$

${\mu }_{max}^2-{\mu }_{min}^2=2a^2+2b^2-(a^2+b^2+2ab)=a^2+b^2-2ab=(a-b{)}^2.[D]$