Question

Let $M$ and $m$ respectively denote the maximum and the minimum values of $[f(\theta)]^{2}$, where $f(\theta)=\sqrt{a^{2} \cos ^{2} \theta+b^{2} \sin ^{2} \theta}$ $+\sqrt{a^{2} \sin ^{2} \theta+b^{2} \cos ^{2} \theta}$. Then $M-m=$

• A. $a^2 + b^2$
• B. $(a -b)^2$
• C. $a^2 b^2$
• D. $(a + b)^2$

Answer 1

The Correct Option is B

If $f(\theta)=\sqrt{a^{2} \cos ^{2} \theta+ b^{2} \sin ^{2} \theta}+\sqrt{a^{2} \sin ^{2} \theta +b^{2} \cos ^{2} \theta}$ $\therefore[f(\theta)]^{2}=a^{2} \cos ^{2} \theta +b^{2} \sin ^{2} \theta +a^{2} \sin ^{2} \theta +b^{2} \cos ^{2} \theta$ $+2 \sqrt{\left(a^{2} \cos ^{2} \theta +b^{2} \sin ^{2} \theta\right)\left(a^{2} \sin ^{2} \theta +b^{2} \cos ^{2} \theta\right)}$ $\because[f(\theta)]^{2}$ will be maximum, if $\sin ^{2} \theta=\cos ^{2} \theta=\frac{1}{2}$ and will be minimum, if either $\sin ^{2} \theta=0$ or $\cos ^{2} \theta=0$ $\therefore M=\left(a^{2}+b^{2}\right)+2 \sqrt{\left(\frac{a^{2}}{2}+\frac{b^{2}}{2}\right)^{2}}$ $=2\left(a^{2}+b^{2}\right)$ and $m=\left(a^{2}+b^{2}\right)+2 \sqrt{a^{2} b^{2}}$ $=a^{2}+b^{2}+2 a b$ $\therefore M-m=2\left(a^{2}+b^{2}\right)-\left[a^{2}+b^{2}+2 a b\right]$ $=a^{2}+b^{2}-2 a b=(a-b)^{2}$