Question

If $f\left(x\right)=\sqrt{a^2{\cos }^{2}x+b^2{\sin }^{2}x}+\sqrt{a^2{\sin }^{2}x+b^2{\cos }^{2}x}$ then the difference between the maximum and minimum values of ${\left\{f\left(x\right)\right\}}^2$ 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 ${(a-b)}^2$

$\left(f\right(x){)}^2=h(x)=a^2+b^2+2\sqrt{(a^2{\sin }^{2}x+b^2{\cos }^{2}x)(a^2{\cos }^{2}x+b^2{\sin }^{2}x)}$

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

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

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

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

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

Now
$h\left(x\right)$ is maximum at $x=\frac{\pi }{4}$

and minimum at $x=\frac{\pi }{2}$

Hence
$h\left(\frac{\pi }{4}\right)=h_{max}=a^2+b^2+2\sqrt{\frac{(a^2-b^2{)}^2+4a^2{b}^2}{4}}$

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

$=a^2+b^2+2\sqrt{\frac{(a^2+b^2{)}^2}{4}}$

$=2(a^2+b^2)$ ...(i)
And
$f\left(\frac{\pi }{2}\right)$ gives
$a^2+b^2+2ab$ ...(ii)

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