Question

If $xϵR$ then maximum value of $R=2(a-x)(x-\sqrt{x^2-b^2})$ is

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

Answer 1

Answer: (C)

$a^2-b^2$

(C) : Let $z=x-\sqrt{x^2-b^2}\therefore \frac{1}{z}=\frac{x+\sqrt{x^2-b^2}}{b^2}$

$z=x-\sqrt{x^2-b^2}$

$\therefore \frac{b^2}{z}=x+\sqrt{x^2-b^2}$

$\therefore z+\frac{b^2}{z}=2x$

and $\frac{b^2}{z}-z=2\sqrt{x^2-b^2}$

Now $2(a-x)(x-\sqrt{x^2-b^2})=(2a-2x)z$ $=(2a-(z+\frac{b^2}{z}))z=2az-z^2-b^2$ $=a^2-b^2-(a^2+z^2-2az)$ $=(a^2-b^2)-{(a-z)}^2\le a^2-b^2$

$\therefore$ Maximum value $=a^2-b^2.$