Question

The maximum value if 2 (a-x)(x + $\sqrt{x^2+b^2})$ is ( x , a, b $\in$ R )

• A. a+b
• B. a-b
• C.
• D.

Answer 1

The correct option is D

put t = x + $\sqrt{x^2+b^2}$ $\to$ $\sqrt{x^2+b^2}$ - x = $\frac{b^2}{t}$ $\Rightarrow$ 2x = t - $\frac{b^2}{t}$

x = $\frac{1}{2}\left(\frac{t^2-b^2}{t}\right)$ , so y = 2 (a-x)t = 2at - $t^2+b^2$

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

y $\le$ $a^2+b^2$