Question

For any real $x$, the expression $2(K-x)[x+\sqrt{x^2+K^2}]$ cannot exceed

• A. $4K^2$
• B. $2K^2$
• C. $8K^2$
• D. None of these

Answer 1

The correct option is B

$2K^2$

Explanation for the correct option:

Step 1. Let $f\left(x\right)=2(K-x)[x+\sqrt{x^2+K^2}]$

$f\text{'}\left(x\right)=2\left[(K-x)\left(1+\frac{1}{2\sqrt{x^2+K^2}}\times 2x\right)+\left(x+\sqrt{x^2+K^2}\times -1\right)\right]$

Step 2. For maximum value, $f\text{'}\left(x\right)=0$

$2\left[(K-x)\left(1+\frac{1}{2\sqrt{x^2+K^2}}\times 2x\right)+\left(x+\sqrt{x^2+K^2}\times -1\right)\right]=0$

$\Rightarrow$ $(K-x)\left[\frac{\sqrt{x^2+K^2}+x}{\sqrt{x^2+K^2}}\right]=\sqrt{x^2+K^2}+x$

$\Rightarrow$ $(K-x)\left(x+\sqrt{x^2+K^2}\right)=x\sqrt{x^2+K^2}+x^2+K^2$

$\Rightarrow$ $x(K-x)+(K-x)\sqrt{x^2+K^2}=x\sqrt{x^2+K^2}+x^2+K^2$

$\Rightarrow$ $\left(K-x-x\right)\sqrt{x^2+K^2}=x^2+K^2+x^2-Kx$

$\Rightarrow$ $\left(K-2x\right)\sqrt{x^2+K^2}=2x^2-Kx+K^2$

Step 3. Squaring both side, we get

$\left(K^2+4x^2-4Kx\right)\left(x^2+K^2\right)={\left(2x^2-Kx+K^2\right)}^2$

$\Rightarrow$$K^2{x}^2+K^4+4x^4+4x^2{K}^2-4K{x}^3-4K^{3}x=4x^4+K^2{x}^2+K^4-4K{x}^3-2K^{3}x+4K^2{x}^2$

$\Rightarrow$ $-4K^{3}x=-2K^{3}x$

$\Rightarrow$ $-2K^{3}x=0$

$\Rightarrow$ $x=0$

At $x=0$,

$\begin{array}{rcl}f\left(x\right)& =& 2K\left(0+\sqrt{K^2}\right)\\ & =& 2K^2\end{array}$

$\therefore$$2(K-x)[x+\sqrt{x^2+K^2}]$ Cannot exceed $\begin{array}{rcl}& & \mathbf{2}{\mathit{K}}^{\mathbf{2}}\end{array}$

Hence, Option ‘B’ is Correct.