Question

For the equation $|x^2+{\left(k\right)}^{1/3}|=16\left|x\right|kϵR$ to have a real solution the value of k can be

• A. $\le 2^{16}$
• B. $\le 2^{18}$
• C. $\le 2^{20}$
• D. $\le 2^{21}$

Answer 1

The correct option is B $\le 2^{18}$

Given $|x^2+{\left(k\right)}^{1/3}|=16\left|x\right|,kϵR,xϵR$

$|x+\frac{k^{1/3}}{x}|=16$

We know that

$|ax+\frac{b}{x}|\ge 2\sqrt{ax.\frac{b}{x}}\ge 2\sqrt{ab}$

$\therefore |x+\frac{k^{1/3}}{x}|\ge 2\sqrt{x.\frac{k^{1/3}}{x}}$

$\therefore 2k^{1/6}\le 16$

$k^{1/6}\le 8$

$\therefore k\le 2^{18}$