Question

If the fractional part of $\frac{1}{x}$ and $x^2$ for some $x\in (\sqrt{2},\sqrt{3})$ are equal then the value of $x^4-\frac{3}{x}$ is

• A. 0
• B. 1
• C. 2
• D. 5

Answer 1

Answer: (D)

$\Theta \left\{\frac{1}{x}\right\}=\left\{x^2\right\}\Rightarrow \frac{1}{x}=x^2-2$
$\Theta x\in (\sqrt{2},\sqrt{3})$
$\Rightarrow x^3-2x-1=0\Rightarrow (x+1)\left(x^2-x-1\right)=0$
$\Theta x\ne -1\therefore x^2-x-1=0$
$\Rightarrow x^2=x+1\Rightarrow x^4=x^2+2x+1$
$\Rightarrow x^4=3x+2$
$\therefore x^4-\frac{3}{x}=3x+2-3\left(x^2-2\right)=8-3\left(x^2-x\right)$
$=8-3(1)=5$