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Q.
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
Relations and Functions - Part 2
Solution:
$ \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^3-2 x-1=0 \Rightarrow(x+1)\left(x^2-x-1\right)=0 $
$\Theta x \neq-1 \therefore x ^2- x -1=0 $
$\Rightarrow x^2=x+1 \Rightarrow x^4=x^2+2 x+1 $
$\Rightarrow x^4=3 x+2$
$\therefore x ^4-\frac{3}{ x }=3 x +2-3\left( x ^2-2\right)=8-3\left( x ^2- x \right)$
$=8-3(1)=5$