Question

Solution of $\left|\frac{x}{x-1}\right|+|x|=\frac{x^2}{|x-1|}$ is $x\in \left\{t\right\}\cup (k,\infty )$ . Find $k+t$

Answer 1

Let $f\left(x\right)=\frac{x}{x-1}$ and $g\left(x\right)=x$
$\therefore f\left(x\right)+g\left(x\right)=\frac{x}{x-1}+x=\frac{x^2}{x-1}$
Using, $|f\left(x\right)|+|g\left(x\right)|=|f\left(x\right)+g\left(x\right)|$
if $f\left(x\right)\cdot g\left(x\right)\ge 0$
$\Rightarrow \frac{x}{x-1}x\ge 0\Rightarrow \frac{x^2}{x-1}\ge 0$
$\Rightarrow x\in \left\{0\right\}\cup (1,\infty )$
Thus $k+t=1$