Question

Solve the system
$x^2+2|x-3|-10<0,$ $\frac{|x^2-4x|+3}{x^2+|x-5|}\ge 1$,

The values of $x$ satisfying both the inequalities is

• A. $x\in [1-\sqrt{5},0)\cup (\frac{1}{2},-1+\sqrt{17})$
• B. $x(1-\sqrt{17},-2/3)\cup [\frac{1}{2},\sqrt{5}-1)$
• C. $x\in (1-\sqrt{17},-2/3)\cup (\frac{1}{2},\sqrt{5}-1]$
• D. None of these.

Answer 1

The correct option is A $x\in [1-\sqrt{5},0)\cup (\frac{1}{2},-1+\sqrt{17})$

$x^2+2|x-3|-10<0,$ .....(1)
$\frac{|x^2-4x|+3}{x^2+|x-5|}\ge 1$ ....(2)
For inequality (1),
When$x<3$
$x^2-2(x-3)-10<0$
$x^2-2x-4<0$
$x\in (1-\sqrt{5},1+\sqrt{5})$
But since, $x<3$
So, $x\in (1-\sqrt{5},3)$
When $x\ge 3$
$x^2+2(x-3)-10<0$
$x^2+2x-16<0$
$x\in (-1-\sqrt{17},-1+\sqrt{17})$
But since, $x\ge 3$
$\Rightarrow x\in [3,-1+\sqrt{17})$
So, the solution of inequality (1) is $(1-\sqrt{5},-1+\sqrt{17})$
For inequality (2),
When $x<0$
$\frac{x^2-4x+3}{x^2-(x-5)}\ge 1$
$x^2-4x+3\ge x^2-x+5$
$\Rightarrow x\le \frac{-2}{3}$
$x\in (-\infty ,\frac{-2}{3}]$
$\Rightarrow x\in (-\infty ,0)$
When $0\le x<4$
$\frac{{-(x}^2-4x)+3}{x^2-(x-5)}\ge 1$
$\Rightarrow 2x^2-5x+2\le 0$
$x\in [\frac{1}{2},4)$
When$4\le x<5$
$\frac{{(x}^2-4x)+3}{x^2-(x-5)}\ge 1$
$\Rightarrow x^2-4x+3\ge x^2-x+5$
$\Rightarrow x\le \frac{-2}{3}$
$x\in (-\infty ,\frac{-2}{3}]$
Hence, no solution.
When$x\ge 5$
$\frac{(x^2-4x)+3}{x^2+(x-5)}\ge 1$
$\Rightarrow x\le \frac{8}{5}$
$x\in (-\infty ,\frac{8}{5}]$
But $x\ge 5$
No solution.
Hence, the solution of inequality (2) is $(-\infty ,0)\cup [\frac{1}{2},4)$
Hence, the common solution of both the inequalities is
$(1-\sqrt{5},0)\cup [\frac{1}{2},-1+\sqrt{17})$