Question

$\left|x-1-x^2\right|\le \left|x^2-3x+4\right|$

Answer 1

$\mid x-1-x^2\mid \le \mid x^2-3x+4\mid$

1. $\mid x-1-x^2\mid \le x^2-3x+4$`

a. $-(x^2-3x+4)\le (x-1-x^2)$

$\Rightarrow 2x-3\le 0$

$\Rightarrow x\le \frac{3}{2}$ - Equation a

b. $x-1-x^2\le x^2-3x+4$

$\Rightarrow 2x^2-4x+5\ge 0$

$D\equiv 16-40<0,a=2>0$ solution is $x\in R$ - Equation b

2. $x^2-3x+4\le -\mid x-1-x^2\mid$

c. $(x-1-x^2)\le -x^2+3x-4$

$\Rightarrow 2x-3\ge 0$

$\Rightarrow x\ge \frac{3}{2}$ -Equation c

d. $(x-1-x^2)\ge x^2-3x+4$

$\Rightarrow 2x^2-4x+5\le 0$

$\Rightarrow D\equiv 16-40<0,a=2>0,2x^2-4x+5>0\forall x\in R$

Not $<0$ - Equation d

Range is $x\in (-\infty ,\frac{3}{2}]$