Question

If $\sqrt{\frac{x^2-4x+3}{x^2-3x+2}}\le 1,$ then the interval(s) in which $x$ can not lie, is/are

• A. $(2,\infty )$
• B. $[3,\infty )$
• C. $(1,2)$
• D. $(-\infty ,1)$

Answer 1

Answer: (A), (C), (D)

The correct options are
A $(2,\infty )$
C $(1,2)$
D $(-\infty ,1)$

For $\sqrt{\frac{x^2-4x+3}{x^2-3x+2}}\le 1$ to be true
$\frac{x^2-4x+3}{x^2-3x+2}\ge 0$
$\Rightarrow \frac{(x-1)(x-3)}{(x-2)(x-1)}\ge 0,x\ne 2,1$
$\therefore x\in (-\infty ,1)\cup (1,2)\cup [3,\infty )\cdots \left(1\right)$

Now,
$\sqrt{\frac{x^2-4x+3}{x^2-3x+2}}\le 1$
Squaring both side, we get
$\frac{(x-1)(x-3)}{(x-2)(x-1)}\le 1$
$\Rightarrow \frac{(x-3)}{(x-2)}\le 1,x\ne 1$
$\Rightarrow \frac{(x-3)}{(x-2)}-1\le 0,x\ne 1$
$\Rightarrow \frac{x-3-x+2}{x-2}\le 0,x\ne 1$
$\Rightarrow \frac{-1}{x-2}\le 0,x\ne 1$
$\Rightarrow \frac{1}{x-2}\ge 0,x\ne 1,2$
$\therefore x\in (2,\infty )\cdots \left(2\right)$
From equation $\left(1\right)$ and $\left(2\right),$ we get
$x\in [3,\infty )$