Question

lf $x$ is real, then $\frac{(x-1)(x-3)}{(x-2)(x-4)}$ lies in the interval

• A. $(1,3)$
• B. $(2,4)$
• C. $(-\infty ,\infty )$
• D. $(3,4)$

Answer 1

The correct option is C $(-\infty ,\infty )$

Let $y=\frac{(x-1)(x-3)}{(x-2)(x-4)}=\frac{x^2-4x+3}{x^2-6x+8}$
$\Rightarrow y(x^2-6x+8)=x^2-4x+3$
$\Rightarrow x^2(y-1)-2(3y-2)x+(8y-3)=0$
Since $x$ is real $D\ge 0$
$\Rightarrow (3y-2{)}^2-(y-1\left)\right(8y-3)\ge 0$
$\Rightarrow 9y^2-12y+4-(8y^2-11y+3)\ge 0\Rightarrow y^2-y+1\ge 0$
Clearly discriminant of quadratic $y^2-y+1$ is negative
Hence $y^2-y+1>0$ for all $y$
Therefore $y\in (-\infty ,\infty )$