Question

If $\frac{x-P}{x^2-3x+2}$ takes all real values for $x\in R$, then the range of $P$ is

• A. $1\le P\le 2$
• B. $1<P<2$
• C. $P<1orP>2$
• D. $P\ge 2$ or $P\le 1$

Answer 1

Answer: (A)

Let $\frac{x-P}{x^2-3x+2}=y$, provided $x^2-3x+2\ne 0$
$\Rightarrow x\ne 1,2$
$\Rightarrow y{x}^2-(3y+1)y+(2y+P)=0$
$\because x\in R$, so $D\ge 0$
$\Rightarrow (3y+1{)}^2-4y(2y+P)\ge 0$
$\Rightarrow y^2+(6-4P)y+1\ge 0$
$\because y\in R$, so $D\le 0$
$\Rightarrow (6-4P{)}^2-4\le 0$
$\Rightarrow (4P-6+2)(4P-6-2)\le 0$
$\Rightarrow (P-1)(P-2)\le 0$
$\Rightarrow P\in [1,2]$