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. $P < 1 or P > 2$
• D. $P \ge 2$ or $P \le 1$

Answer 1

Answer: (A)

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