If frac{x-P}{x^2-3x+2} takes all real values | vedclass

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(B) Let $y = \frac{x-P}{x^2-3x+2}$.For $y$ to take all real values,the equation $yx^2 - (3y+1)x + (2y+P) = 0$ must have real roots for $x$ for all $y

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$1 < P < 2$

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(B) Let $y = \frac{x-P}{x^2-3x+2}$.For $y$ to take all real values,the equation $yx^2 - (3y+1)x + (2y+P) = 0$ must have real roots for $x$ for all $y \in \mathbb{R}$.This implies the discriminant $D \geq 0$ for all $y$.$D = (3y+1)^2 - 4y(2y+P) \geq 0$$9y^2 + 6y + 1 - 8y^2 - 4Py \geq 0$$y^2 + (6-4P)y + 1 \geq 0$.For this quadratic in $y$ to be non-negative for all $y \in \mathbb{R}$,the discriminant of this quadratic must be $\leq 0$.$(6-4P)^2 - 4(1)(1) \leq 0$$(6-4P)^2 \leq 4$$|6-4P| \leq 2$$-2 \leq 6-4P \leq 2$$-8 \leq -4P \leq -4$$1 \leq P \leq 2$.However,if $P=1$,$y = \frac{x-1}{(x-1)(x-2)} = \frac{1}{x-2}$,which cannot take the value $0$. If $P=2$,$y = \frac{x-2}{(x-1)(x-2)} = \frac{1}{x-1}$,which cannot take the value $0$. Thus,$P 
eq 1$ and $P 
eq 2$.Therefore,$1 < P < 2$.

If $\frac{x-P}{x^2-3x+2}$ takes all real values for $x \in \mathbb{R} \setminus \{1, 2\}$,then the range of $P$ is

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