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If $\mathbb{R}-(\alpha, \beta)$ is the range of $\frac{x+3}{(x-1)(x+2)}$, then the sum of the intercepts of the line $\alpha x+\beta y+1=0$ on the coordinate axes is
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Verified Answer
The correct answer is:
$10$
$y=\frac{(x+3)}{(x-1)(x+2)}$
$$
\begin{aligned}
& (x-1)(x+2) y=x+3 \\
& x^2 y+(y-1) x-(2 y+3)=0 \\
& \because x \in \mathrm{R} \\
& \therefore \mathrm{D} \geq 0 \\
& (y-1)^2+4 y(2 y+3) \geq 0 \\
& (9 y+1)(y+1) \geq 0 \\
& y \in(-\infty,-1] \cup\left[-\frac{1}{9}, \infty\right) \Rightarrow y \in \mathrm{R}-\left(-1, \frac{-1}{9}\right) \\
& \alpha=-1 \\
& \beta=\frac{-1}{9} \\
& \alpha x+\beta y+1=0 \\
& -x-\frac{1}{9} y+1=0 \Rightarrow \frac{x}{1}+\frac{y}{9}=1
\end{aligned}
$$
$\therefore$ Sum of intercepts $=1+9=10$
$$
\begin{aligned}
& (x-1)(x+2) y=x+3 \\
& x^2 y+(y-1) x-(2 y+3)=0 \\
& \because x \in \mathrm{R} \\
& \therefore \mathrm{D} \geq 0 \\
& (y-1)^2+4 y(2 y+3) \geq 0 \\
& (9 y+1)(y+1) \geq 0 \\
& y \in(-\infty,-1] \cup\left[-\frac{1}{9}, \infty\right) \Rightarrow y \in \mathrm{R}-\left(-1, \frac{-1}{9}\right) \\
& \alpha=-1 \\
& \beta=\frac{-1}{9} \\
& \alpha x+\beta y+1=0 \\
& -x-\frac{1}{9} y+1=0 \Rightarrow \frac{x}{1}+\frac{y}{9}=1
\end{aligned}
$$
$\therefore$ Sum of intercepts $=1+9=10$
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