If R-(alpha, beta) is the range of frac{x+3}{ | vedclass

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(B) Let $y = \frac{x+3}{(x-1)(x+2)}$.$(x-1)(x+2)y = x+3$$y(x^2+x-2) = x+3$$yx^2 + (y-1)x - (2y+3) = 0$.Since $x \in R$,the discriminant $D \geq 0$:$(y

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$10$

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(B) Let $y = \frac{x+3}{(x-1)(x+2)}$.$(x-1)(x+2)y = x+3$$y(x^2+x-2) = x+3$$yx^2 + (y-1)x - (2y+3) = 0$.Since $x \in R$,the discriminant $D \geq 0$:$(y-1)^2 + 4y(2y+3) \geq 0$$y^2 - 2y + 1 + 8y^2 + 12y \geq 0$$9y^2 + 10y + 1 \geq 0$$(9y+1)(y+1) \geq 0$.The solution is $y \in (-\infty, -1] \cup [-\frac{1}{9}, \infty)$.Thus,the range is $R - (-1, -\frac{1}{9})$.Comparing with $R - (\alpha, \beta)$,we get $\alpha = -1$ and $\beta = -\frac{1}{9}$.The line equation is $-x - \frac{1}{9}y + 1 = 0$,or $x + \frac{1}{9}y = 1$.This can be written as $\frac{x}{1} + \frac{y}{9} = 1$.The intercepts are $1$ and $9$.The sum of the intercepts is $1 + 9 = 10$.

If $R-(\alpha, \beta)$ 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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