Let R-(alpha, beta) be the range of f(x) = fr | vedclass

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(B) Let $y = \frac{x+3}{x^2+x-2}$. Then $y(x^2+x-2) = x+3$,which implies $yx^2 + (y-1)x - (2y+3) = 0$. For $x$ to be a real number,the discriminant $D

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$10$

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(B) Let $y = \frac{x+3}{x^2+x-2}$. Then $y(x^2+x-2) = x+3$,which implies $yx^2 + (y-1)x - (2y+3) = 0$. For $x$ to be a real number,the discriminant $D \geq 0$. $D = (y-1)^2 - 4(y)(-(2y+3)) = y^2 - 2y + 1 + 8y^2 + 12y = 9y^2 + 10y + 1 \geq 0$. Factoring the quadratic,we get $(9y+1)(y+1) \geq 0$. The solution to this inequality 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 equation of the line is $-x - \frac{1}{9}y + 1 = 0$,which simplifies to $x + \frac{1}{9}y = 1$. The $x$-intercept is $1$ and the $y$-intercept is $9$. The sum of the intercepts is $1 + 9 = 10$.

Let $R-(\alpha, \beta)$ be the range of $f(x) = \frac{x+3}{(x-1)(x+2)}$. Then,the sum of the intercepts of the line $\alpha x + \beta y + 1 = 0$ on the coordinate axes is:

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