The set of all real values of the expression | vedclass

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(C) Let $y = \frac{x^2-x+2}{x^2+x-2}$.Rearranging the terms, we get $y(x^2+x-2) = x^2-x+2$.$yx^2 + yx - 2y = x^2 - x + 2$.$(y-1)x^2 + (y+1)x - (2y+2)

Vedclass · Accepted Answer

$(-\infty, -1] \cup \left[\frac{7}{9}, \infty\right)$

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(C) Let $y = \frac{x^2-x+2}{x^2+x-2}$.Rearranging the terms, we get $y(x^2+x-2) = x^2-x+2$.$yx^2 + yx - 2y = x^2 - x + 2$.$(y-1)x^2 + (y+1)x - (2y+2) = 0$.For $x$ to be a real number, the discriminant $D \ge 0$.$D = (y+1)^2 - 4(y-1)(-2y-2) \ge 0$.$(y+1)^2 + 8(y-1)(y+1) \ge 0$.$(y+1)[(y+1) + 8(y-1)] \ge 0$.$(y+1)(y+1+8y-8) \ge 0$.$(y+1)(9y-7) \ge 0$.The critical points are $y = -1$ and $y = \frac{7}{9}$.Testing the intervals, the inequality holds for $y \in (-\infty, -1] \cup \left[\frac{7}{9}, \infty\right)$.Thus, the correct option is $C$.

The set of all real values of the expression $\frac{x^2-x+2}{x^2+x-2}$ for all $x \in R-\{-2, 1\}$ is

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