\(\frac{1}{6}, \frac{1}{4}\)
3, -2
-3, 2
0,1
Correct answer is B
The expression is not defines when y2 - y - 6 = 0 => y2 - 3y + 2y - 6 = 0 => (y-3)(y+2) = 0 => y -3 = 0 or y + 2 = 0 => y = 3 or -2
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