Which of the following values of the variable x, (a)x = 0, (b)x = -3, (c)x = 9, satisfy the inequalities 0 < \(\frac{x + 3}{x - 1}\) < 2?
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Correct Answer: Option C
Explanation:
0 < \(\frac{x + 3}{x - 1}\) < 2
Put x = 0, -3 and 9
0 < \(\frac{9 + 3}{9 - 1}\) \(\leq\) 2
i.e. 0 < 1.5 \(\leq\) 2 (true)
but 0 < \(\frac{0 + 3}{0 - 1}\) \(\leq\) 2
i.e. 0 < -3 \(\leq\) 2 (not true)
-3 \(\leq\) 2
-3 is not greater than 0
0 < \(\frac{x + 3}{x - 1}\) < 2
Put x = 0, -3 and 9
0 < \(\frac{9 + 3}{9 - 1}\) \(\leq\) 2
i.e. 0 < 1.5 \(\leq\) 2 (true)
but 0 < \(\frac{0 + 3}{0 - 1}\) \(\leq\) 2
i.e. 0 < -3 \(\leq\) 2 (not true)
-3 \(\leq\) 2
-3 is not greater than 0