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?

(a), (b), (c)

(c)

None of the choices

All of the above

Correct answer is B

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

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