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.

(a), (b), (c)

B.

(c)

C.

None of the choices

D.

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