First slide

Introduction to sets

Question

$Solution set of the inequality \frac{1}{2^{x} - 1} > \frac{1}{1 - 2^{x - 1}} is$

Easy

A

$(1, \infty)$
B

$(0, \log_{2} (\frac{4}{3}))$
C

$(- 1, \infty)$
D

$(0, \log_{2} (\frac{4}{3})) \cup (1, \infty)$

Solution

$Put 2^{x} = t . Then t > 0 . Now, given inequality becomes$
$\begin{aligned} \frac{1}{t - 1} > \frac{2}{2 - t} \\ \Rightarrow \frac{1}{t - 1} - \frac{2}{2 - t} > 0 \\ \Rightarrow \frac{2 - t - 2 t + 2}{(t - 1) (2 - t)} > 0 \\ \Rightarrow \frac{4 - 3 t}{(t - 1) (2 - t)} > 0 \\ \Rightarrow \frac{(t - \frac{4}{3})}{(t - 1) (t - 2)} > 0 . \end{aligned}$

From sign scheme, we get
$\begin{array}{l} 1 < t < \frac{4}{3} or t > 2 . \\ \Rightarrow 1 < 2^{x} < \frac{4}{3} or 2^{x} > 2 \\ \Rightarrow x \in (0, \log_{2} (\frac{4}{3})) \cup (1, \infty) \end{array}$

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