Solveeit Logo
Login

Question

Mathematics Question on inequalities

Solution set of the inequality 12x−1>11−2x−1\frac{1}{2^{x}-1} > \frac{1}{1-2^{x-1}} is

A

(1,∞)(1, \infty)

B

(0,log2(4/3))(0, log_2(4/3))

C

(−1,∞)(-1, \infty)

D

(0,log2(4/3))∪(1,∞)(0, log_2(4/3)) \cup (1, \infty)

Answer

(0,log2(4/3))∪(1,∞)(0, log_2(4/3)) \cup (1, \infty)

Explanation

Solution

Put 2x=t2^x = t. Then t>0t > 0. Now, given inequality becomes 1t−1>22−t⇒1t−1−22−t>0⇒2−t−2t+2(t−1)(2−t)>0\frac{1}{t-1} > \frac{2}{2-t} \Rightarrow \frac{1}{t-1}-\frac{2}{2-t} >0 \Rightarrow \frac{2-t-2t+2}{\left(t-1\right)\left(2-t\right)} > 0 ⇒4−3t(t−1)(2−t)>0⇒(t−43)(t−1)(t−2)>0.\Rightarrow \frac{4-3t}{\left(t-1\right)\left(2-t\right)} >0 \Rightarrow \frac{\left(t-\frac{4}{3}\right)}{\left(t-1\right)\left(t-2\right)} >0. From sign scheme we get 1<t<4/31 < t < 4/3 or t>2.t > 2. ⇒1<2x<4/3\Rightarrow 1 < 2^{x} < 4/3 or 2x>22^{x} > 2 ⇒x∈(0,log2(4/3))∪(1,∞)\Rightarrow x \in\left(0, log_{2}\left(4/3\right)\right)\cup\left(1, \infty\right)