Solution set of the inequality (1/2x-1) (1/1-2x-1) is

Question

Solution set of the inequality (1/2x-1) > (1/1-2x-1) is (A) (1, ∞) (B) (0, log2(4/3)) (C) (-1, ∞) (D) (0, log2(4/3)) ∪ (1, ∞)

Tardigrade Admin · Accepted Answer

Put 2x = t. Then t > 0. Now, given inequality becomes (1/t-1) > (2/2-t) ⇒ (1/t-1)-(2/2-t) >0 ⇒ (2-t-2t+2/(t-1)(2-t)) > 0 ⇒ (4-3t/(t-1)(2-t)) >0 ⇒ ((t-(4/3))/(t-1)(t-2)) >0.

From sign scheme we get 1 < t < 4/3 or t > 2. ⇒ 1 < 2x < 4/3 or 2x > 2 ⇒ x ∈(0, log2(4/3))∪(1, ∞)

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

$(1, \infty)$

$(0, l o g_{2} (4/3))$

$(- 1, \infty)$

$(0, l o g_{2} (4/3)) \cup (1, \infty)$

Q. Solution set of the inequality 2x−11​>1−2x−11​ is

(1,∞)

(0,log2​(4/3))

(−1,∞)

(0,log2​(4/3))∪(1,∞)

Put 2x=t. Then t>0. Now, given inequality becomes t−11​>2−t2​⇒t−11​−2−t2​>0⇒(t−1)(2−t)2−t−2t+2​>0 ⇒(t−1)(2−t)4−3t​>0⇒(t−1)(t−2)(t−34​)​>0. From sign scheme we get 1<t<4/3 or t>2. ⇒1<2x<4/3 or 2x>2 ⇒x∈(0,log2​(4/3))∪(1,∞)

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

$(1, \infty)$

$(0, l o g_{2} (4/3))$

$(- 1, \infty)$

$(0, l o g_{2} (4/3)) \cup (1, \infty)$