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JEE MainMathematicsEasyInequalities and absolute value

The complete solution set of
∣2x−1x−1∣>2\left| \frac{2x - 1}{x - 1} \right| > 2 is

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2x−1x−1∣>2\frac{2x - 1}{x - 1} \mid > 2
⇒(2x−1)(x−1)>2    ∵∣x−1∣⇒+ve\Rightarrow \frac{(2x - 1)}{(x - 1)} > 2\ \ \ \ \because|x - 1| \Rightarrow \text{+ve}
⇒∣2x−1∣>2∣x−1∣\Rightarrow |2x - 1| > 2|x - 1|
⇒(2x−1)−2∣x−1∣>0\Rightarrow (2x - 1) - 2|x - 1| > 0
Case 1; x<1/2x < 1/2
⇒−2x+1+2x−2>0\Rightarrow - 2x + 1 + 2x - 2 > 0
⇒−1>0⇒\Rightarrow - 1 > 0 \RightarrowNo values.
Case 2: 12≤x<1\frac{1}{2} \leq x < 1
2x−1+2x−2>02x - 1 + 2x - 2 > 0
4x−3>04x - 3 > 0
⇒x>3/4\Rightarrow x > 3/4
⇒x∈(34,1)\Rightarrow x \in \left( \frac{3}{4},1 \right)
Case 3: x≥1x \geq 1
2x−1−2(x−1)>02x - 1 - 2(x - 1) > 0
⇒1>0\Rightarrow 1 > 0 Possible for all values of x>1x > 1
x≠1x \neq 1
⇒x∈(1,  ∞)\Rightarrow x \in (1,\ \ \infty)
Taking union of all cases.
x∈(34,  1)∪(1, ∞)x \in \left( \frac{3}{4},\ \ 1 \right) \cup (1,\ \infty)

Source: Grass-G / M-05 Mathematics Paper 10 October.docx

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