Q2Long answer
x +1 4 4 x − 1 2 x 1 Solution From the first inequality, we have − ≥0 2 x +1 4 2 x −1 ⇒ ≥0 2 x +1 ⇒ (2x – 1 ≥ 0 and 2x + 1 > 0) or (2x – 1 ≤ 0 and 2x + 1 < 0) [Since 2x + 1 ≠ 0) 1 1 1 1 ⇒ (x ≥ and x > – ) or (x ≤ and x < – ) 2 2 2 2 1 1 ⇒ x≥ or x < – 2 2 1 1 ⇒ x ∈ ( – ∞ , – ) ∪ [ , ∞) ... (1) 2 2 6x 1 From the second inequality, we have 4 x − 1 − 2 < 0 8 x +1 ⇒ <0