Solve for x
(i) 2 |x+1| + 2 |x| = 6 and x ∈ Ι
(ii) x 2 + x + 1 + |x – 3| ≤ |x 2 + 2x – 2|
(iii) |2x – 4| – 2|x 2 + x – 3| + 2|x – 1||x + 1| = 0
Text Solution
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(i) –2, 1 (ii) [3, ∞ ) (iii) [–1,1] ∪ [2, ∞ )
Sol. (i) 2 |x+1| + 2 |x| = 6 (x ∈ Ι )
Case – I
x ∈ (– ∞ ,–1]
2 –x–1 + 2 –x = 6 ⇒ 
(2 –x ) = 6 ⇒ x = –2
Case - II
x ∈ [0, ∞ )
2 x+1 + 2 x = 6 ⇒ 3.2 x = 6 ⇒ x = 1
(ii) x 2 + x + 1 + |3 – x| = |x 2 + 2x – 2|
⇒ |x 2 + x + 1| + |x – 3| = |x 2 + 2x – 2|
⇒ (x 2 + x + 1) (x – 3) ≥ 0 ⇒ x ≥ 3
(iii) |2x – 4| – 2|x 2 + x – 3| + 2|x – 1||x + 1| = 0
⇒ |x – 2| + |x 2 –1| = |x 2 + x – 3|
⇒ (x – 1) (x + 1) (x – 2) ≥ 0
x ∈ [–1,1] ∪ [2, ∞ )
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