(i) |x 3 – 1| ≥ 1 – x ⇒ |(x – 1)| (x 2 + x + 1) ≥ 1 – x

Case - I x ≥ 1 ⇒ (x – 1) (x 2 + x + 1) + (x – 1) ≥ 0

(x – 1) (x 2 + x + 2) ≥ 0 ⇒ (x – 1) ≥ 0

x ≥ 1 ⇒ x ∈ [1, ∞ )

case - II x < 1

[– (x – 1) (x 2 + x + 1)] + (x – 1) ≥ 0 ⇒ – (x – 1) [x 2 + x + 1 – 1] ≥ 0

(x – 1) (x 2 + x) ≤ 0 = x (x – 1) (x + 1) ≤ 0

x ∈ (– ∞ , –1] ∪ [0, 1)

Taking Union of both the cases, we get x ∈ (– ∞ , –1] ∪ [0, ∞ ) Ans.

(ii) |(x –2) 2 | ≥ 1 ⇒ (x – 2) 2 ≥ 1

(x – 2 + 1) (x – 2 – 1) ≥ 0 ⇒ (x – 1) (x – 3) ≥ 0

x ∈ (– ∞ , 1] ∪ [3, ∞ ) Ans.

(iii) < 2

Case - I x ≤ –2 ⇒ – 2 < 0 ⇒ < 0

> 0 ⇒ > 0 i.e. x ∈ ∪ (0, ∞ )

i.e. x ∈ (– ∞ , –2]

case - II x > –2

< 2 ⇒ –1 <0 ⇒ > 0

i.e. x ∈ (– ∞ , 0) ∪ (1, ∞ ) (Intersection with the given case )

i.e. x ∈ (–2, 0) ∪ (1, ∞ )

(iv) > 0

Case - I x > 2

Case - II x < 2

x ∈ (2, ∞ ) – 1 > 0 Not possible l

x ∈ (2, ∞ ) Ans.

(v) Squaring

x 2 – 4x + 4 > 4x 2 – 12x + 9

3x 2 – 8x + 5 < 0 ⇒ (x – 1)(3x – 5) < 0 ⇒ x ∈

(vi) case - I : x < –2

–x – 2 – x + 3 < –2x – 1 ⇒ 1 < –1 Not possible

Case - II : –2 ≤ x <

x + 2 – x + 3 < – 2x – 1 ⇒ 2x < –6 ⇒ x < –3 Not possible

case-III : ≤ x < 3

x + 2 – x + 3 < 2x + 1 ⇒ x > 2

case -IV : x ≥ 3

x + 2 + x – 3 < 2x + 1 ⇒ 1 > –1 Hence x ∈ (2, ∞ )

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