The solution of 2 x + 2 | x | ≥ 2 2 is:

( - ∞ , log 2 2 - 1 ] ∪ [ 1 2 , ∞ )

The solution of 2 x + 2 | x | ≥ 2 2 is:

( - ∞ , log 2 2 - 1 ] ∪ [ 1 2 , ∞ )

Solve ( x + 3 ) 5 - x - 1 5 ≥ 244 .

Given, ( x + 3 ) 5 - ( x - 1 ) 5 ≥ 244 . . . ( i ) ( x + 1 + 2 ) 5 - ( x + 1 - 2 ) 5 ≥ 244 Let λ = x + 1 . . . ( i i ) ( λ + 2 ) 5 - ( λ - 2 ) 5 ≥ 244 Using the binomial theorem we get, 2 { c 1 5 . λ 4 . 2 1 + c 3 5 . λ 2 . 2 3 + c 5 5 . λ 0 . 2 5 } ≥ 244 2 { 10 . λ 4 + 80 . λ 2 + 32 } ≥ 244 5 λ 4 + 40 λ 2 + 16 ≥ 61 5 λ 4 + 40 λ 2 - 45 ≥ 0 λ 4 + 8 λ 2 - 9 ≥ 0 Now, let λ 2 = t ... i i i t 2 + 8 t - 9 ≥ 0 ( t + 9 ) ( t - 1 ) ≥ 0 t ≤ - 9 or t ≥ 1 Using equation i i i we get, λ 2 ≤ - 9 (This is not a possible solution.) Hence, λ 2 ≥ 1 λ 2 - 1 ≥ 0 ( λ + 1 ) ( λ - 1 ) ≥ 0 Using equation (ii) we get, ( x + 1 + 1 ) ( x + 1 - 1 ) ≥ 0 ( x + 2 ) ( x ) ≥ 0 x ∈ ( - ∞ , - 2 ] ∪ [ 0 , ∞ ) .

Solve x - 2 - 1 ≥ 3 .

Given, x - 2 - 1 ≥ 3 When x ≥ a , x ∈ ( - ∞ , - a ] ∪ [ a , ∞ ) : a > 0 Using the above concept, we can write, x - 2 - 1 ∈ ( - ∞ , - 3 ] ∪ [ 3 , ∞ ) Adding one on both sides of the equation, we get, x - 2 ∈ ( - ∞ , - 2 ] ∪ [ 4 , ∞ ) . x cannot be negative, so neglecting the negative portion we can write, x - 2 ≥ 4 Using the concept x ≥ a , x ∈ ( - ∞ , - a ] ∪ [ a , ∞ ) : a > 0 . Now, we can write, x - 2 ∈ ( - ∞ , - 4 ] ∪ [ 4 , ∞ ) Adding two on both sides of the equation, we get, x ∈ ( - ∞ , - 2 ] ∪ [ 6 , ∞ ) This is the required solution set for the given equation.