Solve the following equations (where [•] denotes greatest integer function and {•} represent fractional part function and sgn represents signum function)
(i) [x] + |x – 2| ≤ 0 and x ∈ [–1,3]
(ii) [2x] – 2x = [x + 1]
(iii) [x 2 ] + 2 [x] = 3x, 0 ≤ x ≤ 2
Text Solution
Verified by Experts(I) [X]
(i) no solution (ii) –1, – 
(iii) 0,1
Sol. (i) [x] + |x–2| = 0, x ∈ [–1,3)
Case-I x ∈ [–1,0)
–1 + 2 – x = 0 ⇒ x = 1 (reject)
Case-II x ∈ [0,1)
0 + 2 – x = 0 ⇒ x = 2 (reject)
Case-III x ∈ [1,2)
1 + 2 – x = 0 ⇒ x = 3 (reject)
Case-IV x ∈ [2,3)
2 + x – 2 = 0 ⇒ x = 0 (reject)
⇒ x ∈φ
(ii) [2x] – 2x = [x + 1]
⇒ {2x} = – [x] – 1
⇒ – [x]–1 ∈ [0,1)
⇒ [x] ∈ (–2,–1] ⇒ [x] = – 1
⇒ {2x} = 0 ⇒ 2x ∈ I
⇒ {x} = 0 or 
⇒ x = –1 or – 
(iii) [x 2 ] + 2[x] = 3x, 0 ≤ x ≤ 2
Case -I x ∈ [0,1)
⇒ 0 + 2 (0) = 3x ⇒ x = 0
Case- II x ∈ [1, 
)
1 + 2(1) = 3x ⇒ x = 1
Case- III x ∈ [ 
)
2 + 2(1) = 3x ⇒ x = 
(reject)
Case- IV x ∈ [ 
,2)
3 + 2 = 3x ⇒ x = 
(reject)
Case- V x = 2
4 + 2(2) = 3 (2) ( not posible)
⇒ x ∈ {0,1}
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