Solve the following equations
(where [•] denotes greatest integer function and {•} represent fractional part function)
(i) 2[x] + 3{x} = 4x – 1
(ii) 4[x] = x + {x}
(iii) [x] + 2{–x} = 3x
Text Solution
Verified by Experts(III) [X]
(i) x ∈φ
(ii) {0}
(iii) {0,2/5, –1/5}
Sol. (i) 2[x] + 3 {x} = 4x – 1 ⇒ {x} = 1–2[x]
⇒ 0 ≤ 1– 2 [x] < 1 ⇒ [x] ∈ 
⇒ x ∈φ
(ii) Case - I : x ∈ Ι then rc 4x = x
x = 0.
Case-II: x ∉ Ι , then rc 4[x] = [x] + 2{x} ⇒ {x} = 
0 ≤ {x} < 1 ⇒ 0 ≤ [x] < 
∴ [x] = 0 ∴ {x} = 0 ∴ x = 0 + 0 = 0
(iii) [x] + 2 {–x} = 3x
Case-I when x ∈ Ι
x + 0 = 3x ⇒ x = 0
Case - II when x 
Ι
[x] + 2 (1 – {x}) = 3([x] + {x})
⇒ 2 – 2[x] = 5{x}
⇒ {x} = 
[x] | 1 | 0 | –1 |
{x} | 0 | 2/5 | 4/5 |
x | 1 | 2/5 | –1/5 |
Reject | Pass | Pass |
⇒ x ∈ {0, 2/5, –1/5}
──────────────────────────────────────────────────────────────────────────────────────────
Prepare Smarter with CGP Edu
Get practice questions, solutions, and test series in one place.
Write a Review
Share your experience with this question and solution.
Commentary
Send your comment, doubt, correction, or feedback to admin.
Similar Questions
Explore conceptually related problems