Sets, Relations & Functions
Floor and fractional part functions
Grade 11
<p>If roots of the equation \(|x-1| = 2[x] - 3\{x\}\) is \(x_i\), then find \(\left[\dfrac{\sum x_i}{5}\right]\) here (where \([x]\) & \(\{x\}\) denotes integral and fractional part of \(x\))</p>
Step-by-Step Solution
Key Concept: We need to analyze the equation |x-1| = 2[x] - 3{x} by considering cases based on the sign of (x-1) and the value of [x]. Since {x} = x - [x], we can express everything in terms of [x] and the fractional part.
<p><strong>Step 1: Set up the equation with substitution.</strong> Let [x] = n (integer) and {x} = f where 0 ≤ f < 1. Then x = n + f. The equation becomes: |n + f - 1| = 2n - 3f.</p><p><strong>Step 2: Analyze the constraint on the RHS.</strong> Since |n + f - 1| ≥ 0, we need 2n - 3f ≥ 0, which gives f ≤ 2n/3.</p><p><strong>Step 3: Case 1 - When n + f ≥ 1.</strong> Then |n + f - 1| = n + f - 1. The equation becomes: n + f - 1 = 2n - 3f, which simplifies to 4f = n + 1, so f = (n+1)/4. For valid f: 0 ≤ (n+1)/4 < 1, giving -1 ≤ n < 3. Also need (n+1)/4 ≤ 2n/3. This gives 3(n+1) ≤ 8n, so n ≥ 3/5, meaning n ≥ 1. Combined: n ∈ {1, 2}. For n = 1: f = 1/2, x = 3/2. For n = 2: f = 3/4, x = 11/4.</p><p><strong>Step 4: Case 2 - When n + f < 1.</strong> Then |n + f - 1| = 1 - n - f. The equation becomes: 1 - n - f = 2n - 3f, which simplifies to 2f = 3n - 1, so f = (3n-1)/2. For valid f: 0 ≤ (3n-1)/2 < 1, giving 1/3 ≤ n < 1. Since n is an integer, there's no valid n in this range. However, checking n = 0: f = -1/2 (invalid). No solutions from this case.</p><p><strong>Step 5: Verify the solutions.</strong> For x = 3/2: |3/2 - 1| = 1/2 and 2[3/2] - 3{3/2} = 2(1) - 3(1/2) = 2 - 3/2 = 1/2. ✓ For x = 11/4: |11/4 - 1| = 7/4 and 2[11/4] - 3{11/4} = 2(2) - 3(3/4) = 4 - 9/4 = 7/4. ✓</p><p><strong>Step 6: Calculate the final answer.</strong> Sum of roots: Σx_i = 3/2 + 11/4 = 6/4 + 11/4 = 17/4. Then Σx_i/5 = 17/20 = 0.85. Therefore [Σx_i/5] = [0.85] = 0.</p><p><strong>∴ Answer: 0</strong></p>
Correct Answer: 0