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Difficulty: Very hardInequalities and Absolute Value Equations

How many integer values of xx satisfy the inequality ∣x2−4x∣≤3x−6|x^2 - 4x| \leq 3x - 6?

  1. A
    3
  2. 4Answer
  3. C
    5
  4. D
    6
  5. E
    8

Answer

4 integer values satisfy the inequality.
The solution requires breaking the inequality into cases based on the sign of x2−4xx^2 - 4x while enforcing that the right-hand side 3x−63x - 6 must be non-negative (meaning x≥2x \geq 2). Combining the valid sub-intervals 3≤x<43 \leq x < 4 and 4≤x≤64 \leq x \leq 6 yields 3≤x≤63 \leq x \leq 6. The integers satisfying this inequality are 3, 4, 5, and 6, giving a total of 4 integers.

Step-by-Step Solution

1
Establish the domain restriction based on non-negativity.
Since the left side ∣x2−4x∣≥0|x^2 - 4x| \geq 0 for all real xx, the right side must also be non-negative: 3x−6≥0  ⟹  x≥23x - 6 \geq 0 \implies x \geq 2.
An absolute value quantity can never be less than a negative number.
2
Analyze Case 1 where the expression inside the absolute value is non-negative (x2−4x≥0x^2 - 4x \geq 0).
For x≥2x \geq 2, x(x−4)≥0x(x - 4) \geq 0 implies x≥4x \geq 4. The inequality becomes x2−4x≤3x−6  ⟹  x2−7x+6≤0  ⟹  (x−1)(x−6)≤0x^2 - 4x \leq 3x - 6 \implies x^2 - 7x + 6 \leq 0 \implies (x - 1)(x - 6) \leq 0, so 1≤x≤61 \leq x \leq 6. Intersecting with x≥4x \geq 4 yields 4≤x≤64 \leq x \leq 6.
When x≥4x \geq 4, ∣x2−4x∣=x2−4x|x^2 - 4x| = x^2 - 4x.
3
Analyze Case 2 where the expression inside the absolute value is negative (x2−4x<0x^2 - 4x < 0).
For x≥2x \geq 2, x(x−4)<0x(x - 4) < 0 implies 2≤x<42 \leq x < 4. The inequality becomes −(x2−4x)≤3x−6  ⟹  −x2+4x≤3x−6  ⟹  x2−x−6≥0  ⟹  (x−3)(x+2)≥0-(x^2 - 4x) \leq 3x - 6 \implies -x^2 + 4x \leq 3x - 6 \implies x^2 - x - 6 \geq 0 \implies (x - 3)(x + 2) \geq 0, so x≤−2x \leq -2 or x≥3x \geq 3. Intersecting with 2≤x<42 \leq x < 4 yields 3≤x<43 \leq x < 4.
When 2≤x<42 \leq x < 4, ∣x2−4x∣=−(x2−4x)|x^2 - 4x| = -(x^2 - 4x).
4
Combine the valid intervals from both cases and count the integer solutions.
Combining [3,4)[3, 4) and [4,6][4, 6] gives the interval [3,6][3, 6]. The integer values in this range are x=3,4,5,6x = 3, 4, 5, 6. Total count = 4.
The union of the solution intervals yields all real numbers between 3 and 6 inclusive.

Key Concept

Solving Absolute Value Inequalities with Variable RHS
Estimated Time:2m 30s
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Topics

Inequalities and Absolute Value Equations

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