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Zorluk: Çok zorInequalities and Absolute Value Equations

How many integer values of xx satisfy the inequality x24x3x6|x^2 - 4x| \leq 3x - 6?

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

Cevap

4 integer values satisfy the inequality.
The solution requires breaking the inequality into cases based on the sign of x24xx^2 - 4x while enforcing that the right-hand side 3x63x - 6 must be non-negative (meaning x2x \geq 2). Combining the valid sub-intervals 3x<43 \leq x < 4 and 4x64 \leq x \leq 6 yields 3x63 \leq x \leq 6. The integers satisfying this inequality are 3, 4, 5, and 6, giving a total of 4 integers.

Adım Adım Çözüm

1
Establish the domain restriction based on non-negativity.
Since the left side x24x0|x^2 - 4x| \geq 0 for all real xx, the right side must also be non-negative: 3x60    x23x - 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 (x24x0x^2 - 4x \geq 0).
For x2x \geq 2, x(x4)0x(x - 4) \geq 0 implies x4x \geq 4. The inequality becomes x24x3x6    x27x+60    (x1)(x6)0x^2 - 4x \leq 3x - 6 \implies x^2 - 7x + 6 \leq 0 \implies (x - 1)(x - 6) \leq 0, so 1x61 \leq x \leq 6. Intersecting with x4x \geq 4 yields 4x64 \leq x \leq 6.
When x4x \geq 4, x24x=x24x|x^2 - 4x| = x^2 - 4x.
3
Analyze Case 2 where the expression inside the absolute value is negative (x24x<0x^2 - 4x < 0).
For x2x \geq 2, x(x4)<0x(x - 4) < 0 implies 2x<42 \leq x < 4. The inequality becomes (x24x)3x6    x2+4x3x6    x2x60    (x3)(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 x2x \leq -2 or x3x \geq 3. Intersecting with 2x<42 \leq x < 4 yields 3x<43 \leq x < 4.
When 2x<42 \leq x < 4, x24x=(x24x)|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.

Anahtar Kavram

Solving Absolute Value Inequalities with Variable RHS
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