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Difficulty: HardInequalities and Absolute Value Equations

For how many integer values of kk does the equation ∣x−4∣+∣x+2∣=k|x - 4| + |x + 2| = k have at least one real solution xx satisfying ∣x∣≤5|x| \leq 5?

  1. A
    5
  2. B
    6
  3. 7Answer
  4. D
    8
  5. E
    12

Answer

7 integer values of k satisfy the equation.
The sum of absolute values f(x)=∣x−4∣+∣x+2∣f(x) = |x - 4| + |x + 2| represents the total distance from xx to 44 and −2-2 on the number line. On the interval [−2,4][-2, 4], this total distance is constant and equal to 66, which is the minimum possible value. On the restricted domain [−5,5][-5, 5], the maximum distance occurs at the boundary point x=−5x = -5, where f(−5)=∣−9∣+∣−3∣=12f(-5) = |-9| + |-3| = 12. Because f(x)f(x) is continuous, it takes every real value from 66 to 1212 inclusive. The integer values that kk can assume are 6,7,8,9,10,11,6, 7, 8, 9, 10, 11, and 1212, yielding a total of 7 integer values.

Step-by-Step Solution

1
Define the piecewise absolute value function f(x)=∣x−4∣+∣x+2∣f(x) = |x - 4| + |x + 2| and analyze its critical points.
Critical points occur at x=−2x = -2 and x=4x = 4.
Absolute value expressions change definition where their inner terms equal zero.
2
Evaluate f(x)f(x) across the three intervals defining the domain [−5,5][-5, 5].
For x<−2x < -2, f(x)=(4−x)−(x+2)=2−2xf(x) = (4 - x) - (x + 2) = 2 - 2x. For −2≤x≤4-2 \leq x \leq 4, f(x)=(4−x)+(x+2)=6f(x) = (4 - x) + (x + 2) = 6. For x>4x > 4, f(x)=(x−4)+(x+2)=2x−2f(x) = (x - 4) + (x + 2) = 2x - 2.
Removing absolute value bars according to domain regions gives explicit linear formulas.
3
Find the minimum and maximum values of f(x)f(x) on the domain ∣x∣≤5|x| \leq 5 (i.e., −5≤x≤5-5 \leq x \leq 5).
Minimum value is f(x)=6f(x) = 6 for x∈[−2,4]x \in [-2, 4]. Maximum value occurs at x=−5x = -5, giving f(−5)=2−2(−5)=12f(-5) = 2 - 2(-5) = 12. At x=5x = 5, f(5)=2(5)−2=8f(5) = 2(5) - 2 = 8. Range of f(x)f(x) is [6,12][6, 12].
Continuous piecewise functions achieve extreme values at domain boundaries or constant minimum intervals.
4
Count the integer values of kk in the closed interval [6,12][6, 12].
The integers are 6,7,8,9,10,11,126, 7, 8, 9, 10, 11, 12, giving 12−6+1=712 - 6 + 1 = 7 integers.
The number of integers in an inclusive range [a,b][a, b] is given by b−a+1b - a + 1.

Key Concept

Absolute Value Equations and Piecewise Function Ranges
Estimated Time:2m 0s
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Topics

Inequalities and Absolute Value Equations

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