Examkin

Question

Difficulty: Very hardLinear Inequalities and Absolute Value

If xx is a real number that satisfies the inequality ∣∣2x−5∣−7∣≤4||2x - 5| - 7| \le 4, and y=∣3−x∣y = |3 - x|, what is the difference between the maximum possible value and the minimum possible value of yy?

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

Answer

The difference between the maximum possible value and the minimum possible value of yy is 5.
Solving ∣∣2x−5∣−7∣≤4||2x - 5| - 7| \le 4 gives 3≤∣2x−5∣≤113 \le |2x - 5| \le 11, which restricts xx to the disconnected domain [−3,1]∪[4,8][-3, 1] \cup [4, 8]. Evaluating y=∣3−x∣y = |3 - x| across these intervals gives a maximum value of 6 (at x=−3x = -3) and a minimum value of 1 (at x=4x = 4). The difference between the maximum and minimum values is 6−1=56 - 1 = 5.

Step-by-Step Solution

1
Unpack the outer absolute value inequality ∣∣2x−5∣−7∣≤4||2x - 5| - 7| \le 4.
−4≤∣2x−5∣−7≤4  ⟹  3≤∣2x−5∣≤11-4 \le |2x - 5| - 7 \le 4 \implies 3 \le |2x - 5| \le 11.
An absolute value inequality ∣u∣≤a|u| \le a (for a≥0a \ge 0) is equivalent to −a≤u≤a-a \le u \le a.
2
Solve the double inequality 3≤∣2x−5∣≤113 \le |2x - 5| \le 11 by splitting it into two conditions.
Condition 1: ∣2x−5∣≤11  ⟹  −11≤2x−5≤11  ⟹  −3≤x≤8|2x - 5| \le 11 \implies -11 \le 2x - 5 \le 11 \implies -3 \le x \le 8.
Condition 2: ∣2x−5∣≥3  ⟹  2x−5≥3|2x - 5| \ge 3 \implies 2x - 5 \ge 3 or 2x−5≤−3  ⟹  x≥42x - 5 \le -3 \implies x \ge 4 or x≤1x \le 1.
The quantity ∣2x−5∣|2x - 5| must simultaneously satisfy upper and lower absolute value bounds.
3
Intersect Condition 1 and Condition 2 to determine the complete domain of xx.
x∈[−3,1]∪[4,8]x \in [-3, 1] \cup [4, 8].
Values in the open interval (1,4)(1, 4) make ∣2x−5∣<3|2x - 5| < 3 and must be excluded from the domain.
4
Evaluate the range of y=∣3−x∣y = |3 - x| over the valid domain of xx.
On [−3,1][-3, 1], y=3−xy = 3 - x decreases from 3−(−3)=63 - (-3) = 6 to 3−1=23 - 1 = 2, giving y∈[2,6]y \in [2, 6].
On [4,8][4, 8], y=x−3y = x - 3 increases from 4−3=14 - 3 = 1 to 8−3=58 - 3 = 5, giving y∈[1,5]y \in [1, 5].
The complete range of yy is [1,6][1, 6].
Combining the output ranges of both disjoint intervals yields all possible values for yy.
5
Calculate the difference between the maximum and minimum values of yy.
Maximum y=6y = 6, Minimum y=1y = 1, Difference = 6−1=56 - 1 = 5.
Subtracting the minimum value 1 from the maximum value 6 gives the required difference.

Key Concept

Solving compound nested absolute value inequalities and finding the extreme values of a transformed function over disconnected solution intervals.
Estimated Time:2m 0s
Rate this question

Topics

Linear Inequalities and Absolute Value

More questions from this topic