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Difficulty: Very hardLinear Inequalities and Absolute Value

For how many integer values of kk does the inequality ∣2x−k∣+∣x+3∣≤7|2x - k| + |x + 3| \le 7 have at least one real solution xx such that x≥1x \ge 1?

  1. A
    8
  2. B
    9
  3. 10Answer
  4. D
    11
  5. E
    13

Answer

10 integer values of kk satisfy the given condition.
The correct answer is 10 because analyzing the condition x≥1x \ge 1 simplifies the inequality to ∣2x−k∣≤4−x|2x - k| \le 4 - x, requiring x∈[1,4]x \in [1, 4]. The double inequality k−4≤x≤k+43k - 4 \le x \le \frac{k + 4}{3} yields solutions overlapping with [1,4][1, 4] if and only if −1≤k≤8-1 \le k \le 8, which contains exactly 10 integers.

Step-by-Step Solution

1
Simplify the absolute value term ∣x+3∣|x + 3| using the given condition x≥1x \ge 1.
Since x≥1x \ge 1, x+3>0x + 3 > 0, so ∣x+3∣=x+3|x + 3| = x + 3. The inequality becomes ∣2x−k∣+x+3≤7|2x - k| + x + 3 \le 7, which simplifies to ∣2x−k∣≤4−x|2x - k| \le 4 - x.
Establishing the sign of x+3x + 3 allows eliminating one set of absolute value bars.
2
Determine the valid range for xx.
Since ∣2x−k∣≥0|2x - k| \ge 0, it must hold that 4−x≥04 - x \ge 0, which implies x≤4x \le 4. Combined with x≥1x \ge 1, any solution xx must lie in the interval [1,4][1, 4].
An absolute value quantity cannot be less than a negative number.
3
Unwrap the absolute value inequality ∣2x−k∣≤4−x|2x - k| \le 4 - x.
−(4−x)≤2x−k≤4−x-(4 - x) \le 2x - k \le 4 - x. Splitting into two linear inequalities:
1) 2x−k≤4−x  ⟹  3x≤k+4  ⟹  x≤k+432x - k \le 4 - x \implies 3x \le k + 4 \implies x \le \frac{k + 4}{3}.
2) 2x−k≥x−4  ⟹  x≥k−42x - k \ge x - 4 \implies x \ge k - 4.
Thus, k−4≤x≤k+43k - 4 \le x \le \frac{k + 4}{3}.
Rewriting absolute value inequalities as compound inequalities defines explicit bounds on xx in terms of kk.
4
Find the range of kk for which [k−4,k+43][k - 4, \frac{k + 4}{3}] overlaps with [1,4][1, 4].
For an overlapping solution to exist in [1,4][1, 4]:
1) The upper bound k+43\frac{k + 4}{3} must be at least 1: k+43≥1  ⟹  k≥−1\frac{k + 4}{3} \ge 1 \implies k \ge -1.
2) The lower bound k−4k - 4 must be at most 4: k−4≤4  ⟹  k≤8k - 4 \le 4 \implies k \le 8.
Combining these gives −1≤k≤8-1 \le k \le 8.
The solution interval for xx must have a non-empty intersection with the allowed domain [1,4][1, 4].
5
Count the total number of integer values of kk in the interval [−1,8][-1, 8].
The integers are −1,0,1,2,3,4,5,6,7,8-1, 0, 1, 2, 3, 4, 5, 6, 7, 8, giving a total of 8−(−1)+1=108 - (-1) + 1 = 10 integer values.
Counting inclusive integer endpoints gives the total count.

Key Concept

Solving absolute value inequalities involving parameters and restricted variable domains.
Estimated Time:2m 0s
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Topics

Linear Inequalities and Absolute Value

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