Examkin

Soru

Zorluk: Çok zorLinear Inequalities and Absolute Value

For all real numbers xx that satisfy the absolute value inequality ∣3−2x∣−4≤x|3 - 2x| - 4 \le x, the rational expression y=∣x−6∣2−xy = \frac{|x - 6|}{2 - x} is defined. Which of the following intervals represents the complete set of all possible real values of yy?

  1. (−∞,0]∪[197,∞)(-\infty, 0] \cup [\frac{19}{7}, \infty)Cevap
  2. B
    (−∞,−15]∪[197,∞)(-\infty, -\frac{1}{5}] \cup [\frac{19}{7}, \infty)
  3. C
    (−∞,0)∪(1,∞)(-\infty, 0) \cup (1, \infty)
  4. D
    [−15,197][-\frac{1}{5}, \frac{19}{7}]
  5. E
    (−∞,1]∪[127,∞)(-\infty, 1] \cup [\frac{12}{7}, \infty)

Cevap

The complete set of all possible real values of yy is (−∞,0]∪[197,∞)(-\infty, 0] \cup [\frac{19}{7}, \infty).
The correct answer is (−∞,0]∪[197,∞)(-\infty, 0] \cup [\frac{19}{7}, \infty). Solving ∣3−2x∣−4≤x|3 - 2x| - 4 \le x gives −13≤x≤7-\frac{1}{3} \le x \le 7. Excluding x=2x = 2 where the denominator is zero, evaluating y=∣x−6∣2−xy = \frac{|x - 6|}{2 - x} on [−13,2)[-\frac{1}{3}, 2) yields y≥197y \ge \frac{19}{7}, on (2,6)(2, 6) yields y<0y < 0, and on [6,7][6, 7] yields −15≤y≤0-\frac{1}{5} \le y \le 0. Taking the union of these intervals gives (−∞,0]∪[197,∞)(-\infty, 0] \cup [\frac{19}{7}, \infty).

Adım Adım Çözüm

1
Isolate the absolute value expression in the inequality.
∣3−2x∣≤x+4|3 - 2x| \le x + 4
Adding 44 to both sides prepares the inequality for standard double-inequality solving.
2
Set up the compound inequality and solve for xx.
−(x+4)≤3−2x≤x+4-(x + 4) \le 3 - 2x \le x + 4, yielding x≤7x \le 7 from −x−4≤3−2x-x - 4 \le 3 - 2x, and x≥−13x \ge -\frac{1}{3} from 3−2x≤x+43 - 2x \le x + 4 (remembering to flip the inequality sign when dividing by −3-3). Thus, x∈[−13,7]x \in [-\frac{1}{3}, 7].
An absolute value inequality ∣A∣≤B|A| \le B (with B≥0B \ge 0) is equivalent to −B≤A≤B-B \le A \le B.
3
Identify domain restrictions for y=∣x−6∣2−xy = \frac{|x - 6|}{2 - x}.
The expression is undefined at x=2x = 2. Therefore, the domain of xx is divided into three sub-intervals: [−13,2)[-\frac{1}{3}, 2), (2,6)(2, 6), and [6,7][6, 7].
The denominator cannot be zero, and the absolute value ∣x−6∣|x - 6| changes definition at x=6x = 6.
4
Analyze yy on the first interval [−13,2)[-\frac{1}{3}, 2).
For x<6x < 6, ∣x−6∣=6−x|x - 6| = 6 - x. So y=6−x2−x=1+42−xy = \frac{6 - x}{2 - x} = 1 + \frac{4}{2 - x}. As xx increases from −13-\frac{1}{3} towards 22, 2−x2 - x decreases from 73\frac{7}{3} to 0+0^+, so yy increases from 1+47/3=1971 + \frac{4}{7/3} = \frac{19}{7} to +∞+\infty. Hence y∈[197,∞)y \in [\frac{19}{7}, \infty).
As the positive denominator approaches zero from above, the positive fraction grows without bound towards +∞+\infty.
5
Analyze yy on the second interval (2,6)(2, 6).
Here x<6x < 6, so ∣x−6∣=6−x|x - 6| = 6 - x and y=1+42−xy = 1 + \frac{4}{2 - x}. As xx increases from 2+2^+ to 66, 2−x2 - x increases from 0−0^- to −4-4. Thus yy increases from −∞-\infty up to 1+4−4=01 + \frac{4}{-4} = 0. Hence y∈(−∞,0)y \in (-\infty, 0).
As the negative denominator moves away from zero towards −4-4, the expression increases from −∞-\infty to 00.
6
Analyze yy on the third interval [6,7][6, 7].
For x≥6x \ge 6, ∣x−6∣=x−6|x - 6| = x - 6. So y=x−62−x=−1−42−xy = \frac{x - 6}{2 - x} = -1 - \frac{4}{2 - x}. At x=6x = 6, y=0y = 0. At x=7x = 7, y=1−5=−15y = \frac{1}{-5} = -\frac{1}{5}. As xx increases from 66 to 77, yy decreases continuously from 00 down to −15-\frac{1}{5}. Hence y∈[−15,0]y \in [-\frac{1}{5}, 0].
Combining (−∞,0)(-\infty, 0) from the second interval and [−15,0][-\frac{1}{5}, 0] from the third interval gives (−∞,0](-\infty, 0].
7
Combine the ranges from all intervals.
y∈(−∞,0]∪[197,∞)y \in (-\infty, 0] \cup [\frac{19}{7}, \infty).
Taking the union of all output values across the valid domain yields the total range.

Anahtar Kavram

Solving absolute value inequalities and finding the range of rational expressions with absolute values over restricted domains.
Tahmini Süre:3m 0s
Bu soruyu puanla

Konular

Linear Inequalities and Absolute Value

Bu konudan daha fazla soru