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Zorluk: OrtaReal Numbers, Number Line, and Absolute Value

If aa and bb are real numbers such that ∣a−3∣=5|a - 3| = 5 and ∣2b+1∣=9|2b + 1| = 9, what is the minimum possible value of ∣a−b∣|a - b|?

  1. A
    1
  2. B
    2
  3. 3Cevap
  4. D
    4
  5. E
    6

Cevap

The minimum possible value of ∣a−b∣|a - b| is 33.
Solving ∣a−3∣=5|a - 3| = 5 gives two possible values for aa: a=8a = 8 and a=−2a = -2. Solving ∣2b+1∣=9|2b + 1| = 9 gives two possible values for bb: b=4b = 4 and b=−5b = -5. Evaluating the distance ∣a−b∣|a - b| for all four pairs (a,b)(a,b) gives ∣8−4∣=4|8 - 4| = 4, ∣8−(−5)∣=13|8 - (-5)| = 13, ∣−2−4∣=6|-2 - 4| = 6, and ∣−2−(−5)∣=3|-2 - (-5)| = 3. The minimum possible value is 33.

Adım Adım Çözüm

1
Solve the absolute value equation ∣a−3∣=5|a - 3| = 5 for all possible values of aa.
a−3=5  ⟹  a=8a - 3 = 5 \implies a = 8 or a−3=−5  ⟹  a=−2a - 3 = -5 \implies a = -2. Thus, a∈{−2,8}a \in \{-2, 8\}.
An absolute value equation ∣x∣=k|x| = k splits into two linear equations: x=kx = k and x=−kx = -k.
2
Solve the absolute value equation ∣2b+1∣=9|2b + 1| = 9 for all possible values of bb.
2b+1=9  ⟹  2b=8  ⟹  b=42b + 1 = 9 \implies 2b = 8 \implies b = 4 or 2b+1=−9  ⟹  2b=−10  ⟹  b=−52b + 1 = -9 \implies 2b = -10 \implies b = -5. Thus, b∈{−5,4}b \in \{-5, 4\}.
An absolute value equation ∣2b+1∣=9|2b + 1| = 9 has two cases: 2b+1=92b + 1 = 9 and 2b+1=−92b + 1 = -9.
3
Calculate ∣a−b∣|a - b| for all four possible pairs of (a,b)(a, b).
For (8,4):∣8−4∣=4(8, 4): |8 - 4| = 4.
For (8,−5):∣8−(−5)∣=13(8, -5): |8 - (-5)| = 13.
For (−2,4):∣−2−4∣=6(-2, 4): |-2 - 4| = 6.
For (−2,−5):∣−2−(−5)∣=3(-2, -5): |-2 - (-5)| = 3.
To find the minimum possible value of ∣a−b∣|a - b|, every valid combination of aa and bb must be tested.
4
Identify the minimum value among the calculated absolute differences.
The minimum calculated value is 33.
Comparing 4,13,6,4, 13, 6, and 33 yields 33 as the smallest value.

Anahtar Kavram

Solving absolute value equations and finding distances between points on the real number line
Tahmini Süre:1m 30s
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