Question

For all real numbers x the condition |3x - 20| + |3x - 40| = 20 necessarily holds if

• A. \(6 < x < 11\)
• B. \(7 < x < 12\)
• C. \(10 < x < 15\)
• D. \(9 < x < 14\)

Answer 1

The Correct Option is B

The minimum value of the modulus function is zero, which means the minimum value for each of the two modulus functions occurs at \(x =\frac{ 20}{3}\) and \(x = \frac{40}{3}\), respectively.

For values of x less than \(\frac{ 20}{3}\), both functions increase. Similarly, for x greater than \(\frac{40}{3}\), the sum of the functions increases. In the range between \(\frac{ 20}{3}\) and \(\frac{40}{3}\), the values of the functions remain the same, as one function's increase matches the other's decrease.

As a result, the sum of the modulus functions remains at 20 within the range of values \([\frac{ 20}{3}, \frac{40}{3}]\), which can be expressed as \([6.66, 13.33].\)

This condition is satisfied by the option: \(7 < x < 12.\)

Answer 2

Case 1: \(x≥\frac {40}{3}\)
\(|3x - 20| + |3x - 40| = 20\)
\(3x-20 +3x-40 =20\)
\(6x=80\)
\(x=\frac {80}{6}\)
\(x=13.33\)

Case 2: \(\frac {20}{3}≤ x < \frac {40}{3}\)
\(|3x - 20| + |3x - 40| = 20\)
\(3x-20+40-3x =20\)
\(20=20\)
Then, \(x ∈ [\frac {20}{3}, \frac {40}{3}]\)

Case:3 \(x < \frac {20}{3}\)
\(20-3x+40-3x =20\)
\(6x=40\)
\(x=\frac {20}{3}\) (this is not possible)
From case 1, 2 and 3
\(\frac {20}{3} ≤ x ≤ \frac {40}{3}\)
On analyzing all the options, option C satisfies for all x.

So, the correct option is (B): \(7<x<12\)