For all real numbers $ x $, the condition $ |3x - 20| + |3x - 40| = 20 $ necessarily holds if
For all real numbers $ x $, the condition $ |3x - 20| + |3x - 40| = 20 $ necessarily holds if
\(6 < x < 11\)
\(7 < x < 12\)
\(10 < x < 15\)
\(9 < x < 14\)
The Correct Option is B
Approach Solution - 1
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.
Consider the expression:
\[ |3x - 20| + |3x - 40| = 20 \]
Behavior of the Function:
- For values of \( x < \frac{20}{3} \), both expressions inside the modulus decrease, so their sum increases.
- For values of \( x > \frac{40}{3} \), both expressions increase together, so again, the sum increases.
- In the range \( \left[ \frac{20}{3}, \frac{40}{3} \right] \), one function increases while the other decreases, exactly balancing each other.
This means:
\[ \text{The sum stays constant at 20 when } x \in \left[ \frac{20}{3}, \frac{40}{3} \right] \Rightarrow x \in [6.66, 13.33] \]
Correct Option:
Among the given choices, the interval that lies completely within this range is:
\( \boxed{(B)\ 7 < x < 12} \)
Approach Solution -2
Solving the Equation:
\[ |3x - 20| + |3x - 40| = 20 \]
Case 1: \( x \geq \frac{40}{3} \)
In this case, both expressions inside the absolute values are non-negative:
\[ |3x - 20| = 3x - 20, \quad |3x - 40| = 3x - 40 \]
\[ \Rightarrow (3x - 20) + (3x - 40) = 20 \\ \Rightarrow 6x - 60 = 20 \\ \Rightarrow 6x = 80 \\ \Rightarrow x = \frac{80}{6} = \frac{40}{3} \approx 13.33 \]
Case 2: \( \frac{20}{3} \leq x < \frac{40}{3} \)
Here, \(3x - 20\) is positive and \(3x - 40\) is negative:
\[ |3x - 20| = 3x - 20, \quad |3x - 40| = -(3x - 40) = 40 - 3x \]
\[ \Rightarrow (3x - 20) + (40 - 3x) = 20 \\ \Rightarrow 20 = 20 \quad \text{(Always true)} \]
So, the equation holds for all \( x \in \left[ \frac{20}{3}, \frac{40}{3} \right] \)
Case 3: \( x < \frac{20}{3} \)
In this case, both expressions are negative:
\[ |3x - 20| = 20 - 3x, \quad |3x - 40| = 40 - 3x \]
\[ \Rightarrow (20 - 3x) + (40 - 3x) = 20 \\ \Rightarrow 60 - 6x = 20 \\ \Rightarrow 6x = 40 \\ \Rightarrow x = \frac{20}{3} \]
This value is not valid in this case since it contradicts the condition \( x < \frac{20}{3} \). So, no solution in this case.
Conclusion:
Valid values of \( x \) lie in:
\[ \frac{20}{3} \leq x \leq \frac{40}{3} \Rightarrow 6.67 \leq x \leq 13.33 \]
From the given options, the interval (B) \( 7 < x < 12 \) lies completely within the valid solution range.
Final Answer:
Option (B): \( \boxed{7 < x < 12} \)
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