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 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:
\[ |3x - 20| + |3x - 40| = 20 \]
This means:
\[ \text{The sum stays constant at 20 when } x \in \left[ \frac{20}{3}, \frac{40}{3} \right] \Rightarrow x \in [6.66, 13.33] \]
Among the given choices, the interval that lies completely within this range is:
\( \boxed{(B)\ 7 < x < 12} \)
\[ |3x - 20| + |3x - 40| = 20 \]
In this case, both expressions inside the absolute values are non-negative:
\[ |3x - 20| = 3x - 20, \quad |3x - 40| = 3x - 40 \]
\[ \Rightarrow (3x - 20) + (3x - 40) = 20 \\ \Rightarrow 6x - 60 = 20 \\ \Rightarrow 6x = 80 \\ \Rightarrow x = \frac{80}{6} = \frac{40}{3} \approx 13.33 \]
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 \( x \in \left[ \frac{20}{3}, \frac{40}{3} \right] \)
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.
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.
Option (B): \( \boxed{7 < x < 12} \)
LIST I | LIST II | ||
A. | The solution set of the inequality \(-5x > 3, x\in R\), is | I. | \([\frac{20}{7},∞)\) |
B. | The solution set of the inequality is, \(\frac{-7x}{4} ≤ -5, x\in R\) is, | II. | \([\frac{4}{7},∞)\) |
C. | The solution set of the inequality \(7x-4≥0, x\in R\) is, | III. | \((-∞,\frac{7}{5})\) |
D. | The solution set of the inequality \(9x-4 < 4x+3, x\in R\) is, | IV. | \((-∞,-\frac{3}{5})\) |
When $10^{100}$ is divided by 7, the remainder is ?