Which of the following is/are true?
I. |3x - 19| > 5 - x holds for all values of x.
II. There is no value of x for which |2x - 10| < 4 - x holds.
III. -|5x - 23| is negative for all values of x.A. None
B. I only
C. I and II only
D. I and III only
E. I, II and III
I. \(|3x - 19| > 5 - x\) holds for all values of x. When \(x < \frac{19}{3}\) (notice that \(\frac{19}{3} = 6 \frac{1}{3}\)), \(3x - 19 < 0\) thus in this case \(|3x - 19| = -(3x - 19)\), so we'd get \(-(3x - 19) > 5 - x\):
\(-3x + 19 > 5 - x\);
\(14 > 2x\);
\(x < 7\).
Since we consider \(x<6 \frac{1}{3}\) range, then for this case we'd have \(x < 6 \frac{1}{3}\).
When \(x \geq \frac{19}{3}\), \(3x - 19 \geq 0\) thus in this case \(|3x - 19| = 3x - 19\), so we'd get 3x - 19 > 5 - x:
\(3x - 19 > 5 - x\);
\(4x > 24\);
\(x > 6\).
Since we consider \(x \geq 6 \frac{1}{3}\) range, then for this case we'd have \(x \geq 6 \frac{1}{3}\).
\(x < 6 \frac{1}{3}\) and \(x \geq 6 \frac{1}{3}\) give all values of x. So, \(|3x - 19| > 5 - x\) holds for all values of x.
II. There is no value of x for which |2x - 10| < 4 - x holds.Here notice that x cannot be more than 4, because if it is then the RHS (the right hand side) becomes negative and it cannot be greater than the LHS (the left hand side), which is an absolute value and therefore is always more than or equal to 0.
So, x must be less than or equal to 4. Now, when \(x \leq 4\), \(2x - 10 < 0\) and thus 2x - 10 = -(2x - 10). So, we'd get: -(2x - 10) < 4 - x:
\(-2x + 10 < 4 - x\);
\(x > 6\).
\(x \leq 4\) (the range we consider) and \(x > 6\) have no overlap, therefore there is no value of x for which |2x - 10| < 4 - x holds.
III. -|5x - 23| is negative for all values of x.This statement says that \(-|5x - 23| < 0\) for all values of x:
This is clearly wrong because when x = 23/5, \(|5x - 23| = 0\), not greater than 0.
Answer: C.
Hope it's clear.