Match the LIST-I with LIST-II
| LIST-I | LIST-II | ||
| (Inequality) | (Solution Set) | ||
| A. | \( |3x-5|\ge 4 \) | I. | \( (-\infty,-1]\cup\left[\displaystyle\frac{7}{3},\infty\right) \) |
| B. | \( |3x|\ge|6-3x| \) | II. | \( (-\infty,-4]\cup[2,\infty) \) |
| C. | \( |x+1|\ge 3 \) | III. | \( [1,\infty) \) |
| D. | \( |2-3x|\ge 5 \) | IV. | \( \left(-\infty,\displaystyle\frac{1}{3}\right]\cup[3,\infty) \) |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → A-IV, B-III, C-II, D-I
For k > 0, \( |u|\ge k \) means \( u\ge k \) or \( u\le -k \).
A. \( |3x-5|\ge 4 \)
This matches IV. \( \left(-\infty,\frac{1}{3}\right]\cup[3,\infty) \).
\( \displaystyle 3x-5\ge 4\ \Rightarrow\ x\ge 3 \)
\( \displaystyle 3x-5\le -4\ \Rightarrow\ x\le\frac{1}{3} \)
B. \( |3x|\ge|6-3x| \)
This matches III. \( [1,\infty) \).
Both sides are non-negative, so squaring keeps the inequality:
\( \displaystyle 9x^2\ge 36-36x+9x^2 \)
\( \displaystyle 36x\ge 36 \)
\( \displaystyle x\ge 1 \)
C. \( |x+1|\ge 3 \)
This matches II. \( (-\infty,-4]\cup[2,\infty) \).
\( \displaystyle x+1\ge 3\ \Rightarrow\ x\ge 2 \)
\( \displaystyle x+1\le -3\ \Rightarrow\ x\le -4 \)
D. \( |2-3x|\ge 5 \)
This matches I. \( (-\infty,-1]\cup\left[\frac{7}{3},\infty\right) \).
Dividing by −3 reverses the inequality:
\( \displaystyle 2-3x\ge 5\ \Rightarrow\ -3x\ge 3 \)
\( \displaystyle \Rightarrow\ x\le -1 \)
\( \displaystyle 2-3x\le -5\ \Rightarrow\ -3x\le -7 \)
\( \displaystyle \Rightarrow\ x\ge\frac{7}{3} \)
So the matching is A-IV, B-III, C-II, D-I.