Match the LIST-I with LIST-II
| LIST-I (Inequality) | LIST-II (Solution set) | ||
| A. | \( |x-3|\lt1 \) | I. | \( x\in\left(-\displaystyle\frac{2}{5},\displaystyle\frac{8}{5}\right) \) |
| B. | \( |3-5x|\lt5 \) | II. | \( x\in\left(-\infty,-\displaystyle\frac{1}{2}\right) \) |
| C. | \( 2x+1\lt|2x+1| \) | III. | \( x\in\left(-\displaystyle\frac{1}{3},5\right) \) |
| D. | \( |7-3x|\lt8 \) | IV. | \( x\in(2,4) \) |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → A-IV, B-I, C-II, D-III
For \( k\gt0 \), \( |u|\lt k\Leftrightarrow-k\lt u\lt k \).
(A) \( |x-3|\lt1 \)
\( \displaystyle -1\lt x-3\lt1 \)
\( \displaystyle \Rightarrow\ 2\lt x\lt4 \)
This matches (IV).
(B) \( |3-5x|\lt5 \)
\( \displaystyle -5\lt3-5x\lt5 \)
\( \displaystyle -8\lt-5x\lt2 \)
Dividing by −5 reverses the inequalities.
\( \displaystyle -\frac{2}{5}\lt x\lt\frac{8}{5} \)
This matches (I).
(C) \( 2x+1\lt|2x+1| \)
A number is less than its absolute value only when it is negative.
\( \displaystyle 2x+1\lt0 \)
\( \displaystyle \Rightarrow\ x\lt-\frac{1}{2} \)
This matches (II).
(D) \( |7-3x|\lt8 \)
\( \displaystyle -8\lt7-3x\lt8 \)
\( \displaystyle -15\lt-3x\lt1 \)
Dividing by −3 reverses the inequalities.
\( \displaystyle -\frac{1}{3}\lt x\lt5 \)
This matches (III).
So the matching is (A)-(IV), (B)-(I), (C)-(II), (D)-(III).