For the function \( f(x)=|x-1|+|x-2|+|x-3| \)
Match the LIST-I with LIST-II
| LIST-I | LIST-II | ||
| Function \( f(x) \) = | For Interval | ||
| A. | \( -3x+6 \) | I. | \( x\ge 3 \) |
| B. | \( -x+4 \) | II. | \( x\lt 1 \) |
| C. | \( x \) | III. | \( 1\le x\lt 2 \) |
| D. | \( 3x-6 \) | IV. | \( 2\le x\lt 3 \) |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → A - II, B - III, C - IV, D - I
Each modulus opens according to its sign on the interval: \( |x-a|=x-a \) when \( x\ge a \) and \( |x-a|=a-x \) when \( x\lt a \).
(A) \( -3x+6 \)
This matches (II) x < 1.
All three brackets are negative:
\( \displaystyle f(x)=(1-x)+(2-x)+(3-x)=6-3x \)
(B) \( -x+4 \)
This matches (III) 1 ≤ x < 2.
Only \( x-1 \) is non-negative:
\( \displaystyle f(x)=(x-1)+(2-x)+(3-x)=4-x \)
(C) \( x \)
This matches (IV) 2 ≤ x < 3.
Now \( x-1 \) and \( x-2 \) are non-negative:
\( \displaystyle f(x)=(x-1)+(x-2)+(3-x)=x \)
(D) \( 3x-6 \)
This matches (I) x ≥ 3.
All three brackets are non-negative:
\( \displaystyle f(x)=(x-1)+(x-2)+(x-3)=3x-6 \)
So the matching is A - II, B - III, C - IV, D - I.
Shortcut: The coefficient of x equals (number of brackets that are non-negative) − (number that are negative). It is −3 before 1, −1 between 1 and 2, +1 between 2 and 3 and +3 after 3, which gives A-II, B-III, C-IV, D-I directly.