Practice Set 2 – Solutions അകലങ്ങൾ
Three questions, each with four parts, all built on one idea: |x − a| is the distance between x and a on the number line. "= r" means the two points at distance r from a; "< r" means everything closer than r.
- The distance between x and y on the number line is |x − y|; the distance of x from 0 is |x|.
- |x + 2| = |x − (−2)| is the distance between x and −2.
- |x − a| < r ⇔ a − r < x < a + r (closer to a than r).
Question 1 · Solve the equations
Read |x − a| = r as "x is at distance r from a". There are two such points: a + r to the right and a − r to the left. For |x + 2|, write it as |x − (−2)|.
A distance does not say which side. Every equation |x − a| = r with r > 0 has exactly two solutions, a − r and a + r, one on each side of a. (If r = 0 there is only one, x = a.)
Question 2 · Between which numbers?
"Distance from a less than r": mark a, go r to the left and r to the right. x lies strictly between those two points (the end points themselves are not included).
The ends are drawn as hollow circles because "<" excludes them: in (ii) x = 3 gives |3 − 2| = 1, which is not less than 1.
Question 3 · The integers inside
Take each interval from Question 2 and list the whole numbers (positive, negative or 0) strictly inside it. Leave out the end points.
Engineers write tolerances exactly like this: a bolt of length "40 mm ± 0.5 mm" must satisfy |x − 40| ≤ 0.5. A quality checker who knows that x lies between 39.5 and 40.5 can sort bolts in seconds.
Try your own
Check yourself
- |x − a| = r ⇒ x = a + r or x = a − r (two points at distance r from a).
- |x − a| < r ⇒ a − r < x < a + r (an open interval around a).
- For |x + b| use |x − (−b)|: the centre is −b.
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