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Absolute Value solved questions
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CHAPTER 2Absolute Value
2.1 Solve |* + 3|<5.
\x + 3\<5 if and only if -5<x + 3s5.
Answer -8 s jc < 2 [Subtract 3.] In interval notation, the solution is the set [—8, 2].
2.2 Solve |3jt + 2|<l.
|3* + 2|<1 if and only if -1<3* + 2<1, -3<3*<-l [Subtract 2.]
Answer -1< x < - 5 [Divide by 3.] In interval notation, the solution is the set (-1, - 3).
2.3 Solve |5-3*|<2.
|5-3x|<2 if and only if -2<5-3x<2, -7<-3x<-3 [Subtracts.]
Answer | > x > 1 [Divide by —3 and reverse the inequalities.] In interval notation, the solution is the set
(i,3).
2.4 Solve |3*-2|s=l.
Let us solve the negation of the given relation: |3* — 2|<1. This is equivalent to — l<3x — 2<1,
1<3*<3 [Add 2.], ^ < x < l [Divide by 3.]
The points not satisfying this condition correspond to AT such that x < 3 or x>\. Answer
2.5 Solve |3 - x\ = x - 3.
|M| = — u when and only when w^O. So, \3>-x\ = x—3 when and only when 3 — *:£0; that is,
3 s x. Answer
2.6 Solve |3 - *| = 3 - x.
\u\ = u when and only when j/>0. So, |3-*|=3 — x when and only when 3-*>(); that is,
3 s x. Answer
2.7 Solve \2x + 3| = 4.
If c>0, \u\ = c if and only if w = ±c. So, \2x + 3| = 4 when and only when 2^: + 3=±4. There
are two cases: Case 1. 2*+ 3 = 4. 2x = 1, x = | . Case 2. 2 A t + 3 = - 4 . 2x = -7, ac = -|.
So, either x = | or x = — j. AnswerSo, either x = | or x = — j. Answer
2.8 Solve |7-5*| = 1.
|7-5*| = |5*-7|. So, there are two cases: Casel. 5x-7 = l. 5* = 8, *=f. Case 2. 5*-7=-l.
5x = 6, AC = f .
So, either * = | or *=|. Answer
2.9 Solve U/2 + 3|<l.
This inequality is equivalent to -l<jc/2 + 3<l, -4<x/2<-2 [Subtracts.], -8<x<-4 [Multi-
ply by 2.] Answer
2.10 Solve |l/*-2|<4.
This inequality is equivalent to —4<1/* — 2<4, -2<l/*<6 [Add 2.] When we multiply by x, there
are two cases: Casel. *>0. -2*<1<6*, x>-\ and g < * , \<x. Case 2. *<0. -2x>\>
6x, x<—\ and !>*, x< — \.
So, either x<— \ or \<x. Answer
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