CHAPTER 2
Solve |l + 3/A-|>2.
This breaks up into two cases: Case 1. l + 3/x>2. 3/x>l [Hence, x>0.], 3>x. Case 2. 1 +
ilx<-2. 3/x<-3 [Hence, *<0.], 3>-3x [Reverse < to >.], -Kje [Reverse > to <.].
So, either 0
2 -10|<6.
This is equivalent to -6
So, either 2sjc<4 or —4s*<—2. Answer
Solve |2*-3| = |* + 2|.
There are two cases: Case 1. 2*-3 = .v + 2. jc-3 = 2, A-=5. Case 2. 2x - 3 = -(jt + 2). 2x - 3 =
-x-2, 3x-3 = -2, 3.x = 1, x=\.
So, either A-= 5 or x=j. Answer
Solve 2x-l = \x + l\.
Since an absolute value is never negative. 2.v-laO. There are two cases: Case 1. x + 7>0. 2x — l =
A-+ 7, A--1 = 7, A-= 8. Case 2. x + 7<0. 2* - 1 =-(A-+ 7), 2*-l = -jc-7, 3x - 1 =-7, 3x = -6,
je=-2. But then, 2jc-l = -5<0.
So, the only solution is x = 8. Answer
Solve |2*-3|<|x + 2|.
This is equivalent to -\x + 2\ <2x -3< |x + 2|. There are two cases: Case 1. A: + 2>0. -(x + 2)<
2Ar-3
2*-3>;t + 2, -x-2>2^-3>A: + 2, l>3jc and A:>5, j>je and x>5 [impossible]. So, j
Solve \2x - 5| = -4.
There is no solution since an absolute value cannot be negative.
Solve 0<|3* + l|
x < — | [Divide by 3.] The inequality 0 < \3x + l| excludes the case where 0 = |3* + 1|, that is, where
*--*.
Answer All A: for which — 5 < A- < -1 except jc = — 3.
The well-known triangle inequality asserts that |« + U|S|M| + |U|. Prove by mathematical induction that, for
n >2, |u, + H 2 + • • • + u n \ < |u,| + |u z | + • • • + |M,,|.
The case n = 2 is the triangle inequality. Assume the result true for some n. By the triangle inequality
and the inductive hypothesis,
|u, + « 2 + • • • + u n + w n + 1 | s |u, + u 2 + ••• + «„! + k + 1 | s (|M,| + |u z | + • • • + | M J) + |u,, + 1 |
and, therefore, the result also holds for n + 1.
Prove |M — v\ > | \u\ — \v\ \.
\u\ = \u + (u-v)\^\v\ + \u-v\ [Triangleinequality.] Hence, \u - v\ a \u\ - \v\. Similarly, |i>-u|s
|y|-|w|. But, \v - u\ = \u - v\. So, \u - v\ a (maximum of |u|-|y| and |u| - |M|) = | |u| - \v\ \.
Solve |*-l|<|x-2|.
Analytic solution. The given equation is equivalent to -|A- -2|0.
-(x-2)x-l>
x-2, -x+2>x-l>x-2, 3>2x, \>x. Thus, the solution consists of all A-such that A-<|.
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2.11
2.12
2.13
2.14
2.15
2.16
2.17
2.18
2.19
2.20
Solve |l + 3/A-|>2.
This breaks up into two cases: Case 1. l + 3/x>2. 3/x>l [Hence, x>0.], 3>x. Case 2. 1 +
ilx<-2. 3/x<-3 [Hence, *<0.], 3>-3x [Reverse < to >.], -Kje [Reverse > to <.].
So, either 0
2 -10|<6.
This is equivalent to -6
So, either 2sjc<4 or —4s*<—2. Answer
Solve |2*-3| = |* + 2|.
There are two cases: Case 1. 2*-3 = .v + 2. jc-3 = 2, A-=5. Case 2. 2x - 3 = -(jt + 2). 2x - 3 =
-x-2, 3x-3 = -2, 3.x = 1, x=\.
So, either A-= 5 or x=j. Answer
Solve 2x-l = \x + l\.
Since an absolute value is never negative. 2.v-laO. There are two cases: Case 1. x + 7>0. 2x — l =
A-+ 7, A--1 = 7, A-= 8. Case 2. x + 7<0. 2* - 1 =-(A-+ 7), 2*-l = -jc-7, 3x - 1 =-7, 3x = -6,
je=-2. But then, 2jc-l = -5<0.
So, the only solution is x = 8. Answer
Solve |2*-3|<|x + 2|.
This is equivalent to -\x + 2\ <2x -3< |x + 2|. There are two cases: Case 1. A: + 2>0. -(x + 2)<
2Ar-3
2*-3>;t + 2, -x-2>2^-3>A: + 2, l>3jc and A:>5, j>je and x>5 [impossible]. So, j
Solve \2x - 5| = -4.
There is no solution since an absolute value cannot be negative.
Solve 0<|3* + l|
x < — | [Divide by 3.] The inequality 0 < \3x + l| excludes the case where 0 = |3* + 1|, that is, where
*--*.
Answer All A: for which — 5 < A- < -1 except jc = — 3.
The well-known triangle inequality asserts that |« + U|S|M| + |U|. Prove by mathematical induction that, for
n >2, |u, + H 2 + • • • + u n \ < |u,| + |u z | + • • • + |M,,|.
The case n = 2 is the triangle inequality. Assume the result true for some n. By the triangle inequality
and the inductive hypothesis,
|u, + « 2 + • • • + u n + w n + 1 | s |u, + u 2 + ••• + «„! + k + 1 | s (|M,| + |u z | + • • • + | M J) + |u,, + 1 |
and, therefore, the result also holds for n + 1.
Prove |M — v\ > | \u\ — \v\ \.
\u\ = \u + (u-v)\^\v\ + \u-v\ [Triangleinequality.] Hence, \u - v\ a \u\ - \v\. Similarly, |i>-u|s
|y|-|w|. But, \v - u\ = \u - v\. So, \u - v\ a (maximum of |u|-|y| and |u| - |M|) = | |u| - \v\ \.
Solve |*-l|<|x-2|.
Analytic solution. The given equation is equivalent to -|A- -2|
-(x-2)
x-2, -x+2>x-l>x-2, 3>2x, \>x. Thus, the solution consists of all A-such that A-<|.
6
2.11
2.12
2.13
2.14
2.15
2.16
2.17
2.18
2.19
2.20