Student’s Manual
Essential Mathematics for
Economic Analysis
th
4 edition
Knut Sydsæter
Peter Hammond
Arne Strøm
For further supporting resources please visit:
www.mymathlab.com/global
, Preface
This student’s solutions manual accompanies Essential Mathematics for Economic Analysis (4th edition, FT
Prentice Hall, 2012). Its main purpose is to provide more detailed solutions to the problems marked ⊂⊃ in the
SM
text. The answers provided in this Manual should be used in combination with any shorter answers provided
in the main text. There are a few cases where only part of the answer is set out in detail, because the rest
follows the same pattern.
We would appreciate suggestions for improvements from our readers, as well as help in weeding out
inaccuracies and errors.
Oslo and Coventry, July 2012
Knut Sydsæter ()
Peter Hammond ()
Arne Strøm ()
Contents
1 Introductory Topics I: Algebra . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1
2 Introductory Topics II: Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4
3 Introductory Topics III: Miscellaneous . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6
4 Functions of One Variable . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9
5 Properties of Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13
6 Differentiation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14
7 Derivatives in Use . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18
8 Single-Variable Optimization . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22
9 Integration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26
10 Interest Rates and Present Values . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34
11 Functions of Many Variables . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36
12 Tools for Comparative Statics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38
13 Multivariable Optimization . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 43
14 Constrained Optimization . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51
15 Matrix and Vector Algebra . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 61
16 Determinants and Inverse Matrices . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 64
17 Linear Programming . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 70
© Knut Sydsæter, Peter Hammond, and Arne Strøm 2012
, CHAPTER 1 INTRODUCTORY TOPICS I: ALGEBRA 1
Chapter 1 Introductory Topics I: Algebra
1.3
5. (a) (2t −1)(t 2 −2t +1) = 2t (t 2 −2t +1)−(t 2 −2t +1) = 2t 3 −4t 2 +2t −t 2 +2t −1 = 2t 3 −5t 2 +4t −1
(b) (a + 1)2 + (a − 1)2 − 2(a + 1)(a − 1) = a 2 + 2a + 1 + a 2 − 2a + 1 − 2a 2 + 2 = 4. Alternatively,
apply the quadratic identity x 2 + y 2 − 2xy = (x − y)2 with x = a + 1 and y = a − 1 to obtain
(a + 1)2 + (a − 1)2 − 2(a + 1)(a − 1) = [(a + 1) − (a − 1)]2 = 22 = 4.
(c) (x + y + z)2 = (x + y + z)(x + y + z) = x(x + y + z) + y(x + y + z) + z(x + y + z) =
x 2 + xy + xz + yx + y 2 + yz + zx + zy + z2 = x 2 + y 2 + z2 + 2xy + 2xz + 2yz (d) With a = x + y + z
and b = x − y − z, (x + y + z)2 − (x − y − z)2 = a 2 − b2 = (a + b)(a − b) = 2x(2y + 2z) = 4x(y + z).
13. (a) a 2 + 4ab + 4b2 = (a + 2b)2 by the first quadratic identity. (d) 9z2 − 16w2 = (3z − 4w)(3z + 4w),
according to the difference-of-squares formula. (e) − 15 x 2 + 2xy − 5y 2 = − 15 (x 2 − 10xy + 25y 2 ) =
− 15 (x − 5y)2 (f) a 4 − b4 = (a 2 − b2 )(a 2 + b2 ), using the difference-of-squares formula. Since
a 2 − b2 = (a − b)(a + b), the answer in the book follows.
1.4
1 1 x+2 x−2 x+2−x+2 4
5. (a) − = − = = 2
x−2 x+2 (x − 2)(x + 2) (x + 2)(x − 2) (x − 2)(x + 2) x −4
(b) Since 4x + 2 = 2(2x + 1) and 4x 2 − 1 = (2x + 1)(2x − 1), the lowest common denominator (LCD)
is 2(2x + 1)(2x − 1). Then
6x + 25 6x 2 + x − 2 (6x + 25)(2x − 1) − 2(6x 2 + x − 2) 42x − 21 21
− = = =
4x + 2 4x − 1
2 2(2x + 1)(2x − 1) 2(2x + 1)(2x − 1) 2(2x + 1)
18b2 a 18b2 − a(a − 3b) + 2(a 2 − 9b2 ) a(a + 3b) a
(c) 2 − +2= = =
a − 9b 2 a + 3b (a + 3b)(a − 3b) (a + 3b)(a − 3b) a − 3b
1 1 (a + 2) − a 2 1
(d) − = = =
8ab 8b(a + 2) 8ab(a + 2) 8ab(a + 2) 4ab(a + 2)
2t − t 2 5t 2t t (2 − t) 3t −t (t − 2) 3t −3t 2
(e) · − = · = · =
t +2 t −2 t −2 t +2 t −2 t +2 t −2 t +2
a 1 − 2a 1
a− 1
a 1 − 1
(f) = 1 2 = 4a − 2, so 2 − 2a
= 2 − (4a − 2) = 4 − 4a = 4(1 − a)
0.25 4
0.25
2 1 2(x + 1) + x − 3x(x + 1) 2 − 3x 2
6. (a) + −3= =
x x+1 x(x + 1) x(x + 1)
t t t (2t − 1) − t (2t + 1) −2t
(b) − = = 2
2t + 1 2t − 1 (2t + 1)(2t − 1) 4t − 1
3x 4x 2x − 1 3x(x − 2) + 4x(x + 2) − (2x − 1) 7x 2 + 1
(c) − − = = 2
x+2 2−x (x − 2)(x + 2) (x − 2)(x + 2) x −4
1 1 1 1 1 1 1 1
+ + xy − 2 − 2 · x2y2
x y x y y+x x2 y x2 y y2 − x2
(d) = = = x + y (e) = = 2
1 1 1 1 1 1 1 y + x2
· xy + + · x 2y2
xy xy x2 y2 x2 y2
(f) To clear the fractions within both the numerator and denominator, multiply both by xy to get
a(y − x) y−x
=
a(y + x) y+x
© Knut Sydsæter, Peter Hammond, and Arne Strøm 2012
,2 CHAPTER 1 INTRODUCTORY TOPICS I: ALGEBRA
1
1 −2
1 −2
8. (a) 1
4 − 1
5 = 5
20 − 4
20 = 1
20 , so 4 − 5 = 20 = 202 = 400
n n·n n(n − 1) − n2n2 −n
(b) n − =n− = n − = =
1 1 n−1 n−1 n−1
1− 1− ·n
n n
1 1 1 1 1 u
(c) Let u = x p−q . Then + = + = + =1
1 + x p−q 1 + x q−p 1 + u 1 + 1/u 1+u 1+u
1 1
+ (x 2 − 1)
x − 1 x2 − 1 (x + 1) + 1 x+2 1
(d) = 3 = =
2 x − x − 2x + 2 (x + 2)(x − 2x + 1)
2 (x − 1)2
x− (x 2 − 1)
x+1
1 1
− 2
1 1 x − (x + h)
2 2 −2xh − h 2
(x + h) 2 x −2x − h
(e) − 2 = = 2 , so = 2
(x + h) 2 x x (x + h)
2 2 x (x + h) 2 h x (x + h)2
10x 2 2x
(f) Multiplying denominator and numerator by x 2 − 1 = (x + 1)(x − 1) yields = .
5x(x − 1) x−1
1.5
5. The answers given in the main text for each
√ respective
√ part
√ emerge
√ after multiplying
√ both numerator
√ √
and denominator by the following: (a) 7 − 5 (b) 5 − 3 (c) 3 + 2 (d) x y − y x
√ √ √
(e) x + h + x (f) 1 − x + 1.
2
12. (a) (2x )2 = 22x = 2x if and only if 2x = x 2 , or if and only if x = 0 or x = 2. (b) Correct because
a p−q = a p /a q . (c) Correct because a −p = 1/a p . (d) 51/x = 1/5x = 5−x if and only if 1/x = −x
or −x 2 = 1, so there is no real x that satisfies the equation. (e) Put u = a x and v = a y , which reduces
the equation to uv = u + v, or 0 = uv − u − v = (u − 1)(v − 1) − 1. This is true only for special values
of u and v and so for special values of x and y. In particular, the equation is false when x = y = 1.
√ √
(f) Putting u = x and v = y reduces the equation to 2u · 2v = 2uv , which holds if and only if
uv = u + v, as in (e) above.
1.6
3x + 1 3x + 1 3x + 1 − 2(2x + 4) −x − 7
4. (a) 2 < has the same solutions as − 2 > 0, or > 0, or >0
2x + 4 2x + 4 2x + 4 2x + 4
A sign diagram reveals that the inequality is satisfied for −7 < x < −2. A serious error is to multiply the
inequality by 2x + 4, without checking the sign of 2x + 4. If 2x + 4 < 0, mulitiplying by this number
will reverse the inequality sign. (It might be a good idea to test the inequality for some values of x. For
example, for x = 0 it is not true. What about x = −5?)
120 480 − 3n
(b) The inequality is equivalent to ≤ 0.75, or ≤ 0. A sign diagram reveals that the
n 4n
inequality is satisfied for n < 0 and for n ≥ 160. (Note that for n = 0 the inequality makes no sense. For
n = 160, we have equality.) (c) Easy: g(g − 2) ≤ 0 etc. (d) Note that p2 − 4p + 4 = (p − 2)2 , and
p+1
the inequality reduces to ≥ 0. The fraction makes no sense if p = 2. The conclusion follows.
(p − 2)2
−n − 2 −n − 2 − 2n − 8 −3n − 10
(e) The inequality is equivalent to − 2 > 0, i.e. > 0, or > 0, etc.
n+4 n+4 n+4
(f) See the text and use a sign diagram. (Don’t cancel x 2 . If you do, x = 0 appears as a false solution.)
© Knut Sydsæter, Peter Hammond, and Arne Strøm 2012
, CHAPTER 1 INTRODUCTORY TOPICS I: ALGEBRA 3
5. (a) Use a sign diagram. (b) The inequality is not satisfied for x = 1. If x = 1, it is obviously satisfied
if and only x + 4 > 0, i.e. x > −4 (because (x − 1)2 is positive when x = 1). (c) Use a sign diagram.
(d) The inequality is not satisfied for x = 1/5. If x = 1/5, it is obviously satisfied for x < 1.
(e) Use a sign diagram. (Note that (5x − 1)11 has the same sign as 5x − 1.)
3x − 1 3x − 1 −(1 + x 2 )
(f) > x + 3 if and only if − (x + 3) > 0, i.e. > 0, so x < 0. (1 + x 2 is
x x x
x−3 x−3 −2x(x + 2)
always positive.) (g) > 2x − 1 if and only if − (2x − 1) < 0, i.e. < 0.
x+3 x+3 x+3
Then use a sign diagram. (h) x 2 − 4x + 4 = (x − 2)2 , which is 0 for x = 2, and strictly positive for
x = 2. (i) x 3 + 2x 2 + x = x(x 2 + 2x + 1) = x(x + 1)2 . Since (x + 1)2 is always ≥ 0, we see that
x 3 + 2x 2 + x ≤ 0 if and only if x ≤ 0.
Review Problems for Chapter 1
5. (a) (2x)4 = 24 x 4 = 16x 4 (b) 2−1 − 4−1 = 1
2 − 1
4 = 41 , so (2−1 − 4−1 )−1 = 4.
(c) Cancel the common factor 4x 2 yz2 . (d) −(−ab3 )−3 = −(−1)−3 a −3 b−9 = a −3 b−9 , so
a 5 · a 3 · a −2 a6
[−(−ab3 )−3 (a 6 b6 )2 ]3 = [a −3 b−9 a 12 b12 ]3 = [a 9 b3 ]3 = a 27 b9 (e) −3 · a 6
= 3
= a3
3 3 −3 a a
x 3 8 −3 x 8 −3 x
(f) · −2 = · −2 = = (x 5 )−3 = x −15
2 x 8 x x −2
√ √ √ √ √ √ √ √
9. All are straightforward,
√ except (c), (g), and (h): (c) − 3 3 − 6 = −3 + 3 6 = −3 + 3 3 2
= −3 + 3 2 (g) (1 + x + x 2 + x 3 )(1 − x) = (1 + x + x 2 + x 3 ) − (1 + x + x 2 + x 3 )x = 1 − x 4
(h) (1 + x)4 = (1 + x)2 (1 + x)2 = (1 + 2x + x 2 )(1 + 2x + x 2 ) and so on.
12. (a) and (b) are easy. (c) ax + ay + 2x + 2y = a(x + y) + 2(x + y) = (a + 2)(x + y)
(d) 2x 2 − 5yz + 10xz − xy = 2x 2 + 10xz − (xy + 5yz) = 2x(x + 5z) − y(x + 5z) = (2x − y)(x + 5z)
(e) p2 − q 2 + p − q = (p − q)(p + q) + (p − q) = (p − q)(p + q + 1) (f) u3 + v 3 − u2 v − v 2 u =
u2 (u − v) + v 2 (v − u) = (u2 − v 2 )(u − v) = (u + v)(u − v)(u − v) = (u + v)(u − v)2 .
s s s(2s + 1) − s(2s − 1) 2s
16. (a) − = = 2
2s − 1 2s + 1 (2s − 1)(2s + 1) 4s − 1
x 1−x 24 −x(x + 3) − (1 − x)(x − 3) − 24 −7(x + 3) −7
(b) − − = = =
3−x x + 3 x2 − 9 (x − 3)(x + 3) (x − 3)(x + 3) x−3
y−x y−x 1
(c) Multiplying numerator and denominator by x 2 y 2 yields 2 = = .
y −x 2 (y − x)(y + x) x+y
17. (a) Cancel the factor 25ab. (b) x 2 − y 2 = (x + y)(x − y). Cancel x + y. (c) The fraction can be
(2a − 3b)2 2a − 3b 4x − x 3 x(2 − x)(2 + x) x(2 + x)
written as = . (d) = =
(2a − 3b)(2a + 3b) 2a + 3b 4 − 4x + x 2 (2 − x) 2 2−x
25. Let each side have length s, and let the area be K. Then K is the sum of the areas of the triangles
ABP , BCP , and CAP in Fig. SM1.R.25, which equals 21 sh1 + 21 sh2 + 21 sh3 = K. It follows that
h1 + h2 + h3 = 2K/s, which is independent of where P is placed.
© Knut Sydsæter, Peter Hammond, and Arne Strøm 2012
Essential Mathematics for
Economic Analysis
th
4 edition
Knut Sydsæter
Peter Hammond
Arne Strøm
For further supporting resources please visit:
www.mymathlab.com/global
, Preface
This student’s solutions manual accompanies Essential Mathematics for Economic Analysis (4th edition, FT
Prentice Hall, 2012). Its main purpose is to provide more detailed solutions to the problems marked ⊂⊃ in the
SM
text. The answers provided in this Manual should be used in combination with any shorter answers provided
in the main text. There are a few cases where only part of the answer is set out in detail, because the rest
follows the same pattern.
We would appreciate suggestions for improvements from our readers, as well as help in weeding out
inaccuracies and errors.
Oslo and Coventry, July 2012
Knut Sydsæter ()
Peter Hammond ()
Arne Strøm ()
Contents
1 Introductory Topics I: Algebra . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1
2 Introductory Topics II: Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4
3 Introductory Topics III: Miscellaneous . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6
4 Functions of One Variable . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9
5 Properties of Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13
6 Differentiation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14
7 Derivatives in Use . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18
8 Single-Variable Optimization . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22
9 Integration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26
10 Interest Rates and Present Values . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34
11 Functions of Many Variables . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36
12 Tools for Comparative Statics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38
13 Multivariable Optimization . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 43
14 Constrained Optimization . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51
15 Matrix and Vector Algebra . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 61
16 Determinants and Inverse Matrices . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 64
17 Linear Programming . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 70
© Knut Sydsæter, Peter Hammond, and Arne Strøm 2012
, CHAPTER 1 INTRODUCTORY TOPICS I: ALGEBRA 1
Chapter 1 Introductory Topics I: Algebra
1.3
5. (a) (2t −1)(t 2 −2t +1) = 2t (t 2 −2t +1)−(t 2 −2t +1) = 2t 3 −4t 2 +2t −t 2 +2t −1 = 2t 3 −5t 2 +4t −1
(b) (a + 1)2 + (a − 1)2 − 2(a + 1)(a − 1) = a 2 + 2a + 1 + a 2 − 2a + 1 − 2a 2 + 2 = 4. Alternatively,
apply the quadratic identity x 2 + y 2 − 2xy = (x − y)2 with x = a + 1 and y = a − 1 to obtain
(a + 1)2 + (a − 1)2 − 2(a + 1)(a − 1) = [(a + 1) − (a − 1)]2 = 22 = 4.
(c) (x + y + z)2 = (x + y + z)(x + y + z) = x(x + y + z) + y(x + y + z) + z(x + y + z) =
x 2 + xy + xz + yx + y 2 + yz + zx + zy + z2 = x 2 + y 2 + z2 + 2xy + 2xz + 2yz (d) With a = x + y + z
and b = x − y − z, (x + y + z)2 − (x − y − z)2 = a 2 − b2 = (a + b)(a − b) = 2x(2y + 2z) = 4x(y + z).
13. (a) a 2 + 4ab + 4b2 = (a + 2b)2 by the first quadratic identity. (d) 9z2 − 16w2 = (3z − 4w)(3z + 4w),
according to the difference-of-squares formula. (e) − 15 x 2 + 2xy − 5y 2 = − 15 (x 2 − 10xy + 25y 2 ) =
− 15 (x − 5y)2 (f) a 4 − b4 = (a 2 − b2 )(a 2 + b2 ), using the difference-of-squares formula. Since
a 2 − b2 = (a − b)(a + b), the answer in the book follows.
1.4
1 1 x+2 x−2 x+2−x+2 4
5. (a) − = − = = 2
x−2 x+2 (x − 2)(x + 2) (x + 2)(x − 2) (x − 2)(x + 2) x −4
(b) Since 4x + 2 = 2(2x + 1) and 4x 2 − 1 = (2x + 1)(2x − 1), the lowest common denominator (LCD)
is 2(2x + 1)(2x − 1). Then
6x + 25 6x 2 + x − 2 (6x + 25)(2x − 1) − 2(6x 2 + x − 2) 42x − 21 21
− = = =
4x + 2 4x − 1
2 2(2x + 1)(2x − 1) 2(2x + 1)(2x − 1) 2(2x + 1)
18b2 a 18b2 − a(a − 3b) + 2(a 2 − 9b2 ) a(a + 3b) a
(c) 2 − +2= = =
a − 9b 2 a + 3b (a + 3b)(a − 3b) (a + 3b)(a − 3b) a − 3b
1 1 (a + 2) − a 2 1
(d) − = = =
8ab 8b(a + 2) 8ab(a + 2) 8ab(a + 2) 4ab(a + 2)
2t − t 2 5t 2t t (2 − t) 3t −t (t − 2) 3t −3t 2
(e) · − = · = · =
t +2 t −2 t −2 t +2 t −2 t +2 t −2 t +2
a 1 − 2a 1
a− 1
a 1 − 1
(f) = 1 2 = 4a − 2, so 2 − 2a
= 2 − (4a − 2) = 4 − 4a = 4(1 − a)
0.25 4
0.25
2 1 2(x + 1) + x − 3x(x + 1) 2 − 3x 2
6. (a) + −3= =
x x+1 x(x + 1) x(x + 1)
t t t (2t − 1) − t (2t + 1) −2t
(b) − = = 2
2t + 1 2t − 1 (2t + 1)(2t − 1) 4t − 1
3x 4x 2x − 1 3x(x − 2) + 4x(x + 2) − (2x − 1) 7x 2 + 1
(c) − − = = 2
x+2 2−x (x − 2)(x + 2) (x − 2)(x + 2) x −4
1 1 1 1 1 1 1 1
+ + xy − 2 − 2 · x2y2
x y x y y+x x2 y x2 y y2 − x2
(d) = = = x + y (e) = = 2
1 1 1 1 1 1 1 y + x2
· xy + + · x 2y2
xy xy x2 y2 x2 y2
(f) To clear the fractions within both the numerator and denominator, multiply both by xy to get
a(y − x) y−x
=
a(y + x) y+x
© Knut Sydsæter, Peter Hammond, and Arne Strøm 2012
,2 CHAPTER 1 INTRODUCTORY TOPICS I: ALGEBRA
1
1 −2
1 −2
8. (a) 1
4 − 1
5 = 5
20 − 4
20 = 1
20 , so 4 − 5 = 20 = 202 = 400
n n·n n(n − 1) − n2n2 −n
(b) n − =n− = n − = =
1 1 n−1 n−1 n−1
1− 1− ·n
n n
1 1 1 1 1 u
(c) Let u = x p−q . Then + = + = + =1
1 + x p−q 1 + x q−p 1 + u 1 + 1/u 1+u 1+u
1 1
+ (x 2 − 1)
x − 1 x2 − 1 (x + 1) + 1 x+2 1
(d) = 3 = =
2 x − x − 2x + 2 (x + 2)(x − 2x + 1)
2 (x − 1)2
x− (x 2 − 1)
x+1
1 1
− 2
1 1 x − (x + h)
2 2 −2xh − h 2
(x + h) 2 x −2x − h
(e) − 2 = = 2 , so = 2
(x + h) 2 x x (x + h)
2 2 x (x + h) 2 h x (x + h)2
10x 2 2x
(f) Multiplying denominator and numerator by x 2 − 1 = (x + 1)(x − 1) yields = .
5x(x − 1) x−1
1.5
5. The answers given in the main text for each
√ respective
√ part
√ emerge
√ after multiplying
√ both numerator
√ √
and denominator by the following: (a) 7 − 5 (b) 5 − 3 (c) 3 + 2 (d) x y − y x
√ √ √
(e) x + h + x (f) 1 − x + 1.
2
12. (a) (2x )2 = 22x = 2x if and only if 2x = x 2 , or if and only if x = 0 or x = 2. (b) Correct because
a p−q = a p /a q . (c) Correct because a −p = 1/a p . (d) 51/x = 1/5x = 5−x if and only if 1/x = −x
or −x 2 = 1, so there is no real x that satisfies the equation. (e) Put u = a x and v = a y , which reduces
the equation to uv = u + v, or 0 = uv − u − v = (u − 1)(v − 1) − 1. This is true only for special values
of u and v and so for special values of x and y. In particular, the equation is false when x = y = 1.
√ √
(f) Putting u = x and v = y reduces the equation to 2u · 2v = 2uv , which holds if and only if
uv = u + v, as in (e) above.
1.6
3x + 1 3x + 1 3x + 1 − 2(2x + 4) −x − 7
4. (a) 2 < has the same solutions as − 2 > 0, or > 0, or >0
2x + 4 2x + 4 2x + 4 2x + 4
A sign diagram reveals that the inequality is satisfied for −7 < x < −2. A serious error is to multiply the
inequality by 2x + 4, without checking the sign of 2x + 4. If 2x + 4 < 0, mulitiplying by this number
will reverse the inequality sign. (It might be a good idea to test the inequality for some values of x. For
example, for x = 0 it is not true. What about x = −5?)
120 480 − 3n
(b) The inequality is equivalent to ≤ 0.75, or ≤ 0. A sign diagram reveals that the
n 4n
inequality is satisfied for n < 0 and for n ≥ 160. (Note that for n = 0 the inequality makes no sense. For
n = 160, we have equality.) (c) Easy: g(g − 2) ≤ 0 etc. (d) Note that p2 − 4p + 4 = (p − 2)2 , and
p+1
the inequality reduces to ≥ 0. The fraction makes no sense if p = 2. The conclusion follows.
(p − 2)2
−n − 2 −n − 2 − 2n − 8 −3n − 10
(e) The inequality is equivalent to − 2 > 0, i.e. > 0, or > 0, etc.
n+4 n+4 n+4
(f) See the text and use a sign diagram. (Don’t cancel x 2 . If you do, x = 0 appears as a false solution.)
© Knut Sydsæter, Peter Hammond, and Arne Strøm 2012
, CHAPTER 1 INTRODUCTORY TOPICS I: ALGEBRA 3
5. (a) Use a sign diagram. (b) The inequality is not satisfied for x = 1. If x = 1, it is obviously satisfied
if and only x + 4 > 0, i.e. x > −4 (because (x − 1)2 is positive when x = 1). (c) Use a sign diagram.
(d) The inequality is not satisfied for x = 1/5. If x = 1/5, it is obviously satisfied for x < 1.
(e) Use a sign diagram. (Note that (5x − 1)11 has the same sign as 5x − 1.)
3x − 1 3x − 1 −(1 + x 2 )
(f) > x + 3 if and only if − (x + 3) > 0, i.e. > 0, so x < 0. (1 + x 2 is
x x x
x−3 x−3 −2x(x + 2)
always positive.) (g) > 2x − 1 if and only if − (2x − 1) < 0, i.e. < 0.
x+3 x+3 x+3
Then use a sign diagram. (h) x 2 − 4x + 4 = (x − 2)2 , which is 0 for x = 2, and strictly positive for
x = 2. (i) x 3 + 2x 2 + x = x(x 2 + 2x + 1) = x(x + 1)2 . Since (x + 1)2 is always ≥ 0, we see that
x 3 + 2x 2 + x ≤ 0 if and only if x ≤ 0.
Review Problems for Chapter 1
5. (a) (2x)4 = 24 x 4 = 16x 4 (b) 2−1 − 4−1 = 1
2 − 1
4 = 41 , so (2−1 − 4−1 )−1 = 4.
(c) Cancel the common factor 4x 2 yz2 . (d) −(−ab3 )−3 = −(−1)−3 a −3 b−9 = a −3 b−9 , so
a 5 · a 3 · a −2 a6
[−(−ab3 )−3 (a 6 b6 )2 ]3 = [a −3 b−9 a 12 b12 ]3 = [a 9 b3 ]3 = a 27 b9 (e) −3 · a 6
= 3
= a3
3 3 −3 a a
x 3 8 −3 x 8 −3 x
(f) · −2 = · −2 = = (x 5 )−3 = x −15
2 x 8 x x −2
√ √ √ √ √ √ √ √
9. All are straightforward,
√ except (c), (g), and (h): (c) − 3 3 − 6 = −3 + 3 6 = −3 + 3 3 2
= −3 + 3 2 (g) (1 + x + x 2 + x 3 )(1 − x) = (1 + x + x 2 + x 3 ) − (1 + x + x 2 + x 3 )x = 1 − x 4
(h) (1 + x)4 = (1 + x)2 (1 + x)2 = (1 + 2x + x 2 )(1 + 2x + x 2 ) and so on.
12. (a) and (b) are easy. (c) ax + ay + 2x + 2y = a(x + y) + 2(x + y) = (a + 2)(x + y)
(d) 2x 2 − 5yz + 10xz − xy = 2x 2 + 10xz − (xy + 5yz) = 2x(x + 5z) − y(x + 5z) = (2x − y)(x + 5z)
(e) p2 − q 2 + p − q = (p − q)(p + q) + (p − q) = (p − q)(p + q + 1) (f) u3 + v 3 − u2 v − v 2 u =
u2 (u − v) + v 2 (v − u) = (u2 − v 2 )(u − v) = (u + v)(u − v)(u − v) = (u + v)(u − v)2 .
s s s(2s + 1) − s(2s − 1) 2s
16. (a) − = = 2
2s − 1 2s + 1 (2s − 1)(2s + 1) 4s − 1
x 1−x 24 −x(x + 3) − (1 − x)(x − 3) − 24 −7(x + 3) −7
(b) − − = = =
3−x x + 3 x2 − 9 (x − 3)(x + 3) (x − 3)(x + 3) x−3
y−x y−x 1
(c) Multiplying numerator and denominator by x 2 y 2 yields 2 = = .
y −x 2 (y − x)(y + x) x+y
17. (a) Cancel the factor 25ab. (b) x 2 − y 2 = (x + y)(x − y). Cancel x + y. (c) The fraction can be
(2a − 3b)2 2a − 3b 4x − x 3 x(2 − x)(2 + x) x(2 + x)
written as = . (d) = =
(2a − 3b)(2a + 3b) 2a + 3b 4 − 4x + x 2 (2 − x) 2 2−x
25. Let each side have length s, and let the area be K. Then K is the sum of the areas of the triangles
ABP , BCP , and CAP in Fig. SM1.R.25, which equals 21 sh1 + 21 sh2 + 21 sh3 = K. It follows that
h1 + h2 + h3 = 2K/s, which is independent of where P is placed.
© Knut Sydsæter, Peter Hammond, and Arne Strøm 2012
