Essential Mathematics for
Economic Analysis
th
4 edition
Knut Sydsæter
Peter Hammond
Arne Strøm
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, 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 (knutsy@econ.uio.no)
Peter Hammond (hammond@stanford.edu)
Arne Strøm (arne.strom@econ.uio.no)
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
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.)
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 .
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.
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