Test Bank For Thomas calculus 11th edition solution manual All Chapters Included

CHAPTER 1 PRELIMINARIES

1.1 REAL NUMBERS AND THE REAL LINE
"
1. Executing long division, 9 œ 0.1, 2
9 œ 0.2, 3
9 œ 0.3, 8
9 œ 0.8, 9
9 œ 0.9

"
2. Executing long division, 11 œ 0.09, 2
11 œ 0.18, 3
11 œ 0.27, 9
11 œ 0.81, 11
11 œ 0.99

3. NT = necessarily true, NNT = Not necessarily true. Given: 2 < x < 6.
a) NNT. 5 is a counter example.
b) NT. 2 < x < 6 Ê 2
2 < x
2 < 6
2 Ê 0 < x
2 < 2.
c) NT. 2 < x < 6 Ê 2/2 < x/2 < 6/2 Ê 1 < x < 3.
d) NT. 2 < x < 6 Ê 1/2 > 1/x > 1/6 Ê 1/6 < 1/x < 1/2.
e) NT. 2 < x < 6 Ê 1/2 > 1/x > 1/6 Ê 1/6 < 1/x < 1/2 Ê 6(1/6) < 6(1/x) < 6(1/2) Ê 1 < 6/x < 3.
f) NT. 2 < x < 6 Ê x < 6 Ê (x
4) < 2 and 2 < x < 6 Ê x > 2 Ê
x <
2 Ê
x + 4 < 2 Ê
(x
4) < 2.
The pair of inequalities (x
4) < 2 and
(x
4) < 2 Ê | x
4 | < 2.
g) NT. 2 < x < 6 Ê
2 >
x >
6 Ê
6 <
x <
2. But
2 < 2. So
6 <
x <
2 < 2 or
6 <
x < 2.
h) NT. 2 < x < 6 Ê
1(2) >
1(x) <
1(6) Ê
6 <
x <
2

4. NT = necessarily true, NNT = Not necessarily true. Given:
1 < y
5 < 1.
a) NT.
1 < y
5 < 1 Ê
1 + 5 < y
5 + 5 < 1 + 5 Ê 4 < y < 6.
b) NNT. y = 5 is a counter example. (Actually, never true given that 4  y  6)
c) NT. From a),
1 < y
5 < 1, Ê 4 < y < 6 Ê y > 4.
d) NT. From a),
1 < y
5 < 1, Ê 4 < y < 6 Ê y < 6.
e) NT.
1 < y
5 < 1 Ê
1 + 1 < y
5 + 1 < 1 + 1 Ê 0 < y
4 < 2.
f) NT.
1 < y
5 < 1 Ê (1/2)(
1 + 5) < (1/2)(y
5 + 5) < (1/2)(1 + 5) Ê 2 < y/2 < 3.
g) NT. From a), 4 < y < 6 Ê 1/4 > 1/y > 1/6 Ê 1/6 < 1/y < 1/4.
h) NT.
1 < y
5 < 1 Ê y
5 >
1 Ê y > 4 Ê
y <
4 Ê
y + 5 < 1 Ê
(y
5) < 1.
Also,
1 < y
5 < 1 Ê y
5 < 1. The pair of inequalities
(y
5) < 1 and (y
5) < 1 Ê | y
5 | < 1.


5.
2x  4 Ê x 
2

6. 8
3x 5 Ê
3x
3 Ê x Ÿ 1 ïïïïïïïïïñqqqqqqqqp x
1

7. 5x
$ Ÿ (
3x Ê 8x Ÿ 10 Ê x Ÿ 5
4


8. 3(2
x)  2(3  x) Ê 6
3x  6  2x
Ê 0  5x Ê 0  x ïïïïïïïïïðqqqqqqqqp x
0

"
9. 2x
# 7x  7
6 Ê
"#
7
6 5x
Ê "
5
ˆ
10 ‰
6 x or
"
3 x

6
x 3x
4
10. 4  2 Ê 12
2x  12x
16
Ê 28  14x Ê 2  x qqqqqqqqqðïïïïïïïïî x
2

,2 Chapter 1 Preliminaries
"
11. 4
5 (x
2)  3 (x
6) Ê 12(x
2)  5(x
6)
Ê 12x
24  5x
30 Ê 7x 
6 or x 
67

12.
x2 5 Ÿ 123x
4 Ê
(4x  20) Ÿ 24  6x
Ê
44 Ÿ 10x Ê
22
5 Ÿ x qqqqqqqqqñïïïïïïïïî x

22/5

13. y œ 3 or y œ
3

14. y
3 œ 7 or y
3 œ
7 Ê y œ 10 or y œ
4

15. 2t  5 œ 4 or 2t  & œ
4 Ê 2t œ
1 or 2t œ
9 Ê t œ
"# or t œ
9#

16. 1
t œ 1 or 1
t œ
1 Ê
t œ ! or
t œ
2 Ê t œ 0 or t œ 2

17. 8
3s œ 9
2 or 8
3s œ
#9 Ê
3s œ
7# or
3s œ
25
# Ê sœ
7
6 or s œ 25
6


18. s
#
1 œ 1 or s
#
1 œ
1 Ê s
# œ 2 or s
# œ ! Ê s œ 4 or s œ 0


19.
2  x  2; solution interval (
2ß 2)

20.
2 Ÿ x Ÿ 2; solution interval [
2ß 2] qqqqñïïïïïïïïñqqqqp x

2 2

21.
3 Ÿ t
1 Ÿ 3 Ê
2 Ÿ t Ÿ 4; solution interval [
2ß 4]

22.
1  t  2  1 Ê
3  t 
1;
solution interval (
3ß
1) qqqqðïïïïïïïïðqqqqp t

3
1

23.
%  3y
7  4 Ê 3  3y  11 Ê 1  y  11
3 ;
solution interval ˆ1ß 11 ‰
3


24.
1  2y  5  " Ê
6  2y 
4 Ê
3  y 
2;
solution interval (
3ß
2) qqqqðïïïïïïïïðqqqqp y

3
2

25.
1 Ÿ z
5
1Ÿ1 Ê 0Ÿ z
5 Ÿ 2 Ê 0 Ÿ z Ÿ 10;
solution interval [0ß 10]

26.
2 Ÿ
1 Ÿ 2 Ê
1 Ÿ
3z
#
3z
# Ÿ 3 Ê
32 Ÿ z Ÿ 2;
solution interval

23 ß 2‘ qqqqñïïïïïïïïñqqqqp z

2/3 2

27.
"#  3
"
x  "
# Ê
7# 
x" 
5# Ê 7
#  "
x  5
#

Ê 2
7 x 2
5 ; solution interval ˆ 27 ß 25 ‰


"
28.
3  2
x
43 Ê 1 2
x ( Ê 1 x
#  7

Ê 2x 2
7 Ê 2
7  x  2; solution interval ˆ 27 ß 2‰ qqqqðïïïïïïïïðqqqqp x
2/7 2

, Section 1.1 Real Numbers and the Real Line 3

29. 2s 4 or
2s 4 Ê s 2 or s Ÿ
2;
solution intervals (
_ß
2]  [2ß _)

" "
30. s  3 # or
(s  3) # Ê s
5# or
s 7
#
Ê s
5# or s Ÿ
7# ;
solution intervals ˆ
_ß
7# ‘ 

5# ß _‰ ïïïïïïñqqqqqqñïïïïïïî s

7/2
5/2

31. 1
x  1 or
("
x)  1 Ê
x  0 or x  2
Ê x  0 or x  2; solution intervals (
_ß !)  (2ß _)

32. 2
3x  5 or
(2
3x)  5 Ê
3x  3 or 3x  7
Ê x 
1 or x  73 ;
solution intervals (
_ß
1)  ˆ 73 ß _‰ ïïïïïïðqqqqqqðïïïïïïî x

1 7/3

33. r"
# 1 or
ˆ r# 1 ‰ 1 Ê r1 2 or r  1 Ÿ
2
Ê r 1 or r Ÿ
3; solution intervals (
_ß
3]  [1ß _)

34. 3r
5
" 2
5 or
ˆ 3r5
"‰  2
5
Ê 3r
5 
or
3r5 
53 Ê r  37 or r  1
7
5
solution intervals (
_ß ")  ˆ 73 ß _‰ ïïïïïïðqqqqqqðïïïïïïî r
1 7/3

35. x#  # Ê kxk  È2 Ê
È2  x  È2 ;
solution interval Š
È2ß È2‹ qqqqqqðïïïïïïðqqqqqqp x

È # È#


36. 4 Ÿ x# Ê 2 Ÿ kxk Ê x 2 or x Ÿ
2;
solution interval (
_ß
2]  [2ß _) ïïïïïïñqqqqqqñïïïïïïî r

2 2

37. 4  x#  9 Ê 2  kxk  3 Ê 2  x  3 or 2 
x  3
Ê 2  x  3 or
3  x 
2;
solution intervals (
3ß
2)  (2ß 3) qqqqðïïïïðqqqqðïïïïðqqqp x

3
2 2 3

" " " " " " " "
38. 9  x#  4 Ê 3  kxk  # Ê 3 x # or 3 
x  #
" "
Ê 3 x or
#"  x 
3" ;
#
solution intervals ˆ
"# ß
3" ‰  ˆ 3" ß #" ‰ qqqqðïïïïðqqqqðïïïïðqqqp x

1/2
1/3 1/3 1/2

39. (x
1)#  4 Ê kx
1k  2 Ê
2  x
1  2
Ê
1  x  3; solution interval (
"ß $) qqqqqqðïïïïïïïïðqqqqp x

1 3

40. (x  3)#  # Ê kx  3k  È2
Ê
È2  x  3  È2 or
3
È2  x 
3  È2 ;
solution interval Š
3
È2ß
3  È2‹ qqqqqqðïïïïïïïïðqqqqp x

3
È #
3  È #

, 4 Chapter 1 Preliminaries

Ê ˆx
12 ‰ <
2
41. x#
x  0 Ê x#
x + 1
4 < 1
4
1
4 ʹx
1
2 ¹< 1
2 Ê
12 < x
1
2 < 1
2 Ê 0 < x < 1.
So the solution is the interval (0ß 1)


42. x#
x
2 0 Ê x#
x + 1
4
9
4 Ê ¹x
1
2 ¹ 3
2 Ê x
1
2
3
2 or
ˆx
12 ‰ 3
2 Ê x 2 or x Ÿ
1.
The solution interval is (
_ß
1]  [2ß _)

43. True if a 0; False if a  0.

44. kx
1k œ 1
x Í k
(x
1)k œ 1
x Í 1
x 0 Í xŸ1

45. (1) ka  bk œ (a  b) or ka  bk œ
(a  b);
both squared equal (a  b)#
(2) ab Ÿ kabk œ kak kbk
(3) kak œ a or kak œ
a, so kak# œ a# ; likewise, kbk# œ b#
(4) x# Ÿ y# implies Èx# Ÿ Èy# or x Ÿ y for all nonnegative real numbers x and y. Let x œ ka  bk and
y œ kak  kbk so that ka  bk# Ÿ akak  kbkb# Ê ka  bk Ÿ kak  kbk .

46. If a 0 and b 0, then ab 0 and kabk œ ab œ kak kbk .
If a  0 and b  0, then ab  0 and kabk œ ab œ (
a)(
b) œ kak kbk .
If a 0 and b  0, then ab Ÿ 0 and kabk œ
(ab) œ (a)(
b) œ kak kbk .
If a  0 and b 0, then ab Ÿ 0 and kabk œ
(ab) œ (
a)(b) œ kak kbk .

47.
3 Ÿ x Ÿ 3 and x 
"# Ê
"
#  x Ÿ 3.

48. Graph of kxk  kyk Ÿ 1 is the interior
of “diamond-shaped" region.




49. Let $ be a real number > 0 and f(x) = 2x + 1. Suppose that | x
1 | < $ . Then | x
1 | < $ Ê 2| x
1 | < 2$ Ê
| 2x
# | < 2$ Ê | (2x + 1)
3 | < 2$ Ê | f(x)
f(1) | < 2$

50. Let % > 0 be any positive number and f(x) = 2x + 3. Suppose that | x
0 | < % /2. Then 2| x
0 | < % and
| 2x + 3
3 | < %. But f(x) = 2x + 3 and f(0) = 3. Thus | f(x)
f(0) | < %.

51. Consider: i) a > 0; ii) a < 0; iii) a = 0.
i) For a > 0, | a | œ a by definition. Now, a > 0 Ê
a < 0. Let
a = b. By definition, | b | œ
b. Since b =
a,
|
a | œ
(
a) œ a and | a | œ |
a | œ a.
ii) For a < 0, | a | œ
a. Now, a < 0 Ê
a > 0. Let
a œ b. By definition, | b | œ b and thus |
a| œ
a. So again
| a | œ |
a|.
iii) By definition | 0 | œ 0 and since
0 œ 0, |
0 | œ 0. Thus, by i), ii), and iii) | a | œ |
a | for any real number.