Summary of contents Part 1 Foundation topics F.1 F.2 F.3 F.4 F.5 F.6 F.7 F.8 F.9 F.10 F.11 F.12 F.13
Arithmetic Introduction to algebra Expressions and equations Graphs Linear equations Polynomial equations Binomials Partial fractions Trigonometry Functions Trigonometric and exponential functions Differentiation Integration
3 63 101 127 161 177 195 223 243 267 287 315 353
Part II 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29
Complex numbers 1 Complex numbers 2 Hyperbolic functions Determinants Matrices Vectors Differentiation Differentiation applications Tangents, normals and curvature Sequences Series 1 Series 2 Curves and curve fitting Partial differentiation 1 Partial differentiation 2 Integration 1 Integration 2 Reduction formulas Integration applications 1 Integration applications 2 Integration applications 3 Approximate integration Polar coordinate systems Multiple integrals First-order differential equations Second-order differential equations Introduction to Laplace transforms Data handling and statistics Probability
385 412 437 459 489 519 544 562 583 605 639 663 688 731 752 768 794 821 834 851 873 902 921 943 968 1004 1027 1045 1076
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Quiz F.1
Frames
1
Place the appropriate symbol < or > between each of the following pairs of numbers: (a) 3 2 (b) 8 13 (c) 25 0
1
to
4
Find the value of each of the following: (a) 13 þ 9 3 2 5 (b) ð13 þ 9Þ ð3 2Þ 5
5
to
12
Round each number to the nearest 10, 100 and 1000: (a) 1354 (b) 2501 (c) 2452 (d) 23 625
13
to
15
Write each of the following as a product of prime factors: (a) 170 (b) 455 (c) 9075 (d) 1140
19
to
22
Find the HCF and the LCM of each pair of numbers: (a) 84, 88 (b) 105, 66
23
to
24
28
to
36
37
to
46
47
to
48
49
to
52
56
to
65
66
to
67
12 Convert each of the following to fractional form in lowest terms: (a) 0:8 (b) 2:8 (c) 3:3_ 2_ (d) 5:5
68
to
73
13 Write each of the following in abbreviated form: (a) 1.010101 . . . (b) 9.2456456456 . . .
70
to
71
14 Write each of the following as a number raised to a power: 3 8 (a) 36 33 (b) 43 25 (c) 92 (d) 70
78
to
89
2 3 4 5 6
7
8
9
Reduce each of the following fractions to their lowest terms: 12 144 49 64 (b) (c) (d) (a) 18 21 14 4 Evaluate the following: 3 2 11 5 (b) (a) 7 3 30 6
(c)
3 4 þ 7 13
(d)
5 4 16 3
Write the following proportions as ratios: 1 1 3 of C (a) of A, of B and 2 5 10 1 1 1 (b) of P, of Q , of R and the remainder S 4 3 5 Complete the following: 4 (b) 48% of 50 ¼ (a) ¼ % 5 9 ¼ % (d) 15% of 25 ¼ (c) 14
10 Round each of the following decimal numbers, first to 3 significant figures and then to 2 decimal places: (a) 21.355 (b) 0.02456 . (c) 0 3105 (d) 5134.555 11 Convert each of the following to decimal form to 3 decimal places: 4 7 9 28 (b) (c) (d) (a) 15 13 5 13
4
Frames
15 Find the value of each of the following to 3 dp: pffiffiffi 1 1 1 (a) 153 (b) 5 5 (d) ð 9Þ2 (c) ð 27Þ3
90
to
94
16 Write each of the following as a single decimal number: (b) 16:1105 10 2 (a) 3:2044 103
95
to
98
17 Write each of the following in standard form: (a) 134.65 (b) 0.002401
99
to
101
18 Write each of the following in preferred standard form: (b) 9.3304 (a) 16:1105 10 2
102
to
104
105
to
108
20 Express the following numbers in denary form: (b) 456:7218 (a) 1011:012 : (c) 123 2912 (d) CA1:B2216
112
to
126
21 Convert 15:60510 to the equivalent octal, binary, duodecimal and hexadecimal forms.
127
to
149
19 In each of the following the numbers have been obtained by measurement. Evaluate each calculation to the appropriate level of accuracy: (a) 11:4 0:0013 5:44 8:810 1:01 0:00335 (b) 9:12 6:342
5
Partial fractions
227
A¼6
7
Because 16 28 ¼ Að 2Þ þ Bð0Þ so that 12 ¼ 2A giving A ¼ 6. Therefore: 8x 28 6 2 ¼ þ ðx 2Þðx 4Þ x 2 x 4 the required partial fraction breakdown. This example has demonstrated the basic process whereby the partial fractions of a given rational expression can be obtained. There is, however, one important proviso that has not been mentioned: To effect the partial fraction breakdown of a rational algebraic expression it is necessary for the degree of the numerator to be less than the degree of the denominator. If, in the original algebraic rational expression, the degree of the numerator is not less than the degree of the denominator then we divide out by long division. This gives a polynomial with a rational remainder where the remainder has a numerator with degree less than the denominator. The remainder can then be broken down into its partial fractions. In the following frames we consider some examples of this type.
8
Example 1 2
Express
x þ 3x 10 in partial fractions. x2 2x 3
The first consideration is . . . . . . . . . . . .
Is the numerator of lower degree than the denominator?
9
No, it is not, so we have to divide out by long division: 1 2
2
x 2x 3 x þ3x 10 x2 2x 3 5x 7
;
x2 þ 3x 10 5x 7 ¼1þ 2 2 x 2x 3 x 2x 3
Now we factorize the denominator into its prime factors, which gives . . . . . . . . . . . .
ðx þ 1Þðx 3Þ ;
x2 þ 3x 10 5x 7 ¼1þ 2 x 2x 3 ðx þ 1Þðx 3Þ
The remaining fraction will give partial fractions of the form: 5x 7 A B ¼ þ ðx þ 1Þðx 3Þ x þ 1 x 3 Multiplying both sides by the denominator ðx þ 1Þðx 3Þ: 5x 7 ¼ . . . . . . . . . . . .
10
228
Programme F.8
11
Aðx 3Þ þ Bðx þ 1Þ 5x 7 Aðx 3Þ þ Bðx þ 1Þ is an identity, since the RHS is the LHS merely written in a different form. Therefore the statement is true for any value of x we choose to substitute. As was said previously, it is convenient to select a value for x that makes one of the brackets zero. So if we put x ¼ 3 in both sides of the identity, we get . . . . . . . . . . . .
12
15 7 ¼ Að0Þ þ Bð4Þ i.e.
8 ¼ 4B
;B¼2
Similarly, if we substitute x ¼ 1, we get . . . . . . . . . . . .
13
5 7 ¼ Að 4Þ þ Bð0Þ ; 12 ¼ 4A ;A¼3 5x 7 3 2 ; ¼ þ ðx þ 1Þðx 3Þ x þ 1 x 3 So, collecting our results together: x2 þ 3x 10 5x 7 5x 7 ¼1þ 2 ¼1þ x2 2x 3 x 2x 3 ðx þ 1Þðx 3Þ 3 2 þ ¼1þ xþ1 x 3 Example 2 Express
2x2 þ 18x þ 31 in partial fractions. x2 þ 5x þ 6
The first step is . . . . . . . . . . . .
14
to divide the numerator by the denominator since the numerator is not of lower degree than that of the denominator. ;
2x2 þ 18x þ 31 8x þ 19 ¼2þ 2 . x2 þ 5x þ 6 x þ 5x þ 6
Now we attend to
8x þ 19 . x2 þ 5x þ 6
Factorizing the denominator, we have . . . . . . . . . . . .
15
8x þ 19 ðx þ 2Þðx þ 3Þ so the form of the partial fractions will be . . . . . . . . . . . .
Partial fractions
229
16
A B þ xþ2 xþ3 i.e.
8x þ 19 A B ¼ þ ðx þ 2Þðx þ 3Þ x þ 2 x þ 3
You can now multiply both sides by the denominator ðx þ 2Þðx þ 3Þ and finish it off: 2x2 þ 18x þ 31 ¼ ............ x2 þ 5x þ 6
2þ
17
3 5 þ xþ2 xþ3 So move on to the next frame
18
Example 3 3x3 x2 13x 13 in partial fractions. Express x2 x 6 Applying the rules, we first divide out: 3x3 x2 13x 13 ¼ ............ x2 x 6
3x þ 2 þ
Now we attend to
7x 1 in the normal way. x2 x 6
3x þ 2 þ
3 4 þ xþ2 x 3
Because 7x 1 A B ¼ þ ðx þ 2Þðx 3Þ x þ 2 x 3 x¼3 x ¼ 2
; 7x 1 ¼ Aðx 3Þ þ Bðx þ 2Þ 20 ¼ Að0Þ þ Bð5Þ ;B¼4 15 ¼ Að 5Þ þ Bð0Þ ;A¼3
Remembering to include the polynomial part: ;
19
7x 1 x 6
x2
3x3 x2 13x 13 3 4 ¼ 3x þ 2 þ þ x2 x 6 xþ2 x 3
Now one more entirely on your own just like the last one. Example 4 2x3 þ 3x2 54x þ 50 in partial fractions. x2 þ 2x 24 Work right through it: then check with the next frame. Express
2x3 þ 3x2 54x þ 50 ¼ ............ x2 þ 2x 24
Finish it off
20
230
Programme F.8
21
2x 1 Ăž
1 5 x 4 xĂž6 2x 1
Here it is: 2
3
2
x Ăž 2x 24 2x Ăž 3x 54x Ăž 50 2x3 Ăž 4x2 48x x2 6x Ăž 50 x2 2x Ăž 24 4x Ăž 26 2x3 Ăž 3x2 54x Ăž 50 4x 26 Âź 2x 1 2 2 x Ăž 2x 24 x Ăž 2x 24 4x 26 A B Âź Ăž ; 4x 26 Âź AĂ°x Ăž 6Ă&#x17E; Ăž BĂ°x 4Ă&#x17E; Ă°x 4Ă&#x17E;Ă°x Ăž 6Ă&#x17E; x 4 x Ăž 6 xÂź4 10 Âź AĂ°10Ă&#x17E; Ăž BĂ°0Ă&#x17E; ; A Âź 1 x Âź 6 50 Âź AĂ°0Ă&#x17E; Ăž BĂ° 10Ă&#x17E; ;BÂź5 2x3 Ăž 3x2 54x Ăž 50 1 5 Âź 2x 1 Ăž ; x2 Ăž 2x 24 x 4 xĂž6
;
Âź 2x 1 Ăž
1 5 x 4 xĂž6
At this point let us pause and summarize the main facts so far on the breaking into partial fractions of rational algebraic expressions with denominators in the form of a product of two simple factors
Review summary 22
1 To effect the partial fraction breakdown of a rational algebraic expression it is necessary for the degree of the numerator to be less than the degree of the denominator. In such an expression whose denominator can be expressed as a product of simple prime factors, each of the form ax Ăž b: (a) Write the rational expression with the denominator given as a product of its prime factors. A where A is (b) Each factor then gives rise to a partial fraction of the form ax Ăž b a constant whose value is to be determined. (c) Add the partial fractions together to form a single algebraic fraction whose numerator contains the unknown constants and whose denominator is identical to that of the original expression. (d) Equate the numerator so obtained with the numerator of the original algebraic fraction. (e) By substituting appropriate values of x in this equation determine the values of the unknown constants. 2 If, in the original algebraic rational expression, the degree of the numerator is not less than the degree of the denominator then we divide out by long division. This gives a polynomial with a rational remainder where the remainder has a numerator with degree less than the denominator. The remainder can then be broken down into its partial fractions.
Partial fractions
231
Review exercise Express the following in partial fractions: xþ7 1 x2 7x þ 10 3 2
3x2 8x 63 x2 3x 10 10x þ 37 x2 þ 3x 28
4
2x2 þ 6x 35 x2 x 12
1
xþ7 xþ7 A B ¼ ¼ þ x2 7x þ 10 ðx 2Þðx 5Þ x 2 x 5 ; x þ 7 ¼ Aðx 5Þ þ Bðx 2Þ x¼5 12 ¼ Að0Þ þ Bð3Þ x¼2 9 ¼ Að 3Þ þ Bð0Þ xþ7 4 3 ¼ x2 7x þ 10 x 5 x 2
2
;B¼4 ; A ¼ 3
10x þ 37 10x þ 37 A B ¼ ¼ þ x2 þ 3x 28 ðx 4Þðx þ 7Þ x 4 x þ 7 ; 10x þ 37 ¼ Aðx þ 7Þ þ Bðx 4Þ x ¼ 7 33 ¼ Að0Þ þ Bð 11Þ x¼4 77 ¼ Að11Þ þ Bð0Þ 10x þ 37 7 3 ¼ þ ; 2 x þ 3x 28 x 4 x þ 7
3
;B¼3 ;A¼7
3x2 8x 63 x 33 ¼3þ x2 3x 10 ðx þ 2Þðx 5Þ x 33 A B ¼ þ ðx þ 2Þðx 5Þ x þ 2 x 5 ; x 33 ¼ Aðx 5Þ þ Bðx þ 2Þ x¼5 28 ¼ Að0Þ þ Bð7Þ ; B ¼ 4 x ¼ 2
35 ¼ Að 7Þ þ Bð0Þ
;A¼5
2
; 4
23
3x 8x 63 5 4 ¼3þ x2 3x 10 xþ2 x 5
2x2 þ 6x 35 8x 11 ¼2þ x2 x 12 ðx þ 3Þðx 4Þ 8x 11 A B ¼ þ ðx þ 3Þðx 4Þ x þ 3 x 4 8x 11 ¼ Aðx 4Þ þ Bðx þ 3Þ x¼4 21 ¼ Að0Þ þ Bð7Þ ;B¼3 x ¼ 3 2
;
35 ¼ Að 7Þ þ Bð0Þ
;A¼5
2x þ 6x 35 5 3 ¼2þ þ x2 x 12 xþ3 x 4
24
240
Programme F.8
Test exercise F.8 48
Express each of the following in partial fractions: 1 2
x 14 x2 10x þ 24 13x 7 11x þ 3
10x2
1
to
7
1
to
7
3
4x2 þ 9x 73 x2 þ x 20
8
to
21
4
6x2 þ 19x 11 ðx þ 1Þðx2 þ 5x 2Þ
25
to
31
5
10x 13 32
to
34
35
to
36
ð2x 3Þ2 6
3x2 34x þ 97 ðx 5Þ3
7
9x2 16x þ 34 ðx þ 4Þð2x 3Þð3x þ 1Þ
37
to
40
8
8x2 þ 27x þ 13 x3 þ 4x2 þ x 6
41
to
42
Further problems F.8 49
Frames
Express each of the following in partial fractions: 1
7x þ 36 x2 þ 12x þ 32
2
5x 2 x2 3x 28
3
xþ7 x2 7x þ 10
4
3x 9 x2 3x 18
5
7x 9 2x2 7x 15
6
14x 6x2 x 2
8
7x 7 6x2 þ 11x þ 3
7
13x 7 11x þ 3
10x2 9
18x þ 20
10
2
ð5x þ 2Þ2
ð3x þ 4Þ 11
12x 16
12
2
75x2 þ 35x 4
14
ð5x þ 2Þ3 15
5x2 13x þ 5 ðx 2Þ3
ð4x 5Þ 13
35x þ 17
64x2 148x þ 78 ð4x 5Þ3
8x2 þ x 3
16 2
ðx þ 2Þðx 1Þ
4x2 24x þ 11 ðx þ 2Þðx 3Þ2