Unit 1 Chapter-3 Partial Fractions, Algebraic ... 101/Unit 1 Chapter 3 (Surds... · 1 1 Unit 1...
Transcript of Unit 1 Chapter-3 Partial Fractions, Algebraic ... 101/Unit 1 Chapter 3 (Surds... · 1 1 Unit 1...
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1
Unit 1
Chapter-3
Partial Fractions, Algebraic Relationships, Surds, Indices, Logarithms
3.1 Partial Fractions:
A fraction of the form 1
733
2
x
x where the degree of the numerator is less than the degree of
the denominator is referred to as a proper fraction. If the degree of the numerator is greater
than or equal to the degree of the denominator the fraction is referred to as an improper
fraction.
An improper fraction of the form 1
12
2
x
xmay be written as
1
21
1
2
1
1
1
1222
2
2
2
xxx
x
x
x
This process is called expressing or decomposing a single fraction as a sum of two (or
several) separate fractions or Partial fractions.
Example1
Consider the rational expression 11
22
x
x
x.
This expression can be expressed in the compact form as 11
1122
2
xx
xxx= 11
232
2
xx
xx.
Example 2
Consider the expression 42
3
xx
x .
This expression can be expressed as the sum of two Partial fractions as
42
3
xx
x= 42
x
B
x
A.
Here the values of A and B are to be determined in the following way,
42
3
xx
x= 42
x
B
x
A.
243 xBxAx
2
2
Substituting 2x gives 6
5A and substituting 4x gives
6
1B . So
42
3
xx
x=
46
1
26
5
xx.
3.1.1 Supplementary Problems
1))1)(1(
3
xx 2)
)1)(4(
3
xx 3)
)2)(2(
1
xx
x
4) )2)(12(
3
xx 5)
xx
x
)1(
3
6)
)23)(1(
12
xx
x
7) )3)(2)(1(
3
xxx
x 8)
)1)(3(2
422
xxx
xx 9)
)12)(13(
12
xx
x
10) )12)(1(
5 2
xxx
xx 11)
)2)(12(
31
xx
x 12)
)2)(1(
)1(
xx
xx
13) )3)(1(
42
xxx
x
Multiple choice Exercise
Type 1
1) If ,3
2
1))(1(
xx
q
cxx
px the value of p and q are:
(a) 1,2 qp (b) 1,2 qp (c) 2,1 qp (d) 1,1 qp (e) 1,1 qp
2. If 062 pxx has equal roots and pp ,0 is :
(a) 48 (b)0 (c) 6 (d)3 (e) 24
3. If 1)(4 22 qxpxx , the values of p and q are:
(a) 2,5 qp (b) 2,1 qp (c) 5,2 qp (d) 5,1 qp (e) 1,0 qp
3.2 Algebraic Relationships
Solving a quadratic equation 02 cbxax .
3
3
Let 02
22 a
cx
a
bx
0442
22
2
2
22
a
c
a
b
a
bx
a
bx
a
c
a
b
a
bx
2
22
42
2
2
4
4
2 a
acb
a
bx
a
acb
a
bx
2
4
2
2 .
Hence the two values for x are a
acb
a
bx
2
4
2
2 and
a
acb
a
bx
2
4
2
2
Note: i) when 042 acb , x has two distinct roots.
ii) when 042 acb , x has two equal roots.
iii) when 042 acb , x has no real roots.
Example:
Solve; 0162 x
162 x
16 x
142
x
14 x
But a1 , when a is a real number.
Note:
The number 1 is not a real number. It is considered as an imaginary number and denoted
with the letter i. i.e. 1i . So, in the above example the solutions of the given equation
are ix 4 and ix 4 .
4
4
3.2.1 Solve the following quadratic equations by completing the square
1) 0462 2 xx 2) 0842 xx 3) 0372 2 xx
4) 022 axx 5) 022 baxx 6) 02 cbxax
Determine the nature of the roots of the following equation but do not solve the equations.
7) 0962 xx 8) 01062 xx 9) 0352 2 xx
10) 0243 2 xx 11) 09124 2 xx 12) 09124 2 xx
13) For what values of K is 169 2 kxx a perfect square?
14) The roots of 0123 2 kxx are equal .Find K.
15)Find a if 052 axx has equal roots.
3.3 Surds
3.3.1 General rules of Surds
A number which can be expressed as a fraction of integers (assuming the denominator is
never 0) is called a rational number. Examples of rational numbers are 5
4,
2
5 and 2. A
number which cannot be expressed as a fraction of two integers is called an irrational
number. Example of irrational numbers are 3 7,2 and Π. An irrational number involving a
root is called a Surd.
3.3.2 Multiplication of Surds:
baba
For example:
i. 636123123
ii. 864232232
3.3.3 Division of Surds:
b
aba
For example:
5
5
i. 6362
72272
ii. 395
45545
3.3.4 Addition and Subtraction of Surds:
baba
baba
Example (1):
Simplify (a) 75212243 and (b) 32850
(a)
317
3103239
325234381
325234381
75212243
(b)
211
242225
21624225
32850
6
6
3.3.5 Rationalization of the Denominator
For example: 4
4
)()())((
37)3()7()37)(37(
59)5()9()59)(59(
22
22
22
nmnmnmnm
Example (2):
Simplify 3
5
3
35
3
3
3
5
3
5
Example (3):
Simplify 37
4
3737
)37(4
37
37
37
4
37
4
Example (4)
Find, without using tables or calculators, the value of 23
1
23
1
7
6
29
2323
)23)(23(
)23()23(
23
1
23
1
3.3.6 Supplementary Problems
1. Simplify each of the following:
i. 27
ii. 243
iii. 9818
iv. 11217528
v. 81255
vi. 27327
vii. 28
22463112
viii. 21
112
ix. 73
1
73
1
7
7
2. Express the following in the form of cba :
i. 2)23(
ii. 2)323(
iii. 2)13(
iv. 3223
3223
v. 2332
2332
vi. 27
27
3. Rationalize the denominators of the following, giving your answer in the
simplest form possible:
i. 35
21
35
21
ii. 3253
5237
iii. 27
14
4. Find the value of each of the following:
i. 53
1
53
1
ii. 832
1850
iii. 22 )51(
1
)51(
1
8
8
3.4 Indices
If a positive integer a is multiplied by itself three times .We get 3a , i.e. 3aaaa .Here a
is called the base and 3, the index or power .Thus 4a means the 4th power of a, In general, na means the power of a , where n is any positive index of the positive integer a.
3.4.1 Rules of indices
There are several important rules to remember when dealing with indices.
If a, b, m and n are positive integers, then
(1) nmnm aaa e.g. 1385 333
(2) nmnm aaa e.g. 11314 555
(3) mnnm aa )( e.g.
1052 5)5(
(4) mmm baba )( e.g.
555 )23(23
(5)
m
mm
b
aba
e.g.
4
44
3
535
(6) 10 a e.g. 150
(7) n
n
aa
1
e.g. 3
3
5
15
(8) nn aa
1
e.g. 33
1
88
(9) mnn
m
aa )( e.g. 233
2
)8(8
Example 1
Evaluate: (i) 32 (ii) 3
1
8 (iii) 4
3
16 (iv) 2
3
25
i. 8
1
2
12
3
3
ii. 288 33
1
iii. 821616 33
44
3
iv. 125
1
5
1
25
1
25
125
3
3
2
32
3
9
9
Example 2
Evaluate: (i) 2
1
5
2
3
1
aaa (ii) 423ba (iii) 21
15 23 aaa
i.
30
7
30
151210
2
1
5
2
3
1
2
1
5
2
3
1
a
a
a
aaa
ii.
812
4243
423
ba
ba
ba
iii.
30
17
2
1
5
2
3
1
2
1
5
2
3
1
2
115 23
a
a
aaa
aaa
3.4.2 Solving Exponential Equations
Example 3
Solve the following exponential equations: (i) 322 x (ii) 25.04 1 x
i.
5
22
322
5
x
x
x
ii.
2
11
44
4
14
25.04
11
1
1
x
x
x
x
x
Example 4
Solve the equation: xxx 2122 332
xxxx
xxx
2122222
2122
33
332
Let xy 2
18
1
0118
0178
188
2
2
ory
yy
yy
yyy
When 8
1y
3
22
8
12
3
x
x
x
10
10
When 1y
)(12 leinadmissibx
3
32
2
1
8
12 x
Hence 3x
Example 5
If 2793 2 yx and8
142 yx
, calculate the value of x and y.
28
142
12793.
2
yx
yx
From (1):
334
33
333
333
34
34
322
yx
yx
yx
yx
From (2):
432
22
222
2
122
32
32
3
2
yx
yx
yx
yx
(3) – (4): 1
66
y
y
Substitute y = 1 into (3):
1
34
x
x
1 x and 1y
11
11
3.4.3 Supplementary Problems
1. Evaluate each of the following without using a calculator:
i. 17
ii. 017
iii. 2
3
49
iv. 3
2
8
v. 5
3
243
vi. 4
1
81
vii. 3
4
27
1
viii.
2
4
1
ix. 2
1
576
1
x. 3
4
512
xi. 13
2
48
xii. 34
1
4625
1
2. Simplify each of the following giving your answer in index form:
i. 6
1
3
1
2
1
aaa
ii. 243 aaa
iii. 64 412 aa
iv. 2
3
2
5
416
aa
v.
15
5
2
3
1
ba
vi.
24
8
3
4
1
ba
vii. aa 3
viii. aa 8 7
ix. 3 24 3 aaa
x. 9 36 2 xx aa
xi. 33
3
64
8
b
a
xii.
3
42
2
9
ba
ab
3. Solve the following equations:
i. 813 x
ii. 1255 x
iii. 832 x
iv. 8
12 x
v. 2
116 x
12
12
vi. 49
17 x
vii. 15 x
viii. 34 273 xx
ix. 634 2 xx
x. 121 842 xxx
xi. 122 2793 xxx
xii. 125
145 13 xx
4. By using appropriate substitution, or otherwise, solve the following equations:
i. 1222 22 xx
ii. 3
1333 22 tt
iii. 01292 212 xx
iv. xxx 3393 312
v. xx 3439
vi. 24225 xx
vii. 033289 1 xx
viii. xx 3246813 22
5. Solve the following pairs of simultaneous equations:
i. 3437,497 yxyx
ii. 32 375,75 abab
iii. 82,2433 52 yxyx
iv. 16
12,6255 242 yxyx
v. 16
182,27813 yxyx
13
13
3.5 Logarithms
For any number y such that xay (a>0 and a≠1), the logarithm of y to the base a is defined
to be x and is denoted by yalog .
Thus
For example,
2100log10100
481log381
10
2
3
4
Note: The logarithm of 1 to any base is 0, i.e. 01log a
The logarithm of a number to a base of the same number is 1, i.e. 1log aa
The logarithm of a negative number is not defined.
Example 1
Find the value of (i) 64log2 (ii) 3log9 (iii) 9
1log 3 (iv) 25.0log8
i. Let x64log 2
6
22
264
6
x
x
x
ii. Let x3log 9
2
1
12
33
93
21
x
x
x
x
iii. Let x9
1log 3
2
33
39
39
1
2
1
x
x
x
x
iv. Let x25.0log8
xay means that yalog =x
14
14
3
2
23
22
24
1
825.0
32
3
x
x
x
x
x
Example 2
Find the logarithm of the following to the base indicated in brackets: (i) 27 (3) (ii) 64 (8)
(iii) 1000 (10) (iv) 0.25 (2)
i. 327log
327
3
3
ii. 264log
864
8
2
iii. 31000log
101000
10
3
iv. 225.0log
225.0
2
2
15
15
3.5.1 Laws of Logarithms
Proof: (1) Let xma log and yna log
xam and
yan
Multiply m by n: yx aanm
nmyxmn
amn
aaa
yx
logloglog
(2) Let xma log and yna log
xam and
yan
Divide m by n: yx aanm
nmyxn
m
an
m
aaa
yx
logloglog
(3) Let xma log
mnxnm
am
am
a
n
a
xnn
x
loglog
Example 3
Without using tables, evaluate 5log2 2
41log-70log
35
41log 10101010 :
(1) nmmn aaa logloglog
(2) nmn
maaa logloglog
(3) mnm a
n
a loglog
16
16
2
10log2
10log
100log
5 41
270
35
41log
5log2 2
41log-70log
35
41log
10
2
10
10
2
10
10101010
Example 4
Given that 2log24log 1010 p , Calculate the value of p without using tables or
Calculators.
2log24log 1010 p
)4(log 2
10 p =2
22 104 p
2p =
4100
P= 5
Since p cannot be -5 because 10log (-5) is not defined, p=5
3.5.2 Supplementary Problems
Write the following in logarithmic from:
1. 2552 2. 1120 3. 34373
4. 3
13 1
5. 32
8
1 6. 36216
Write the following in index form:
7. 38log 2 8. 4625log 5 9. 12
1log 2
17
17
10. 01log 9 11. 236
1log 6 12. 31000log10
Solve the following equations:
13. x1log 3 14. x5
2 2log 15. x5
1log 5
16. x25.0log 4 17. x128log 2 18. x7log 7
19. x5log5
20. 2
1log 4 x 21. 3log 5 x
Simply the following logarithms:
22. 5log10 23. 64log 8 24. 3
5 5log 25. 5125log 5
26. 4
1log16 27.
x5log 5 28. 12log 3 4log 3 29. 5log 7 15log 7
30. 3 5log 6 25log 6 31. 5log3log21log 222
32. 4log310log5log2 333 33. 17log2
3
3
5log2
4
17log
17
81log
2
110101010
If 6309.2log 3 and 456.15log 3 , evaluate the following without the use of calculators or
logarithm tables:
34. 10log 3 35. 15log 3 36. 25log 3
37. 5log 3 38. 5.2log 3 39.
3
13log 3
40. 8
1log 3 41. 100log 3 42. 12log 3
43. 12.0log 3 44. 08.0log 3 45. 25log 3
48. Evaluate the following without using calculators:
(1) 5log32log4log2 101010
18
18
(2) 6log7log9log914log27log 1010101010
(3) 32log2
116log
2
38log
3
2222
(4) 6
5log3
25
24log
9
10log 444
(5) 52log91log175log 101010
(6) 105
8log4
9log
7
32log
15
4log 6666
3.5.3 Common logarithms
Example 1
Solve the Equation 7.43 x
7.43 x
Taking logarithms to base 10,
7.4lg3lg x
7.4lg3lg x
x = 3lg
7.4lg = 1.409
Example 2
Given that x2
x3 = 18 , find x correct to two decimal places .
x2x3 =18
x6 =18
Taking logarithms to base 10,
x6lg = 18lg
18lg6lg x
19
19
x =6lg
18lg = 1.61(correct to 2 decimal places).
3.5.4 Supplementary Problems
1. Solve the following equations, giving your answer correct to 3 significant figures , where
necessary :
(1) 235 37 x
(2) 74.054 x (3)
22 51005 xx
(4) 0127772 xx
Simplify 5lg6lg
125lg8lg27lg
10. Given that ba 3 ,
(1) Find the value of b)
3
1( in terms of a.
(2)Find the value of a and b if 14353 22 bb
2.4 Summary
Surds:
(1) abba (2) b
aba
(3) ba and ba are conjugate surds. The product of Conjugate surds is a rational
number.
Indices:
(1) nmnm aaa (2) nmnm aaa (3) mnnm aa )(
(4) 10 a (5) nn
aa 1
(6) nn aa
1
(7) mnn mn
m
aaa )(
Logarithms
(1) xya log means ya x .
(2) )(logloglog nmnm aaa
20
20
(3) n
mnm aaa logloglog
(4) mnm a
n
a loglog
(5) 01log a
(6) 1log aa