partial fractions - solutions.pdf
TRANSCRIPT
8/17/2019 Partial Fractions - Solutions.pdf
http://slidepdf.com/reader/full/partial-fractions-solutionspdf 1/4
AQA Core 4 Algebra
1 of 4 18/03/13 © MEI
Section 2: Partial fractions
Solutions to Exercise
1. i)
1
1 1
1 1
x x x x
x x
Putting 1
Putting 1 1
1 1 1
1 1
x x x
ii)
14
2 1 3 2 1 3
14 3 2 1
x
x x x x
x x x
Putting 42 7 6
Putting 2 2
2
14 2 6
2 1 3 2 1 3
x
x x x x
2.
3 2
1 2 2 3 1 2 2 3
3 2 2 2 3 1 2 3 1 2
x C
x x x x x x
x x x x x C x x
Putting 1
3
5 15
Putting 4
3
4 3
Putting 5 5
2 2 4
2
C C
3 2 1 4 2
1 2 2 3 3 1 3 2 2 3
x
x x x x x x
3. i)
7
1 2 1 2
7 2 1
x
x x x x
x x x
Putting 6 3 2
Putting
9 3 3
7 3 2
1 2 2 1
x
x x x x
ii)
2
1 1
4 2 2 2 2
1 2 2
x x
x x x x x
x x x
Putting 3
4
3 4
Putting 1
4
1 4
8/17/2019 Partial Fractions - Solutions.pdf
http://slidepdf.com/reader/full/partial-fractions-solutionspdf 2/4
AQA C4 Algebra 2 Exercise solutions
2 of 4 18/03/13 © MEI
1 3 1
4 4 2 4 2
x
x x x
iii)
2
2
9
1 3 2 1 1 3 2 1
9 3 2 1 1 2 1 1 3
x x
C
x x x x x x
x x x x x x C x x
Putting 8 4 2
Putting 84 28 3
Putting 7
2 4 4
1C C
2
9
2 3 1
1 3 2 1 1 3 2 1
x x
x x x x x x
4. i)
2 2
2
1 1 1
2 1
x
x x x
x x
Putting 2
Equating coefficients of x
2
2 2 2
1 1 1
x
x x x
ii)
2 2
3 2
2 3 2 3 2 3
3 2 2 3
x
x x x
x x
Putting 3
4
Equating coefficients of x 3 1
2
3 2 4 1
2 3 2 3 2 3
x
x x x
5. i)
2 2
2
4
1 3 1 1 3 1 1
4 1 1 3 1 1 3
C
x x x x x
x x x C x
Putting 3 9
9
Putting 4 2 2
C C
Equating coefficients of x²
0 3
3 9
3
2
9
3 2
1 3 1 1 3 1 1x x x x
8/17/2019 Partial Fractions - Solutions.pdf
http://slidepdf.com/reader/full/partial-fractions-solutionspdf 3/4
AQA C4 Algebra 2 Exercise solutions
3 of 4 18/03/13 © MEI
ii)
2 2
2
13 7
2 3 2 2 3
13 7 2 3 3 2
x C
x x x x x
x x x x C x
Putting 1 1
Putting 8
C
Equating coefficients of x² 8
C
2
13 7 8 1 8
2 3 2 2 3
x
x x x x x
6. i) The degree of the numerator and denominator are both 2, so there
will be a constant term.
2
2
2 6 7
2 2 1 2 2 1
2 6 7 2 2 1 2 1 2
x x C
x x x x
x x x x x C x
Putting 3 3 1
Putting 1 3
2 2 2
3C C
Equating coefficients of x² 2 1
2
2 6 7 1 3
1
2 2 1 2 2 1
x x
x x x x
ii) The degree of the numerator is 3 and the degree of the denominator is 2, so
there will be a linear expression.
3 2
3 2 2
2 4 1
1 1
2 4 1 1 1 1
x x x C D
x
x x x x
x x x x x x x C x Dx
Putting 1
C
Putting 2 2
D D
Equating coefficients of x³
Equating coefficients of x² 3
3 2
2 4 1 1 2
2 3
1 1
x x x
x
x x x x
iii) The degree of the numerator and denominator are both 3, so there will be a
constant term.
3 2
2 2
3 2 2
2
3 9 2
1 4 1 1 4
3 9 2 1 4 1 4
4 1
x x x
C D
x x x x x
x x x x x x x
C x D x
Putting
15 5 3
C C
Putting
50 25 2
D D
Equating coefficients of x³