Date post: | 30-Jun-2015 |
Category: |
Technology |
Upload: | shmaths |
View: | 3,923 times |
Download: | 6 times |
DHS H2 MATHEMATICS : a revision package
NEXT
At the end of this revision, you should be able to:
Express a given algebraic fraction in terms of partial fractions
Use the cover-up rule
Express an improper algebraic fraction as aproper fraction before expressing it in partialfractions
NEXT
A complex algebraic fraction can be expressed asthe sum of 2 or more simpler fractionspartial
, known frac
astions
9 4
(3 5)( 2)
x
x x
3
3 5x
2
- 2x
the complex algebraic fraction its partial fractions
NEXT
The number and the form of the partial fractions of an F( )
algebraic fraction depends factor son i the n G( : G( )
)x
xx
2 2
F( ) 2 5Consider
G( ) 1 2 3
x x
x x x x
a non-repeated linear factor of the form ax b a repeated linear factor, i.e. ( 2) is repeated twicex
2a non-repeated quadratic factor of the form ax bx c
NEXT
In G( ), there is a:x
For every in G( ),
there will be a parti
non-repea
al fracti
ted l
on o
inear fac
f the f
tor
orm
xA
ax b
ax b
10 4
( 1)(3 2 )2
x
x x
3 2 2 1x
A
x
B
(2 (8
10 4
(3 2 1) 1 1 ))x x
x
x
23 1 12 8 1
B
x
A
x
C
x
NEXT
2
2For a in G( ), there'll
be partial fractions
rep
of
eated linear fa
the form
t
c or ax b xA B
ax b ax b
2
5 1
(2 1)
x
x
22 1 (2 1)
A B
x x
2(2 1
5 1
(7 4))x
x
x
22 1 (2 1 (7 4))
A
x
B
x
C
x
2
2For every in G( ),
there'll be a partial fraction of
quadratic fac
the form
tor xAx B
ax b
b
x c
ax x c
2(2 3 () 1)
x
xx
22 3 1
Ax B
x
C
x
22
1
5xx
2 2 5
C
x
A B
x
x
x
D
Note that x2 is a repeated linear factor (i.e. x repeated twice)
Note that x2 is a repeated linear factor (i.e. x repeated twice)
NEXT
We can determine these numerator constants by:
After writing out the form of the partial fractions, weneed to find the constants in the numerators of thesepartial fractions. For example,
2 2(2 3)( 1) 2 3 1
x Ax B C
x x x x
We want to find what values are A, B and C.
We want to find what values are A, B and C.
coveusi r-ung the p r ,ule like terms in on bothcompa sidr es, oring x suitablesubstitu values ting of x NEXT
The is used to find the numerator constant of a partial fraction corresponding to a non-repeated linear factor in G( )
cov
. F
er-up ru
or exam ,
le
plex
COVER-UP RULE COMPARING SUBSTITUTION
2 2
non-repeatedlinear factor
2 2
(( 1 1) 1) 1
x Bx C
x x
A
x x
We can use the cover-up rule to find as it isthe numerator of the partial fraction corresponding to t non-repeated linear factohe r 1.
A
x NEXT
The root of 1 is 1 (since if 1 0 1)...x x x
COVER-UP RULE COMPARING SUBSTITUTION
2 2
2 2
( 1)( 1) 1 1
x Bx C
x x x
A
x
NEXT
We "cover up" 1 in the algebraic fraction...x
Sub the root 1 into algebraic fraction to find :x A
2
2 1 22
1 1A
2 2
2 2
( 1)( 1) 1 1
A Bx C
xx x
x
x
Now to find and , we either compare like terms in , or substitute in suitable values of . But first, we "combine" the partial fractions on the RHS:
B C xx
COVER-UP RULE COMPARING SUBSTITUTION
2 2
2 2 2
( 1)( 1) 1 1
x
x x x x
Bx C
NEXT
2
2 2( 1)( 1) ( 1
2 2 2( 1) ( )( 1
)( 1)
)x x Bx C x
x x x x
...and equate the numerators on both sides:22 2 2( 1) ( )( 1)x x Bx C x
If you are quick, skip this step and go to the next step (i.e. equate the numerators)...
If you are quick, skip this step and go to the next step (i.e. equate the numerators)...
COVER-UP RULE COMPARING SUBSTITUTION
22 2 2( 1) ( )( 1)x x Bx C x
2Comparing the terms,x2 2 2( 0 (LHS) 2 (RHS))2 0 x x BB x
Comparing the constant terms,( 2 (LHS) 2 (RHS2 )2 )CC
2B
0C
We can now compare like terms in on both sides:x
NEXT
COVER-UP RULE COMPARING SUBSTITUTION
22 2 2( 1) ( )( 1)x x Bx C x
Sub 0 :x 22(0) 2 2(0 1) ( (0) )(0 1)B C
2 2 C
We can also substitute suitable values of to find and .
xB C
0C
Sub 2 :x 6 10 2B
2B
It is actually faster here to compare like terms in x to find B and C. So always try to use the faster method.
It is actually faster here to compare like terms in x to find B and C. So always try to use the faster method.
NEXT
This makes B disappear, leaving only C.This makes B disappear, leaving only C.
COVER-UP RULE COMPARING SUBSTITUTION
And finally, write out the partial fractions with thefound values for the numerator constants:
NEXT
2 2
2 2 2 2
( 1)( 1) 1 1
x x
x x x x
To check whether your partial fractions are correct, "combine" back the partial fractions on the RHS to see if you could obtain the algebraic fraction on the LHS.
To check whether your partial fractions are correct, "combine" back the partial fractions on the RHS to see if you could obtain the algebraic fraction on the LHS.
COVER-UP RULE rewind
COMPARING SUBSTITUTION
The can also be used to find the numerator of the partial fraction which denominator is the highest power of a repeated linear factor in G
c
( ). For example,
over-up rule
x
2
non-repeated repeated highest powerlinear factor linear factor of the repeated
linear
2 2
factor
11
2 1 12 1
x x A B
x xx x
C
x
We can use the cover-up rule to find (i.e. numerator of partial fraction corresponding to non-repeated linear factor)...
A
...and (i.e. numerator of partial fraction which denominator is the highest power of the repeated linear factor).
CNEXT
COVER-UP RULE rewind
COMPARING SUBSTITUTION
2
non-repeated repeated highest powerlinear factor linear factor of the repeated
linear
2 2
factor
11
2 1 12 1
x x A B
x xx x
C
x
The root of 1 is 1 (since if 1 0 1)...x x x
We "cover up" 1 in the algebraic fraction...x
Sub the root 1 into algebraic fraction to find C:x
211 1 13
1 2C
NEXT
The cover-up rule is actually a shortcut for substituting values of x. Can you see out why?
The cover-up rule is actually a shortcut for substituting values of x. Can you see out why?
COVER-UP RULE rewind
COMPARING SUBSTITUTION
2
2
11 3
( 2)( 1) 2 1 ( 1)
x x A B
x x x x x
Now find (using cover-up method) and (by comparing like terms or substituting suitable values):
A B
(1) 1, 2
(2) 1, 2
(3) 1, 2
(4) 1, 2
A B
A B
A B
A B
Click on correct
answer to proceed
COVER-UP RULE rewind
COMPARING SUBSTITUTION
2
2
11 3
( 2)( 1) 2 1 ( 1)
x x A B
x x x x x
Now find (using cover-up method) and (comparing like terms or substituting suitable values):
A B
(1) 1, 2
(2) 1, 2
(3) 1, 2
(4) 1, 2
A B
A B
A B
A B
Yes, this is the correct answer!
Yes, this is the correct answer!
NEXT
...can be expressed in partial fractions straightaway.
not fully factA) If its denominator is , we needto factorise it f
oi
risedrst:
22( 1)
4
( 3) 3 1
x A x
x
B C
x x x
2
4 4
( 3)( 1) ( 3)( 1)( 1) 3 1 1
x x A B C
x x x x x x x x
NEXT
not fully factorised
fully factorised
...can be expressed in partial fractions straightaway.
impropB) If the algebraic fraction is , we needto make it proper fi
errst:
is improper if degree of deF(
gree of G(x))
F(x) .G( )
x
x
For example, the following are improper:3 1
1
x
x
degree 3
degree 1 2 1
x
x
degree 1
degree 1 II
NEXT
F( )We can make an improper fraction proper
G( )using either:
x
x
Long division - if degree F( ) G( )x x
Splitting the numerator - if degree F( ) G( )x x
NEXT
LONG DIVISION SPLITTING NUMERATOR
3 23 4 5
1 2
x x x
x x
3 2
2
3 4 5
2
x x x
x x
2 3 2
3 2
2
2
2 3 4 5
(3 3 6 )
2 2 5
(2 2 4
1
2
)
3
4
x x x x x
x x x
x x
x x
x
x
( 1)( 23
42
)
1x
x xx
5 73 2 (cover-up rule)
3( 1) 3( 2)x
x x
properfraction now
properfraction now
improper fraction(deg of numerator
> deg of
denominator)
improper fraction(deg of numerator
> deg of
denominator)
NEXT
LONG DIVISION SPLITTING NUMERATOR
23 1
3 1
x
x x
2
2
3 1
2 3
x
x x
2
2
2
2 3
3 6 03 1xx x
x x
10 6
33 1
x
x x
"introduce" denominator in the
numerator
"introduce" denominator in the
numerator
improper fraction(deg of numerator = deg of denominator)
improper fraction(deg of numerator = deg of denominator)
split the numerator to create proper fractionsplit the numerator to create proper fraction
7 13 (cover-up rule)
3 1x x
NEXT
Which of the above fractions cannot be expressed in partial fractions straightaway:
2 2
2 2
2
Consider the following fractions:
2 5 2 5(A) (B)
( 4 5)( 1) (x 4 3)( 1)
1 9(C) (D)
( 1)( 3) ( 1)
x x
x x x x x
x x
x x x x
A, B B, C B, D C, DClick on correct
answer to proceed
Which of the above fractions cannot be expressed in partial fractions straightaway:
2 2
2 2
2
Consider the following fractions:
2 5 2 5(A) (B)
( 4 5)( 1) (x 4 3)( 1)
1 9(C) (D)
( 1)( 3) ( 1)
x x
x x x x x
x x
x x x x
A, B B, C B, D C, D
Bingo! Bingo!
NEXT
F( )To express in partial fractions, we need to:
G( )
x
x
determine the form of the partial fractions by looking at the factors in G( )x
determine the constants in the numerators of the partial fractions by:(A) using cover-up rule,(B) comparing like terms in , or(C) substituting suitable values of
xx
F( )first make sure is proper and G( ) is fully factorised
G( )
xx
x
Complete this assignment and submit to your tutor(your tutor will set a dateline)
Express the following in partial fractions:
2
2
5(A)
( 2)(3 1)
4 7(B)
(2 )(1 )
3 23 45(C)
( 3)
x
x x
x
x x
x x
x x