Example 2 Factorise y  27x 3  12x .
Separate the common factors 3 and x from the two terms, rewrite
the resulting two terms as difference of two squares and then
factorise.
2
y  3x 9x 2  4  3x 3x  2 2  3x3x  23x  2 .
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



Algebra
Example 3 Factorise 7 x 2  196 .
The binomial theorem
n
This theorem is used to expand expression of the form  p  q  ,
where n is a positive integer.
 p  qn  nC0 p n q 0  nC1 p n1q1  nC2 p n2 q 2  ...... nCn p 0 q n
Separate the common factor 7 from the two terms, rewrite the
resulting two terms as difference of two squares and then
factorise.
2
7 x 2  196  7 x 2  28  7 x 2  2 7   7 x  2 7 x  2 7 .


Example 1 Expand 2 x  3 .
2x  3
 C0 2x  3  C1 2x  3  C2 2x  3
4
4
0
3
4
1
4
2
2
 4C3 2x  3  4C4 2x  3
1
 




Example 4 Factorise x 4  9 over Q.
4
4

3
0
4
 16 x 4  96 x 3  216 x 2  216 x  81
Rewrite the binomial as difference of two squares and then
factorise.
 
  
  3  x  3  x  3 x  3 x

x4  9  x2
2
  x 2

2
 32  x 2  3 x 2  3
2
2

3
Example 2 Find the coefficient of the term containing x 6 in the
2
Example 5 Factorise 2x  3  5 over R.
9
expansion of 3x  a  , where a is a constant.
Do not expand. Express the polynomial as difference of two
squares and then factorise.
The required term is 9 C3 3x   a  .
6
3
The coefficient is 9 C3 3  a   61236a 3 .
6
3
Factorisation of polynomial functions
Not all polynomials can be factorised analytically (by algebraic
methods). The following methods are applicable to those
polynomials that have factors involving rational numbers and in
some cases, surds.
Common factors Look for common factors and separate them
from the terms of the polynomial by means of the distributive
law in reverse. This is always the first step in all factorisations.
Example Factorise f x   25 x 3  10 x 2 
5
x.
2
5
1
x  55xxx  52xx  5  x . All
2
2
three terms have common factors 5 and x.
1
1


Hence f x   5x 5xx  2 x    5x 5x 2  2 x   .
2
2


f x   25 x 3  10 x 2 
Difference of two squares Rewrite the polynomial in the form
a 2  b 2 that can be factorised as a  b a  b  .
Note: Polynomials in the form of sum of two squares cannot be
factorised.
1
.
25
The polynomial has no common factors.
Example 1 Factorise 36 x 2 
2
1
1 
1
1 
2
 6 x       6 x   6 x   .
25
5 
5
5 
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36 x 2 
2x  32  5  2x  32  
5
  2x  3  5 2x  3  5 .
2
Quadratic trinomials For trinomials that have linear factors
involving rational numbers, the quickest method is by trial and
error.
x 2  bx  c  x  px  q , by trial and error, find numbers p
and q such that pq  c AND p  q  b .
ax 2  bx  c  mx  p nx  q  , by trial and error, find numbers
m, n, p and q such that mn  a , pq  c AND mq  np  b .
Note: Check the value of b 2  4c in the first case and b 2  4ac
in the second case before trying. If the value is negative, the
trinomial has no linear factors. If the value is zero, the two linear
factors are the same.
Example 1 Factorise 3x 2  2 x  1 .
This trinomial cannot be factorised because
2
b 2  4ac   2  431  8 , a negative value.
Example 2 Factorise f x  2x 2  4x  6 .
The trinomial has a common factor of 2, hence
f x   2 x 2  2x  3 .
For the resulting trinomial inside the brackets,
2
b 2  4c   2  4 3  16 , a positive value; it can be
factorised.
f x  2 x 2  2x  3  2x  3x  1 , by trial and error.




Algebra
1
Example 3 Factorise y  2rx 2  28rx  98r .

Example 4 Factorise x 4  x 2  16 over R.

y  2rx 2  28rx  98r  2r x 2  14x  49
Common factors.
For the resulting trinomial, b 2  4c   142  449  0 , hence
the two linear factors are the same.
2
 y  2r x  7 .
Example 4 Factorise x 4  x 2  12 completely over Q.
   x   12
2
Rewrite the polynomial as x 2



 x2  4 x2  3


2
Add 9x 2 to complete the square.
 

 x2


2
 8 x 2  16  9 x 2




 x 2  4  3x   x 2  3x  4 x 2  3x  4 .
The two quadratic factors do not have linear factors,
b 2  4ac  0 .
2
2
Note: Examples 3 and 4 show two methods in completing the
Difference of two squares
square for polynomials of the form x 2
 
2
Do not expand. Replace x  3 by y, then factorise by trial and
error.
x  32  2x  3  35  y 2  2 y  35   y  7 y  5
 x  3  7x  3  5  x  4x  8
Completing the square If you fail to find the linear factors of a
quadratic trinomial by trial and error, try completing the square.
Sum/difference of two cubes Both forms can be factorised to a
linear factor and a quadratic factor. The quadratic factor cannot
be factorised further.

 a  ba
a 3  b 3  a  b a 2  ab  b 2
a b
3
3
2
 ab  b


The two terms in the cubic function have a common factor of 2.
2x 3  54  2 x 3  27  2 x 3  33  2x  3 x 2  3x  9 .

 7   x  3  7 x  3  7  .
2
2


3
3


1 3
 3
2 x 2  3x  6  2 x 2  x  3  2 x 2  x         3
2
2


 2 2
 4


2
2



3
 3   3
 2 x 2  x         3  2 x 
2


 4    4 

 


3
57 
  
4
16 
2
2
2



3   57  
3
57  

 2 x    
 2 x   
  x 

4   4  
4
4  






3
57 
3
57 
.
x  
x 



4
4 
4
4 

3
57 


4
4 
Example 3 Factorise x  2 x  7 over R.
2
 

625x 3  40  5 125x 3  8  5 5x  23

3



3
Example 3 Factorise 1000 x 3  5 2 over R.
 5
 10 x  5  10 x   10 x  5    5  


 10 x  5 100 x  10 5 x  5
 510 x  5 20 x  2 5 x  1 .
3
1000 x 3  5 2  10 x  
3
3
2
2
2
2
Example 4 Factorise 2x  1  1 over R.
 x2 1  6 x2 1  6
Do not expand, use ‘the sum of two cubes’.
2x  13  1  2x  13  13
2 2
2
2
2
2
2
2



 x  1  6 x  1  6 
2
2
2
2



  x 2   1  6   x 2   1  6  

 

 

  x  1  6  x  1  6  x  1  6  x  1  6  .





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
 55x  25x  25x  2 2  55x  2 25x 2  10x  4 .
2
   2x   1  1  7
 x   2x   1  6  x  1   6 
2

Example 2 Factorise 625 x 3  40 over R.
Example 2 Factorise 2 x 2  3x  6 over R.
2


Example 1 Factorise 2 x 3  54 over R.
2
x 4  2x 2  7  x 2
2
2
6 6
x 2  6x  2  x 2  6x  
 
 2
 2   2 
 x 2  6x  9  9  2  x 2  6x  9  7
4
 Bx 2  C .
and subtract a x 2 term to complete the square.
Example 1 Factorise x 2  6 x  2 over R.
2
2
If B 2  4C  0 , follow the method shown in example 3, i.e. add
and subtract a constant to complete the square.
If B 2  4C  0 , follow the method shown in example 4, i.e. add
Example 5 Factorise x  3  2x  3  35 .
 x  3 

 x 2  16  9 x 2  9 x 2
Trial and error
 x  2x  2 x 2  3
2
2
x 4  x 2  16  x 2
3

 2x  1  12x  1  12x  1  12

2




 2x 4x  4x  1  2x  1  1  2 x 4 x 2  6 x  3 .
2
Algebra
2
Example 5 Factorise x 6  64 over R.
The factor theorem is best used for polynomials with linear
factors of rational coefficients. The value(s) of  is found by
trial and error. The possible values of  for trying depend on
the first and last coefficients of
a. factor.of .a n
.
Px  a0 x n  a1 x n1  ...... an1 x  an ,   
a. factor.of .a 0
Treat x 6  64 as difference of two squares.
   2   x  2 x  2 
 x  2x  2x  4x  2x  2x  4 .
x 6  64  x 3
2
3 2
3
2
3
3
3
2
If all these  values give P   0 , it does not necessarily
mean that Px  has no linear factors, because the coefficients of
the linear factor(s) may be irrational.
If x  64 is treated as difference of two cubes, then
6
   2   x  2  x   2 x  2  
 x  2x  2x   4 x  16 
 x  2x  2x   4 x  16  4 x  4 x 
 x  2x  2x   8 x  16  4 x 
 x  2x  2x  4  2 x  
 x  2x  2x  4  2xx  4  2x .
x 6  64  x 2
3
2 3
2
2
2 2
2 2
2
2 2
2
2
2 2
2
2
2
2
2
2 2
2
Example 1 Use the factor theorem to find a linear factor of
3x 3  2 x 2  7 x  2 , then find the quadratic factor and hence all
the linear factors.
2
Let Px  3x 3  2 x 2  7 x  2 . The possible values of  for
1,2
1 2
testing are   
, i.e.    , ,1,2 .
1,3
3 3
2
2
2
P1  31  21  71  2  0 .
3
Grouping The four terms in a cubic polynomial are grouped into
two and two. The two groups are then factorised. Correct
grouping gives a common factor in the two groups.
Example 1 Factorise 2 x 3  3x 2  32 x  48 over R.
3
2
P 1  3 1  2 1  7 1  2  0 , x  1 is a factor.
Divide Px  by x  1 to find the quadratic factor.
3x 2  5 x  2
x  1 3x 3  2 x 2  7 x  2

Group the first two terms and the last two.
2x 3  3x 2  32x  48  2x 3  3x 2  32x  48


 x 2x  3  162x  3  2x  3x  16
 2x  3x  4   2x  3x  4x  4 .
2
2
 3x 3  3x 2
 5x 2  7 x
2
2

  5x 2  5x
2
0
Hence Px  x  1 3x 2  5x  2  x  13x  1x  2 .

Example 2 Factorise 8x  2 x  x  1 over R.
2

 
 
 
8x 3  2x 2  x  1  8x 3  1  2x 2  x  2x  13  2x 2  x


3
 2x  12x  2x1  12  x2x  1

 2x  14x
2

 2x  1 4x 2  2x  1  x2x  1
2



 2x  1  x  2x  1 4x 2  x  1 .

2 x  2
  2 x  2
Note: It is also possible to proceed by grouping the first and the
third terms; and the second and the last terms.
3



Another possible outcome:
3
2
1
 1  1
 1
 1
P    3    2    7    2  0 ,  x  is a
3
3
3
3
3

 





factor.
3x 2  3x  6
x  13  3x 3  2 x 2  7 x  2
The factor theorem
Consider a polynomial Px  that has x   as a linear factor.
Then Px   x   Qx  , where Qx  is a polynomial one

 3x 3  x 2
 3x 2  7 x

  3x 2  x
degree lower than Px  , obtained by expanding and comparing
coefficients, or dividing Px  by x   .
Replacing x by  in Px   x   Qx  ,
P      Q  . Hence P   0 .
Conversely, for any polynomial Px  , if P   0 , then x  
is a factor of Px  . This statement is known as the factor
theorem, and can be used to find the linear factors of a
polynomial if other methods failed.
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

6 x  2
  6 x  2
0
1 2
1


Hence Px    x   3x  3x  6  3 x   x 2  x  2
3
3


1

 3 x  x  2x  1 . It is equivalent to the previous result.
3





Algebra
3
Example 2 Given x  1 is a factor of 3x 4  x 3  9 x 2  9 x  2 ,
find the cubic factor Qx  such that
3x 4  x 3  9x 2  9x  2  x  1Qx .
Let Qx  ax 3  bx 2  cx  d , then
(i) R  P 5  2 5   5  5 5  11  1339
4


Expand to obtain 3x  x  9 x  9 x  2
3
2
 ax 4  a  bx 3  b  cx 2  c  d x  d .
Compare the coefficients on both sides, a  3 , a  b  1 ,
b  c  9 , c  d  9 and d  2 . Hence b  2 , c  7 .
Qx  3x 3  2x 2  7 x  2 .
This example illustrates an alternative method in finding Qx 
to long division shown in the last example.
Example 3 Use the factor theorem to find the linear factors of
x 4  x 3  x 2  3x  2 .
Let Px  x 4  x 3  x 2  3x  2 . Test   1,2 .
P 1   1   1   1  3 1  2  0
4
3
2
4
3
2
P1  1  1  1  31  2  0 , x  1 is a factor.
Hence Px   x  1Qx  .
Use long division or comparing coefficients to find
Qx  x 3  x  2 .


Hence Px   x  1 x  x  2 . Use the factor theorem on
3
Qx  x  x  2 to find its linear factor. Test   1,2 .
3

3
13
3 3 3
3
(ii) R  P   2      5   11 
4
2 2 2
2
(iii)
4
3
R  P2a  22a  2a  52a  11  32a 4  8a 3  10a  11
Example 2 Given that x  2 is a factor of 3x 3  px 2  qx  2
and the remainder is 20 when the cubic polynomial is divided
by x  2 . Find the values of p and q.
Use the factor theorem and the remainder theorem to set up two
simultaneous equations for p and q.
Let Px   3x 3  px 2  qx  2 .
x  2 is a factor of Px  , P2  0 ,
32  p2  q2  2  0 ,
2 p  q  11 ……(1)
The remainder is 20 when divided by x  2 , P 2  20 ,
3
2
3 2  p 2  q 2  2  20 ,
3
2 p  q  3 ……(2)
2
Solve eqs (1) and (2) for p and q.
(1) + (2), 4 p  8 ,  p  2
(1)  (2), 2q  14 , q  7
Exponential (index) laws
3
Q1  1  1  2  0 , x  1 is a factor of Qx  .
Hence Px   x  1x  1T x  . T x  is quadratic and found by
long division of Px  by the expansion of x  1x  1 , or
comparison of coefficients as discussed in example 2,
T x   x 2  x  2 .
1. a m  a n  a m n
1
1
am
2.
 a  n ; a m   m ; n  a mn
n
a
a
a
3.

4.
 Px  x  1x  1 x  x  2 .
2
3
4
3x 4  x 3  9x 2  9x  2  x  1 ax3  bx 2  cx  d .
4
Example 1 Find the remainder when Px  2x 4  x 3  5x  11
is divided by (i) x  5 , (ii) 2 x  3 , (iii) x  2a .
a 
m n
 a mn
abn  a n b n
1
 
q
p
p
p
The remainder theorem When a polynomial Px  is divided
Good to know: a 0  1 , a p  a , a p  a or a q .
by a linear binomial x   , the remainder can be found quickly
without actually carrying out the division.
ab2n  b 3n , and express with positive
Example
1
Simplify
n
Since
Quotient
Qx 
ab 4  b 4 n
indices.
Dividend (Polynomial)
P x 
x   
q
 
ab2n  b 4n
Divisor
R
Remainder
ab 
4 n
 Px   x   Qx   R .


    
When x  , P         Qx   R  R ,

    

 
i.e. the remainder R  P  when Px  is divided by x   .
 
b
5n




a 2n b 2n  b 4n b 2n a 2n  b 2n
 4n n
a n b 4 n  b 5n
b a  bn

     b a  b a  b 


a  b 

2

b 2 n a n  b n
an  bn

an  bn
.
b 2n
2
2n
n
n
n
n
n
n
This is known as the remainder theorem.
© Copyright itute.com 2005
Algebra
4
e 2 x 1  4e x 1  3e
.
e x 1  e
Example 2 Simplify

Logarithm laws





 

x 2
e 2 x 1  4e x 1  3e e e 2 x  4e x  3 e 0 e
 4 ex  3


e x 1  e
e ex 1
ex 1

e
x




 3 ex 1
 e x  3 for x  0 .
ex 1


e 2 x 1  4e x 1  3e
is
e x 1  e
undefined for x  0 whilst e x  3 is defined for all x,  they
cannot be equal at x  0 .
p  2p
Example 3 Simplify
5 p  10 p
5
2
p  2p
3
2
3
2
5 p  10 p
1
2
p

3
2
 p  2
1
2
5p
p

 p  2
.
1
2
3 1

2 2
log a p
log a b
Law 4 shows the relationship between log a p and log b p .
log a p  0 for 0  p  1 ; log a p  0 for p  1 .

5
4. log b p 
Good to remember: log a 1  0 ; log a a  1 ;
log a p is undefined for p  0 ;
3
2
3
2
 p
1
2. log a p  log a q  log a   ;  log a q  log a  
q
q
3. log a p n  n log a p
Note: It is necessary to state that x  0 , 
5
2
For p, q  0 ,
1. log a p  log a q  log a  pq
For even n, log a p n is defined for all p  R whilst n log a p is
defined only for p  0 ,
p
.
5
log a p n  n log a p for p  0 , and
log a p n  n log a p for p  0 .

Example 4 Simplify f n  100 n  2  10 n1  100
that f 2  1000 f 0 .

3
2
. Show
The following graphs of y  log e x 2 and y  2 log e x illustrate
the point.
   2  10  10  10 
 10   210 10   10   10  10    10  10 
100
n
 2  10 n 1  100
n 2

3
2
3
2 2
n

n
 10 2
3
2 2
n
3
2 2
n
n
3

 f n   10 n  10 .

f 2  10
3
  11 ;
 10   110   10  11
f 0  10 0  10
3
3
3
2
Example 5 Simplify
3
3
6  3 2x 5 y 2
6 x y 
6
6  3 2x 5 y 2
6 x y 
6
4


2
3
 6
 6 x y


5
1
3


, express in positive indices.
2
3
1
1
3

1
2
3
2


2
3

2 3 x y
2
1
3
 23 x
1
3
© Copyright itute.com 2005
5
2
2 3 23 x3 y 3
2  3x y 
2
3 23 x 3 y 3
1
3

6 2x 5 y 2
 10 3  11 3  1000 f 0 .
1
3
y 
6
1
3
2
1
2  33 y 3
x
1
3
.
Algebra
5
 1 
Evaluate 5 log 2   .
 32 
Example 1
Example 6 Change 5 x to base 10 and base e.
 1 
 1 
5 log 2    5 log 2  5   5 log 2 2 5  5  5 log 2 2  25 .
 32 
2 




Example 2 Simplify 2 log10 3x 2 y  3 log10 2xy 2 .
 
 
   log 2 xy 
9x y   log 8x y   log  9x y   log  9 x  .
4
2
3
2
3
6
10
 8x y 
10
4
 8y 
Your calculator has only log (i.e. log 10 ) and ln (i.e. log e ).
log 10 10
log e 10
 2.0959 , or log 3 10 
 2.0959 .
log 10 3
log e 3
Example 4 Show that 3 log 4 x  2 log 8 x 
5
log 2 x .
6
Change both logarithms on the left side of the identity to base 2.
3 log 2 x 2 log 2 x
3 log 4 x  2 log 8 x 

log 2 4
log 2 8
3 log 2 x
log 2 2 2

2 log 2 x
log 2 2 3

y  e x  x  log e y .
6
10
log 3 10 
y  10 x  x  log10 y ;
10
4
y  x2  x   y ;
y  Sinx  x  Sin1  y  ;
2 3
Example 3 Evaluate log 3 10 .

Equivalent relations
Examples are:
2
2 log 10 3x 2 y  3 log 10 2 xy 2  log 10 3x 2 y
 log 10
5 x  10 x log10 5  10 0.6990x ; 5 x  e x loge 5  e1.6094x .
3
2
log 2 x  log 2 x
2
3
In each of the above cases, both left and right statements give
exactly the same relationship between x and y, i.e. they are
equivalent. Try to plot the graphs of a pair of equivalent
relations. They are the same plot. The left relation uses y as the
subject, and the right relation uses x. In the last two examples,
the left relations are expressed in index (exponential) form
whilst the right relations are in logarithm form.
Inverse relations
y  e x and x  log e y are equivalent relations, but
y  e x and x  e y are inverse relations, and so are
y  log e x and x  log e y . (Read each relation carefully)
In inverse relations, the x and y-coordinates of all the points are
interchanged.
5
3 2
    log 2 x  log 2 x.
6
2 3
Example 5 Show that
1
 log 10 e .
log e 10
Change both sides to a common base b.
log b e
1
.
LHS 

log b 10 log b 10
log b e
RHS 
log b e
.  LHS  RHS .
log b 10
Example 6 Given 10 p  e q , (i) find q in terms of p, (ii) find p
in terms of q.
(i) 10 p  e q , log e 10 p  log e e q , p log e 10  q log e e ,
 q  p log e 10 .
q
 q log 10 e .
log e 10
These two results show the way to change the base of an
exponential function.
(ii) Since q  p log e 10 ,  p 
10 p  e p loge 10 ; e q  10 q log1 0 e . In general, a x  b x logb a .
© Copyright itute.com 2005
y  e x and x  e y
y  log e x and x  log e y
Algebra
6
y  e x and x  e y are inverses of each other. Express
x  e y with y as the subject, y  log e x .
y  log e x and x  log e y are inverses of each other. Express
x  log e y with y as the subject, y  e x .
Hence y  e x and y  log e x are inverses of each other, as
discussed previously in Functions and Graphs.
Relations and functions
A relation is a set of ordered pairs (points). A function is a
relation such that no two points have the same x-coordinate.
Use the vertical line test to determine whether a relation is a
function (cuts through only one point) or not (cuts through more
than one point). If it is a function, then it is either a many-to-one
function or a one-to-one function. If it is not a function, then it
is either a many-to-many relation or a one-to-many relation.
Other examples of inverse pairs are:
  
y  Sin(x) for  ,  and y  Sin1 x ;
 2 2
y  x 2 for x  0 and y  x ;
y  x 2 for x  0 and y   x .
Inverse functions
Every relation has an inverse that may or may not be a function.
If a relation is a one-to-many relation, then its inverse is a manyto-one function. If a relation is a one-to-one function, then its
inverse is also a function (a one-to-one function). If a relation is
a many-to-many relation or many-to-one function, then its
inverse is not a function.
Use the horizontal line test to determine whether the inverse is
a function (cuts through only one point) or not (cuts through
more than one point).
Example 1 The following two graphs show the original relation
2
y  x  1 that is a many-to-one function, and its inverse
2
x   y  1 that is not a function.
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Algebra
7
  
Example 2 The relations y  Sin(x) for  ,  , y  x 2 for
 2 2
2
x  0 and y  x for x  0 are one-to-one functions,  their
inverses are also one-to-one functions.
If a relation is a function, function notations can be used to
represent it, e.g. y  x 2 for x  0 , f : R   R, f x  x 2 . Since
its inverse y   x is also a function, use f
inverse function, f
1
: R   R, f
1
x   
1
to denote
x.
Example 3 Restrict the domain of y  cos x to produce a oneto-one function. Use function notations to represent it and its
inverse function.
There are an infinite number of possible restrictions of the
domain, e.g. 0,   . Let the function with this restriction be
g : 0,    R, g x   cos x .
The following graphs show that g and g 1 are both one-to-one
functions. The graph of g
line y  x.
1
Example 4 Given a function with equation
1
5
y  log e 3x  1  , find the domain, range and equation of
2
2
its inverse with y as the subject.
The given function is defined for 3x  1  0 , i.e. x 
1
.  its
3
1 
domain is  ,   . Its range is R. Hence the domain and range
3 
1 
of its inverse are R and  ,   respectively.
3 
1
5
1
5
The inverse of y  log e 3x  1  is x  log e 3 y  1  .
2
2
2
2
5 1
Transpose to make y the subject. x   log e 3 y  1 ,
2 2
2 x  5  log e 3 y  1 , the equivalent is 3 y  1  e 2 x5 ,
1
1
1
3 y  e 2 x5  1 , y  e 2 x 5  1 or e 2 x 5  .
3
3
3


Example 5 Given f : R  R, f x  3e 2 x  1 . Find f
1
.
is obtained by reflecting g in the
f is a one-to-one function,  f
graphing f is y  3e
2x
1
exists. The equation for
 1 . f has  1,  as its range,  the
domain of f 1 is  1,  . The graph of f
in the line y  x .
1
is the reflection of f
g has a range of  1,1 , this becomes the domain of g 1 .
The inverse of y  cos x is x  cos y , the equivalent of
x  cos y is y  cos1 x .
 g 1 :  1,1  R, g 1 x  cos1 x .
The equation of f
subject.
1
is x  3e 2 y  1 . Transpose to make y the
x  1 2y
 e , the equivalent is
3
1
 x  1
 x  1
2 y  log e 
 , y  log e 
.
3
2


 3 
1
 x 1
Hence f 1 :  1,   R, f 1 x   log e 
.
2
 3 
x  3e 2 y  1 , x  1  3e 2 y ,
© Copyright itute.com 2005
Algebra
8
Example 6 Given j :  1,5  R, j x  
2
 3 , find the
x 1
domain, range, asymptote(s) and equation of its inverse.
Consider the given function j: Domain  1,5 . Vertical
asymptote x  1 . Note that horizontal asymptote does not exist
because the function has an end point at x  5 ,
2
2
10
10 
. Range  ,   . Equation y 
 3.
y
3
x 1
5 1
3
3

10 
The inverse of j: Domain  ,   . Range  1,5 . Horizontal
3

2
asymptote y  1 . Equation x 
 3 . Make y the subject
y 1
2
2
2
by transposition. x  3 
, y 1
, y
1 .
y 1
x3
x3
Both h and k have the same horizontal asymptote y  2 and
 the same range  2,  .  h 1 and k 1 have the same
vertical asymptote x  2 and domain  2,  . The inverse of
1
1
 2 . The equivalent relation to
y
 2 is x 
2
 y  12
x  1
1
x
 2 is obtained by transposition to make y the
 y  12
subject.
x
1
 y  1
y 1 
2
2, x2
1
x2
1
 y  1
2
1
, y
x2
,
Example 7 Split up y 
1
k 1 :  2,   R, k 1 x  
Example 8
x  12
 2 into two one-to-one
functions h and k, and find h 1 and k 1 .
1
,
x2
1 .
Hence h 1 :  2,   R, h 1 x   
and
 y  12 
1
1,
x2
1
1 .
x2
Find the domain and equation of the inverse of
2
8
1
1
y   x    1 , where x   .
3
9
3
function to domain  1,  to form one-to-one function k.
 8 
The given function has an end point   ,2  , and a turning
 3 
1 
point  ,1 that is the lowest point of the function. Hence the
3 
range is 1,   , not 2,  .  domain of the inverse is 1,   .
Hence h :  ,1  R, hx  
Equation of the inverse is x 
y
1
x  12
 2 is a many-to-one function and it is symmetrical
about the vertical asymptote x  1 . Restrict the function to
domain  ,1 to form one-to-one function h. Restrict the
and
k :  1,   R, k x  
1
2
x  12
1
x  12
2,
2.
The graphs of h 1 and k 1 are obtained by reflecting the graphs
of h and k in the line y  x .
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2
1
1
 y    1 . Make y the
9
3
2
1
1
1

subject, x  1   y   , 9x  1   y   ,
9
3
3

1
1
1
y    9x  1 , y   3 x  1 , y  3 x  1  .
3
3
3
Note: (1) The inverse is not a function. (2) The inverse has an
8

end point  2,  .
3

Algebra
9
Example 2 Solve 5e 3 x  2  6 .
Inverse functions undo each other
When a one-to-one function f and its inverse function f
used to form composite function f
1
 f or f  f
1
1
are
, then
f 1  f x  x , f  f 1 x  x , i.e. they undo each other.
Example 1 Given f x  x  2 , find f
undo each other.
1
3
x . Show that they
Example 3 Solve 8e x  2  3e 2 x .
Equation of f is y  x  2 ,  equation of f
3
i.e. y  x  2 ,  f
3
f
1
 f x   f
1
1
x  
3
 f x  3 f x   2  3 x  23

is x   y  2 ,
1
3
x 2.
  x
 
f  f 1 x  f f 1 x   f 1 x  2 
3
5e 3 x  2  6 , e 3 x  2  1.2 , the equivalent is 3x  2  log e 1.2 ,
1
3x  log e 1.2  2 ,  x  log e 1.2  2 in exact form or
3
x  0.6059 .
3
3
2 x
 x.
e x  2 3 x  2 2  x  3 2 x 3
 , e  , the equivalent
 , e
8
8
e 2 x 8
1
3
3
is 2 x  log e   , x  log e   in exact form or x  0.4904 .
2
8
8
8e x  2  3e 2 x ,
Alternative method: 8e x  2  3e 2 x , 8e x  2  3e 2 x  0 ,


Example 2 Given g x   log e x , state g 1 x  , find g 1  g x
e 2 x 8e 2 x  3  0 . Since e 2 x  0 , 8e 2 x  3  0 , e 2 x 
g 1 x   e x .
Example 4 Solve 3e 2 x  4e x  1  0 .
and g  g 1 x , and simplify.
g 1  g x   g 1 g x   e g  x   e loge x  x .
gg
1
x  g g x  log e e
1
x
 
3 ex
3e
x.
x
2
 
 4 ex 1 0 ,

e
 loge  x 12
  e  x 1


2
e
 x 1 
log e 

 x 1 


 e

1
loge  x 12 loge  x 12 2
1
2
[factorise by trial and error]
either 3e  1  0 or e  1  0 ,
1
1
 e x  or e x  1 , i.e. x  log e   or x  0 .
3
3
e 2 loge  x 1loge  x 1 .
2
2 loge  x 1loge  x 12

1 ex 1  0 ,
x
Example 3 Simplify
3
etc.
8
1
2
1
2
 x 1  
 x 1  2


   e loge  x 1     e 2 loge  x 1  










x 1
.
x 1
x
Example 5 Solve 3e 2 x  4e x  1  0 .
This equation has no real solutions for x, because all three terms
are greater than 0,  sum > 0.
Example 6 Solve e 2 x  4e x  4  0 .
e 
x 2
 
Solving exponential equations algebraically
 4 e x  4  0 cannot be factorised over Q, set of rational
numbers,  use the quadratic formula to obtain
Example 1 Solve the following equations.
(a) 4 x  8 (b) 10 2 x 3  0.01 (c) 4 x 1  10 x 1 .
ex 
(a) 4
x
 8 , 2 
2 x
 2 3 , 2 2 x  23 ,  2 x  3 , x 
(b) 10 2 x 3  0.01 , 10 2 x 3 
3
.
2
1
, 10 2 x 3  10 2 ,
10 2
1
 2 x  3  2 , x  .
2
(c) 4 x 1  10 x 1 , 4 and 10 cannot be changed to the same base.
Take the log (to base 10 is simpler than to base e in this case) of
both sides of the equation. log 10 4 x1  log10 10 x 1 ,
x  1 log10 4  x  1 , x  10.6021  x  1 ,
0.6021 x  0.6021  x  1 , x  4.0259 .
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  4 
 42  41 4
,
21

e x  2  2 2 . Since e x  0 ,

 e x  2  2 2 , hence x  log e 2  2 2 in exact form or
x  1.5745 .
Example 7 Find k such that the curve y  3e kx2 passes
through  2,3 .
When x  2 , y  3 , 3  3e 2k  2 , e 2k  2  1 ,
2k  2  0 , k  1 .
Algebra
10
Example 8 Find a and b such that the curve y  aebx  1 passes
through the points  1,2 and 1,4  .
log x 243  2.5 , the equivalent is
Use the two points to set up two simultaneous equations:
 1,2  aeb  1  2 ,  aeb  1 ………(1)
3 
1
5 2
1,4  aeb  1  4 ,  aeb  3 ………….(2)
Eq.(2)/eq.(1),
ae b 3 2b
 , e  3 , 2b  log e 3 ,
ae b 1
1
log e 3  log e 3 ……(3)
2
Substitute eq.(3) in eq.(2), ae log
3
a 
 3.
3
35  x 2.5 ,
243  x 2.5 ,
 x 2.5 , 3 2.5  x 2.5 , x  3
Example 3 Solve log x 5x  3 , where x  0 .
log x 5x  3 , log x 5  log x x  3 , log x 5  1  3 , log x 5  2 , the
b 
e
Example 2 Solve log x 243  2.5 .
3
equivalent is x 2  5 , x  5 .
3, a 3 3,
Example 4 Solve log 10 x  log 10 x  1  1 .
Example 9 Find b and c such that the curve y  2e bx  c passes
through the points 2,6 and 4,10  .
2,6  2e 2b  c  6 ,  2e 2b  6  c ………...(1)
4,10  2e 4b  c  10 ,  2e 4b  10  c ……..(2)
10  c
2e 4b 10  c
,  e 2b 
………(3)

2b
6c
6c
2e
 10  c 
Substitute eq.(3) in eq.(1), 2
6c ,
 6c 
 x 
log 10 x  log 10 x  1  1 , log 10 
  1 , the equivalent is
 x  1
x
10
 10 1 ,  x  10x  1 , x  10 x  10 , 9 x  10 , x   .
x 1
9
Example 5 Solve log 10
Eq.(2)/eq.(1),
 35  x  log  35  x  1 .
10
 35  x  log  35  x  1 ,
log  35  x  35  x   1 ,
the equivalent is  35  x  35  x   10 ,
log 10
2
2
210  c  6  c , 6  c  210  c  0 ,
10
10
1
6  c2  26  c  8  0 .
Factorise to obtain 6  c   46  c   2  0 .
35  x 2  10 , x 2  25 , x  5 .
Hence either 6  c  4  0 , i.e. c  2 , or 6  c  2  0 , i.e.
c  8 . The second result is not possible because it leads to an
Example 6 Solve 2 log e x  2  log e x  1  0 .
impossibility 2e 2b  2 . c  2 ……(4)
Substitute eq.(4) in eq.(1), 2e 2b  4 , e 2b  2 , 2b  log e 2 ,
1
b  log e 2  log e 2 .
2
2 log e x  2  log e x  1  0 , log e
 x  2 2
x 1
x  22
x 1
0,
 1 , x  2  x  1 , x 2  4 x  4  x  1 ,
2
5  25  12
,
2
5  13
5  13
i.e. x 
or
. Only the first solution is correct
2
2
because x  2 for 2 log e x  2 to be defined.
x 2  5x  3  0 ,  x 
Solving logarithmic equations algebraically
Example 1 Solve the following equations.


1
(i) log 2 x  5 (ii) log 2 x 2  1  3 (iii) log 2    1.5
x
Example 7
Find a.
(i) log 2 x  5 , the equivalent is x  2 5 , x  32 .


(ii) log 2 x 2  1  3 , the equivalent is x 2  1  2 3 , x 2  9 ,
x  3 .
1
1
(iii) log 2    1.5 , the equivalent is  21.5 , x  2 1.5 or 2
x
x
or 0.3536
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3

2
 2,1 is a point on the curve
 2,1  1  2 log e 1  a 2 ,
1
1  2a  e 2 , a 
1
2
 e  1.
y  2 log e 1  ax .
1
 log e 1  2a  ,
2
Algebra
11
Example 8 Find a and b such that y  log e ax  b  1 passes
 3
through the points 0,1 and 1,  .
 2
0,1  1  log e
b  1 ,  log e b  0 ,
a
b  1 , b  1 ……(1)
1
 3 3
1,    log e a  b  1 , log e a  b  ,
2
2
2


 a  b  e ……(2)
Substitute eq.(1) in eq.(2), a  e  1 .
1
a  b  e2 ,
 3 

a  cos 1 
 2  6



Algebraic solution of equations involving circular functions
It is prudent to consider the domain when solving an equation
involving one or more circular functions.
In this section only equations involving only one of the circular
functions are considered.
In solving such equations algebraically, always transpose an
equation to make sinkx , coskx or tankx the subject of the
equation.
 2 




,  , , 2 
6
6 6
6
11
  11
  11

11
,  , ,
.
 
  
,  , ,
2
6
3
6 6
6
3 3 3
2
Example 3 Find the exact coordinates of the intersections of
y  3 sin2 x   1 and y  3 cos2 x   1 , where   x   .
Solve the two equations simultaneously to find the intersections.
Example 1 Solve 3 sin2 x   2 , where 0  x  6 .
y  3 cos2 x   1 ……(2)
y  3 sin2 x   1 ………(1)
The domain is   x   , or 2  2 x  2 .
The domain is 0  x  6 , or 0  2 x  12 .
2
3 sin2 x   2 , sin2 x    .
3
2
Since sin 2 x  has a negative value,  , then the ‘angle’ 2 x
3
must be in the third or fourth quadrant.
sin 2 x 
3
1

, tan2 x  
.
cos2 x  3
3
1
Since tan2 x  has a positive value,
, the ‘angle’ 2 x must be
3
in the first or third quadrant.
Substitute eq.(1) in eq.(2), 3 sin2 x   1  3 cos2 x   1 ,
3 sin 2 x   3 cos2 x  ,
2x
a
a
a
2
a  sin 1    0.7297 (Calculator)
3
2 x    0.7297 , 2  0.7297 , 3  0.7297 , 4  0.7297
2 x  3.8713 , 5.5535, 10.1545, 11.8366
 x  1.9357 , 2.7767, 5.0772, 5.9183
 
Example 2 Find the exact solution(s) of 2 3 cos   3  0 ,
2
where 4    4 .
The domain is 4    4 , or  2 

2
 2 .
3
 
 
.
2 3 cos   3  0 , cos  
2 2
2

3
 
Since cos  has a positive value,
,  the ‘angle’
must
2
2
2
 
be in the first or fourth quadrant.
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 1  

a  tan 1 

 3 6
2 x  2 

,  
 
,
,

,
6
6
11
11
5  7
5  7
, 
, x  
, 
,
.
2x  
, ,
,
12
6
6 6 6
12 12 12
Substitute each of these (2x) values in eq.(1),
5
 11 
1
y  3 sin 
  1  3   1  ,
2
 6 
2
1
 5 
 1
y  3 sin 
  1  3    1   ,
2
 6 
 2
5
 
1
y  3 sin   1  3   1  ,
6
2
2
 
 
1
 7 
 1
y  3 sin
  1  3    1   . The coordinates are
2
 6 
 2
 11 5   5 1    5 
 7 1 
,  ,  ,  and 
,  .
,  , 

 12 2 
 12 2   12 2   12 2 
6
6
Algebra
12
Example 4 Solve sin 2 x  sinx cosx for x, where
0 x

2
. Note: sin 2 x  is the proper notation for sinx  .
2
Caution: Do not divide both sides of the equation by sin x  ,
because sin x   0 is a possibility. See below.
sin 2 x  sinx cosx , sin 2 x  sinx cosx  0 , common
factor, sinx sinx   cosx   0 .
 either sin x   0 , x  0 ,
sin x   cosx 
0

or sinx   cosx   0 ,
for cosx   0 ,
cosx 
cosx 
 tanx   1  0 , tanx   1, x 

4
.
The two possible solutions for x are 0,

.
4
 t 
Example 5 Find t when x  10 sin   12 is a minimum,
 12 
where 0  t  48 .
t
 4 .
12
 t 
Minimum of x is 22 , it occurs when sin   1 ,
 12 
t 3 7
,
. t  18 , 42.
 
12 2
2
The domain is 0  t  48 , or 0 
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Algebra
13