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Title: GCE Further Mathematics (6360) Further Pure Unit 2 (MFP2) Textbook
Description: Design for As/A level student in mathematics and further pure mathematics,also for student in sciences,engineering,economics
Description: Design for As/A level student in mathematics and further pure mathematics,also for student in sciences,engineering,economics
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GCE Further Mathematics
(6360)
Further Pure Unit 2 (MFP2)
Textbook
Version: 1
...
1
1
...
3
1
...
5
Introduction
The general complex number
The modulus and argument of a complex number
The polar form of a complex number
Addition, subtraction and multiplication of complex numbers
of the form x + iy
1
...
7 Products and quotients of complex numbers in their polar form
1
...
9 Further consideration of |z2 – z1| and arg(z2 –z1)
1
...
1
4
...
3
4
...
5
4
...
7
10
11
13
14
15
38
Chapter 3: Summation of finite series
3
...
2
3
...
4
9
21
Chapter 2: Roots of polynomial equations
2
...
2
2
...
4
2
...
6
2
...
8
5
5
6
8
De Moivre’s theorem
Using de Moivre’s theorem to evaluate powers of complex numbers
Application of de Moivre’s theorem in establishing trigonometric identities
Exponential form of a complex number
The cube roots of unity
The nth roots of unity
The roots of z n , where is a non-real number
54
56
60
69
71
74
77
continued overleaf
2
MFP2 Textbook– A-level Further Mathematics – 6360
Further Pure 2: Contents (continued)
Chapter 5: Inverse trigonometrical functions
5
...
2
5
...
4
5
...
1 Definitions of hyperbolic functions
6
...
3 Graphs of hyperbolic functions
6
...
5 Osborne’s rule
6
...
7 Integration of hyperbolic functions
6
...
9 Logarithmic form of inverse hyperbolic functions
6
...
11 Integrals which integrate to inverse hyperbolic functions
6
...
1 Introduction
7
...
3 Area of surface of revolution
132
133
137
Answers to the exercises in Further Pure 2
143
3
MFP2 Textbook– A-level Further Mathematics – 6360
Chapter 1: Complex Numbers
1
...
2
The general complex number
1
...
4
The polar form of a complex number
1
...
6
The conjugate of a complex number and the division of complex numbers of the
form x iy
1
...
8
Equating real and imaginary parts
1
...
10 Loci on Argand diagrams
This chapter introduces the idea of a complex number
...
4
MFP2 Textbook– A-level Further Mathematics – 6360
1
...
For example, if you use a calculator to evaluate 64 you get an error message
...
As
1 cannot be evaluated, a symbol is used to denote it – the symbol used is i
...
1
...
The term x iy is a complex number with x being the real part and y the
imaginary part
...
The set of real numbers,
(with which you are familiar), is really a subset of the set of complex numbers,
...
5
MFP2 Textbook– A-level Further Mathematics – 6360
1
...
The point P(x, y) in the plane of coordinates with axes Ox
and Oy represents the complex number x iy and the number is uniquely represented by that
point
...
y
P(x, y)
r
θ
O
x
If the complex number x iy is denoted by z, and hence z x iy, z (‘mod zed’) is defined
as the distance from the origin O to the point P representing z
...
The modulus of a complex number z
is given by z x 2 y 2
The argument of z, arg z, is defined as the angle between the line OP and the positive x-axis –
usually in the range (π, –π)
...
6
MFP2 Textbook– A-level Further Mathematics – 6360
Example 1
...
1
Find the modulus and argument of the complex number 1 3i
...
3
3
2 and tan 3 3
...
O
1
x
3
Note that when tan 3, θ could equal 2π or π
...
This is why you need to be careful when evaluating the
argument of a complex number
...
Find the modulus and argument of each of the following complex numbers:
(a) 1 i,
(b) 3i,
(c) 4,
(d) 3 i
...
2
...
Give your answers for arg z in radians to two decimal places
...
4
The polar form of a complex number
y
In the diagram alongside, x r cos and
y r sin
...
This is
called the polar, or modulus–argument,
form of a complex number
...
Exercise 1B
1
...
2
...
3
3
8
x
MFP2 Textbook– A-level Further Mathematics – 6360
1
...
Consider the example below
...
5
...
Solution
(a) z1 z2 (3 4i) (1 2i)
4 2i
...
(since i 2 1)
(b) z1 z2 (3 4i) (1 2i)
2 6i
...
Find z1 z2 and z1 z2 when:
(a) z1 1 2i and z2 2 i,
(b) z1 2 6i and z2 1 2i
...
6
The conjugate of a complex number and the division of complex
numbers of the form x + iy
The conjugate of a complex number z x iy (usually denoted by z* or z ) is the complex
number z* x iy
...
On an
Argand diagram, the point representing the complex number z* is the reflection of the point
representing z in the x-axis
...
z z* z
2
Division of two complex numbers demands a little more care than their addition or
multiplication – and usually requires the use of the complex conjugate
...
6
...
2
Solution
3 4i (3 4i)(1 2i)
1 2i (1 2i)(1 2i)
3 4i 6i 8i2
1 2i 2i 4i
5 10i
5
1 2i
...
e
...
For the sets of complex numbers z1 and z2 , find z1 where
2
(a) z1 4 2i and z2 2 i,
(b) z1 2 6i and z2 1 2i
...
7
Products and quotients of complex numbers in their polar form
If two complex numbers are given in polar form they can be multiplied and divided without
having to rewrite them in the form x iy
...
7
...
6
6
3
3
Solution
6 cos π cos π i sin π cos π i sin π cos π i sin π sin π
3
6
3
6
3
6
3
6
6 cos π cos π sin π sin π i sin π cos π cos π sin π
3
6
3
6
3
6
3
6
z1 z2 2 cos π i sin π 3 cos π i sin π
3
3
6
6
2
Using the identities:
cos( A B ) cos A cos B sin A sin B
6 cos π π i sin π π
3 6
3 6
sin( A B) sin A cos B cos A sin B
6 cos π i sin π
...
Exercise 1E
z1
if z1 2 cos π i sin π and z2 3 cos π i sin π
...
(a) Find
11
MFP2 Textbook– A-level Further Mathematics – 6360
Example 1
...
2
If z1 r1 , 1 and z2 r2 , 2 , show that z1 z2 r1r2 cos(1 2 ) i sin(1 2 )
...
If z1 (r1 , 1 ) and z2 (r2 , 2 ) then z1 z2 (r1r2 , 1 2 ) – with the
proviso that 2π may have to be added to, or subtracted from, 1 2 if
1 2 is outside the permitted range for
There is a corresponding result for division – you could try to prove it for yourself
...
8
Equating real and imaginary parts
z
Going back to Example 1
...
1, z1 can be simplified by another method
...
Then,
1 2i
(1 2i)(a ib) 3 4i
a 2b i(b 2a) 3 4i
...
Suppose we let a ib
Thus b 2 and a 1, giving 1 2i as the answer to a ib as in Example 1
...
1
...
It also illustrates the concept of equating real and imaginary parts
...
8
...
Solution
Let z (a ib), then z* (a ib)
...
Multiplying out, 3a 4ia 3ib 4i 2b a ia ib i 2b 13 2i
...
Equating real and imaginary parts,
2a 3b 13,
5a 4b 2
...
Hence, z 2 3i
...
If z1 3, 2π
3
π
and z 2, 6 , find, in polar form, the complex numbers
2
z2
...
Find the complex number satisfying each of these equations:
(a) z1 z2 ,
z
(b) z1 ,
2
(a) (1 i) z 2 3i,
(c) z12 ,
(d) z13 ,
(e)
(b) ( z i)(3 i) 7i 11,
13
(c) z i 2 z* 1
...
9
Further consideration of z2 z1 and arg( z2 z1)
Section 1
...
Consider now the complex number z z2 z1 , where z1 x1 iy1 and z2 x2 iy2
...
y
A ( x1 , y1 )
B ( x2 , y 2 )
C
O
x
Then z z2 z1 ( x2 x1 ) i( y2 y1 ) and is represented by the point C ( x2 x1 , y2 y1 )
...
From this it follows that
1
2
z2 z1 OC ( x2 x1 ) 2 ( y2 y1 ) 2 ,
that is to say z2 z1 is the length AB in the Argand diagram
...
This in turn is the angle between
AB and the positive x direction
...
Find z2 z1 and arg( z2 z1 ) in
(a) z1 2 3i, z2 7 5i,
(b) z1 1 3i, z2 4 i,
(c) z1 1 2i, z2 4 5i
...
10 Loci on Argand diagrams
A locus is a path traced out by a point subjected to certain restrictions
...
Consider the simplest case first, when the point P represents the complex number z such that
z k
...
z k represents a circle with centre O and radius k
If instead z z1 k , where z1 is a fixed complex number represented by the point A on an
Argand diagram, then (from Section 1
...
It
follows that P must lie on a circle with centre A and radius k
...
The locus of P is therefore the region on and
within the circle with centre A and radius k
...
Again, using the result of Section 1
...
Hence, the locus
of P is a straight line
...
y
All the loci considered so far have been
related to distances – there are also
simple loci in Argand diagrams
involving angles
...
15
P
x
MFP2 Textbook– A-level Further Mathematics – 6360
This condition implies that the angle between OP and Ox is fixed ( ) so that the locus of P is
a straight line
...
In exactly the same way as before, the locus of a point P satisfying arg( z z1 ) , where z1
is a fixed complex number represented by the point A, is a line through A
...
Finally, consider the locus of any point P satisfying arg( z z1 )
...
y
A
β
α
O
x
16
MFP2 Textbook– A-level Further Mathematics – 6360
Exercise 1H
1
...
4
2
...
6
2
3
...
4
4
...
17
MFP2 Textbook– A-level Further Mathematics – 6360
Miscellaneous exercises 1
1
...
[AQA June 2001]
2
...
(a) Find z in the form a ib, where a and b are real
...
(c) Find the values of z and z z*
...
The complex number z satisfies the equation
zz* 3z 2 z* 2i,
where z* denotes the complex conjugate of z
...
[AQA March 2000]
4
...
where z
[AQA Specimen]
5
...
(b) Mark the point P on C at which z is a minimum
...
(c) Mark the point Q on C at which arg z is a maximum
...
[NEAB June 1998]
18
MFP2 Textbook– A-level Further Mathematics – 6360
6
...
4
(b) Indicate, by shading, the region for which
z 2 3i 3 and arg z 1 π
...
The complex number z is defined by
z 1 3i
...
(ii) Find the modulus and argument of z, giving your answer for the argument in
the form pπ where 1 p 1
...
The complex number
12
z2 is defined by z2 z z1
...
6
(ii) Mark on an Argand diagram the points P and P2 which represent z1 and z2 ,
1
respectively
...
1
[AQA June 2000]
19
MFP2 Textbook– A-level Further Mathematics – 6360
8
...
3
(b) The complex number z is such that
0 arg z 1 2π
3
π arg z 3 π
...
(ii) Mark on this diagram the point A belonging to R at which z has its least possible
value
...
(i) Calculate the value of z A
...
[AQA March 1999]
9
...
3i
Express each of z and w in the form a ib, where a and b are real
...
Give each modulus in an exact surd form and each argument in radians
between π and π
...
Find the exact length of PQ and hence, or
otherwise, show that the triangle OPQ is right-angled
...
1
Introduction
2
...
3
Cubic equations
2
...
5
Cubic equations with related roots
2
...
7
Polynomial equations of degree n
2
...
When you have completed it, you will:
know how to solve any quadratic equation;
know that there is a relationship between the number of real roots and form of a
polynomial equation, and be able to sketch graphs;
know the relationship between the roots of a cubic equation and its coefficients;
be able to form cubic equations with related roots;
know how to extend these results to polynomials of higher degree;
know that complex conjugates are roots of polynomials with real coefficients
...
1
Introduction
You should have already met the idea of a polynomial equation
...
Similarly, a
polynomial equation of degree 3 has x 3 as the highest power of x and is called a cubic
equation; one with x 4 as the highest power of x is called a quartic equation
...
2
...
However, even if this section is familiar to you it provides a suitable
base from which to move on to equations of higher degree
...
There are normally two ways of solving a quadratic equation – by
factorizing and, in cases where this is impossible, by the quadratic formula
...
e
...
For example, a sketch of part
of y x 2 2 x 8 is shown below
...
22
MFP2 Textbook– A-level Further Mathematics – 6360
A sketch of part of the curve y x 2 4 x 4 is shown below
...
The equation x 2 4 x 4 0 may be written as
( x 2) 2 0 and x 2, a repeated root
...
A sketch of
part of the curve y x 2 4 x 5 is shown below
...
2
Certainly, x 2 4 x 5 will not factorize so the quadratic formula x b b 4ac has to
2a
be used to solve this equation
...
It follows that the equation x 2 4 x 5 0 does have
Chapter1, this becomes
2
two roots, but they are both complex numbers
...
You may also have observed that whether a quadratic equation has real or complex roots
depends on the value of the discriminant b 2 4ac
...
Solve the equations
(a) x 2 6 x 10 0,
2
...
Cubic equations
As mentioned in the introduction to this chapter, equations of the form ax 3 bx 2 cx d 0
are called cubic equations
...
The reason for this is that cubic curves are continuous – they do
not have asymptotes or any other form of discontinuity
...
Hence the curve must cross the line y 0 at least once
...
24
MFP2 Textbook– A-level Further Mathematics – 6360
A typical cubic equation, y ax 3 bx 2 cx d with a 0, can look like any of the sketches
below
...
x
O
y
O
y
x
O
x
In each of the two sketch graphs above, the curve crosses the line y 0 just once, indicating
just one real root
...
Example 2
...
1
(a) Find the roots of the cubic equation x3 3x 2 x 3 0
...
y
Solution
(a)
4
If f ( x) x3 3x 2 x 3,
then f (1) 1 3 1 3 0
...
f ( x) ( x 1)( x 2 4 x 3)
( x 1)( x 3)( x 1)
...
25
2
(b)
-5
0
-2
0
5
x
MFP2 Textbook– A-level Further Mathematics – 6360
Example 2
...
2
Find the roots of the cubic equation x3 4 x 2 x 26 0
...
Then f (2) 8 16 2 26 0
...
The quadratic in this expression has no simple roots, so using the quadratic formula on
x 2 6 x 13 0,
2
x b b 4ac
2a
6 36 52
2
6 4i
2
3 2i
...
Exercise 2B
1
...
26
MFP2 Textbook– A-level Further Mathematics – 6360
2
...
Note that the factor a is required to ensure that the coefficients of
x 3 are the same, so making the equations identical
...
The two sides are identical so the coefficients of x 2 and x can be compared, and also the
number terms,
b a( )
c a( )
d a
...
Exercise 2C
1
...
2
...
If 3, 7 and 5, state
2
the cubic equation
...
5
Cubic equations with related roots
The example below shows how you can find equations whose roots are related to the roots of
a given equation without having to find the actual roots
...
Example 2
...
1
The cubic equation x3 3 x 2 4 0 has roots , and
...
Solution: method 1
From the given equation,
3
0
4
...
From which the equation of the cubic must be
x3 6 x 2 0 x 32 0
or
x 3 6 x 2 32 0
...
( 2)( 2)( 2) 2 4 8
4 0 12 8
0
...
28
MFP2 Textbook– A-level Further Mathematics – 6360
(c)
1 1 1 1
0 0
...
4
4
1 1 1 1 1
...
29
MFP2 Textbook– A-level Further Mathematics – 6360
The second method of finding the cubic equations in Example 2
...
1 is shown below
...
Solution: method 2
(a) As the roots are to be 2 , 2 and 2 , it follows that, if X 2 x, then a cubic equation in
X must have roots which are twice the roots of the cubic equation in x
...
(b) In this case, if you put X x 2 in x3 3 x 2 4 0, then any root of an equation in X
must be 2 less than the corresponding root of the cubic in x
...
(c) In this case you use the substitution X 1 or x 1
...
X
X
3
On multiplying by X , this gives
1 3X 4 X 3 0
or
4 X 3 3X 1 0
as before
...
The cubic equation x3 x 2 4 x 7 0 has roots , and
...
2
...
3
...
30
MFP2 Textbook– A-level Further Mathematics – 6360
2
...
So,
2
2 2 , or
2 2 for three numbers , and
2
This result is well worth remembering – it is frequently needed in questions involving the
symmetric properties of roots of a cubic equation
...
6
...
Find the cubic equations with
(a) roots , and , (b) roots 2 , 2 and 2
...
]
Solution
(a)
5
6
1
2 1 5 5
...
Hence the cubic equation is x3 6 x 2 5 x 1 0
...
2
2 2 2
...
Thus 2 2 2 2
2
2 36 (2 1 5) 46
...
Hence the cubic equation is x3 13x 2 46 x 1 0
...
7
Polynomial equations of degree n
The ideas covered so far on quadratic and cubic equations can be extended to equations of any
degree
...
Suppose the equation ax n bx n 1 cx n 2 dx n 3 k 0 has n roots , , , then
b
a ,
c
a ,
d
a ,
(1) n k
until, finally, the product of the n roots
...
In practice, you are unlikely to meet equations of degree higher than 4 so this section
concludes with an example using a quartic equation
...
7
...
Write down
(a) , (b)
...
2
Solution
4
2 2
...
2
2
(c) Now (
(a)
2 2 2 2 2(
2
...
6 can be extended to any number of letters
...
32
MFP2 Textbook– A-level Further Mathematics – 6360
Exercise 2E
1
...
(a) Find the equation with roots , , and
...
2
...
Using the ideas from
Chapter 1, if p and q are real,
f ( p iq ) a ( p iq ) n b( p iq ) n 1 k
u iv ,
where u and v are real
...
But i raised to an
odd power is the same as i raised to an odd power multiplied by 1 , and odd powers of i
comprise the imaginary part of f ( p iq )
...
Now if p iq is a root of f ( x) 0, it follows that u iv 0 and so u 0 and v 0
...
If a polynomial equation has real coefficients and if p iq,
where p and q are real, is a root of the polynomial, then its
complex conjugate, p iq, is also a root of the equation
It is very important to note that the coefficients in f ( x) 0 must be real
...
33
MFP2 Textbook– A-level Further Mathematics – 6360
Example 2
...
1
The cubic equation x3 3 x 2 x k 0, where k is real, has one root equal to 2 i
...
Solution
As the coefficients of the cubic equation are real, it follows that 2 i is also a root
...
To find k,
k (2 i )(2 i )(1) 5,
k 5
...
8
...
Find the other three
roots
...
Hence x (1 2i) x (1 2i) is a quadratic factor of the quartic
...
Hence x 2 2 x 5 is a factor of x 4 2 x 3 14 x 15
...
Comparing the coefficients of x3,
2 a2
a 4
...
Hence the quartic equation may be written as
( x 2 2 x 5)( x 2 4 x 3) 0
( x 2 2 x 5)( x 3)( x 1) 0,
and the four roots are 1 2i, 1 2i, 3 and 1
...
A cubic equation has real coefficients
...
Find the cubic
equation in the form x 3 ax 2 bx c 0
...
The cubic equation x3 2 x 2 9 x 18 0 has one root equal to 3i
...
3
...
Find the other
three roots
...
The equation
x 3 3x 2 px 4 0,
where p is a constant, has roots , and , where 0
...
(b) Find the value of p
...
The numbers , and satisfy the equations
2 2 2 22
11
...
(b) The numbers , and are also the roots of the equation
x 3 px 2 qx r 0,
where p, q and r are real
...
(ii) Calculate the product of the three roots
...
[AQA June 2000]
3
...
2 x3 3x 4 0
(a) Write down the values of , and
...
[AQA March 2000]
4
...
7 x 3 8 x 2 23x 30 0
(a) Write down the value of
...
[AQA Specimen]
36
MFP2 Textbook– A-level Further Mathematics – 6360
5
...
(a) Given that 3, write down the value of p
...
(c) One of the two non-real roots of the cubic equation is 3 4i
...
(ii) Find the value of r
...
(a) Prove that when a polynomial f x is divided by x a, the remainder is f a
...
When g x is divided by x i, where i 1, the
remainder is 3
...
(ii) Show that when g x is divided by 2 x i, the remainder is 6i
...
1
Introduction
3
...
3
Summation of series by the method of induction
3
...
When you have completed it, you will:
know new methods of summing series;
know which method is appropriate for the summation of a particular series;
understand an important method known as the method of induction;
be able to apply the method of induction in circumstances other than in the summation of
series
...
1
Introduction
You should already be familiar with the idea of a series – a series is the sum of the terms of a
sequence
...
For instance, the sum of an arithmetic progression is a series
...
Thus,
2 5 8 11 14
is a series of 5 terms, in arithmetic progression, with common difference 3
...
Instead of adding a fixed number to find
the next consecutive number in the series, you multiply by a fixed number (called the
common ratio)
...
A finite series is a series with a finite number of terms
...
39
MFP2 Textbook– A-level Further Mathematics – 6360
3
...
1 2 2 3 3 4
n n 1
In others you have to show that the sum of a series is a specific number or a given expression
...
show that 1 1 1
1 2 2 3 3 4
n 1
n n 1
The method of differences is usually used when the sum of the series is not given
...
The aim in the method of differences is to express
1
is the r th term of the first series
the r th term, which will be a function of r (just as
r r 1
above), as the difference of two expressions in r of the same form
...
If
you can express ur in this way, it can be seen that setting r 1 and then r 2 gives
u1 u2 f 1 f 2 f 2 f 3
f 1 f 3
...
The left hand side of this expression is the required sum of the series,
n
ur
...
40
MFP2 Textbook– A-level Further Mathematics – 6360
Example 3
...
1
Find the sum of the series
1 1 1
1
...
Nor is
the answer given
...
1
...
The only
As above, the r th term, ur , is given by
r
r r 1
1
sensible way to do this is to express
in partial fractions
...
r r 1 r r 1
Then, 1 A(r 1) Br
...
Comparing the constant
terms, A 1
...
r r 1 r r 1
Hence, in this case the f r mentioned previously would be 1 , with f r 1 1 , and so
r
r 1
on
...
r 1
Because the 1 , 1 , etc
...
Hence,
n
1
1
r (r 1) 1 n 1
r 1
(n 1) 1
n 1
n
...
2
...
Hence find
2
2
n
r 3
...
2
2
2
2
r 2 r 1 r 1 r 2 r 2 r 1 r 1
2 2
2
r r 2 r 1 r 2r 1
r 2 r 2 2r 1 r 2 2r 1
r 2 4r
4r 3
...
Listing the terms in columns, as in Example 3
...
1,
2 3 1 2
3 4 2 3
r 1
4 13
12 22 02 12
r2
4 23
r 3
4 33
r n 1
4 n 1 n 1 n 2 n 2 n 1
rn
4 n3
2
2
2
2
2
2
2
2
3
2
2
2
n 2 n 1 n 1 n 2
...
r 1
Summing the right hand side, all the terms cancel out except those shaded in the scheme
above, so the sum is n 2 n 1 02 12
...
2
Hence,
n
1
r 3 4 n2 n 1
2
, as required
...
2
...
2
...
Some series may be such that a term in one line cancels
with a term on a line two rows below it
...
2
...
1 5 3 7 5 9
2n 1 2n 3
Solution
As in Example 3
...
1, the way forward is to express
1
Let
1
2n 1 2n 3
in partial fractions
...
2r 1 2r 3
2r 1 2r 3
Multiplying both sides by 2r 1 2r 3 ,
1 A 2r 3 B 2r 1
...
Comparing the constant terms,
1 3 A B
...
Thus,
4
4
1
2r 1 2r 3
1 1
1
...
4 2n 1 4 2n 3
1
1
1
1
4 2n 5 4 2n 1
MFP2 Textbook– A-level Further Mathematics – 6360
There will be two terms left at the beginning of the series when the columns are added,
1 1 and 1 1
...
Therefore, addition gives
1
1
1
1
1 1 1
1
1 1 1 1
1 5 3 7 5 9
2n 1 2n 3 4 1 4 3 4 2n 1 4 2n 3
1 1 1 1 1
4 3 2 n 1 2n 3
1 4 2n 3 2n 1
4 3 2n 1 2n 3
4 n 1
1 4
4 3 2n 1 2n 3
n 1
...
(a) Simplify r r 1 r 1 r
...
r 1
2
...
r (r 1)(r 2) (r 1)(r 2)(r 3) r (r 1)(r 2)(r 3)
n
(b) Hence sum the series
1
r(r 1)(r 2)(r 3)
...
(a) Show that r 1 r 1 6r 2 2
...
r 1
44
MFP2 Textbook– A-level Further Mathematics – 6360
3
...
Suppose you have to show that the sum of n terms of a series is S(n)
...
You may think that this rather begs the
question but it must be understood that the result is assumed to be true for only one value of
n, namely n k
...
Finally, it is
demonstrated that the result is true for n 1
...
By building up the result, it can be said that the summation result is
true for all positive integers n
...
For convenience, and comparison, the examples worked in Section 3
...
Example 3
...
1
Show that
n
r
r 1
1
n
...
1 2 2 3 3 4
k k 1 k 1
Adding the next term to both sides,
1 1 1
1
1
1
...
Hence, if the result is true for n k , it
is true for n k 1
...
Therefore the result is true for all positive integers by induction
...
3
...
r 1
Solution
Assume that the result is true for n k , that is to say
k
r 3 1 k 2 k 1
...
Hence, if the result is true for n k , it
is true for n k 1
...
Therefore the result is true for all positive integers by induction
...
Prove the following results by the method of induction:
(a) 1 2 2 3 3 4 n n 1 1 n n 1 n 2
...
6
(c)
n
1
r r 2 6 n n 1 2n 7
...
r 1
47
MFP2 Textbook– A-level Further Mathematics – 6360
3
...
This chapter concludes with a look at its use in three other
connections – sequences, divisibility and de Moivre’s theorem for positive integers
...
4
...
n 1
Prove by induction that for all n 1, un 2 n 1
...
2 1
Then, using the relationship given,
uk 1 3 2
uk
3
3
2
k 1
2 1
2k 1
2 2k 1
2k 1 1
3 2k 1 1 2 2k 1
2k 1 1
3 2k 1 3 2k 1 2
2k 1 1
2 2 1
k 1
2k 1 1
k 2
2 k 1 1
2 1
2 k 1 1
...
Hence, if the result is true for
11
n k , it is true for n k 1
...
Therefore the
2 1
result is true for all positive integers n 1 by induction
...
4
...
Solution
The best approach is a little different to that used so far
...
2 k 1
2 k 1
When n k 1 the expression is 3 7
...
This expression is equal to 32( k 1) 32 k
32 k 2 32 k
32 k 32 32 k
32 k 32 1
8 32 k
...
In other words, if the result is true for n k , it is true for
n k 1
...
Hence, 32 n 7 is divisible by 8
for all positive integers n by induction
...
4
...
Prove by induction that for integers n 1, cos i sin cos n i sin n
...
Multiplying both sides by cos i sin ,
cos i sin k cos i sin cos k i sin k cos i sin
k 1
cos i sin cos k cos i sin k cos i sin cos k i2 sin k sin
cos k cos sin k sin i sin k cos cos k sin
cos k 1 i sin k 1 ,
which is of the same form but with k 1 replacing k
...
But when k 1,
cos i sin
1
true for all positive integers n by induction
...
Therefore the result is
i 2 1
MFP2 Textbook– A-level Further Mathematics – 6360
Exercise 3C
1
...
(b) 12n 2 5n1 is divisible by 7
...
[Hint: use the formula for differentiating a product]
dx
(d) x n 1 is divisible by x 1
...
Use the identity
1
1 1
r r 1 r r 1
n
to show that
r r1 1 n n 1
...
(a) Use the identity
4r 3 r 2 r 1 r 1 r 2
2
n
to show that
4r
3
2
n 2 n 1
...
[AQA June 2000]
3
...
4 4
[AQA March 1999]
4
...
r 1
[NEAB June 1998]
5
...
[AQA Specimen]
6
...
[AEB June 1997]
51
MFP2 Textbook– A-level Further Mathematics – 6360
7
...
n
n 1
...
Verify the identity
2r 1 2 r 1
2
...
n 1
r 2
[AEB January 1998]
9
...
(a) Verify that f r f r 1 Ar for some integer A, stating the value of A
...
2
2
r 1
[AEB January 2000]
10
...
r 1
(a) By setting n 1, or otherwise, determine the value of A
...
(c) Deduce the sum of the infinite series
1
4 7
3n 2
...
1
De Moivre’s theorem
4
...
3
Application of de Moivre’s theorem in establishing trigonometric identities
4
...
5
The cube roots of unity
4
...
7
The roots of z n , where is a non-real number
This chapter introduces de Moivre’s theorem and many of its applications
...
53
MFP2 Textbook– A-level Further Mathematics – 6360
4
...
4), you saw a very important result known as de Moivre’s theorem
...
De Moivre’s theorem holds not only when n is a positive integer, but also when it is negative
and even when it is fractional
...
Then k is a positive integer and
cos i sin n cos i sin k
1
cos i sin k
1
...
In order to remove i from
the denominator of the expression above, the numerator and denominator are multiplied by
the complex conjugate of the denominator, in this case cos k i sin k
...
p
where p and q are integers, then
q
q
p
p
pθ
pθ
cos q θ i sin q θ cos q q i sin q q
cos pθ i sin pθ
...
p
Taking the q th root of both sides,
p
p
p
cos i sin cos i sin q
...
A simple example will illustrate this
...
1
2
But cos π i sin π 1
cos π 1 and sin π 0
and
1 i
...
There are, in fact, q different values of cos π i sin π q and this will be
shown in section 4
...
cos i sin
n
cos n i sin n
for positive and negative integers, and fractional values of n
55
MFP2 Textbook– A-level Further Mathematics – 6360
4
...
The method for doing this will be illustrated through examples
...
2
...
6
6
Solution
It would, of course, be possible to multiply cos π i sin π by itself three times, but this would
6
6
be laborious and time consuming – even more so had the power been greater than 3
...
56
MFP2 Textbook– A-level Further Mathematics – 6360
Example 4
...
2
Find
3 i
10
in the form a ib
...
De Moivre’s
theorem could provide an alternative method but it can be used only for complex numbers in
the form cos i sin , and 3 i is not in this form
...
4) can be used to express it in polar form
...
r
2
12 2 and tan 1 so that π
...
57
x
MFP2 Textbook– A-level Further Mathematics – 6360
Example 4
...
3
3
Simplify cos π i sin π
...
6
6
π
π
π
cos π i sin 6 cos 6 i sin 6
6
cos 3π i sin 3π
6
6
cos π i sin π
2
2
3
3
cos π i sin π
2
2
i
...
It is very
n
important to realise that this is a deduction from de Moivre’s theorem and it must not be
quoted as the theorem
...
2
...
2, 2 3
Solution
r
The complex number 2 2 3 i is represented
by the point whose Cartesian coordinates are
2, 2 3 on the Argand diagram shown here
...
Thus
2
3
1
2 2 3 i
3
2 2 3 i
3
3
4 cos 2π i sin 2π
3
3
43 cos 3 2π i sin 3 2π
3
3
1 cos 2π i sin 2π
64
1 1 0
64
1
...
Prove that cos i sin cos n i sin n
...
Express each of the following in the form a ib :
(a)
(d) 1 i
(g)
6
3 3i
5
(b)
(e)
cos 3 i sin 3
2 2i
cos π i sin π
5
5
4
10
(c) cos π i sin π
4
4
1
(f)
1
9
59
3i
5
2
MFP2 Textbook– A-level Further Mathematics – 6360
4
...
The same
principles are used whichever identity is required
...
3
...
Solution
There are several ways of establishing this result
...
Similarly, the expansion of
cos 2 can be used to give cos 3 in terms of cos
...
cos 3 i sin 3 cos i sin
3
cos3 3cos 2 i sin 3cos i sin i sin
2
3
using the binomial expansion of p q 3
cos3 3i cos 2 sin 3cos sin 2 i sin 3
using i 2 1
...
Note that this equation will also give sin 3 by equating the imaginary parts of both sides of
the equation
...
3
...
Solution
tan 4 sin 4 so expressions for sin 4 and cos 4 in terms of sin and cos must be
cos 4
established to start with
...
Now,
tan 4 sin 4
cos 4
3
3
4 cos sin 2 4 cos sin
...
tan 4
sin 2 sin 4
1 6
cos 2 cos 4
But tan sin so
cos
3
tan 4 4 tan 2 4 tan
...
Express sin 3 in terms of sin
...
Express tan 3 in terms of tan
...
Express sin 5 in terms of sin
...
Show that cos 6 32 cos6 48cos 4 18cos 2 1
...
De
Moivre’s theorem can be used to express powers of sin , cos and tan in terms of sines,
cosines and tangents of multiple angles
...
Suppose z cos i sin
...
So,
Adding,
and subtracting,
z cos i sin
1 cos i sin
...
z
z
If z cos i sin
z 1 2 cos
z
1 2i sin
z
z
Also,
z n cos i sin cos n i sin n
n
n
z n 1 cos i sin
n
z
cos n i sin n
cos n i sin n
...
zn
If z cos i sin ,
z n 1 2 cos n
zn
z n 1 2i sin n
zn
A common mistake is to omit the i in 2i sin n , so make a point of remembering this result
carefully
...
3
...
16
Solution
z cos i sin
...
z
z
z
1
1
1
32 cos z 5z 5 10 z 10
z
z
z
1
1
1
z 5 z 10 z
...
z
5
Hence
32 cos 2 cos 5 5 2 cos 3 10 2 cos
cos5 1 cos 5 5cos 3 10 cos ,
16
as required
...
1
cos 16 cos 5 5cos 3 10 cos
5
1 sin 5 5sin 3 10sin c,
16 5
3
Example 4
...
4
(a) Show that cos3 sin 3 1 3sin 2 sin 6
32
(b) Evaluate
π
2
0
cos3 sin 3 d
...
MFP2 Textbook– A-level Further Mathematics – 6360
Solution
2sin z 1
...
z6
z2
Now z 6 1 2i sin 6 and z 2 12 2i sin 2
...
Dividing both sides by 64i,
cos3 sin 3 1 sin 6 3 sin 2
32
32
1 3sin 2 sin 6 ,
32
(b)
π
2
0
cos3 sin 3 1
32
π
2
0
as required
...
32 3 12
This section concludes with an example which uses the ideas introduced here and extends into
other areas of mathematics
...
3
...
(b) Show that the roots of the equation 16 x 4 20 x 2 5 0 are cos rπ for r = 1, 3, 7 and 9
...
(c) Deduce that cos 2
10
10 16
Solution
(a)
Using the ideas introduced at the beginning of this section,
cos 5 i sin 5 cos i sin
...
2
3
4
5
Not every term of this expression has to be simplified
...
The real
part of the right-hand side of the equation comprises those terms with even powers if i in
them, since i 2 1 and is real
...
66
2
2
using cos sin 1
MFP2 Textbook– A-level Further Mathematics – 6360
(b) Now when cos 5 0, either cos 0 or 16 cos 4 20 cos 2 5 0
...
5 π , 3π , 5π , 7π , 9π , 11π , 13π ,
But if cos 5 0,
2 2 2 2 2 2
2
so that
π , 3π , 5π , 7π , 9π , 11π , 13π ,
...
10
10
10
10
10
Now cos 5π cos π 0 and π is, of course, a root of cos 0, so that the roots of the
10
2
2
4
equation 16 x 20 x 5 0 are cos π , cos 3π , cos 7π and cos 9π
...
10
Thus the four roots of the quartic equation 16 x 4 20 x 2 5 0 can be written as cos π
10
3π
...
7), the product of the roots of the quartic
equation 16 x 4 20 x 2 5 0 is 5 so that
16
cos π cos π cos 3π cos 3π 5
...
10
10 16
67
MFP2 Textbook– A-level Further Mathematics – 6360
Exercise 4C
1
...
Prove the following results:
(a) cos 4 8cos 4 8cos 2 1
(b) sin 5 16sin 5 20sin 3 5sin
(c) sin 6 sin 32 cos5 32 cos3 6 cos
(d) tan 3 3 tan tan
1 3 tan 2
3
68
MFP2 Textbook– A-level Further Mathematics – 6360
4
...
These are given by
2 n2
cos 1
...
2! 4! 6!
2n 2 !
2
4
6
2 n 1
sin
...
3! 5! 7!
2n 1!
3
and
5
7
There is also a series for e x given by
2
3
4
n 1
e x 1 x x x x
...
2! 3! 4!
n 1!
If i is substituted for x in the series for e x ,
e
i
i 2 i 3 i 4
...
1 i
2!
3!
4!
n 1!
1 i i
...
It is also important to note that if z cos i sin , then
z n cos i sin
n
cos n i sin n
e ni ,
and if z r cos i sin , then z rei and z n r n e ni
...
Another result can be derived from the exponential form of a complex number:
ei cos i sin
...
Adding these
ei ei cos isin cos isin
2cos ,
or
Subtracting gives
i
i
cos e e
...
2i
or
i
i
cos e e
2
i
i
sin e e
2i
Example 4
...
1
Express 2 2i in the form rei
...
Hence,
y
r 22 2 8,
2
1
2
2 π,
2
4
and
tan
so that
2 2i= 8 e
O
θ
r
πi
4
...
Express the following in the form rei :
(a) 1 i
b) 3 i
(c) 3 3i
(d) 2 3 2i
70
MFP2 Textbook– A-level Further Mathematics – 6360
4
...
They must,
therefore, satisfy the equation z 3 1 0
...
Factorising,
z 3 1 z 1 z 2 z 1 0
...
The other
two come from the quadratic equation z 2 z 1 0
...
It can also be shown that if w is a root of z 3 1,
then w2 is also a root – in fact, the other root
...
Thus the three cube roots of 1 are 1, w and w2 , where w and w2 are non-real
...
2
It doesn’t matter whether w is labelled as 1 i 3 or as 1 i 3 because each is the square
2
2
of the other
...
2
If w 1 i 3 , then w2 1 i 3
...
Take w 1 i 3 ; w can be represented
2
2
by the point whose Cartesian coordinates are
1 3
2 , 2 on an Argand diagram
...
Thus,
2
2
4πi
2πi
The other root is w e 3 e 3 and can also be written as
2πi
3
...
y
Plotting the three cube roots of unity on an
Argand diagram shows three points equally
spaced (at intervals of 2π ) round a circle of
3
2π
3
radius 1 as shown in the diagram alongside
...
5
...
Solution
w 1 w w because w 1 ,
w w 1 w w
...
72
1,0
x
MFP2 Textbook– A-level Further Mathematics – 6360
Example 4
...
2
Show that
1 1 1 0
...
So the denominators of the left-hand side of the
equation can be replaced to simplify to 1 2 1 1
...
Exercise 4E
1
...
6
The nth roots of unity
The equation z n 1 clearly has at least one root, namely z 1, but it actually has many more,
most of which (if not all) are complex
...
To find the remaining roots, the right-hand side of the equation z n 1 has to be examined
...
But also, 1 e2πi because
e 2πi cos 2π i sin 2π 1 i0 1
...
Substituting the
right-hand side of the equation z n 1 by this term gives z n e2 kπi
...
Different integer values of k will give rise to different roots, as
shown below
...
n
n
n 0, 1, 2, , n 1 gives the n distinct roots of the equation z n 1
...
2 n 1 πi
e n
2 nπi
e n
Similarly, if k is set equal to n 1,
the same root as that given by k 1, and so on
...
All the roots lie on the circle
z 1 because the modulus of every root is 1
...
In other words, the
n n n
n
roots are represented by n points equally spaced
around the unit circle at angles of 2π starting at
n
1, 0 – the point representing the real root z 1
...
6
...
Solution
z 6 1 e 2 kπi
ze
Therefore
e
Hence the roots are
k 0,
k 1,
2 kπi
6
kπi
3
k 0, 1, 2, 3, 4, 5
...
y
e
2πi
3
πi
e3
–1
1
x
e
4πi
3
e
5πi
3
Two further points are worth noting
...
In example 4
...
1, the roots would be
given as z e
kπi
3
for k 0, 1, 2, 3
...
6
...
75
MFP2 Textbook– A-level Further Mathematics – 6360
Of course, there are variations on the above results
...
This looks daunting but if you can
recognise the left-hand side as a geometric progression with common ratio z, it becomes more
straightforward
...
The
6
root to be excluded is the root z 1 because z 1 is indeterminate when z 1
...
2
2
Exercise 4F
1
...
In each case, show the roots on an Argand diagram
...
Solve the equation z 4 z 3 z 2 z 1 0
...
Solve the equation 1 2 z 4 z 2 8 z 3 0
...
By considering the roots of z 5 1, show that cos 2π cos 4π cos 6π cos 8π 1
...
7
The roots of zn=α where α is a non-real number
Every complex number of the form a ib can be written in the form rei , where r is real and
lies in an interval of 2π (usually from 0 to 2π or from π to π)
...
Now
using e
because e
ei 2πi ei e 2πi
pq
ei
Similarly
e p eq
2πi
1
...
z n rei 2 kπi
So,
and, taking the nth root of both sides,
1 i 2 kπi
n
z r ne
1
r ne
i 2 kπ
n
k 0,1, 2, 3, , n 1
...
All lie on the circle z r n
and are equally spaced around the circle at
1 i
2π
...
n
The equation z n , where rei ,
has roots z
1 i 2 kπ
r ne n
1 2 π
n
n
r e
k 0, 1, 2, , n 1
77
1 iπ
n n
r e
MFP2 Textbook– A-level Further Mathematics – 6360
Example 4
...
1
Find the three roots of the equation z 3 2 2i
...
y
From the diagram alongside,
2, 2
r 22 22 8, tan 1, and π
...
z 3 8e
πi 2 kπi
4
θ
x
...
12
So the roots are
πi
k 0,
z 2 e12
k 1,
z 2 e 12
k 2,
z 2 e 12
9πi
7πi
or 2 e 12
...
12
12
17πi
The roots can also be written
This chapter closes with one further example of the use of the principles discussed
...
7
...
5
Solution
At first sight, it is tempting to use the binomial expansion on z 1 but this generates a
5
quartic equation (the terms in z 5 cancel) which would be difficult to solve
...
78
MFP2 Textbook– A-level Further Mathematics – 6360
Taking the fifth root of each side,
2 kπi
z 1 e 5 z
k 1, 2, 3, 4
...
Solving the equation for z,
or
2 kπi
1 z e 5 1
1
...
The term e 5 can be
written as cos 2kπ i sin 2kπ making the denominator have the form p iq
...
As p would then equal cos 2kπ 1 and q would equal
5
2kπ , this would be a rather cumbersome method
...
1
z
Thus,
e
z
So
e
e
kπi
5
e
kπi
5
1
e
kπi
5
e
kπi
5
e
kπi
5
e
(for reasons which
,
2 kπi kπi
5
5
e
i
i
But e e sin so that e
2i
2 kπi
5
kπi
5
kπi
5
kπi
5
...
Solve the following equations:
(b) z 3 1 i
(a) z 4 16i
(d) z 2 1
(e)
z 1
3
(c) z 8 1 3 i
8i
(f)
80
z 1
5
z5
MFP2 Textbook– A-level Further Mathematics – 6360
Miscellaneous exercises 4
1
...
(b) Hence solve the equation z 4 64 0 giving your answers in the form
r cos i sin , where r 0 and π π
...
[AEB June 1996]
2
...
(b) Using your answers to part (a),
(i) show that
3 i
1 i
5
10
(ii) solve the equation
1 3 i,
2 2
z 3 1 i
3 i
giving your answers in the form a ib, where a and b are real numbers to be
determined
to two decimal places
...
(a) By considering z cos i sin and using de Moivre’s theorem, show that
sin 5 sin 16sin 4 20sin 2 5
...
(c) Deduce the exact values of sin π and sin 2π , explaining clearly the reasons for your
5
5
answers
...
(a) Show that the non-real cube roots of unity satisfy the equation
z 2 z 1 0
...
Find the possible values of a
...
(a) Verify that
z 1
is a root of the equation
5
1
...
(c) Mark on an Argand diagram the points corresponding to the five roots of the equation
...
(d) By considering the Argand diagram, find
(i) arg z1 in terms of π,
(ii) z1 in the form a cos π , where a and b are integers to be determined
...
(a) (i) Show that w
2πi
e5
is one of the fifth roots of unity
...
(b) Let p w w4 and q w2 w3 , where w e
2πi
5
...
(ii) Write down the quadratic equation, with integer coefficients, whose roots are p
and q
...
5
5
[NEAB June 1998]
82
MFP2 Textbook– A-level Further Mathematics – 6360
7
...
zn
(ii) Write down the corresponding result for z n 1
...
(ii) By substituting z cos i sin in the above identity, deduce that
cos3 sin 3 1 3sin 2 sin 6
...
(a) (i) Express e 2 e
i
2
in terms of sin
...
i
(b) Derive expressions, in the form ei where π π, for the four non-real roots of
the equation z 6 1
...
(i) Explain why the equation has only five roots in all
...
(iii) Show that the non-real roots are
1 ,
z1 1
1 ,
z2 1
1 ,
z3 1
1 ,
z4 1
where z1 , z2 , z3 and z4 are the non-real roots of the equation z 6 1
...
[AQA March 2000]
83
MFP2 Textbook– A-level Further Mathematics – 6360
9
...
(b) Show that one of the roots of the equation
z 3 2 2i
πi
is 2 e12 , and find the other two roots giving your answers in the form rei , where r
is a surd and π π
...
(d) Find the area of the triangle ABC, giving your answer in surd form
...
Denoting by w, , and the
complex numbers represented by P, A, B and C, respectively, show that
w 2 w 2 w 2
[AQA June 1999]
84
6
...
1
Introduction and revision
5
...
3
Applications to more complex differentiation
5
...
5
Applications to more complex integrals
This chapter revises and extends work on inverse trigonometrical functions
...
85
MFP2 Textbook– A-level Further Mathematics – 6360
5
...
However, in order to present a clear picture, and for the
sake of completeness some revision is included in this section
...
Note that sin 1 y is not cosec y which would
normally be written as (sin y ) 1 when expressed in terms of sine
...
The sketch of y sin x will be
familiar to you and is shown below
...
The graph of y sin 1 x being the inverse, is the
reflection of y sin x in the line y x and a sketch of it is as shown
...
In order to overcome this obstacle, we restrict the range of y to π y π so that the
2
2
1
sketch of y sin x becomes the sketch shown
...
This value is usually called the principal value
...
2
2
Notice that the gradient of y sin 1 x is always greater than zero
...
The sketches of
y cos x and y cos 1 x are shown below
...
Instead we choose the range 0 x π and the sketch is as shown
...
When it comes to tan 1 x we can restrict the range to π and π
...
2
2
The sketch of y tan 1 x is shown below
...
Express in terms of π the values of:
(a) tan 1 1
(d) cos 1 0
(b) cos 1 3
2
(e) tan 1 1
3
88
(c) sin 1 1
2
(f) cos 1 (1)
MFP2 Textbook– A-level Further Mathematics – 6360
5
...
This is due to the fact that the gradient of the
graph of y sin 1 x is always greater than zero as was shown earlier
...
For y cos 1 x using similar working we would arrive at
If y cos 1 x
dy
1
dx
1 x2
89
MFP2 Textbook– A-level Further Mathematics – 6360
If
y tan 1 x
then we write
tan y x ,
and, differentiating implicitly,
dy
1
dx
dy
12
dx sec y
1
1 tan 2 y using sec 2 y 1 tan 2 y
dy
1
dx 1 x 2
sec2 y
or
If y tan 1 x
dy
1
dx 1 x 2
Exercise 5B
1
...
dx
1 x2
90
MFP2 Textbook– A-level Further Mathematics – 6360
5
...
These would include
the function of a function rule, and the product and quotient rules
...
Example 5
...
1
If y sin 1 (2 x 1) , find
dy
...
2
4 x 4 x2
2
2 x x2
1
x x2
Example 5
...
2
Differentiate sin 1 e x
...
3
...
dx
This time we need to use the product rule and the function of a function rule
...
dx
1 4x
Example 5
...
4
1
Differentiate cos x
1 x2
1
If y cos x
1 x2
Then, using the quotient rule,
1 x2 1
2
dy
1 x
dx
1
2
1
cos x 2 1 x
1 x2
1
1 2 x cos x
...
(a) tan 1 3x
(b) cos 1 3 x 1
(c) sin 1 2x
2
...
(a) sin 3 3x
x
(b)
4
...
1 x
2
93
MFP2 Textbook– A-level Further Mathematics – 6360
5
...
However, this does not preclude a question requiring a proof of a result from this
booklet being set
...
The first one is
dx
a 2 x2
This integral requires a substitution
...
Example 5
...
1
Evaluate
2
0
dx
4 x2
We have
2
dx
1 x
1
0 4 x 2 2 tan 2 0
1 tan 1 1 1 tan 1 0
2
2
1 π 0
2 4
π
8
2
Example 5
...
2
Evaluate
3
2
0
dx
9 x2
We have
3
2
0
3
sin 1 x 2
3 0
dx
9 x2
3
1 2
sin sin 1 0
3
sin 1 1 sin 1 0
2
π 0
6
π
6
Exercise 5D
Integrate the following, leaving your answers in terms of π
...
1
3
2dx
1 x2
2
...
1
1
1
2
3dx
1 x
dx
0 1 x2
3
...
3
1
3
95
dx
x 1
2
4
3
dx
25 x 2
MFP2 Textbook– A-level Further Mathematics – 6360
5
...
Most will involve completing the square of
a quadratic expression, a method you will no doubt have used many times before in other
contexts
...
Example 5
...
1
Find
dx
x2 4 x 8
...
The result is
x 2 c
...
5
...
The substitution
MFP2 Textbook– A-level Further Mathematics – 6360
Example 5
...
3
Find
xdx
x4 9
Here the substitution u x 2 transforms the given integral into standard form for if u x 2
1 du
du 2 xdx and we have 2
u2 9
1 1 tan 1 u c
2 3
3
2
1 tan 1 x c
6
3
Finally we give a slightly harder example of an integral which uses
dx
a2 x2
in its solution
...
5
...
Integrals of this type where the numerator is a linear expression in x and the denominator is a
quadratic in x usually integrate to ln p ( x) tan 1 q ( x) where p ( x) and q ( x) are functions of
f ' x dx
x
...
You should have been taught this result when studying the module
Core 3
...
This will be done by means of examples
...
5
...
As in previous examples, we need to complete the square on 4x x 2 , and we write
4 x x2 4 x 2
So that the interval becomes
dx
4 x 2
98
2
2
MFP2 Textbook– A-level Further Mathematics – 6360
The substitution u x 2 simplifies the result to the standard form
du
4 u2
sin 1 u c
2
sin 1
x 2 c
2
Example 5
...
6
Find
dx
1 6 x 3x2
...
In this particular context the result you will need is that
f ' x
f x dx
2 f x c
[This result can be easily verified using the substitution u f x , since then, du f ' x dx
and the integral becomes
du ]
...
5
...
Now the derivative of 7 6x x 2 is 6 2x so we write x as 1 6 2 x 3 and the integral
2
becomes
1 6 2x 3
2
dx
7 6 x x2
or separating the integral into two parts
1 6 2x
2
7 6 x x2
The first integral is of the form
3dx
7 6 x x2
f ' x
f x dx apart from a scaler multiplier, and so integrates to
7 6x x 2 , whilst completing the square on the denominator of the second integral, we
3dx
which integrates to 3sin 1 x 3 using the substitution u x 3
...
Integrate
(a)
1
x 4x 5
(b)
1
2x 4x 5
(b)
2
x
x x 1
(b)
(b)
(c)
1
x x2
(c)
2
(c)
2
2
...
Find
(a)
dx
7 6x x
2
dx
3 2x x
2
dx
x 1 2 x
4
...
1
Definitions of hyperbolic functions
6
...
3
Graphs of hyperbolic functions
6
...
5
Osborne’s rule
6
...
7
Integration of hyperbolic functions
6
...
9
Logarithmic form of inverse hyperbolic functions
6
...
11 Integrals which integrate to inverse hyperbolic functions
6
...
When you have completed it, you will:
know what hyperbolic functions are;
be able to sketch them;
be able to differentiate and integrate them;
have learned some hyperbolic identities;
understand what inverse hyperbolic functions are and how they can be expressed in
alternative forms;
be able to sketch inverse hyperbolic functions;
be able to differentiate inverse hyperbolic functions and recognise integrals which
integrate to them;
be able to solve equations involving hyperbolic functions
...
1
Definitions of hyperbolic functions
It was shown in Chapter 4 that sin x 1 eix e ix and cos x 1 eix +e ix
...
The definitions of sinh x and cosh x (often
called hyperbolic sine and hyperbolic cosine – pronounced ‘shine x’ and
‘cosh x’) are:
sinh x 1 e x e x
2
1 e x e x
cosh x
2
There are four other hyperbolic functions derived from these, just as there are four
trigonometric functions
...
e 1
Again,
1
sinh x
1
x
x
1
2 e e
cosech x
2
...
103
MFP2 Textbook– A-level Further Mathematics – 6360
Exercise 6A
1
...
2
sech x,
(b) coth x,
(c) tanh 1 x,
2
(d) cosech 3x
...
There is no unit for x when evaluating, for example, sinh x
...
3 :
2
2
sinh 2 e e 3
...
3
1
...
3 e e
1
...
2
You can work out these values on a calculator using the e x button
...
It is worth remembering that
0
0
sinh 0 e e 1 1 0;
2
2
0
0
cosh 0 e e 1 1 1
...
Use a calculator to evaluate, to two decimal places:
(a) sinh 0
...
3,
(c) sech 2
...
6),
(e) cosh (– 0
...
3
Graphs of hyperbolic functions
The graphs of hyperbolic functions can be sketched easily by plotting points
...
It would also be worthwhile committing the general shapes of
y sinh x, y cosh x and y tanh x to memory
...
In Section 5
...
1 e
Now, as e 2 x 0 for all values of x, it follows that the numerator in the bracketed expression
above is less than its denominator, so that tanh x 1
...
So the graph of y tanh x has an asymptote at y 1
...
1 e
2 x
As e 0 for all values of x, it follows that the numerator of this fraction is less than its
denominator, from which it can be deduced that tanh x 1
...
So the graph of y tanh x has an asymptote at
x 1
...
y tanh x
y
1
0
x
–1
105
MFP2 Textbook– A-level Further Mathematics – 6360
Exercise 6C
1
...
4
(b) y cosech x,
(c) y coth x
...
For example,
2
2
x x
cosh x e e 1 e2 x 2 e2 x ,
2
4
2
x x
and
sinh x e e 1 e2 x 2 e 2 x
2
4
from which, by subtraction,
2
cosh 2 x sinh 2 x 1 e 2 x 2 e2 x 1 e2 x 2 e2 x
4
4
1
...
sech 2 x 1 tanh 2 x
Or again, dividing both sides by sinh 2 x instead,
cosh 2 x sinh 2 x 1
sinh 2 x sinh 2 x sinh 2 x
coth 2 x 1 cosech 2 x
...
In exactly the same way, expressions for sinh x y , cosh x y and cosh x y can be
worked out
...
Show that
(a) sinh x y sinh x cosh y cosh x sinh y,
(b) cosh x y cosh x cosh y sinh x sinh y
...
The hyperbolic
formulae given above help to find corresponding results for hyperbolic functions
...
cosh x y cosh x cosh y sinh x sinh y,
putting y x,
cosh x x cosh x cosh x sinh x sinh x,
or
cosh 2 x cosh 2 x sinh 2 x
...
sinh 2 x 2sinh x cosh x
cosh 2 x cosh 2 x sinh 2 x
2 cosh 2 x 1
1 2sinh 2 x
Some examples will illustrate extensions of these results
...
4
...
1 tanh x
Solution
tanh 2 x sinh 2 x
cosh 2 x
x
2sinh x cosh 2
...
1 tanh 2 x
Example 6
...
2
Show that cosh 3x 4 cosh 3 x 3cosh x
...
Exercise 6E
1
...
2
...
109
2
[using cosh x sinh x 1]
MFP2 Textbook– A-level Further Mathematics – 6360
6
...
In fact the only differences
are those of sign – for example, whereas cos 2 x sin 2 x 1, the corresponding hyperbolic
identity is cosh 2 x sinh 2 x 1
...
To change a trigonometric function into its corresponding
hyperbolic function, where a product of two sines appears
change the sign of the corresponding hyperbolic term
For example,
because
cos x y cos x cos y sin x sin y
then
cosh x y cosh x cosh y sinh x sinh y
...
However, care must be exercised in using this rule, as the next example shows
...
cos x
It should be noted that Osborne’s rule is only an aid to memory
...
The method shown in Section 6
...
110
MFP2 Textbook– A-level Further Mathematics – 6360
6
...
Just to remind you,
d e kx ke kx
...
For example,
y sinh x
1 e x e x
...
y sinh kx
1 e kx e kx ,
2
then
dy 1
ke kx +ke kx
dx 2
k 1 e kx +e kx
2
k cosh kx
...
The following results
cosh x
should be committed to memory:
y sinh x,
y cosh x,
y tanh x,
dy
cosh x
dx
dy
sinh x
dx
dy
sech 2 x
dx
Generally:
y sinh kx,
y cosh kx,
y tanh kx,
dy
k cosh kx
dx
dy
k sinh kx
dx
dy
k sech 2 kx
dx
Note that the derivatives are very similar to the derivatives of trigonometric functions, except
that whereas d cos x sin x, d cosh x sinh x
...
6
...
2
Solution
y sinh 1 x
2
dy
cosh 1 x 1
dx
2
2
1 cosh 1 x
...
6
...
Solution
y x cosh 2 x cosh 4 3 x
x cosh 2 x cosh 3 x
...
Exercise 6F
1
...
113
MFP2 Textbook– A-level Further Mathematics – 6360
6
...
Just to remind you,
1 kx
kx
e dx k e
...
sinh x dx cosh x c
cosh x dx sinh x c
2
sech x dx tanh x c
Of course, generally sinh kx 1 cosh kx c
...
cosh x
Putting u cosh x,
du sinh x dx
...
tanh x dx ln cosh x c
coth x dx ln sinh x c
Exercise 6G
1
...
114
MFP2 Textbook– A-level Further Mathematics – 6360
6
...
, so there are inverse
hyperbolic functions
...
To find the value of, say, sinh 1 2 using a calculator, you use it in the same way as you would
if it was a trigonometric function (pressing the appropriate buttons for hyperbolic functions)
...
These sketches are shown below
...
y
y tanh
y sinh
1
1
x
y
x
0
x
–1
0
1
x
Note also that y cosh x does not have an inverse
...
If you look at the graph of y cosh x (in Section
6
...
However, if the
domain of y cosh x is restricted to x 0 there will be a one-to-one mapping, and hence an
inverse, and the range for the inverse will be y 0
...
9
Logarithmic form of inverse hyperbolic functions
The inverse hyperbolic functions cosh 1 x, sinh 1 x and tanh 1 x can be expressed as
logarithms
...
2 xe y e 2 y 1
Multiplying by e y ,
0 e 2 y 2 xe y 1
...
Taking the logarithm of each side,
y ln x x 2 1
...
x x2 1
Thus
and
so that
x
1
2
x 1
,
1
ln x x 1 ln
2
x x 1
2
ln x
y ln x
x 1 ,
x 1
...
8, if y 0 then
y cosh 1 x ln x x 2 1
...
This gives y ln x x 2 1 , but as x x 2 1 0 the negative
sign has to be rejected because the logarithm of a negative number is non-real
...
It is straightforward to obtain the logarithmic form of y tanh 1 x if, after writing
sinh y
tanh y x, tanh y is written as
...
9
...
2
Solution
1 1
2
tanh 1 1 1 ln 1
2 2 1 2
3
2
1 ln 1
2 2
1 ln 3
2
ln 3
...
9
...
4
Solution
2
sinh 1 3 ln 3 3 1
4
4
4
ln 3 9 1
16
4
ln 3 25
4
16
ln 3 5
4 4
ln 2
...
Show that sinh 1 x ln x x 2 1
...
Show that tanh 1 x 1 ln 1 x
...
Express the following in logarithmic form:
(a) cosh 1 3 ,
(b) tanh 1 1 ,
(c) sinh 1 5
...
10 Derivatives of inverse hyperbolic functions
As already seen, if y sinh 1 x , then sinh y x
...
cosh y
dx
dy
So that
1
dx cosh y
1
using cosh 2 y sinh 2 y 1
2
1 x
Again, if y sinh 1 x ,
a
sinh y x ,
a
and
cosh y
dy 1
...
2
a x2
The derivatives of cosh 1 x and cosh 1 x , and also tanh 1 x and tanh 1 x are obtained
a
a
in exactly the same way
...
10
...
dx
Solution
y tanh 1 x
tanh y x
sech 2 y
dy
1
dx
dy
12
dx sech y
1 2
1 x
using sech 2 y tanh 2 y 1
...
y sinh 1 x :
y cosh 1 x :
y tanh 1 x :
dy
1
dx
1 x2
dy
1
dx
x2 1
dy
1
dx 1 x 2
Generally,
y sinh 1
x
:
a
y cosh 1
x
:
a
dy
1
2
dx
a x2
dy
1
dx
x2 a2
y tanh 1
x
:
a
dy
a
2 2
dx a x
Example 6
...
2
Differentiate cosh 1 x
...
120
MFP2 Textbook– A-level Further Mathematics – 6360
Example 6
...
3
dy
If y x 2 sinh 1 x , find
when x 2, giving your answer in the form a b ln c
...
2
y x 2 sinh 1 x
2
dy
1
2 x sinh 1 x x 2
...
Exercise 6I
1
...
x
(c) cosh 1 x ,
4
dy
when x 2, giving your answer in the form a ln b, where a
dx
and b are irrational numbers
...
If y x cosh 1 x, find
121
MFP2 Textbook– A-level Further Mathematics – 6360
6
...
10, that:
dx
x
sinh 1 c
a
a x
x
dx
cosh 1 c
2
2
a
x a
2
a
1
The integrals of
2
a x
2
2
x
dx
1
tanh 1 c
2
a
x a
2
1
and
2
x a2
, in particular, help to widen the ability to integrate
...
Solution
px qx r
, or
, where p 0
...
Example 6
...
1
Find
1
2
dx
16 x
2
dx
2
4 x2
sinh 1 x c
...
11
...
Solution
dx
6 2x
2
dx
2 3 x2
1
2
dx
3 x2
1 sinh 1 x c
...
11
...
Solution
In order to evaluate this integral, you must complete the square in the denominator
...
2
Hence,
dx
x2 2 x 3
dx
x 1 4
2
...
2
z 4
x 2x 3
cosh 1 z c
2
cosh 1 x 1 c
...
11
...
Solution
In this case, the integral is split into two by writing one integral with a numerator which is the
derivative of x 2 6 x 10
...
Now,
dx
But
2x 5 2x 6 1
and
2x 5
2
x 6 x 10
dx
2x 6
2
x 6 x 10
I1 I 2 ,
say
...
Because the derivative of x 2 6 x 10 is 2 x 6, then for I1 the substitution z x 2 6 x 10
gives dz 2 x 6 and consequently
dx
I1 dz
z
1
z 2 dz
1
2
z1
2
2 x 2 6 x 10
...
2
So that
I2
dx
x 32 1
...
Therefore, the complete integral is
I1 I 2 2 x 2 6 x 10 sinh 1 x 3 c
...
Evaluate the following integrals:
dx ,
(a)
(b)
(d)
x 9
dx
(g)
2
2
9 x 49
(e)
dx
2
x 4x 5
,
(h)
,
dx
2
x 16
(c)
,
dx
x 1
2
4
dx
2
x 2x 2
,
(f)
dx
2
4 x 25
,
dx
x 2
2
16
,
...
12 Solving equations
You are likely to meet two types of equations involving hyperbolic functions
...
The first type has the form a cosh x b sinh x c, or a similar linear combination of
hyperbolic functions
...
Example 6
...
1
Solve the equation 7 sinh x 5cosh x 1
...
Multiplying throughout by e x ,
e 2 x 6 e x ,
e 2 x e x 6 0
...
x
Hence, e 3 or e 2
...
125
MFP2 Textbook– A-level Further Mathematics – 6360
Example 6
...
2
Solve the equation cosh 2 x 4sinh x 6
...
The reason for this can be seen on substitution
– instead of having an equation involving cosh x and sinh x, the original equation is reduced
to one involving sinh x only
...
This is a quadratic equation in sinh x which factorizes to
sinh x 5 sinh x 1 0
...
31 or x 0
...
Hence,
and
and
The answers can also be expressed in terms of logarithms, using the results from Section 6
...
Note that it is not advisable to use the definitions of sinh x and cosh x when attempting to
solve the equation in Example 6
...
2 – this would generate a quartic equation in e x which
would be difficult to solve
...
Solve the equations:
(a) 4sinh x 3e x 9,
(c) cosh 2 x 3sinh x 5,
(b) 3sinh x 4 cosh x 4,
(d) cosh 2 x 3cosh x 4,
2
(e) tanh x 7sech x 3
...
(a) Express cosh x sinh x in terms of e x
...
cosh x sinh x
[AQA March 1999]
2
...
(b) Hence, or otherwise, solve the equation
8sinh x 3sech x,
leaving your answer in terms of natural logarithms
...
(a) By considering sinh y x
...
(b) Solve the equation
2 cosh 2 5sinh 8 0,
leaving your answers in terms of natural logarithms
...
(a) Show that the equation
14sinh x 10 cosh x 5
can be expressed as
2e 2 x 5e x 12 0
...
[AQA June 2001]
127
MFP2 Textbook– A-level Further Mathematics – 6360
5
...
(b) Hence show that the substitution x cosh t transforms the equation
16 x3 12 x 5
cosh 3t 5
...
Obtain this root, giving your answer in
the form 2 p 2q , where p and q are rational numbers to be found
...
(a) Using the definitions of sinh and cosh in terms of exponentials, show that
2
tanh e2 1
...
[AEB June 1999]
7
...
dx
tanh x tanh x dx
...
[AQA March 2000]
128
MFP2 Textbook– A-level Further Mathematics – 6360
8
...
(b) Given that
cosh x 17 and sinh y 4 ,
8
3
(i) express y in the form ln n, where n is an integer,
(ii) show that one of the possible values of x y is ln12 and find the other possible
value in the form ln a, where a is to be determined
...
(a) State the values of x for which cosh 1 x is defined
...
(i) Show that C has just one stationary point
...
[NEAB March 1998]
10
...
dx
(b) Hence, or otherwise, prove that
d tanh 1 x 1
...
2 1 x
(d) Show that
1
2
0
tanh 1 x dx a ln b 2 ,
1 x2
where a and b are numbers to be determined
...
The diagram shows a region R in the x–y plane bounded by the curve y sinh x, the x-axis
and the line AB which is perpendicular to the x-axis
...
3
2
(b) (i) Show that cosh ln k k 1
...
3
(c) (i) Show that
ln 3
0
sinh 2 x dx 1 sinh ln 9 ln 9
...
[NEAB June 1998]
130
MFP2 Textbook– A-level Further Mathematics – 6360
Chapter 7: Arc Length and Area of Surface of Revolution
7
...
2
Arc length
7
...
When you
have completed it, you will:
know a formula which can be used to evaluate the length of an arc when the equation of
the curve is given in Cartesian form;
know a formula which can be used to evaluate the length of an arc when the equation of
the curve is given in parametric form;
know methods of evaluating a curved surface area of revolution when the equation of the
curve is given in Cartesian or in parametric form
...
1
Introduction
You will probably already be familiar with some formulae
to do with the arc length of a curve and the area of surface
of revolution
...
You will also be familiar with the formula V
b
a
-axis2ll alss
132
πy 2 dx which gives the volume of the solid
MFP2 Textbook– A-level Further Mathematics – 6360
7
...
In the diagram alongside, s is
the arc length between the points P and Q on the curve
y f x
...
δs
O
y
a
b
x
Now, if P and Q are fairly close to each other then the arc length s must be quite short and
PQN be approximately a right-angled triangle
...
x
2
dy
1
dx
s
x
ds
dx
x y
...
dx
dx
Thus,
s
2
b
dy
1 dx
...
Suppose that x and y are both functions of a parameter t
...
2
2
2
ddst ddxt ddyt
ds dx dy
dt dt
dt
dy
dx
s
dt dt dt,
s
t
2
2
2
y
...
The length of arc of a curve in terms of a parameter t
is given by
s
t2
t1
dx
dt
2
2
dy
dt,
dt
where t1 and t2 are the values of the parameter at
each end of the arc
...
134
MFP2 Textbook– A-level Further Mathematics – 6360
Example 7
...
1
Find the length of the curve y cosh x between the points where x 0 and x 2
...
dx
2
Therefore,
dy
1 1 sinh 2 x
dx
cosh 2 x
2
dy
1 cosh x
...
135
MFP2 Textbook– A-level Further Mathematics – 6360
Example 7
...
2
Show that the length of the curve (called a cycloid) given by the equations x a sin ,
y a 1 cos between 0 and 2π is 8a
...
d
Therefore,
dx
d
2
2
2
dy
2
2
2
a 1 cos a sin
d
a 2 1 2 cos cos 2 sin 2
a 2 1 2 cos 1
2
2
(using sin cos 1)
a 2 2 2 cos
2a 2 1 cos
2a 2 2sin 2
2
dx
d
2
2
(using 2 sin 1 cos 2 )
2
dy
4a 2 sin 2
d
2
2a sin
...
136
MFP2 Textbook– A-level Further Mathematics – 6360
7
...
The area of this surface is
known as the ‘curved surface area’ or ‘area of surface of revolution’
...
This arc is rotated about the x-axis by 2π
y
P
radians
...
You can see from the diagram that the curved surface
generated by the rotation is larger than that of the cylinder of
width δs obtained by rotating the point P about the x-axis, but
smaller than the area of the cylinder width δs obtained by
rotating the point Q about the same axis
...
If the actual area generated by the rotation
δs
Q
δy
δx
y
O
a
y
b
of arc PQ about the x-axis is denoted by δA, it follows that
2πy s A 2π y y s
or, dividing by δs,
2πy A 2π y y
...
Therefore,
dA 2πy
ds
A
b
2πy ds
a
2
b
dy
2πy 1 dx
a
dx
(from section 7
...
3
...
is rotated through 2π radians about the x-axis
...
2
...
dx
Hence,
A
2
2
dy
2πy 1 dx
0
dx
2
2π cosh x cosh x dx
0
2
cosh x dx
2π 1 1 cosh 2 x dx
2
2π
2
0
2
0
2
π x sinh 2 x
2 0
π 2 1 sinh 4 π 0 1 sinh 0
2
2
π 2 1 sinh 4
...
3
...
3
138
MFP2 Textbook– A-level Further Mathematics – 6360
Solution
This was the curve used in example 7
...
2 – from there
2
dx
d
2
=
2π
dy
2a sin 2
...
4πa 2
2
2
2π
2
0
2π
(using cos 2 x 2 cos 2 x 1)
2
0
2π
2
0
Now, consider
cos
2
sin d
...
Integrating for A,
2π
A 8πa 2 2 cos 2 cos3
2 3
2 0
3
8πa 2 2 cos π 2 cos3 π 2 cos 0 2 cos 0
3
3
8πa 2 2 2 2 2
3
3
64 πa 2
...
(a) Show that
(i)
d
tanh sech 2 ,
d
(ii)
d sech sech tanh
...
2
dy
tanh 2
...
Find s in terms of and deduce that, for any point on
curve, y e s
...
(a) Using only the definitions of cosh x and sinh x in terms of exponentials,
(i) determine the exact values of cosh and sinh , where ln 9 ,
4
(ii) establish the identities
cosh 2 x 2 cosh 2 x 1
sinh 2 x 2sinh x cosh x
...
Show that
S π ln 9 p
4
for some rational number p whose value you should state
...
(a) (i) Using only the definitions
cosh 1 e e and sinh 1 e e ,
2
2
prove the identity
cosh 2 sinh 2 1
...
(b) A curve C has parametric representation x sech ,
(i) Show that
dx
d
2
y tanh
...
d
(ii) The arc of the curve between the points where 0 and ln 7 is rotated
through one full turn about the x-axis
...
25
[AEB June 1997]
4
...
2
dy
4 cos 2
...
The area of the surface generated is denoted by S
...
[AEB June 1996]
141
MFP2 Textbook– A-level Further Mathematics – 6360
5
...
(a) Show that dx
dt
2
2
2
dy
t2 1
...
Determine
(i) the length of L,
(ii) the area of the surface generated when L is rotated through 2π radians about
the x-axis
[AEB January 1998]
6
...
dx
(b) A curve has the equation y sinh 1 x x 1 x 2
...
Show that
L 5ln 5 12
...
(a)
2, π
4
2
...
16, 2
...
93
(c) 7
...
71
Exercise 1B
1
...
(a) 2 2 i
(b) 3 cos π i sin π
2
2
(d) 2 cos 5π i sin 5π
6
6
(b) 2 2 3 i
Exercise 1C
1
...
(a) 1 6 8i
5
(b) 2 2i
Exercise 1E
1
...
(a) 6 ,
2
(d)
3 5π
(b) ,
2 6
4π
(c) 9 ,
3
π
(e) 2 ,
9 2
27 , 0
(b) 4 2i
(c) 1 1 i
3
(b) 5, 0
...
(a) 2
...
5i
(c) 7
...
98
Exercise 1G
1
...
39, 0
...
(a)
(b) y
y
(c)
2,1
π
4
3
O
y
O
x
O
x
1,0
x
y
2
...
P
4
...
3 2i
y
2
...
1 2i,
z
(b)
O
z*
4 2i
x
4,3
5 5
(c) 1, 1
...
1 1 i
5
y
5
...
2, 3
arg z 1 π
4
z 2 3i 3
O
2, 0
7
...
(a)
y
(b)(ii)
2 2, 7π
12
4, π
6
(iii) 2 10
x
P2
(b)
y
y
A
(c)(i)
...
(a) 14 2i, 1+i
(c)(ii)
(b)(i)
x
3, 0 1, 0
x
1, 0
3 3 3
i
4
4
20, 0
...
32 (ii)
145
10
MFP2 Textbook– A-level Further Mathematics – 6360
Chapter 2
Exercise 2A
1
...
(a) 1, 1, 3
(c) 2, 2 i
(b) 1, 1 i
Exercise 2C
1
...
2 x3 6 x 2 7 x 10 0
Exercise 2D
1&2 (a) x3 3 x 2 36 x 189 0
3
...
(a) 16 x 4 6 x 2 5 x 4 0
(b) 3
Exercise 2F
1
...
2, –3i
3
...
(a) 1, 5
(b) 2
2
...
(a)
0 3
2
2
(b) 2 x3 3 x 2 8 0
4
...
(a) p 3
(b) (i) q 7
(ii)
2 0
(c) (i) – 3
(ii) 75
6
...
(a) 2r
1
2
...
(b) 1 n n 1 2n 1
6
Miscellaneous exercises 3
2
...
(a) 2
10
...
(a) cos15 i sin15
(b) 1
(f) 1 1 3 i
64
(e) 64
(d) 8i
(c) i
(g) 41472 3
Exercise 4B
1
...
3 tan tan
1 3 tan 2
3
...
(a) 1 z 4 1
2
z4
(b) 1 z 7 1
2
z7
(c) 1 z 6 1
2i
z6
(d) 1 z 3 1
2i
z3
Exercise 4D
1
...
(a) –1
(b) 7
(c) 8
Exercise 4F
1
...
cos 2kπ i sin 2kπ ,
5
5
(b) 2 cos 2kπ i sin 2kπ ,
5
5
k 0 1, 2, 3, 4, 5
k 1, 2
3
...
(a) 2e
(c)
1 6 k 1 πi
2 8 e 24
4 k 1 πi
(e) 2e
6
1 8 k 1 πi
2 6 e 12
k 0, 1, 2, 3
(b)
k 1, 2, 3, , 8
k 1, 2, 3
(d) i
(f) 1 1 i cot kπ
2
5
1 k 0, 1, 2
Miscellaneous exercises 4
1
...
41 0
...
81 1
...
60 1
...
(a)
3
...
(b) 1, 3
5
...
(b)(ii) x 2 x 1 0
7
...
(a)(i) 2i sin
2
(c) centre z 1, radius 1
(b)(i) A 1, B 3
(b) e
kπi
3
(c)(i) coefficients of w6 cancel
k 1, 2
(ii) 1
2
150
(iv) 1 5 ,
4
1
1 5
4
x
MFP2 Textbook– A-level Further Mathematics – 6360
πi
9
...
(a) π
4
(b) π
6
(c) π
6
(d) π
2
(e) π
6
(f) π
Exercise 5C
1
...
(a)
3
x tan 1 x
1 x2
(b)
2e x
(b) e x cos 1 2 x
1 4 x2
(c) 2 x sin 1 2 x 3
3
...
(a)
6x 9x
(c)
2
2 x2
8 12 x 4 x 2
1
3sin 4 3x
x
1 9 x2
3
x3
6x
1 x 1 3x
2
a
1 ax b
2
2
2
1
(b)
2 x tan 1 3x 2 1
1 x
2
2
a
2
1 ax b
151
2
1 4x 2
MFP2 Textbook– A-level Further Mathematics – 6360
Exercise 5D
1
...
π
3
...
2
5
5 2
4
...
π
2
Exercise 5E
1
...
(a) ln x 2 2 x 3 2 tan 1 x 1 + c
2
(b) 1 ln x 2 x 1 1 tan 1 2 x 1 + c
2
3
3
3
...
(a) sin 1 x 1 x 2 + c
(b) 3 3 2 x x 2 sin 1 x 1 + c
2
(c)
1 x x 2 3 sin 1 2 x 1 + c
2
5
152
MFP2 Textbook– A-level Further Mathematics – 6360
Chapter 6
Exercise 6A
1
...
(a) 0
...
86
(c) 0
...
54
Exercise 6C
1
...
00
y
y
(b)
1
O
(c)
(e) 1
...
8cosh 4 x 8cosh 2 x 1
Exercise 6F
1
...
(a) 1 sinh 3 x c
3
(c) 1 x cosh 2 x 1 sinh 2 x c
2
4
(d) x tanh x + c
Exercise 6H
3
...
(a)
3
9 x2
(b)
e
(d) e x sinh 1 x
2
...
(a) sinh 1 x c
3
1 sinh 1 3x c
(d)
3
7
(g) sinh 1 x 2 c
(b) cosh 1 x c
4
1 x 1
(e) sinh
c
2
(h) cosh 1 x 1 c
3
(c) 1 sinh 1 2 x c
2
5
1 x 2
(f) cosh
c
4
(b) 0, ln 7
(c) ln
Exercise 6K
1
...
(a) e x
(b) 1
2
...
(b) ln 2, ln 2 5
4
...
(c) 2
2
3
2
4
3
7
...
(a)
(ii) ln cosh x 1 tanh 2 x + c
2
y
(b)(i) ln 3
two roots
1
O
9
...
(d) 1 ln 3 + c
8
11
...
76
155
(ii) ln 3
4
MFP2 Textbook– A-level Further Mathematics – 6360
Chapter 7
Miscellaneous exercises 7
1
...
(a)(i) 97 ,
72
3
...
(b) k 8π,
5
...
(a) 2
65
72
(b) 6305
5184
20 2π
3
(ii) 576 π
5
156
MFP2 Textbook– A-level Further Mathematics – 6360
Copyright © 2014 AQA and its licensors
...
AQA retains the copyright on all its publications
Title: GCE Further Mathematics (6360) Further Pure Unit 2 (MFP2) Textbook
Description: Design for As/A level student in mathematics and further pure mathematics,also for student in sciences,engineering,economics
Description: Design for As/A level student in mathematics and further pure mathematics,also for student in sciences,engineering,economics