ADVANCED HIGHER MATHS
UNIT
1
REVISION AND FORMULAE
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Unit 1 Outcome 1 BINOMIAL and PARTIAL FRACTIONS
r
r
r
r
Expand (a − b)5
a5 − 5a4b + 10a3b² − 10a²b3 + 5ab4 − b5.
Expand (2x − 1)3
8x3 − 12x² + 6x − 1
Expand
(2 x  5 y)
4
 16 x  160 x y  600 x y
4
3
2
2
 1000 xy 3  625 y 4
Expand
(3a  2b)
5
 243a  810a b  1080a b
5
4
3 2
 720a b  240ab  32b
2 3
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5
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Unit 1 Outcome 1 BINOMIAL and PARTIAL FRACTIONS
 6 6 6 6 6 6 6
       
 0  1  2  3  4  5  6
1 6
(x  )
x
2
1
4 1
31
 x  6 x    15 x    20 x  
 x
 x
x
6
3
5
4
5
1
1 1
 15 x    6 x     
 x
 x  x
6
2
15 6 1
 x  6 x  15 x  20  2  4  6
x
x
x
6
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Unit 1 Outcome 1 BINOMIAL and PARTIAL FRACTIONS
1
5x  3
4

.
2
x  3x  10 x  5 x  2
4x  9x  2x  6
x 2  2x
3
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 4x
4x  6
 1 x 2  2x
3
7
 4x  1  
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x x 2
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Unit 1 Outcome 2 DIFFERENTIATION
f(x) function
c
x
2x
xn
(ax +
b)n
f’(x) derivative
0
1
2
nxn−1
an(ax+b)n−1
sin x
cos x
cos x
−sin x
sin(ax + b)
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cos(ax + b)
These are your
standard derivatives
for now!
All of these
were covered
in the
Higher course
acos(ax + b)
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-asin(ax
+ b)
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Unit 1 Outcome 2 DIFFERENTIATION
New Trigonometric Functions
AND
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Unit 1 Outcome 2 DIFFERENTIATION
Product Rule
Quotient Rule
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Unit 1 Outcome 2 DIFFERENTIATION
The six basic trigonometric derivatives
1 - Derivative of sin x.
The derivative of f(x) = sin x is given by
f '(x) = cos x
2 - Derivative of cos x.
The derivative of f(x) = cos x is given by
f '(x) = - sin x
3 - Derivative of tan x.
The derivative of f(x) = tan x is given by
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f '(x) = sec 2 x
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Unit 1 Outcome 2 DIFFERENTIATION
The six basic trigonometric derivatives
4 - Derivative of cot x.
The derivative of f(x) = cot x is given by
f '(x) = - cosec 2 x
5 - Derivative of sec x.
The derivative of f(x) = sec x is given by
f '(x) = sec x tan x
6 - Derivative of cosec x.
The derivative of f(x) = cosec x is given by
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f '(x) = -Stcosec
x cot x
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Unit 1 Outcome 2 DIFFERENTIATION
Higher derivatives
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Unit 1 Outcome 2 DIFFERENTIATION
Motion
v=
dx/
dt
a=
dv/
dt =
d2x/ 2
dt
These are used in the example over the page
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Unit 1 Outcome 2 DIFFERENTIATION
Find velocity and acceleration after
(a) t secs and (b) 4 secs for
particles travelling along a straight
line if:
(i)
x = 2t3 – t2 +2
(ii) x = t2 +8/t
(iii) x = 8t +
6t2 – 2t
12t – 2
88 m/s
46 m/s2
2t – 8/t2
2+ 16/t3
7.5 m/s 2.25 m/s2
8+
et
8+
e4
et
et
m/s
e4
m/s2
55
m/s2
63 m/s
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Unit 1 Outcome 3 INTEGRATION
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Unit 1 Outcome 3 INTEGRATION
+
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Unit 1 Outcome 3 INTEGRATION
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Unit 1 Outcome 3 INTEGRATION
Example 1
Example of rotating the region about x-axis
In the picture shown, a solid is formed by
revolving the curve y = x about the x-axis,
between x = 0 and x = 3.
FIND THE VOLUME
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Unit 1 Outcome 3 INTEGRATION
Example of rotating the region about y-axis
Example 2
Find the volume of the solid obtained by the region
bounded by y=x3, y=8, and x=0 around the y-axis.
y
8
2
3
V    y dy
8
0
x3 y
o
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

3
3(8)

V 
 0
 5




3y

V 
 5

5
3
8



 0
96
V 
5
x
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Unit 1 Outcome 4 PROPERTIES OF FUNCTIONS
Sketch the graph of y 
x3
x2  x  2
.
[You need not find the coordinates of any stationary points.]
Solution
y-axis:
When
3 3
y

2 2
x0
,
The curve cuts the y-axis at
 3
 0, 
 2
x-axis:
When y  0
x3
0
2
x  x2
x 3  0
x3
The curve cuts the x-axis at (3, 0).
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Unit 1 Outcome 4 PROPERTIES OF FUNCTIONS
Vertical Asymptotes:
x  x20
2


x
–21
y

–2
,
Non-Vertical Asymptote:
As x  
y0
This means that
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( x  2)( x  1)  0
x  2
or
x 1
–19
09


y
1
11

x3
x2  x  2
(since the degree of the denominator is higher
than the degree of the numerator).
y  0 is a non-vertical asymptote.
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Unit 1 Outcome 4 PROPERTIES OF FUNCTIONS
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Unit 1 Outcome 4 PROPERTIES OF FUNCTIONS
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Unit 1 Outcome 4 PROPERTIES OF FUNCTIONS
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Unit 1 Outcome 4 PROPERTIES OF FUNCTIONS
Non-Vertical Asymptote
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Unit 1 Outcome 4 PROPERTIES OF FUNCTIONS
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Unit 1 Outcome 4 PROPERTIES OF FUNCTIONS
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Unit 1 Outcome 4 PROPERTIES OF FUNCTIONS
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Unit 1 Outcome 4 PROPERTIES OF FUNCTIONS
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Unit 1 Outcome 4 PROPERTIES OF FUNCTIONS
f(-x)
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Unit 1 Outcome 4 PROPERTIES OF FUNCTIONS
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Unit 1 Outcome 5 GAUSSIAN ELIMINATION
Past Paper 2002 Q1 5 marks
Use Gaussian elimination to solve the following system of equations
x + y + 3z = 2
2x + y + z = 2
3x + 2y + 5z = 5
1 1 3 2 
2 1 1 2


 3 2 5 5 
R2 –2R1
R3 -3R1
1 1 3 2 
0 1 5 2


0 0 1 1 
1 1 3 2 
0 1 5 2


0 1 4 1 R3 –R2
z=1
y = -3
-y –5z = -2
x - 3 +3= 2
-y = -2 +5
x = 2
(x,y,z) = (2, -3, 1)
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Unit 1 Outcome 5 GAUSSIAN ELIMINATION
Past Paper 2003
Q6 6 marks
Use elementary row operations to reduce the following system of
Equations to upper triangle form
x + y + 3z = 1
Hence express x, y and z in terms of
3x + ay + z = 1
parameter a.
x + y + z = -1
Explain what happens when a = 3
1
0

0
1 1 3 1 
1
3 1
1
3 a 1 1  R2 –3R1
0 a  3 8 2




R3
-R1
1 1 1 1
0
0
2 2 R3/-2
z=1
y = 6/(a-3)
1
3 1
x + 6/( a –3) + 3 = 1
a  3 8 2  (a-3)y –8 = -2
x = -2 –6/(a-3)
0
1 1  (a-3)y = 6
a=3 gives z = ¼ from R2 and z = 1 from R3.
Inconsistent
equations!
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Unit 1 Outcome 5 GAUSSIAN ELIMINATION
Past Paper 2005
Q6 6 marks
Use Gaussian elimination to solve the system of equations below when
.
l2
x + y + 2z = 1
2x + ly + z = 0
3x + 3y + 9z = 5
Explain what happens when l = 2
1 1 2 1 
 2 l 1 0 R2 –2R1


 3 3 9 5 R3 -3R1
1
2 1
1
0 l  2 3 2 


0
0
3 2  R3/3
z = 2/3
(l-2)y –2 = -2
y=0
x = 1 –4/3
x = -1/3
l=2 gives R2 = R3. Infinite number of solutions!
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Unit 1 Outcome 5 GAUSSIAN ELIMINATION
Past Paper 2006
Q9 5 marks
Use Gaussian elimination to obtain solutions of the equations
2x - y + 2z = 1
x + y - 2z = 2
x - 2y + 4z = -1
1 1 2 2 
1 2 4 1 R2 –R1


 2 1 2 1  R3 -2R1
1 1 2 2 
0 3 6 3


0 0 0 0 
1 1 2 2 
0 3 6 3


0 3 6 3
z=t
-3y +6t = -3
y = 2t+1
R3-R2
x +2t+1 –2t = 2
x = 1
(x,y,z:x=1, y=2t+1 and z = t)
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