12 october, 2014 st mungo's academy 1 advanced higher maths revision and formulae unit 1
TRANSCRIPT
11 April 2023 St Mungo's Academy 1
ADVANCED HIGHER MATHS
REVISION AND FORMULAE
UNIT 1
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Unit 1 Outcome 1 BINOMIAL and PARTIAL FRACTIONS
Expand (a − b)5 a5 − 5a4b + 10a3b² − 10a²b3 + 5ab4 − b5.
Expand (2x − 1)3 8x3 − 12x² + 6x − 1
rr
r r
4(2 5 )x yExpand 4 3 2 2
3 4
16 160 600
1000 625
x x y x y
xy y
5(3 2 )a bExpand 5 4 3 2
2 3 4 5
243 810 1080
720 240 32
a a b a b
a b ab b
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Unit 1 Outcome 1 BINOMIAL and PARTIAL FRACTIONS
61( )x
x
6 6 6 6 6 6 6
0 1 2 3 4 5 6
2 36 5 4 3
4 5 62
1 1 16 15 20
1 1 115 6
x x x xx x x
x xx x x
6 4 22 4 6
15 6 16 15 20x x x
x x x
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Unit 1 Outcome 1 BINOMIAL and PARTIAL FRACTIONS
12x
45x
3 2
2
4 9 2 62
x x xx x
4 1x 2
4 62
xx x
3 74 1
2x
x x
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f(x) functionc
x
2x
xn
(ax + b)n
sin x
cos x
sin(ax + b)
cos(ax + b)
0
f’(x) derivative
1
2
nxn−1
an(ax+b)n−1
cos x
−sin x
acos(ax + b)
-asin(ax + b)
These are your standard derivatives
for now!
All of these were covered
in the Higher course
Unit 1 Outcome 2 DIFFERENTIATION
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New Trigonometric Functions
AND
Unit 1 Outcome 2 DIFFERENTIATION
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Unit 1 Outcome 2 DIFFERENTIATION
Product Rule
Quotient Rule
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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
f '(x) = sec 2 x
The six basic trigonometric derivatives
Unit 1 Outcome 2 DIFFERENTIATION
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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 x6 - Derivative of cosec x.
The derivative of f(x) = cosec x is given by
f '(x) = - cosec x cot x
The six basic trigonometric derivatives
Unit 1 Outcome 2 DIFFERENTIATION
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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/dt2
These are used in the example over the page
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Find velocity and acceleration after (a) t secs and (b) 4 secs forparticles travelling along a straight
line if:
(i) x = 2t3 – t2 +2
(ii) x = t2 +8/t
(iii) x = 8t + et
88 m/s
6t2 – 2t 12t – 2
46 m/s2
2t – 8/t2
7.5 m/s
2+ 16/t3
2.25 m/s2
e4 m/s2
et 8 + et
8+ e4 m/s
63 m/s 55 m/s2
Unit 1 Outcome 2 DIFFERENTIATION
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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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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
Example 1
Unit 1 Outcome 3 INTEGRATION
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Find the volume of the solid obtained by the region bounded by y=x3, y=8, and x=0 around the y-axis.
o
8
x
y
3 yx
Example of rotating the region about y-axisExample 2
28
3
0
V y dy
85
3
0
3
5
yV
5
33(8)0
5V
96
5V
Unit 1 Outcome 3 INTEGRATION
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Sketch the graph of 2
32
xx
xy
.[You need not find the coordinates of any stationary points.]
0x2
3
2
3
y
2
3,0
Solutiony-axis: When
The curve cuts the y-axis at
02
32
xx
x03 x 3x
0yx-axis: When
,
The curve cuts the x-axis at (3, 0).
Unit 1 Outcome 4 PROPERTIES OF FUNCTIONS
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This means that is a non-vertical asymptote.
022 xx0)1)(2( xxVertical Asymptotes:
2x 1x or
x –21 –2 –19 09 1 11
y
2
32
xx
xy
Non-Vertical Asymptote:
x 0yAs
,
(since the degree of the denominator is higher than the degree of the numerator).
0y
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
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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f(-x)
Unit 1 Outcome 4 PROPERTIES OF FUNCTIONS
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Unit 1 Outcome 4 PROPERTIES OF FUNCTIONS
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Past Paper 2002 Q1 5 marks
Use Gaussian elimination to solve the following system of equations
x + y + 3z = 22x + y + z = 23x + 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 1 4 1
R3 –R2
1 1 3 2
0 1 5 2
0 0 1 1
z = 1
-y –5z = -2
-y = -2 +5
y = -3
x - 3 + 3 = 2
x = 2
(x,y,z) = (2, -3, 1)
Unit 1 Outcome 5 GAUSSIAN ELIMINATION
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Past Paper 2003 Q6 6 marksUse elementary row operations to reduce the following system of Equations to upper triangle form
x + y + 3z = 13x + ay + z = 1x + y + z = -1
1 1 3 1
3 1 1
1 1 1 1
a
R2 –3R1
R3 -R1
1 1 3 1
0 3 8 2
0 0 2 2
a
R3/-2
z = 1
(a-3)y –8 = -2
(a-3)y = 6
y = 6/(a-3)
x + 6/( a –3) + 3 = 1
x = -2 –6/(a-3)
a=3 gives z = ¼ from R2 and z = 1 from R3. Inconsistent equations!
1 1 3 1
0 3 8 2
0 0 1 1
a
Hence express x, y and z in terms of parameter a.Explain what happens when a = 3
Unit 1 Outcome 5 GAUSSIAN ELIMINATION
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Past Paper 2005 Q6 6 marks
x + y + 2z = 12x + y + z = 03x + 3y + 9z = 5
1 1 2 1
2 1 0
3 3 9 5
R2 –2R1
R3 -3R1
1 1 2 1
0 2 3 2
0 0 3 2
R3/3
z = 2/3
(-2)y –2 = -2
y = 0
x = 1 –4/3
x = -1/3
=2 gives R2 = R3. Infinite number of solutions!
Explain what happens when = 2
Use Gaussian elimination to solve the system of equations below when . 2
Unit 1 Outcome 5 GAUSSIAN ELIMINATION
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Past Paper 2006 Q9 5 marks
2x - y + 2z = 1x + y - 2z = 2x - 2y + 4z = -1
1 1 2 2
1 2 4 1
2 1 2 1
R2 –R1
R3 -2R1
1 1 2 2
0 3 6 3
0 3 6 3
R3-R2
z = t
-3y +6t = -3
y = 2t+1
x +2t+1 –2t = 2
x = 1
x,y,z:x=1, y=2t+1 and z = t)
Use Gaussian elimination to obtain solutions of the equations
1 1 2 2
0 3 6 3
0 0 0 0
Unit 1 Outcome 5 GAUSSIAN ELIMINATION