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CORE 2
1
Summary Notes
Indices
· Rules of Indices
x
a
´ x
x
a
¸ x
a b
(x )
b
= x
a+b
b
= x
a–b
= x
Solve 3
x
Solve 9 =
· x
x
0
= 1
3
–a
=
1
x
·
1
n
2x
= 3
2x =
· x
2
3 ´ (3x)
Let y = 3
–3
x =
–
n
x
+ 3 = 0
x
– 10 ´ 3 + 3 = 0
x
2
3y – 10y + 3 = 0
– 3
a
– 10 ´ 3
2
1
27
· x = x
m
n
2x + 1
ab
(3y – 1)(y – 3) = 0
3
y =
2
1
x
gives 3 =
3
1
x= – 1
3
x
y = 3 gives 3 = 3
x= 1
= ( n x )m = n x m
Graphs and Transformations
ASYMPTOTES
Straight lines that are approached by a graph which never actually meets them.
Vertical Asymptote
x=2
y
1
y=
+ 1
x– 2
5
4
3
2
1
– 5– 4– 3– 2– 1
– 1
1
2
3
4
5
x
– 2
Horizontal Asymptote
y=1
– 3
– 4
– 5
éaù
TRANSLATION - to find the equation of a graph after a translation of ê ú you replace x
ëbû
by (x-a) and y by (y - b)
é3 ù
e.g. The graph of y = x2 -1 is translated through ê ú . Write down the equation of
ë - 2û
the graph formed.
y - b = f(x-a)
(y + 2) = (x-3)2 – 1
y
y = x2 -1
2
5
y = x – 6x + 6
or
4
y = x2 – 6x + 6
3
2
y = f(x-a) +b
1
– 5– 4– 3– 2– 1
– 1
– 2
– 3
1
– 4
– 5
1
2
3
4
5
x
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y = -f(x)
REFLECTING
Reflection in the x-axis, replace y with –y
Reflection in the y-axis, replace x with – x
y = f(-x)
STRETCHING
1
x
k
1
Stretch of factor k in the y direction replace y by
y
k
Stretch of factor k in the x direction replace x by
1
y = f( x )
k
y = kf(x)
2
e.g. Describe a stretch that will transform y = x + x -1 to the
graph of y = 4x2 + 2x -1
4x2 + 2x -1 = (2x)2 + (2x) -1
So x has been replaced by 2x.
Stretch of scale factor ½ in the x direction
3
Sequences and Series 1
· A sequence can be defined by the nth term such as un = n2 +1
u1 = 12 +1
u2 = 22 +1
u3 = 32 +1
u4 = 42 +1
=2
=5
=10
=17
· An INDUCTIVE definition defines a sequence by giving the first term and a rule to find
the next terms.
u2 =7
u3 =15
Un+1 = 2un + 1 u1 =3
· Some sequences get closer and closer to a value called the LIMIT – these are known
as CONVERGING sequences
e.g The sequence defined by Un+1 = 0.2un + 2
u1 =3 converges to a limit l
Find the value of l.
l must satisfy the equation I = 0.2l + 2
0.8l = 2
l = 2 ÷ 0.8
= 2.5
ARITHMETIC SEQUENCE
Each term is found by adding a fixed number (COMMON DIFFERENCE) to the
previous one.
u 2 = u 1+ d
u3 = u2 + d
u1 = a
a is the first term ,d is the common difference the sequence is
a, a + d , a + 2d, a + 3d,……….
Un = a + (n - 1)d
SUM of the first n terms of an AP (Arithmetic progression)
1
1
Sn = n ( 2a + (n-1)d ) or
n(a + l) where l is the last term
2
2
1
· The sum of the first n positive integers is n(n + 1)
2
5
· SIGMA notation
3
Si
3
3
3
3
3
= 0 + 1 +2 + 3 +4 +5
i=0
2
3
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20
Show that
S (3n – 1)
= 610
r=1
= 2 + 5 + 8 + .········59
a = 2
d = 3
20
(2 ´ 2 + (20 – 1) ´ 3)
2
S20 =
= 10 ´ 61
= 610
4
Geometric Sequences
· A geometric sequence is one where each term is found by MULTIPLYING the
previous term by a fixed number (COMMON RATIO)
· The nth term of a geometric sequence a, ar, ar2, ar3, …is arn-1
a is the first term
r is the common ratio
· The sum of a geometric sequence a + ar + ar2 + ar3 ….+ arn-1 is a geometric
a(r n - 1)
series S n =
r -1
a
· If -1 < r < 1 then the sum to infinity is
1- r
Find the sum to inifinty of the series 1 +
2
4
8
+
+
3
9
27
Geometric Series first term = 1 common ratio =
Sum to infinity =
5
1
2
13
2
3
=3
Binomial Expansion
· The number of ways of arranging n objects of which r are one type and (n-r) are
ænö
n!
where n! = n(n-1)(n-2)….x 3 x 2 x 1
another is given by çç ÷÷ =
è r ø r!(n - r)!
n
·
(1 + x)
n
(a + b)
æ n ö
æ n ö 2
æ n ö 3
n
= 1 + ç ÷ x + ç ÷ x + ç ÷ x + .········.x
è 1 ø
è 2 ø
è 3 ø
= a
n
+
æ n ö n–1 1
æ n ö n–2 2
æ n ö 1 n–1
n
b + ç ÷ a
b + .··········. ç
+ b
ç ÷ a
÷ ab
è 1 ø
è 2 ø
è n– 1 ø
Find the coefficient of x3 in the expansion of (2+3x)9
Note 3 + 6 = 9 (n)
n=9 , r=3
æ9ö
çç ÷÷(3x)3 (2)6
è3ø
= 84 × 27 x3 × 64 = 145152 x3
3
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6
Trigonometry
· Exact Values – LEARN
30
45
sin 45 =
√2
1
sin 30 =
1
60
1
2
2
2
cos 30 =
1
cos 45 =
2
45
1
tan 30 =
tan 45 = 1
3
2
2
cos 60 =
√3
60
1
1
3
3
2
2
COSINE RULE
2
2
2
a = b + c – 2bc cos A
2
sin 60 =
30
1
tan 60 =
1
2
3
B
2
b = a + c – 2ac cos B
a
c
2
2
2
c = a + b – 2ab cos A
C
A
b
SINE RULE
a
b
c
=
=
sinA sinB sinC
AREA of a TRIANGLE
1
1
1
ab sinC = bc sinA = ac sinB
2
2
2
e.g. Find the area of triangle PQR
Finding angle P
R
8
9c
m
8
2
2
= 9 + 12
cm
2
cos P =
2
– 2 ´ 9 ´ 12 ´ cos P
2
9 + 12 – 8
2
= 0·745
2 ´ 9 ´ 12
P
Q
12 cm
Area =
P = 41·8·····.
1
´ 9 ´ 12 ´ sin 41·8·····.
2
2
= 35·99991 36 cm =
= 36 cm
2
RADIANS and ARCS
360 º = 2π radians
Try to avoid rounding until you reach your final
answer
180 = p radians
45 =
p
4
4
90 =
radians
p
radians 60 =
2
p
3
30 =
p
6
radians
radians
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LENGTH OF AN ARC l = rθ
l
r
AREA of SECTOR = ½ r2θ
r
7
Trigonometry
y
1
y = sin θ
– 270
– 180
– 90
90
180
360 x
270
– 1
sin (­θ) = ­ sin θ y = cos θ
sin (180 – θ) = sin θ
sin (π – θ) = sin θ (in radians)
y
1
– 270
– 180
– 90
90
180
360 x
270
– 1
cos (­θ) = cos θ cos (180 – θ) = ­ cos θ
cos ( π – θ) = ­ cos θ (in radians)
y = tan θ
y
10
tan x =
sinx
cos x
5
– 270
– 180
– 90
90
180
270
– 5
– 10
Tan (­θ) = ­ tan θ
2
2
sin x + cosx = 1
LEARN THIS IDENTITY
SOLVING EQUATIONS
e.g. Solve the equation 2sin2 x = 3 cos x for 0 < x < 2 π
2
2 sin x = 3 cosx
2
2(1 – cos x) = 3 cos x
2
2cos x + 3cosx – 2 = 0
(2cos x – 1)(cos x + 2) = 0
cos x = -2 has no solution
5
360
x
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cos x = -½
x=
p
3
or 2p -
p 5p
=
3
3
PROVING IDENTITIES
2
Show that (sin x + cos x ) = 1 + 2sinx cos x
2
2
LHS: (sin x + cos x)
2
= (sin x + 2 sinx cos x + cos x)
2
2
= sin x + cos x + 2 sinx cos x
= 1 + 2 sin x cos x
LHS = RHS
TRANSFORMATIONS
Stretch in the y direction scale factor a y = a sin θ
y = sin x
y = 4 sin x
y
5
4
3
2
1
– 180
– 90
–
–
–
–
90
1
2
3
4
180
Stretch in the θ direction scale factor
270
1
b
x
360
y = cos bθ
y = cos θ y = cos 2θ
y
2
1
– 180
– 90
90
180
270
360
– 1
é- c ù
Translation ê ú
ë0 û
y = sin (θ + c)
y = sin x
y = sin (x+30)
2
y
1
– 180
– 90
90
– 1
– 2
6
180
A
270
360
x
x
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é0ù
Translation ê ú y = cosθ + d
ëdû
y = cos θ – 180
8
– 90
y
A
5
4
3
2
1
y = cosθ +2
90
– 1
180
270
x
360
Exponentials and Logarithms
· A function of the form y = ax is an exponential function
· The graph of y = ax is positive for all values
of x and passes through the point (0,1)
y = 3
–x
y = 4
x
y = 2
x
y
16
14
12
· A Logarithm is the inverse of an
exponential function
10
8
6
4
y = a
x
Þ x = loga y
2
– 3
e.g log 32 = 5 because 25 = 32
2
LEARN THE FOLLOWING
loga a = 1
loga a
x
loga 1 = 0
= x
a
log x
= x
Laws of Logarithms
loga m + loga n = loga mn
æ m ö
loga m – loga n = loga ç
÷
è n ø
kloga m = loga m
E.g
k
Solve logx 4 + logx 16 = 3
logx 64 = 3
x
3
= 64
x = 4
7
– 2
– 1
1
2
3
x
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ax =b can be solved by taking logs of both sides
· An equation of the form
9
Differentiation and Integration
dy
n
n–1
= nx
· If y = x then
dx
n+1
x
n
+ c for all values of n except n = – 1
x dx =
·
n + 1
ò
e.g
ò
1 + x
x
2
=
ò
x
1
–2
3
1
+ x2 = 2x2 +
5
2 2
x + c
5
TRAPEZIUM RULE
The trapezium rule gives an approximation of the area under a graph.
If the gap between the ordinates is h then
Area = ½ h (end ordinates + twice ‘sum of interior ordinates’)
Use the trapezium rule 4 strips to estimate the area under the graph of y = 1 + x
2
from x = 0 to x = 2
x
0
0.5 1 1.5 y
y
1
√1.25
√2
√3.25
2
1
Area =
½ x 0.5 ( 1 + √3.25 + 2(√1.25 + √2))
= 1.966917
= 1.97 (3sf)
– 1
1
– 1
8
2
3
x