Summary Mathematics A Level - A* student's notebook (pure)

Circles Revision :




notation : (x-h)2 (y k)" + - = rz
,
where (h , k) is the centre Circumcentre : the centre is at the point where all the perpendicular bisectors




or : x2 +
y2 + Dx +
Ey + F = 0
,
where (-) is the centre of a
triangles sides meet



radius
:" -
F

.
7 V


Mixed exercise 6) 2)(x S)" (y -
+ + 2) = 30 *




xQ(11, 12) 7
a)((3 6)



I
,
c (5 -2)
, distance to 10 0) ,


x( xp(13, 6) b)
2 (11 3)) = 10

sa
+ -




radius (2 d) + + =
Veg
-
P(5 ,

2)(x 3) (y 6)" 102
- + -
=




d) (13 -

3) + (6 6)2
- = 102 radiusis50origineSee
10, 12) A
2
102 = 18

. P lies on the circle I

3) x2 + 3x +
yz + by = 3x -




2y -
7 16) (x b) (y S)"
-
+ - = 17 -
<
B1z
-

1
6)(x k)) 3k)2 (6 5)
a) x2 y2 + Sy (y lines
·



+ + 0x 7 0 + 13 mx 12
y
+ =
= : = + ,

-1
- - -




D
centre :
(-) =
(0 4)
,
-




passes
(3 0) , (x -

6)2 + ( -

mx + 12
-


S)2 =
17



radius :
(2 -7 a)(3 k)2 (3k)2 - + = 13 x2 -
12x + 36 + m2x2 -

14mx + 49 -
17 =
0




=
Vg 9 6k + k + 9k2
-
= 13
X m 2x2 + -
12x -
14mx + 68 = 0 18) (x 3)3 +-



(y 3)2
+ =
52


b) = 3
10k2 6k 4 =
0
14m)2
-




( (1 + m2)
3x
-




12 4 68)0 + c
x2 y2 + Sy y
- .
=
7
-




0
.
-



+ + =



(k 1)(5k 2)
- + = 0


When x = 0
, k= 1 k =
-




2 196m2 + 336m + 144 -

4/68 68m2))o +
(x 3)2 - +
(3x + c +
3) = 52


y2 8y + +7 = 0


(X 1) + (y 3)2 13 196m >
(y z)(y 1) b)k 1 272m2 336m + 144 272)0
=
=
- -
+ -

0
-




+ + =
,




*
7 = -
y =
y
-




,
-
76m2 +
336m 12870
(0 1)
-




10, 7) &
-
-



,




8) y
= 2x -
8 -

19m2 + 84 -

3270
+3x2 + 3x + 3xc -
34 + c + 6 =
0

c) y = 0 at Xaxis

10 -8)
x 7
,

>
-
X
m = 4 m =
f discriminant) (3 3c)2 4 .3
+ - .
( -

34 + ch + 6s)
+ = 0
14 , 0)
x2 1 4x + 12 9 + 18c + 9c2 + 442 13c2 78278
7 y =
- -
-



,
-
=




c(2 4) X[

-
-




x +1
has real solutions,
therefore (2y 42 60c 45170
=
, +
no -
-




doesn't cross X axis

=
X
c =

+ ,
s




7)3x -




y
-
9 = 0
y
= 3x -

9 radius =
Nuz)J +
2
17) AB gradient :
E = -3
It
1:
y
=

=x -
Fo
gradl , = -
k(z y :
=

Ex 1
m(4 , 4)
+
x2 +
px +
yz +
4y = 20 =




y =
=x +c

(3 3)v -




(x 2) +
,

x + px (3x g)2 4(3x 9) 4)
20
(y (: [x 8
+ + =
+ 20
-

=
-




4
+
y
=
-




4 = +




Xi
x2 + px + 9x2 - S4x + 8) + 12x -
36 =
20

33
s =




10x2 + px 42x + 25
c( -
2
, 3) (( -
2
,
2)
A(3 7)
= 0
-




,
1,
Uso
·

radius
don't intersect , so discriminant /O KAl = KCBI =




(p 42)2 - -
4 .

10 .
250
Mo(4, 4)
L
E (3 + 10 -

14 -
35 + 2 -


6
So
B(5, 1) Fy
(3 2) + + =
V+y)2 5) a re a ABC
(p 42)2 -
-




1000/0
: ·




2 ?
20 units
-




y 14y
=



p2 84p + 1764 -
10000
-
+ 49 + 25 =
49 y 2y
+ - + 1




p2 -


84p + 764/ 24 =

12g equation :
(x 2)2
+ +
(y z)2 -
=
50


p =
0 at 42+10 To & 42-10 To y
= 2



: 42-10 To < p(42 + 10 To

, Vectors
Scaler /how many times we
Starting
wanti s




Er
vector
position

Challenge :
: ↳


y
= 5 -




5x =
pi + qj + rk



1B) = xi + y = = i =

(g) i (g)m ( %) =
=




(8, 5)
B(x, y) S
Ex
y
-

=




(Yc Y)" (My K )
2

distance of (x )
25x 50x
z
+ S vector -X + +

y
-
= = -




. .
,

(3, 0)

from the origin x
y2 + z
-x" -
= +
=




e
.
g
A (S 11 8) B(- 3 1 6) to find angle to axis :

99
an


=
.



, ,
X = , ,



34x 5
ou X

0 xcoordinate
-

=
-




=
B(,) or

le #S
3) (n 1)
+ 6) + =
+ (8 -
Cos
-
length
Vestor
of



B =


Zi E +
(os0
=
distance = 15



-99 (osc
=
A :
Zi + 3j 4k
Shapes
-




:



AB zi
Sj 7k
trapezium
=

parallel
-
+
: 1 pair side
Y

0 B :
Hi -

2j + 3k Rhombus : 4S a m e sides

P) -

6 3 2) Q(4 -2 0
Kite : 2 pairs sides , , ,
of same
,



S 2 Parallelogram 2 pairs same sides
5j r
·
6 Gi
:
-


: +
=
10i -




Sj
-
2k
T >
-




PQ 3i + 2k

=
-




=
j
Ng
-




·
B R >
-




&R Zi 4j + 3k
555
17)
=
+
-




X

>

Loso
-




PR = i +
3j + Sk

- Vis Eg
>
-




AB P


Z B =
256 Mu -
X
35 =
29 + 14- 229 My .
sos x 82 = 100 .
10


F = Vu3 &
X = 78 549 . . . .




= 78 .
50 (1d p ) .
.




Ex 12B.




7)a = ti +
2j + 3k

1) Hith
AB = -

Gi +
4j + 3k (s) a =

(t)b (4)c = =

(2)
(a) = 7 49 = 3 + 22 + 33 IABI =
To | a) (b) v74 k1 V3
5)
=
=
=

:




000x 36 3
= 0 98 69 600x
=
=
t 16
.
.
=
= .




-




= Oy 62 3 a by 90
=
= .




13)
A() B((8) (2) , ,
8 =
35 .

5002 = 1460




14)hi
a) AB =
4j -
k c) Scalene
B = bi -



3j + 21 1s)a(j) B(j)c(z) B
17 = 14 + 9 -
2 14 9
. .

CosA

= = 4i
j k
X =
74 498 ...
3
AB(g)
.



+ -




A
- = Hi
/Bl = 7 (ABl = 3
a T
3j
-




~
11 75 7
E() IAG) V7
4S =
7
=


T7
lBi N7
IB1
=
S
7 45
S
&
B(z)
=

1B3 vin
I VS
=




E
=

B 7
c ·
.
3 Mu Sin88 635.


.




1521 =
5 = S 4083
. ...




?
=> S 41. Units

, 53 = 9) 25 is irrational
Sectors
3 = negation : 23 is rational, so can be written in the form
vector
a


3b2 = a & which is in its simplest form .
unit Vector : =
a -
magnitude
So at must be a multiple of 3, so a is a
multiple q =2
of 3 .
Ista =
3c
,
a2 = gc2 a
3 = 2 AS :
Mixed exercise 11


9c = 3b2 a3 = 263 5) Sa + kb & Sa + 26 are parallel find .
K




(x5a) =
3c2 = b2
a multiple of 2 AB
If they're multiples of
both is 3i
So b must be a multiple of 3 .
3
,
a ,
= +
5j
B T3
then fraction is not in its simplest form ,
so i s irrational Sa a must be eve n be 8 = x5 7) C
At = Gi +
3j
54
X =

3 -
if a = 2k a3 Sk = = -
AB Ac +

L Ms
,


xk = 2

8) there integer solution to
x2-y2 8k3 2b3 A + 6i +

8k
L 3i 5j
is no so
3j
= = = - -




=
2
4k3 = b3 = 3i -




2j
Negation ,
there is a Solution to x2 gz = .
2
k =

5
sob must be eve n .
If both a & b a re even, -

x2 534 . 5
-




yz = 2 the fraction is not in its
simplest form, BAC = 13 =
45 + 34 -
2 . ·


Cos A

So contradiction .E is irrational .

(x + y)(x y) (35 5)
°
- = 2
integer solutions :
=
32 5 .
.




A
(1 2)
10)
,



x+
y
= 2 (- , 2)- n =
5 11) (3i kj) + = 355
12) a
x -


y = 1
Mo N
1 5k 355
-




t ax
X+ I x 2 n
y
-

y
+ =
=
-
+
=


-



1 y
=

=
X
(2 y) y
-
=



a
2 1
-




X x
y
= =

y
- -
- -




2 2y
1
q 1
= 3 = 32 k
-
+
=

45
-




+ =

1
x +
y = 1
(x + 1) 2X = - +
y
X =
-

-




2 1 +
x =
y +
y 2
k
-

=
y
-
-



+ -
2
a)a)b =
36
B
1+ 2 *
1 2 2y =
- -




y
=
y
354
-
-




k =
-
1 -




zy = 2
y
=
- X
zy = 1
b) no least the rational number

y
=
zX


·
-




there are no solutions to the equation ,
so there
negation : there is a least the rational number .

is a contradiction

n
= where
they are integers


5) prove that IV is irrational m
= so min contradicts that there
N
,

: a + Xb

negation : a ss u m e 3 is rational z where is a least the rational number
o = b +



the
fraction is in its simplest terms



22 8b2 =
442 13) F
,
= (4i -



5j)N >
R 3


t -




(2) =
F2
(pi qj)
,

Fc = +
2
So b'must must be a multiple of 2
, JF
,

multiples of the
if they're both 2

3 9)
,
18 4" below
1/3-j) O =

Resultant
.




fraction is not in its simplest form , So
:

a)
Gi +
pi-Sj-q) =
Birt; horizontal
8b2 a3
contradiction (2)" is irrational
Gi 3j
=
R = -




6) 2+ 3g /l

+
X3 =



t = 0
2
X
=
=




(4) IRI a
xj q 3 =



at (3i + 4j) 2 To
-




So is a multiple of .
8 u =
=




t = 2
v =
(1Si3j)(a) =f) 4 2
+ = x3


if a t must be be S + 3 x 4 + 3xx
1 +
-
is even p = =
-


even a =
,

15 4
+
3q =
p
- -
-




=
let 2k
(2k) :it
a = a
ms
=
,

a 11
3q
2
5 p
-



a X + =
4k q
1-
-



+ =
-




S +
-

=

q
- =



p+ 3q = 1

, Proof by contradiction :
If a number is rational,


Assume opposite is true it be written in the
form -
·


can



Maths & b with
where
integers
·

a a re no


·

Find the part that doesn't make sense common factors
·

Statement 'hence it cannot be true'

You cannot represent an irrational number


Practice :




a + 4b = 4/a5

&b 2
g
Prove that & is irrational
assume are positive integers
.
.


....




=> assume it is rational where &b
a = 2n + 1 ,
a no common factors .




Nanb E
q
2n + 1 + 4b = 4 =
a = 2k


(a 4b)2 + = 16ab 2n + 1 + 4b = 4bn
+b
(2k) = 2b2

En y + + b = Xb
b + 2 = so be is even so
,
is b


a + Sab + 16b2 = 16ab both &
a b are even


al - Sab + 16b2 = 0 a
2 = 2b2 So have common factor

(a 4b) -
=
0 ↳ So a is even
,
so a is even
↳
against contradiction
a = 4b


a = 2(zb)
hence a is eve n
,
& there's a contradiction




Prove
by contradiction log 27 is irrational .




=> ass u me logz7 is rational ,
a&b no common
factors
=
logz7 b




blogz7 =
a



So cannot be true be LHS
always even


logz76 = a
& RHS is
always even.




= this cannot unless a = b = 0



log2" log73 -
=
0
-


both prime numbers (2&7)