Difficult Questions Explained for
Cambridge IGCSE
Additional Mathematics
(0606)
1st edition, for examination until 2027
© r/IGCSE Resources 2024
0606_m22_qp_22
144- 24h + h2
=
144
APQR
(n2 24n + 144)
-
=
-
Cross-sectional
X
area
height (n)
(h2 24h 144)n
-
[ (84xSin60
area of ABC =
In
=
+
(h3-2442 +
Area of ABC
↑
1653
144h)
Final Answer
PQR3 PQO length
ratio of
-
12
(1-ratio
of areas
Ratio of Areas
144-24h + h2
144-244 + he
144
=
Area
I
of 4
Solve for PR
Area of ABC (165)
Maximum !
#I : differentiate the volume equation with
=
(342
(3-48h +
-
&
respect to h
1446 -
48h + 144)
↑
=
0n
=
n
=
12
4
↑ solve this
(6(12)
use 2nd
to
find max
=
-
second derivative
Iderivative
= (6(1) 48) -
U
=
-
48)
4 final answer
=
883
-
0606_m23_qp_22
Triangle POR is
Congruent to
triangle QOR
PRQ
&
-20
-
2Q
=
Ero-Area
of Sector
Mr - Circle
ic(a)
2 -
(y)"20
Find y in terms of $
-
use
cosine/sine
rule
COSINE
yz
=
az + a2
-
SINE
2a2 (os(n
-
20)
Substitute
y2 into ic(a) 2 & (y)2@
↑C (9)
- (ar
sind sinn-20)
A
-
-
!
+ a2 -zarcos
(
-
201) 24
Final answer
=
y
↑
192-292-2a2cos(i-20)
↑
sing (Sin (r-201)
asin (i- 20)
Sind
Substitute y into
=
y
ic
a
2
-
(y) 20
Raz-(asinCr-20)
Maz
92 Sinc (m-2Q)
-
Sin &
0606_w02_qp_01
i
Up
.
=
50j
+ t
(20i 10j)
+
rq 80i + 20j
+
=
t( 10i + 30j)
-
Y
r = a + tv
where
t is the final position Vector
is
a
t is time
,
rp
=
=
↑
q
50j
+
2(20i + 10j)
Velocity Vector
is
v
ii. Substitute 2 into t 1200 -> 1400 (2 hours)
Original position vector
iii. When lif P3 &
to each other
50j + t(20i + 10j)
40i + 70j
meet , they will be equal
.
80i + 20j + t( 10i + 30j)
-
-
50j + 20ti + 10 +j
=
80i + 20j +
80i + 20j + 2( 10i + 30j)
20 ti + j(50 + 10t)
=
60i + 80j
Y(50 10t) y(20 30t)
+
Distance between Q3P so Fq-Up
60i + 80;40i
=
1
70j
20i + 10j
↳ Find
=
50 20
=
-
30
magnitude of vector by using
Pythagoras
20
185
+
=
IO
->
Final Answer !
20 +Y =
Y(80 10t)
-
20t + 10t = 80
20 + 30t
30t
30t 10t
-
20t
t
=
=
3 t
=
3
hence
meet
=
-
+
=
50 + 10t
=
10ti + 30t
i(80 10t) + j(20 + 30t)
=
-
-
,
.
and
a don't
80
8
0606_w02_qp_01.
i
-
r
.
Sing
pr
=
r
⑳
S
-
R
-
r
R -r
8
So
=
Snor
R ↓
(sino 1)
R
=
+
=
ii. Convert 30- radians
iii. R
tano
3r
=
T
%0
·
30
=
X
R
I
=
r(sing)
R 3r
=
Area of circle
:
Mr
15
5
3r
=
5
isms
x0
W
r
=
tan(E) 5
=
15-5
Xer
Finding y
R2 O
:
:
(3r)20
:
gr20
=
A
garden
~
9r2()
⑳
=
↳ Final
.
Answer
Perimeter : Add all
-L
M o
z=
-
M
=
73
-
Y
=
-
-
E E
-
90 in radians
55 + 55 +
=
27
.
.
E
5xz
=
R
10R
3
79
1 decimal place !
27 8
22 =
y
the sides
55
Er
L
~
~
-
55
55
S
=
-
:
↳ substitute
Fraction roses
Y
↑
&
2
x
55
x=
substitute
R 3r
=
tan(z)
Area of Sector : R /
20
/4
=
=
m
don't
forget the
unit w
0606_s03_qp_01
t
=
1 5
.
6 00 - 7 : 30
:
1 5 hours !
.
45
--------145
-
4525
-
90km
45
↑
Final Position Vector
Use CONT
-
r = a + tv
where
t is the final position Vector
a
is
Original position vector
t is time
%
v
⑪iii
Velocity Vector
is
Lifeboat
Let Original Liner position
Reiner
=
=
36ti
Lifeboat original position
0 0
=
.
↑ Lifeboat
t 1 5
=
.
36(1 5) i
-
.
45 4552j
=
54 + 0j
-
+
.
3 Lifeboat to meet
the Reiner
=
Rlifeboat
542 + 0j = (4582 + 1 5x) j( 5y 45r)
+
-
.
.
54
4522 + 1 5x
=
-
.
455
54
-
=
1 5x
.
45E
=
1 5
0
=
36
-
=
1
.
45
5y
=
452
-
1
.
5y
45
K
.
x
+
302
y
=
302
This is velocity rector
,
to convert to
speed
,
use
.
Pythagoras theorum
(36 302)" + (302)2
-
=
42 9km/h
.
↑
1 5x4 + 1
.
[(4582 + 1 5x) j(1 5y
=
For Liner
54
455-455
(45i 45825) + 1 5(xi + yj)
=
.
=
:
this is the final Answer !
.
-
.
5y]
45(2)
0606_w03_qp_01
ii
.
AO + AB + BC + CO
AO BC
=
AB CO
=
2A0 + 2CO
(8 9)
,
(4 , 2)
(0 0)
,
(6 3)
A0
=
AO
=
CO
,
Co
2 10
62 + 32
=
=
(pythagoras
62 + 22
35
Perimeter = 2 (200) + 2 (355)
=
4 To + G
26 1 units
=
.
i
y
C
.
By Ex intersect
3x 15
=
-
=
C
the point where line
:
:
3x 15
3x
-
2x
Ex
x
=
-
=
Parallel to y
x
15 =
x
=
G
(3x-15 y
coordinate
or
32
6
=
En y)
Line equation
x+ C
=
,
for AB
6)
!(2)
y
+ c
6- (2)
find y
=
equation equal
c= 5
to
B
5 + 15
3x
3#
20
(603)
20 +
:
:
+
=
6
y Ex + 53
:
-
2x
Ex
E
=
=
y
=
9
2(2) + C
=
C
2x + 10
let-2x + 10 equal to
Ex intersection of
,
those lines is point D.
2x + 10
10
=
10
=
Ex
2x + 2x
E
(4 2)
4 x D
:
=
=
=
A cords
-
-
x= 8
y
=
-
C = 10
x
y =(8) + 5
=
2x + C
=
6+ 4
-
=
m
-
3x 15
-
-
* Substitute
3x 15
-
((6)
=
C
=
y 3x 15 intersect.
=
=
y
Because B is where
=
y
Let ABline
c
(x 5
y
x+ 5
=
mx + c
=
ME
=
.
x
=
,
y
:
to
=
use A cords to find line
A (2
=
:
for AD.
x is found substitute the
=
,
Perpendicular to
Find line equation
y Ex
D
they have same gradient (2)
15
=
Ex hence
:
equation.
15
value into either equations
Y
:
y
=
* After
B the line Bis on is
B : (8 , 9) #
y 2(4) y 2#
=
=
,
0606_w03_qp_01
It ra
-
N2r
2r
-
i
Mr + 2x +
.
2x
=
5
125-
=
125
Total Peri
N
-tr
2
x=
15
-
Er T
-
4
EMr2
=
+
2rx
125cM
2
+
Sr
Er
I
125 - E-ir
4
=
=
A
[Rr2
+
(125
-
3
-
r)r +
125r--Mr
125r-T Er
Hence , Shown
&
-
.
n
Er
Mr +
=
r
*
Final Answer !
#
Area
[
,
you can split
.
Pythagoras theorum
L
into rectangle area
A
To find the maximum let
the triangle in half to use
#
W
Substitute
2x=
= 125-M-Er
To find h ,
h?
M
N
diameter of circle
2
triangle
semicircle
=
equation with respect to r
S
rectangle
A
ii. Differentiate the Area
Mr
A 0
:
=
125-Mr-Er O
125 Mr Er
=
+
=
n
=
n
=
r2
2
of triangle E h
:
125
r
=
(i + E)
2
r = 18
.
8 CM
>
ir
=
(*r)(r)
=
↑
Final Answer !
0606_s04_qp_01
5y + x
y
=
to find
2
y
yx z
m, =
-
=
(ax + 3)inx
U
+
=
=
gradient equation at x 1
u
Eg
=
V
v
A
y 3)xk
y= ak
(ax +
Line is perpendicular to normal
gradient of tangent
=
in
Mc
5
=
y (ax + 3)inx
+
axn
a
=
=
=
a(l) + 3
+
aln()
_
a + 3 + aln())
o
=
Crosses x-axis y 0
atgradient
0 = (ax + 3) Inz
gradient at tangent is parallel so you can
Inx 0
equate them to each other
=
=
,
=
x = e0
x
=
1
at x = 1
,
:
a+3
a
=
=
5
5 3
-
a=2
↑
Final answer !
:
0606_s04_qp_01
carl'
&
E
-
CI
(
S
72)
~
-T117
El
2
·
-
-
C
Any
37
11
Wi
Wu
-
Wi
,
·
-
13
frog PP
E
IIIIIIII
⑪ ·X
i
ii. Integrate y
gradient equal is diffrentiated version ofY
So integrate
- to find what the equation of the lineis then use P-2
=
.
1 1
3 5-p
,
&y
-
=
dx
y
as
((0x 1)dx ((0x 1)ax
-2
Substitute 2 points (2 3 5) 3 (5 1 4) to find
.
,
A
=
2
2(2)
A
g
3 5(8)
a+ 8
-
=
a + 50c
-
42
=
=
=
-
,
1 4
.
+
C
=
+
1 4 (50) = a + 50)
a + 82
=
.
+
-
+ p +5
-
2
=
=
-
-
2+ 3 +
10
#3 25+
=
P
2+5
-
p2
70
2p2
42
p2
=
P
=
-
Final Answer !
a = 20#
T
1y
-
20
10
10 #
↳ Final Answer !
28 8 = a
+
+
↑
10 + 10
2018
-
P2
=
=
(
-
P2 + p2
70 = a + 50c
-
1 4
28
a + 8(1)
x
+
10
28
Y
10 p (3 2)
=
1#
=
-
P2 -10
a + 80
c =
=
-
+
=
p2 -10
.
28
=
-
=
-
-x] []
C3 A
-
+ C
=
.
.
,
- 1.4
1
+
.
3 5
limits to find the value of
P
=
03 5
+
=
+ 1
4)
0606_w04_qp_01
4y
3x + 1
=
4y
x=
-
y
1
s substitute
into
28x
27Y
3
xy
=
-
mx + C
=
m
To Find the Area , Use the shoe
-I
=
lace method
=
=
42-y
y
-
28
-
3
27y
4
4-yly -27y
4y2 -y -112y- 28
=
3
+
-
4yz 32y
-
4y
-
=
=
=
+ 28 = 0
Y=1
Y57
M,
=
=
=
+C
c
3
Ex 3
+
=
-
4x + 32
3y + 4x
y
=
=
32
4x
12x + 4x 32
=
16x
=
(9 , 7)
=
gradienta midpoint
Mc
=
-
Midpoint (5 4)
=
,
32
x= 2
y 4(2)
=
perpendicular bisector
=
3y
y
=
S
,
,
--Area =
/
Area
=
9 Hz =
to Find the
Find
20
+C
3(4x)+ 4x 32
1
P(1 , 1)Q
-
Y
20
,
=
-
T.
-
4+
27y
4y2 113y 28 27y(3)
4y2 113y By + 28 0
x
=
c
=
P(1 1)Q(9 7)R(2 8)
(5 , 4)
( + ) y 28 ( ) (7y4 ( 5)(b)
+
:
=
Area
=
Area
=
17 72 2 a 14 8)
+
+
21501
25 units
-
-
-
0606_w04_qp_01
i
20000e-0 05CI0N
.
.
=
ii.
10 because
1970-1968 10
=
12130
20000e-0 05n
.
e-0 05n
.
e
=
=
=
year difference
.
2000
2000
20090
it
-0 05n
.
=
- 0 05n = In
.
(it)
(In t)
,
n=
- 0 05
.
46 1 years
n=
.
n = 47
years
1960 + 47
=
-
rounded up !
2007
↳ Final answer !
(b) (e + 34 y
=
(34)(3)
3y 2
-
3y
-
2
+2
-
=
=
2
-
=
2
=
8x3)
8x Y
3Y
=
Ey 34
3y
2 =
zy
index > 10g
-
32 67
=
-
+
2
=
Y
log(6)
form
0606_s05_qp_01
Find the perimeter 3 equal it to the
Length
r+
x
Circle
Square
2Tr
r
=
2
=
-
+
(simplify
4
-
T
2(i + 4)x
4 = 0
-
Th
2(i + 4)x
-
4
=
0
1 - 2x
2(n + 4)x
I
Formula
2(i + 4)x
:
Mr2 + x2
circle
u
L
square
substitute
(r = (
S
[ 4x
1+
A
+
1 + 4x + 4x2
Th
1 + 4x + 4x2 +
T
x2
7
=
A
make denominator
x2T
+
2
M+ 4
.
(x + 4)x2 4x + 1
(4) 4)(x(4)" 4( )
+
-
+ 1
T
4
+ 42
4
-
=
-
(1th
=
4
Substitute into A
A
R((l 2x)(l-2x)) +2
+
=
=
x =
M(1 -2x)2 + x
*
2(i 4(x
=
2- 4x
V=
-
(R + 4)x-4
_
4x
2T
Area
.
Differentiate A and equal it to 0
.
A
2
=
li
Th
x2i
same
=
0 14
.
a
iii) find
2(i + 4)x
Th
1 + 4x + (4 +
Th
R(x2
&A 2
=
-
4
27 + 8
Th
> 0 , hence
minimum point
0606_w05_qp_01
Find
Position Vector
RE
=
Ri
=
RE
Rs
=
(i + 12j)
+
F3S
of
:
+(3i + 2j)
Y + 12j + 3ti + 2 + j
(3t + 1)i
(12 + 2t)5
+
(85i + 5j) + t( Si + kj)
-
=
Rs
=
Rs
=
RF
Si + 5j +
-
5 + i + k+ ]
(85 - St)i + (5 + kt)j
=
Rs
Solve for K 8 +
,
3t + 1 = 85
-
3t + St 85
=
8t
t
t
=
=
=
5t
-
1
5 + kt
=
12 + 2t
kt 2t 12
-
=
-
84
k (10 5)
8
10 51
2(10 5)
-
.
.
k =
=
.
21
-
.
10 5
5
=
7
7
7 + 21
10 5
.
1=
25 Final
Answer !
0606_w05_qp_01
(a) 3COSX
=
3COS
2cos(34)
Stan I
( )
8
=
3 COS2x
=
Osinx
200s (2)
=
cos(29)
=
identifies
=
=
O sin x
3sinx + Osinc-3
Sinx
=
0
=
V
E
2
:
cos(2) -S T quadrants
=
,
S
-(5)
T46
.
(19 47)
180
x=
160 52
-
A
TY6
=
x=
0
T
x, 19 47
=
=
cost ( + (
=
x = Sin
1
-
Basic angle
↑
3(1-Sin2x)
+
.
&
I
!
Possible Answers
T
[
.
2 554/6
=
y
=
y
5 2
=
SRCREJECT]
4xy26
Ly
y
y
=
=
=
7
,
= 3
=
or
5 5
.
0606_s07_qp_01
i
differentiate the y equation
.
Integrate Y equation with limits O to
ii.
j
.
and let it equal to 0
My
3COSK
-
da
3 COS1
8
4 Since
-
=
4 Since
=
=
&S
=
y
y
=
=
=
=
(2)
0 644
.
3 Sinx + 4 COSX
3 sin(0 644) + 42s(0 644)
.
.
=
.
=
5 00
.
maximum coordinate
=
(0 644 5 00)
.
,
.
4sin(*)) (3cos(0) 4sin (0)
' 4) ( 0)
3cos(E) +
0 +
=
4 +3
=
7 units2
↑
y 4 99
y
+
tank
x = tant
x
[-3coS 4 sin]
=
4 Sinc = 0
-
3 coss
3 sine + 4 cos du
-
3+
Final Answer !
-
+
0606_w07_qp_01
i
.
use the Cosine rule to find BC
BC2
BC
=
BC
=
ii
AB2 + ACI-2ABAC COS (A)
=
Find arc length BC
.
.
BC
10(0 8)
=
.
G
AB2 + ACE - 2ABAC COS (A)
(10(2 + (10)2 2(10x10) Cos (0 8)
ro
=
=
BD
=
BC
BD
=
7 79
radius of circle Centre
-
-
point B
.
(Solve in Calculator in RADIAN MODE]
.
DC
=
0 + 100-200 cos (0 8)
ro
=
.
=
r
7 7883
=
7 79
.
.
=
to find O use since rule !
7 79
.
iii. Area
Sin A
=
(Sector CAB-triangle CAB)
( 20 t (7 79) (1 17)
Sector DBC +
Sector DBC
=
+
=
.
.
A
.
=
7 79
b
10
.
10
(Sin08) Sin Sin"10(%)
8
1 17 rad
.
35 50
=
SinO
Sin0 8
Sino
=
=
.
Sector CAB
Triangle
=
[r20
=
40
CAB =
=
[(10) (0 8)
=
.
.
DC rO
=
rsino
=
Closin
0 8
=
(7 79) (1 17)
.
.
.
=
9 11
.
=
35 87
.
Area
=
=
35 50 + (40
.
39 G unit?
-
35 87)
Perimeter
=
BC + DB + DC
=
8 + 7 79 + 9 11
.
.
.
=
24 9
.
.
0606_s08_qp_01
i
N
.
Y
C=
=
li cos(45) =
↳
845
↑
j
=
ii.
=
.
Let the
Wsnip (2i 3j) + (t)(15i + 15 ;)
==
Usub
=
+
=
a + tv
=
(479 27j) + t (25j)
15i + 155
=
Position Vector
r
r
r
=
a + tv
=
(2i 35) + (1 5)(15i
=
(2i 3j)
+
.
+
+
+ 15
-
24 + 35
a + tv
=
position vector of ship 3 Submarine equal
to each other.
15
S
Sin (43)
=
V
+
15 it + 155t
+
(2 + 15t)i + (3 15t)j
%)
2 + 15t
(22 Si + 22 5j)
=
=
=
47i 275 + 25jt
-
47i + (25 +
-
27)j
47
.
.
15t 47-2
=
r =
iii
24 54 + 25 59
.
.
15t
=
45
.
Position Vector
r
r
IV
=
=
=
a + tv
t
=
45
15
a + tv
(2i 35) + (t)((Si + 15 ; )
Itlitute
+
NaY
Esub 255
.
=
ship
25
or
into
position rector of either
.
submarine
=
&
snip 15i 15i
=
↑
+
Relative velocity
=
Tsub
Ehip-sub
=
15 + 155
=
1-105
-
255
R
=
(479 27j) + 3(255)
=
(47 27j) + 755
=
-
-
47i + 485
0606_s08_qp_01
i
Diffrentiate S In (t + 1) and let t 0
=
=
.
SI
a
=
t
t+ 1
=
·
=
ii
Fl
1 m/S
Diffrentiate S
.
and le + t
=
=
In (t-2)
-
(n(t + 1) + n16
4
z)-t
=
substitute t
=
cit-4
iii
=
=
-
d
i
zo
=
-i
.
v
-
4
=
-
(2t 4)
*
x2
-
2(2t 4)
-
-
( 1(t 1))
+
-
-
-
+
1(t + 1)
2
= acceleration
no
=
-
2(2(4)
=
-
4) 1(4 1)
+
+
I
iV
.
I
2t
-
-
4
+
0
=
t + 1
↓
↳
I
2t
-
t+ 1
2t
-
=
t
t
v
t+1
4
2t
4
1 +4
=
=
-
S
1 .......
/
1
.
23
4
-m
0
S4-Ss
S
In (3 + 1) S
=
S4
S4
Sy
=
=
In (4)
In (4-2) (n(4 + 1) + 1n16
-
(In(2) (n(5) 1n16
=
+
-
In
=
2
Ind + 1n16
-
Inx16
S4
In
S4-Ss
In
S4
=
=
,
194
In (4)
=
-
In
0606_w08_qp_01
#1 : integrate day to
find S
dx2
S4e-2x da
ze
=
-
ze
-2x
2(0)
C = 3 = (e
C
=
-
=
a 0
=
2(0)
3
=
e
y
=
5dx
+
2x
+ 5x +
2x
-
y
2x
-
-
+ c
(substitute
ze
e
=
-
+
-
+ 5x + c
2x
-
e
+ 5x + c
(substitute (2 e")
,
33)
-
C
5
4
e"
2(2)
-
e
=
-
(e
2()
-
+
5(2) + C
+
22
U
:
z2
v
:
x+
:
dy
10)
y
10
=
C
& Final Answer !
(2+ 1) (2x)
(x + 1)(x + 1)
V2
2x2 + 2x
=
(x2)(1)
-
x2
y
y
x2 + 2x + 1
=
z
-
at
22 + 2x
dy
0=
I
x2 + 2x + 1
di
x2 + 2x(0 0)
4)
2
+ 1)
+
,
0
=
x
2x
,
x2 + 2x 0
↑ Final Answer !
=
x=
-
2 , x= 0
substitute into
-
-
y
=
0
curve equation for y
y
=
-
4
:
10#
-
shoelace method :
MNO
=
M : (1 0
1
,
-
gradient
-
+ 3x
N : (0 , 4)
3
=
VU' -UV'
e
z
=
where x
dx
-
y
x= 1 ,
a+
V :
U 2x
x+ 1
:
2x
c
=
(ii) Find coordinate M
0
=
-
=
-
(i) Stationary point
ce4
=
,
M (6 0)
:
,
(x 1)
-
-
A
Ex ,
+
a
-
0 : (0 0)
-I
=
+
=
axis so y 0
En +x GM (1 0)
+
=
:
-
2
,
y
=
-
4
&y where x
=
-2
at
=
yaxis
O ,
y
so
x= 0
=
1
-
O
I
to &
↑ -
-
4 O
O -o
-
-4
=[CE)
=
da
gradient
=
,
Find Coordinate of N :
x=
=
-
4
N (0 , -4)
:
Area unit
0606_s09_qp_01
REMOVED
:
or
:
it cos (45)
=
o
i
=
j
j
=
=
:
Sin (45) 1002
a + +v
10
16i + 28j + t(10i +
(ii) Position Vector
t
:
=
=
4i + 8j
10t
.
.
4i + 8j + 2(10i + 10j)
4i + 8j + 20i + 20;
=
19i + 34j + t (8i + 6j)
=
19i + Sti + 34j + 6tj
-
8t
19 + 8t
28 + 10t
19-16
10t 6t
=
=
-
=
t
& 1 5h
=
10 00 + 12 00
16i + 28j
10j)
=
t
V : 10i + 10j
=
a + +v
+ (34 +
6t)j
(16 + 10t)i + (28 + 10t)j (19 + 8t)i
a + tv
(2hours)
R
=
16i + 28 j + 10ti + 10tj
16 + 10t
a :
:
=
=
=
:
16i + 28 j
j
Velocity vector 10i + 10 ;
R
Q : at 12 00
191 + 34j
P : at 12 : 00
velocity vector
M
r = a + +V
(iv)
100
North East
(i) Speed
12 00 + 1 5
:
.
r
=
r=
r=
34 + 6t
34 28
-
= t 3
=
13 : 30
=
.
r = a ++ V
=
=
3(8i
6j)
(19i 34j)
19i + 34j + 12i + 9j
31i + 43j
+
+
Position Vector
Time of meet
& Final
=
=
Answer !
+
31i + 43j
13 : 30
0606_s09_qp_01
1 5
.
4
i
y
.
=
ST
2T
a sin bx + C
z
with
=2 iv) integratelimitssin()
find
It
ii. Period =
↳
where a = amplitude
so
ii
amplitude
Let
.
y
=
Itsin equal
=
1 + sin 1 S
=
.
sin
=
Period
1
S
R-I
&
1 5-1
.
to
y
=
=
GR
1 5
.
A
+
x-
=
TC
= sin" 0 5
x
=
US
3(5)x 3(55)
=
3x =
x
Ex 3
A
B
=
=
cos(
i
-co
Final Answer !!
-
E-3cos
2n + 35
=
↑
minus Area of rectangle (2iX 1 5)
.
=
2
5
(2)
-
x=
=
Sin
[x 3((os * )]
.
=
under the graph :
area
A
c
to
3 B as
=
24 + 353
=
35
↑
-
-
37
T
Final Answer !
0606_w09_qp_01
iii )
.
2 coty + 3 cosecy 0
2 Closey 1) + 3 cosec y
(a) tan O + cot O
tan O +
tano
-
tanzo + 1
=
2
cosecry- 2 + 3 cosecy
2
cosecy + 3cosecy 2
tano
Sec2O
tano
↓
=
:
Costo
2x2 + 3x
x
cososing
coseco Seco
(b)
+anx
=
y
Sino
COSO
=
RHS !
Sin x
=
COS I
=2
-
x=
cosecy
3 sin x
2
-
=
x
=
0
Sin y
=
cosecy
=
3 Sinx
-
COST
=
2
Sin
0
x
x=
=
,
x = 180
,
u
=
:
y
x
360 70 5
.
*
-
.
.
,
x = 70 5 , 180 , 289 5
↑
y
70 5
x = 70 5
.
y
cos"(5)
x=
=
.
Final Answer !!
y
-
2
A
-
C
T
=
0 180 360
-2
So
C
=
Sin"(-2)
=
M+
=
-
=
=
S
=
x =
0
-2
siny
(cosc -3) 0
0
sin
cos 3 0
Sin "(O)
cosa
t
Sin
0
0
=
=
I
REJECT
Sin x
=
2
siny =I
3 sin x
=
-
Let Cosec
I
=
=
289 5
.
=
T
,
2r
E I
,
Final Answer !!
-
T
0606_s10_qp_11
i
.
)dyo Vu'-UV'
=
di
V2
Inxx(22)
x2
x (2)
+
x
V = x2
V
=
x
-
2n
-
ii )
(2)
.
=
-
-
(x2)2
2x2
-
=
y
0
0
-
y
x
=
-In e
Y =
3x2 (1
-
s
=
coordinate
es
1
=
22
x(1-21na)
1- 2inx
=
23
x4
VU' -UV 23(-E)-1-21nz
=
-
-
3(In 2(nx)
-
=
=
2 + 3 + 6(nx
61nx + 5
=
x44
x6
(3x2)
(23)2
V2
⑫( 2
,
(ei)2
xt
21nx)
(et ze)
Final Answer !
in(et)
=
:
&
-
·
ass ( vi
-
=
x=
x-2xInz
Inx(a)
1 - 21nx
=
2x(nx
=
=
O
Inx(2x)
2x(nx
=
In x
In x(2n)
(x
=
=
Inx
0
=
u 1nx
u
-
=
x4
x4
↑ Final
Answer !
iii
.
y O minimum
acY0
a
maximum
6 In x + 5
=
x4
s
O
6inx + 5
Ginn
=
-
Inx
=
-
0
5
5
e
x=
=
e-70
,
maximum
point !
0606_s10_qp_11
NEXT PAGE
integrate &y
i )
.
dx
find
curve
ii. )
+o
equation
J6cos(2x E)dx
A
-
6cos(2x + 2)
=
0
cos(2x + 1)
=
0
+
y
Gsin(2x + E)
=
(2x + E) COS (0)
+ c
-
=
2
y
=
S
3sin(2x + =) + c
A
⑨
O
i
T
Substitute
5
5
=
-
(F 5)
Cos" (0)
,
3sin(2(i) E)
+
3sin(2(i) E)
+
+ C
=
C
C = 5
y
↑
=
3sin(2x + 2) + 5#
I
=
2
E =2
2x +
2x = 0
x= 0
1 3
2x
3 E
2x +
=
=
-
iz
x =
Final Answer !!
2x +
E 5
2x
5T
=
=
x= T
-
C
0606_s10_qp_11
iii) Find gradient at x
6cos(2(y) E)
+
M
,
=
3 by substituting intoa
G
6 > Gradient of tangent
=
-
gradient of normal
=
i
-
Mz
=
Y
-(
=
=
x +
get a set of coordinates substitute
To
,
y
y
=
5
=
+
-
-
jx + c(3
j(3)
+ c
](3) C
c
5 7
5+
=
y
↑
=
-
+
5
5
=
y
3sin(2(3) E)
=
=
+
x+ 5
+T
Final Answer !
,
5)
= into curve equation .
0606_s10_qp_12
i
) 48 x
=
.
:
Il
solve for t
48
=
48
+
12
(2E
12
(23)
(n(2t + 3)
e"
=
e"
t
12((n(2t 3)
)d and
-
let -
and
+3
=
le+ t = 1
12
(5)
2t + 3
3
=
=
2t
e" 3 - Final
24
Final Answer !
m/s
-
=
Answer !
2
& and let
ii )
.
12(27 3)
+
d
24
dt
2t + 3
-
24(2t
24(2t + 3) (2)
=
-
48(2t + 3)
+
3)
=
-
-
1
Final Answer !
48(2 + 3)
=
m/s
0606_w10_qp_11
XPOA XPOE
↑
An
·)
1
P
PB
=
ii ) Area of Sector BPD-2 triangles
.
4 cm
20
4 cm
tancE) =
PA
of sector
PA =
Area of
=
PO
=
PB
=
=
=
(12) 2 (5)
=
2472
triangle Ebh
((45)(4)
=
=
8
453
Sin() 50
=
[20
45
PC = PO + OC
PO
=
=+
PC
1 4 radius
,
an"(E)
Area
A
40
sinY)
8 cM
=
242
-
2(85)
85
=
.
↑
4 cm + 8 cm
12 cm
47 69cm2
init
Final Answer !
4 + 4 + 2(12
12 (5)
-
48) + 47
8 + 24 85 + 47
-
=
30 7 cm
.
=
44
* Final
Answer !!
0606_w10_qp_11
(
i )
.
1 + x
(1(1 x)
+
d
dx
1 + x
-
2x(1 + x)
U
I'
V
=
V
=
↑ Final
/cix)
=
-
(2(1
&(1
-
+
+
2 +
2(1 + x)
+
=
2
1 +x
-
+
x)
2
=
x
z
(1 + x)
A =2
B
=
-
+
1(1 + x)
=
2x
:
1 + x
-
2) (O 1 -
1
-
(1 + x) =
ja
(1 x)
=
insert limits :
-
I
=
O
+
x(l + x)
1 +x
2x
du
-
2x(2(1 x) 2) (1 + x) =(2)(1 3)
-
2x
=
+
1(I + x)
x)
I (1 x)
dx
2
UV' + VU
( + x)
1 +x
x
du
2
=
I
+
/cix)
2x
=
2x
dx
Answer !
2x
ii. )
x
- dx
2(1 + x)* + c
=
ScErl-y a
iii.
dx
1 Final Answer !
↑
Final Answer !
+
d
0606_w10_qp_13
1) Substitute
x= 0
=
=
50
A220 + Be -
=
A + B = 58
=
dx
y Ae2 + Be-
into
O
2Ae"-Be-"
10)
2(10) e
I
2022
=
2"
-
0
2
40e
40e
=
=
=
0
2
-
20
C
x (Ae2
Be
+
-
x)
e
=
2Ae
210
-
Solve
-
B
=
1
Be- =
3
,
+
-
B
=
A + B
=
20
=
20
x
=
To find
into Y
equation
y substitute 2
,
+
)
=
y
y 47 6
=
20
.
>
-
50
=
1n
10e2 40e -
Final Answer !
(In2 47 6)
,
3A
2
3x In 2
Simultaneously
-
x
e3x
-20
Elimination Method
2A
-
2
2
40
=
-
20
-
2H
=
2Ae-Be-"when x 0
2A
ii.) 94
By 50
.
30
iii )
.
A = 10
Substitute A 10 into A + B 50
=
=
10 + B
=
50
B 40
if
a 20e2 -40e
any 0 minimum &
0 , maximu
a
=
40222 + 40e 20
-
minimum
0606_S22_qp_22
I
S
E
3
2T
-
1
Find areaPQOR 2x2i 4R
2
Area under the graph
A+ B
j
A
=
cos
do
2 + Sin (2r)
=
5
-
=
B
?
at since
=
(o
+ Sin
=
2T
2
=
27 2
-
Write A B as ratio &
:
(2 2 2)
(o))
1
Find A
=
2T
j
+
*
+
2
~
M
↑
2T
It Cos 2
Let
(x
-
=
=
M-1
:
3 B
k
A + B = 47
-
:
27
=
3
+
Find B
4
=
=
-
1
Final Answer !
equal to 1 to find intersects (3)
coss
Sin
Area
+ sin-(E SinE)
+
=
M-1
Area of A
=
A=2
-
of
Rectangle
(M - 2)
simplify
0606 _ S21_QP_22
Al
=
Al
=
(5 11)2 + 74 6)2
-
-
274#
Midpoint AC
:
(5111 4
+
-
A
B
.
·
=
(8 1) y 1#
,
ii. )
BD
=
-
:
4-6
5-11
:
y -1
= -
(x
-
8)
B
3
!
midpoint AC [8 1]
,
s
Perpendicular gradient
:
C
D
perpendicular bisector of AC
gradient Ac
BD
E
=
=- =
BD : y =
use
.
5
----------
D
FB
ED
:
:
(5)
( - 3)
3
x+
2 1 y Ex 39
gradient
to see
"
Final Answer !!
translation
+
-
>
=
+
bi )
.
range of f" (a)
f" (x)
bii. )
0
2x-1
y
=
3x
2x
Final Answer !
-
>
-
1
3ay
=
2x2 3xy
2
=
-
,
b
1 = 0
-
-
a
domain of f(x)
=
3y
=
b = b2
-
,
C
=
-
1
49C
2a
-
( 3y) =
-
( 3y)z
-
-
4(2)) 1)
-
2(2)
=
3y = 4 +
4
f " (x)
=
8 (f (x)0)
-
3x +
ax2 + 8
4
↑ Final Answer !!