2013 jc2 h2 maths rev g solutions pure maths ii
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ACJC 2013 JC2 H2 Mathematics REVISION SET G
COMPLETE SOLUTIONS
Pure Maths Part II
VECTORS
1.
PJC07/1/6 1l :
1
3 1
7 0
p
r ,
(i) Given 1q and 4p , since C lies on 1l ,
4 1
3 1
7 0
OC
and
4 1
0 1
4 0
AC OC OA
2AC l
4
4
1
0 0
2
12 8 9 7OC i j k
(ii) 2AB q i k
Given acute angle between 1l and
2l is 60 ,
2
1
1 0cos60
0 2
2 4
q
q
22 4 2q q 2q
SAJC/97/1/8
1
TJC07/1/4 By Ratio theorem,
1
2OM (a + c) and
1
2OP (5b – 3a)
1OS OP OM 2
(5b – 3a) +
1
2
(a + c)
= 1
22
a + 5
2
b +
1
2
c
C, S and B are collinear OS c + 1 b
Since a, b and c are non-parallel vectors, 1 5 1
2 0; 1 ;2 2 2
1 3,
4 8
1
8OS (5b + 3c) (shown)
N1996/2/15
3.
1l :
1 1
2 1 ,
0 1
r
2l :
2 3
1 1 ,
7 5
r
3
1
If 1 2 3 ----(1)
and 2 1 ----(2)
and 7 5 ----(3),
we have from (1) & (2), 3 and 2 which satisfies (3) since
R.H.S of (3) = 7 + 5( 2) = 3 = L.H.S of (3) .
andl
2 intersect and the coordinates of is (4, 1, 3). l E
2
1
3 1
1 1
5 1 9 3 3cos
35 3 35 3 35
27 2 2sin 1 cos 1
35 35
6
sin , where is the shortest dist from to and 2
10
2 2140 4 2
35
.
pp A l AE
AE
p
ACJC 2004
5 (i) 0 1 1
1 2 2 (2 3) (1 4 3) 5
1 1 3
Therefore the line l lies on 1
5 (ii) 2 2 0
1 3 4
4 3 1
AB OB OA
1
2 2 2 2 2 21
0 1
4 2
1 3 11cos
2380 4 1 1 2 3
AB
AB
d n
d n
1 11cos 44.5185
238
Hence angle between line AB and plane 1 90 44.5185 45.4815 45.5o
Alternatively, Let be the acute angle between the line and the plane
1
2 2 2 2 2 21
0 1
4 2
1 3 11sin
2380 4 1 1 2 3
AB
AB
d n
d n
1 11sin 45.5
238
5 (iii)
(iv)
Normal of plane 2 , 2
1 1 8 4
2 2 2 2 1
3 1 4 2
n
Equation of plane 2 ,
8 2 8 8 4
2 1 2 2 2 or 1 1
4 4 4 4 2
r r r
(Accept parametric form)
The 3 planes do not have a common point (or line) of intersection.
6
1l :
1 3
5 2 ,
12 2
r = and 2l :
1 8
5 11 ,
12 6
r =
Consider
3 8 34 2
2 11 34 17 2
2 6 17 1
A normal to 1Π is
2
2
1
. 1
2 1 2 2
2 5 2 2 4
1 12 1 1
Π
: r r
6 Method 1
Line passing through Q and perpendicular to 1Π ,
4 2
: 0 2 ,
8 1
l
r
Since R lies on l,
4 2
2
8
OR
for some
At point of intersection of l and 1Π ,
4 2 2
2 2 4
8 1
43
8 4 4 8 4
43
83
203
OR
(shown)
Method 2
RQ = Projection of PQ
onto the normal = 1 1
1 1
PQ
n n
n n
3 2 2
85 2 23
4 1 18
3
43
24
234 4 1 4 4 1
1
8 43 3
8 83 3
2043 3
4
0
8
OR OQ RQ
(shown)
2 possible scenarios
1 : 2 2 4Π x y z
2 : 2 5Π x ay bz
3 : 3 7Π x y z
Direction vector of line of intersection of 1Π & 3Π =
1 2 5
3 2 3
1 1 4
5 2
3 0 10 3 4 0 3 4 10
4
a a b a b
b
1
Q (4,0,8)
R (1,5,12)P
2Π
1Π
3Π
7
Given 1
2 4 2
: . 3 2 , 1 and 5
0
OA OB
p q
r
(i) Equation of line AX : r =
4 2
1 3
0p
where .
X is the point of intersection between line AX and plane 1 , therefore
4 2 2
1 3 3 2
0p
8 4 3 9 2 1 . Therefore
2
2OX
p
If OX makes an angle of 45o with z-axis
2
2 0
2 0
1cos 45
8
p
p
2
2
18
28
pp
p
2 2p
Since p > 0, we have 2 2p
(ii) 2
1
: 1 1
1
r .
2
6AB OB OA
q p
Plane 2 parallel to vector 2normal of plane AB AB .
2 1
6 1 0
1q p
2 6 0q p 2 2 4q
(iii) 3
2
: 3 0
0
r
Perpendicular distance
= length of projection of OX onto the normal of both planes
=
2 2
2 3
02 2 2
13 13
X
O
Alternative method:
1
2
: . 3 2
0
r
2 22 2
2
. 3
0 2 2
132 3 2 3
r
Therefore the perpendicular distance is 2
13.
(iv) 1
2
: . 3 2
0
r and 4
1 2
: 3 3
2 0
r .
Since plane 4 is parallel to -2i +3j which is the normal of 1 4 1
normal of 4 =
1 2 6
3 3 4
2 0 3
4
6
Equation of : 4 0
3
r
To find common point of intersection, solve the following equations:
1
2
4
: 2 3 2
: 1
: 6 4 3 0
x y
x y z
x y z
the position vector of the point of intersection is 1
7 12 3011
i j k .
8 Sub , ,0 into 1 and 2.
1
2
: 1( ) 3( ) (0) 8
: 3( ) 1( ) (0) 0
a
b
Using GC, 1, 3
1 3 3
3 1 3
8
b a
a b
a b
1
1 3
: r 3 3 ,
0 8
b a
l a b
1 3
3 3
0 8
b a
a b
5
2
2
+
4
1
0
1
4
Sub a b and 1
4 , we have
11 4 5 4
4
13 4 2
4
a
a
2, 2a b
Since the 3 planes have no point in common, l1 cannot intersect 3.
1
1 1
: r 3 1 ,
0 1
l
Condition 1: 1,3,0 is not on 3. Hence ( 1) (3) (0) 11
4
p q
p
Condition 2:
1
p
q
and
1
1
1
are perpendicular.
1 1
1 0 1 0 1
1
.p p q p q
q
COMPLEX NUMBERS
1
2(a)
3 3
2 2
3 i 3 i
1 i 1 iz
p p
3
22
3 1
1 p
2
8
1 p
2z 2
82
1 p
24 1 3 or 3 (rejected)p p
3
2
3 iarg arg
1 iz
p
3 2
arg 3 i arg 1 3i
3arg 3 i 2arg 1 3i 7
3 26 3 6
For arg( )z ,
7 5arg( ) 2
6 6z
(Shown)
2(b) 5 2z 2i i
2e 2e , 0, 1, 2k
k
By De Moivre’s Theorem,
21
510i
2 e , 0, 1, 2
k
z k
3 31 1 1 1 1
5 5 5 510 10 10 10 10i i i i
i2 e , 2 e ,2 e ,2 e ,2 ez
1 1 3 1 1 1 3
10 5 10 5 10 10 5 10 5i i i i
2 e , 2 e , 2 , 2 e , 2 ez
3(a) 2z
π 5πarg π
6 6z
5πi
62ez
4 ii ea bz 4 4 ii e ea bz
5πi 4
4 6 i2 e e ea b
4e 2
4ln 2
a
a
5πi 4
6ie e
20π 2π
6 3
b
b
3(b)
2 +i
i 2 +3 3 8 8e 2e
k
kw w
i3
i
i3
π π0, 2e 2 cos isin 1 i 3
3 3
1, 2e 2 cos π i sin π 2
π π 1, 2e 2 cos isin 1 i 3
3 3
k w
k w
k w
2,1 i 3,1 i 3w
4 3 8 8 0w w w 31 8 0w w
1, 2, 1 i 3, 1 i 3w
For 4 3iz 8i 8 0z z , let iw z . i, 2i, 3 i, 3 iz
4(a)
2
2
42i 2i 4 0
2i 2i 4( 4)(1) 2i 12i 3
2 2
z z zz
z
Since 1 2 1Re( ) Re( ), 3 iz z z
In trigonometric form, 1
π π2 cos isin
6 6z
Using de Moivre’s theorem, 1
π π2 cos isin
6 6
n n n nz
For 1
nz to real,
πsin 0
6
n
.
Thus smallest possible positive integer is 6.
4(b) (i) i1 e
i i i2 2 2e [e e ]
i2
i2
e cos isin cos isin2 2 2 2
2cos e2
(ii) Let the fourth roots be z. Then
π π
i i 2 π4 4 48 2 8 2 i 16e 16e
k
z
π πi
16 22e
k
z
1, 0 , 1, 2k
(iii) Let 2z w . Then π π
i16 22 2e
k
w
π π π π
i i16 2 32 4π π
2 1 e 4cos e32 4
k kk
w
, 1, 0 , 1, 2k
Thus when 1k ,
π π 9πi i
32 4 32π π 9π4cos e 4cos e
32 4 32w
.
The other roots are
π 7πi i
32 32π 7π4cos e ,4cos e
32 32
and
15πi
3215π4cos e
32
.
5(i) Given 2 8iz , we can rewrite it as
2 i2 28
k
z e
.
12 i i
2 2 4 8e 2 2ek k
z
, where 0,1k .
i.e. i
4 2 2ez
or 3
i42 2e
.
Converting to Cartesian form,1 i
2 22
z
or 1 i
2 22
Hence 2 2iz or 2 2i .
5(ii) Given 2 4 4 2i 0w w , using formula for solving using quadratic formula,
12 8i
2w .
Applying the result in (i), 1
2 2 2i2
w or 1
2 2 2i2
(note that there should only be two solutions for w)
Hence 1 iw or 3 i .
5(iii)
5(iv) The two loci are parallel to each other but are not the same line, so they do not
intersect. They therefore have no points in common.
6
Minimum value of 2
22 2 3 2 2 2z i = 2.
1 2cos 3 2sin i = 2 3 2 i.4 4
z
7
2cos8
a
Hence 2 cos sin i8 8
z a a
= 2 2cos cos 2cos sin i8 8 8 8
= 1 cos sin i4 4
(by double angle formula)
= 1 1
1 i2 2
.
From right angle triangle in diagram,
1sin
128tan18 2 1cos 1
8 2
a
a
.
Hence 3
tan tan cot 2 18 2 8 8
.
8(i) 2 cos 1 1 2 sinz i
1 2 cos sinz i i
1 2z i since cos sin 1i
The locus of P is a circle with radius 2 and centre (1, 1). (shown)
Method 2
Let z x iy , 2 cos 1 1 2 sinx iy i
Compare real and imaginary parts, we get
2 cos 1x and 1 2 siny
11cos and sin
2 2
yx
Since 2 2cos sin 1 ,
2 2
2 21 11 1 1 2
2 2
x yx y
The locus of P is a circle with radius 2 and centre (1, 1). (shown)
x (2,0) O
(i)
(ii)
(1,0)
7
8
a
8(ii)
8(iii) Min. value of 6 4z i
226 1 4 1 2 74 2 units
8(iv) 2sin 0.16515 rad
74
5
tan 0.62025 rad7
Max. arg 6 4 0.455 radz i (3sf)
9(a) Given w* = z − 2i, z = w* + 2i, we obtain |w|2 = w* + 2i + 6.
Let w = ia b .
2 2 i 2i + 6 6 i 2a b a b a b
Comparing real and imaginary parts,
2 − b = 0 b = 2
2
2
4 6
2 0
2 1 0
2 or 1(N.A.)
a a
a a
a a
a
Therefore w = 2 + 2i.
Im(z)
Re(z)
(1, -1)
(-6, 4)
9(b)
arg(a) = 4
a = 1 + i
10
10 (i) Minimum value will be the perpendicular distance from origin to the line segment
CB. Let the point of intersection of this perpendicular and CB be D. By observing
the right angle triangle formed by O, C, and D, OD = | a | sin 4
=
| |
2
a.
10 (ii) The locus of z satisfying both relations is the line segment CB, excluding C. The
argument of the point represented by B is arg( )2
a
, while the argument of the
point represented by C is arg( )a .
Hence the range of values is arg( ) arg( ) arg( )2
a z a
.
11(i) Locus of P is a half-line from and excluding the point representing the complex
number1 2i that makes an angle of with the positive real axis.
11(ii)
11(iii)
5EF and radius 2EG
2sin
5 0.287
Locus of P meets the locus of Q more than once when 0.287 0.287 .
DIFFERENTIATION
1 2 26 2 21 0x xy y
Differentiate with respect to x
d d 2 6 4 0
d d
d4 6 6 2
d
d 6 2
d 4 6
3
2 3
y yx y x y
x x
yy x y x
x
y y x
x y x
y x
y x
2(a) 4e
1tan
xyx
0d
de
1
1 1tan
2
yx
yx
x
x
x
yx
x
yx
1tan
2e
1
1
d
d
)1(
)1(e2
2tan 1
xx
xyx
2(b) xxy )12( )12ln(ln xxy
)12ln(12
21
d
d
x
x
x
yx
y
)12ln(
12
2
d
dx
x
xy
x
y
)12ln(
12
2)12( x
x
xx x
3(a) 2
2
2
2
2
ln 2
cosd(ln( ))
2 (ln co )
2 sin 1
cos
12 ta
d
n
s x
x
dx xx d
x x
x x
x x
x
x
3(b) 2
1
2
2 2
2 2
(1 2 ) (2 )( 2)
2 (1 2tan (
(
))
21 21 )
1 2
1 2 4 2
(1 2 ) (2 )
5
5 15
1
x x
x x
xx
x
x x
x x
x x
d
dx
4.
2
2 2 2
3
2 3 2 2
f ( ) e
f '( ) 3 e (2 e ) 1 e (3 2 ) 1
x
x x x
x x x
x x x x x x
For all values of x , we have
22 2 0, e 0, (3 2 ) 0xx x , and so22 2 e (3 2 ) 0xx x .
Hence f '( ) 1 0x , and so 23f ( ) exx x x is strictly increasing for all values of
x .
DIFFERENTIATION and its APPLICATIONS
1. ln lny xy x
Differentiate implicitly w.r.t x,
1 d dln
d d
1y yxy x x y
y x xx
1 d dln ln
d d
y yy x x y x
y x x
1 dln ln
d
yx x y y x
y x
2 2d (1 ln ) (1 ln )
d 1 ln 1 ln
y y x y x
x xy x y
For the tangent to be parallel to the y–axis, d
0d
x
y
2
1 ln0
(1 ln )
y
y x
1 ln 0y ey
Hence from xyy x , we have ee xx e e 0xx
Hence x = 1.32, and since the tangent parallel to the y-axis takes the form x c ,
hence equation of tangent is 1.32x
2. 2 1
2dy
x at y atdx t
,
Equation of tangent is 212 ( )y ap x ap
p
2py x ap
Equation of normal is 22 ( )y ap p x ap 3 2y px ap ap
(ii) A = (9a, 6a) 29 3a ap p
Equation of normal is 23 (3) 2 (3) 3 33y x a a x a
Since Q lies on the normal, sub 2( ,2 )aq aq into eqn of normal to get:
22 3 33aq aq a 23 2 33 0q q
113 or
3q q
Coordinates of Q are 121 22
,9 3
a a
(iii) Since 3
( , )2
T a a lies on the tangent, it satisfies the equation
2py x ap
23
2ap a ap
22 3 2 0p p
1(2 1)( 2) 0 or 2
2p p p
Points B and C are 1
( , ) and (4 , 4 )4
a a a a
(Gradient of TB) × (gradient of TC) =1
2 ( ) 12
Thus TB is perpendicular to TC
3 By Pythagoras’ Theorem,
22 2
2
15 15
30 (shown)
r h
r h h
1515 h
3(i) 2 345
3V h h
2
2
d d30
d d
d20 30 5 5
d
d0.050929 0.0509 cm/min (3.s.f)
d
V hh h
t t
h
t
h
t
Alternative
2 2
2
d90 3 30
d 3
d d d
d d d
d 120
d 30 5 5
d0.050929 0.0509 cm/min (3.s.f)
d
Vh h h h
h
h h V
t V t
h
t
h
t
3(ii)
2
12 2
30
d 130 30 2
d 2
r h h
rh h h
t
2
d d d
d d d
15 5dWhen 5, 0.050929
d 30 5 5
d0.045552 0.0456 cm/min (3.s.f)
d
r r h
t h t
rh
t
r
t
4(i) Applying the cosine rule to ASB, we have
2 2 2 2 2 2 2 2 2 2 2(3 2) ( 2 ) ( 3 ) 2 2 3 cos .s s s s
Rearranging,
2 2 2
2 2 2 2
4 9 1 6cos ,
2 4 9 4 9
s s s
s s s s
which upon squaring both sides gives the required result.
4(ii) 2 2
2 2 2 2
( 6)1
( 4)( 9) 4 9
s P Q
s s s s
So
2 2 2 2 2 2( 6) ( 4)( 9) ( 9) ( 4).s s s P s Q s
Comparing coefficients gives
9 4 0
1,
P Q
P Q
4 9
and .5 5
P Q
Thus
2 2
1 4 9f ( ) 1
5 4 9s
s s
and
2 2 2 2
2 9 4f '( ) .
5 ( 9) ( 4)
ss
s s
4(iii) When f '( ) 0s , we have
2 2 2 2
2 2 2 2
2 2 2 2
2 2
2 9 40
5 ( 9) ( 4)
9 40 (as 0)
( 9) ( 4)
9 4
( 9) ( 4)
3 2(as all quantities are positive)
9 4
s
s s
ss s
s s
s s
Thus 2 6,s giving 6s as 0.s 2
2 2 2 2 2 2 2 2
2 9 4 2 5( 30)f '( ) .
5 ( 9) ( 4) 5 ( 9) ( 4)
s s ss
s s s s
This implies that
s 6
6 6
f '( )s 0 0 0
f ( )s __
Since 2cos
is decreasing for 0,2
, when f is minimum, 2cos
is
minimum and this implies is maximum. Hence the soccer player should shoot
when 6.s
Alternative solution : use graph
5
Let x be the distance between A and P. Then 2 2 28 64PW x x
Cost of laying the pipe, 260000 64 45000 12C x x
d0
d
C
x
2
6000045000 0
64
x
x
2
2
2 2
2
60000 45000 64
4 3 64
16 9 64
576
7
576 576 (reject as 0)
7 7
x x
x x
x x
x
x x
To show that C is minimum when 576
7x
Method 1: First Derivative Test
x 576
7x
576
7x 576
7x
2
d 6000045000
d 64
C x
x x
< 0 0 > 0
C is a minimum when 576
7x .
Method 2: Second Derivative Test
22
2 2
2 2
2 2
3 32 22 2
6000060000 64
d 64
d 64
60000 64 60000 38400000 for all
64 64
xx
C x
x x
x xx
x x
C is a minimum when 576
7x .
Land
W
P B A
8 Ocean
x 12 – x
BINOMIAL EXPANSION and MACLAURIN’S SERIES
1
111 222
2
3
3cos 2 4 3 cos 2 4 1
4
1 3
1 1 3 32 2cos 2 1
2 2 4 2! 4
1 3 5
32 2 2
3! 4
1
2
xx x x
x xx
x
2 2
3
2 3 2 3
2 3 27 1351 1
2 8 128 1024
1 3 229 633 1 3 229 6331
2 8 128 1024 2 16 256 2048
x x xx
x x x x x x
Replace x with x ,
21 1
2 2
1
2
1 3
2 2
2cos 2 4 3 1 4 3
2
1 2 4 3
1 2 1 3 229 633Thus
2 16 256 20484 3
xx x x
x x
xx x x
x
2(i)
1144
314 4
16
214 16 2 16
2 23 31 164 8192 32 4096
16 2 1
2 1 ...
2 1 ... 2 ...
x
x x
x
x x x x
This expansion is valid for 16
1 16 16.x x
2(ii)
14
22 2 231
32 4096
2 4 3 23 3132 32 4096
23 27132 32 4096
2 23 10132 4096
16 3 2 3 3 ...
2 6 9 ...
2 ...
2 (up to term in )
t t t t t t
t t t t t
t t
t t t
2(iii) Replacing t with –t in the above expansion, 142 23 101
32 409616 3 2t t t t
3 sin 2 sin cos 2 cos sin 2
4 4 4
cos cos
x x x
x x
1 1cos 2 sin 2
2 2
cos
x x
x
2
2
1 4 11 (2 )
22 2
12
xx
x
2
2
1 4 11 (2 )
22 2
12
xx
x
12
211 2 2 1
22
xx x
221
1 2 2 1 ...22
xx x
2 21 11 2 2
22x x x
21 31 2
22x x
4(a)
2
2
2 3 2
3 2
3 2
πtan cos2 10
4
tan 1cos2 10
1 tan
Given that is sufficiently small,
1 41 10
1 2
1 1 2 1 10 1
1 1 2 2 10 10
2 12 10 2 0
6 5 1 0 (Shown)
x x x
xx x
x
x
x xx
x
x x x x x
x x x x x x
x x x
x x x
Solving the cubic equation with GC,
4.95(N.A.) or 1.22(N.A.) or 0.166 (3 s.f.)x x x
4(b)
1
2 1
2
222
2
22 2
22
22 2
1 tan
1 tan
d 1Differentiate w.r.t. , 2
d 1
d d 2Differentiate w.r.t. , 2 2 1 2
d d 1
d d (Shown)
d d 1
y x
y x
yx y
x x
y y xx y x x
x x x
y y xy
x x x
22 2 23 2 2
43 2 2 2
2 2
32
Differentiate w.r.t. ,
1 4 1d d d d d2
d d d d d 1
1 4
1
x
x x xy y y y yy
x x x x x x
x x
x
2 3
2 3
2 3
2 3
d 1 d 1 d 5When 0, 1, , , .
d 2 d 4 d 8
511 8412 2 6
1 1 51
2 8 48
y y yx y
x x x
y x x x
x x x
1
11
2
2 2
2 2 2
2
1 tan
1
1 tan 1
1 3
1 1 1 2 21 1
2 8 2 2
1 3 1 1 11
2 8 2 4 8
11
2
x
x
x x
x x x x
x x x x x
x x
5 2 22 2 2
2 4
( 2)( 3)1 1 ( 2)
2 2 2! 2
3 1
4
x x x
x x
5(i) Since x is sufficiently small for 3x and higher powers of x to be neglected,
2
2
22
22
2
2
sin 2sec
2sin
cos
2
12
2 12
2 1
2 2
x x
xx
xx
xx
x x
x x
OR
2
2
2
2
sin 2sec
sin 2(1 tan )
2(1 )
2 2
x x
x x
x x
x x
5(ii) 1 0 2 3
1
2
1 2 3 4
2 2 2 2 2
3 4 1 1
4 8
1 1 4
2
rr
r
6(i) 1
2
4 2 2 12
xx
2 331 11 11 2 2 22 2
2 1 ...2 2 2! 2 3! 2
x x x
2 31 1 12 ...
2 16 64
Expansion is valid if 1 2 2 2.2
x x x
xx x
6(ii) d:
dx
de 3cos3
d
y yx
x .
d:
dx
22
2
d de e 9sin3 9 e 1
dd
y y yy yx
xx
22
2
d d9e 9 0 Shown .
d d
yy y
x x
d:
dx
3 2
3 2
d d d d2 9e 0
d dd d
yy y y y
x xx x
.
When 0x , 0y , d
3d
y
x ,
2
2
d9
d
y
x ,
3
3
d27
d
y
x .
2 33 9 27...
1! 2! 3!y x x x
2 39 9
3 ...2 2
x x x .
6(iii)
2 1eln 2 1 ln 1 sin 3
1 sin 3
x
x xx
2 3 2 3
2 1 4 4 4 2 2
1 1 1 1 12 2 2 2 ... 2
2 16 64 4 8
x x x
x x x x x x
2 3 2 3
2 3
4 4 ln 1 sin 3
1 1 9 92 ... 3 ...
4 8 2 2
17 352 2 ...
4 8
x x
x x x x x x
x x x
2 10.1 0.1
2 3
0 0
e 17 35ln d 2 2 d
1 sin 3 4 8
0.19131 to 5 dp
x
x x x x xx
INTEGRATION and ITS APPLICATION
1 1 2cos dx x x
2 21 2
4
2 31 2
4
21 2 4
2cos d
2 2 1
1 4cos d
2 4 1
1cos 1
2 2
x x xx x
x
x xx x
x
xx x C
2 (i) tant x 2 2sec 1
dtx t
dx
= 2 2
2
1 1
11
1
dtt t
t
= 2
1
1 2dt
t
= 11tan 2
2t c
= 11tan 2 tan
2x c
(ii)
volume is same as
Exact volume =
2
420
1d
1 sinx
x
41
0
1
1tan 2 tan
2
tan 22
x
3 Method 1:
sin 2 cos dx x x
1sin 3 sin d
2
1 cos3cos
2 3
1 cos3cos
2 3
x x x
xx C
xx C
Method 2:
sin 2 cos dx x x
12
3
[f ( )]2sin cos d [use f ( )[f ( )] d ]
1
2cos
3
nn x
x x x x x x cn
x C
(i)
d dcos 1, 2cos 2
d d
x yt t
t t
d d d 2cos 2
d d d cos 1
y y t t
x t x t
When d
0,d
y
x
3 3 1 1 3cos 2 0 2 , , ,
2 2 4 4 4 42 2t t t x
At point A, 1
, 142
x y
1y is the equation of the tangent to the curve at point A.
Or
2
12
1 siny
x
2
1
1 siny
x
y =2
Since 0 t , the maximum and minimum values of y (i.e. sin 2y t ) is 1 and -1. The
y-coordinate of point A is 1 and since the tangent to this max pt is a horizontal line
(d
0,d
y
x ), therefore the equation of the tangent to the curve at point A is y = 1.
(ii)
1
42
4
4
44
Area 1 d
3 1 sin 2 cos 1 d
4 2
3 1 sin 2 cos sin 2 d
4 2
3 1 1 cos3 cos 2 cos
4 2 3 22
3 1 2 1 1
4 3 22 3 2
3 1 2 2
4 6 3
y x
t t t
t t t t
x tx
4 22 3131)31)(31(
102
x
CBx
x
A
xx
x
By cover-up rule, when 1 3
1 Ax
)31)(()31( 1102 2 xCBxxx
When x = 0, C = 3
When x = 1, B = 1
xx
x
xx
xx
xd
31
3
31
1d
)31)(31(
1021
0
2
1
0
2
y
x
A
34ln
6
1
3tan34ln6
14ln
3
1
3tan331ln6
131ln
3
1
1
1
0
12
xxx
xx exex
coscos )(sind
d
xxxe
xxe
x
x
d )cossin2(
d 2sin
cos
cos
xxxe x d ))(cossin(2 cos
cxe
ceex
xexex
x
xx
xx
)cos1(2
2)(cos2
d )(sin)(cos2
cos
coscos
coscos
5 Equation of Tangent: y = 2x – 2
When x = 3, y = 4
Area )4)(2(
2
1d )1(
3
1
2 xx
43
3
1
3
x
x
3
8 or 2.67 (3 s.f.)
Volume
8
0
24
0
2
d 1)4()3(d 2
2yyy
y
8
0
24
0
23
23642
34
y
yyy
y
3
40
6 2 2
2
1 2 (1 )1 1
1 (1 )
x x
x x x
x x
2 2
1 1 d d1 2 1 (1 )
1 ln 1
(1 )
xx x
x x x x
x Cx
2 2 2 2
2
1e sin(e ) d (2e )sin(e ) d
2
cos(e )
2
x x x x
x
x x
C
1 1
2
1
1 2 2
1 2
1sin d sin . d
11
sin ( 2 ).(1 ) d2
sin 1
x x x x x xx
x x x x x
x x x C
2
2
2
2 2
2 1
d2 3
2 31 d
2 32 2 1
1 d2 3 ( 1) 2
1 1ln 2 3 tan
2 2
xx
x xx
xx x
xx
x x xx
x x x C
7(a)
(i)
Total area of all n rectangles, A
3 3
3 3
33 3 3 3
3 13
1
2
4
1 1 1 1 20 2 2 2
1 3 1 1 2 ... 2
1 12 2 ... 2 1 2 3 ... ( 1)
1 12
1 ( 1)2 (
4
n
r
n n n n n
n
n n n n
nn n
n rn n
nn
n
2
2
2
2 2
1 1)
12 2 1
4
1 1 1 9 1 12 (shown)
4 2 4 24 4
n nn
n nn n
(ii) Area under the curve 2
2
9 1 1 9lim unit
4 2 44n n n
(iii) For
2
9 1 1 90.99
4 2 44n n
2
1 10.0225
2 4n n
Using GC,
0.512 or 21.7n n
Since n , least n = 22
(b)(i) By GC, ( 0.340,1.162) and (0.340,1.162)
(ii) Since both curves are symmetrical about the y-axis,
(I)
Volume required 2
20.34001 0.52
20 0.34001
12π 2 cos(5 ) d 1 d
4x x x
x
32.61 unit or 0.832π (3 s.f.) by GC
(II) 2
2 1
2 1
2 cos(5 )
5 cos (2 )
1cos (2 )
5
y x
x y
x y
2
2
2
11
4
11
4
1
4( 1)
yx
yx
xy
Volume required 1.162 1.162 1
0 1
1 1π d cos (2 ) d
4( 1) 5y y y
y
30.567 unit or 0.180π (3 s.f.) by GC 8(a)
2 2 2
2
2
2 1
1 1 6 1d d d
1 3 6 1 3 1 3
1 1 1 ln 1 3 d
6 3 1/ 3
1 3 ln 1 3 tan 3
6 3
x xx x x
x x x
x xx
x x C
(b) 2 2 22de e 2 e
d
x x xx xx
Using by parts with 22' 1 2 e ,xv x u x ,
2 2
2 2
2 2
2 2
2
2
2 1 d 1 2 d
d
1
2
x x
x x
x x
x x e x x x e x
x e xe x
x e e C
9(i) 6/5 1/2 6/5
0 0 1/2
1/2 6/52 2
0 1/2
2 1 d 1 2 d 2 1 d
37
50
x x x x x x
x x x x
(ii) d2sin , 2cos
d
xx
1
0 0
6 3sin
5 5
x
x
26/5
0
21
0
2
0
2
0
2
0 0
00
2
1
1 d4
4sin 31 2cos d where sin
4 5
4 4sin2cos d
4
1 sin 2cos d
2cos d cos 2 1 d
1sin 2 sin cos
2
sin cos sin 1 sin
3 4 3sin
5 5 5
a
a
a
a a
aa
xx
a
a a a a a a
1
3 (Since sin )
5
12 3sin (shown)
25 5
a
(iii)
(a)
26/5
0
26/5 6/5
0 0
6/5
1
0
1
1 2
3Area of 1 2 1 d
4 5
32 1 d 1 d
5 4
37 3 12 3sin
50 5 25 5
37 18 12 3sin from (i) and (ii)
50 25 25 5
23 3sin units
50 5
xR x x
xx x x
x
(iii)
(b)
Volume generated
y
x
Idea: + –
2
1 4/52
4/5 3/5
22/5
3/5
3 3
44 1 d d
2 5
1 d
5 2
8513.56 units (to 3 s.f) by GC or units
750
yy y y
yy
11 (i) Total Area of rectangles
11 2
01e e e e
n
n n n
n
1
1
e 11
e 1
n
n
n
n
1
e 1
e 1
nn(shown).
10(i) 2 2
2 21
24
3
x y
2 29 4 16x y
(ii)
(iii) Required volume
=
1.08729 1.08729 22
2
0 0
16 9 3d d
4 4
x xx x
x
= 9.487 (3 d.p.)
y
y = 0
x
-2 2
x = -2 x = 2
(1.09, 1.16)
(-1.09, -1.16)
(ii) Actual Area =
11
00
e d e e 1
x xx
Actual Area > Total Area of rectangles
1
e 1e 1
e 1
nn
1
e 1 1 nn
1 1
e 1 n
n (shown)
e (1) xy
e(2)y
x
Point of intersection is 1, e
Volume = 2e 4
2
1 e
eln d d
y y yy
= 4.99 (to 2 d.p)
12 (i) Let 1 y x , then
d 1
d 2
y
x x .
4 3
1 2
3
2
3
2
1 1 d 2 1 d1
12 1 d
2 ln
2 3 ln 3 2 2 ln 2
22 2ln
3
x y yyx
yy
y y
(ii)
Let
d 1ln(1 ),
d
vu x
x x
4
1
4 4
1 1
4
1
1ln(1 ) d
1 12 ln(1 ) 2 d
1 2
14ln 3 2ln 2 d
1
24ln 3 2ln 2 2 2ln
3
6ln 3 4ln 2 2
x xx
x x x xx x
xx
13
14(a)
2 d
2 4d
xx t t
t . when 0, 0x t . when
2
, 08 4
x t t
.
2
8
0
4
0
44
0 0
Area d
sin 4 d
4 cos 4cos d
y x
t t t
t t t t
4
0
24sin
2
2 24 0
2 2
22 2
2
t
15 (i)
4 4
2 2
2
2
3 1
1 1
1( 1
1
1tan
3
x xx x
x x
x xx
x x x C
d d
) d
(ii) Let tanu x
2 2 2
2
dsec tan 1 1
d
1d d
1
ux x u
x
x uu
When
, tan 1
4 4
0, tan 0 0
x u
x u
4144
20 0
3 1 10
1
tan d du1
1=[ tan ]
3
1 21 tan 1
3 4 3
ux x
u
u u u
(iii) 4 44 4
04
2 4tan d 2 tan d 2( )
4 3 2 3x x x x
A parametric 2tan , secy x , where 0 2 .
(iv) When 1, tan 1 , 24
y x
Area of region R =
2
1
2 dy x
24
0
2 24
0
2 24
0
4 24
0
2 tan 2sec tan d
4 tan sec d
4 tan (tan 1)d
4 (tan tan )d (shown)
4 24 4
0 0
24
0
40
4 (tan )d 4 (tan )d
24( ) 4 (sec 1)d
4 3
84[tan ]
3
8 44[1 ]
3 4 3
(v) 1
2 2
0
(2) 2 2 dyV x y
2 2 24
0
8 2 (sec ) sec d 13.4
16 (b)
(i) 2d( sin 2 )
dx x
x =
22 sin 2 2 cos2x x x x
(ii) 2 2(2 cos 2 2 sin 2 )d sin 2x x x x x x x
2 2
2
2
2
(2 cos 2 ) d 2 sin 2 d sin 2
[ cos 2 cos 2 d ] sin 2
1[ cos 2 sin 2 ] sin 2
2
1cos 2 sin 2 sin 2
2
x x x x x x x x
x x x x x x
x x x x x C
x x x x x C
2 21 1 1(2 cos 2 ) d cos 2 sin 2 sin 2
2 4 2x x x x x x x x C
17(i)
2
3 10
1
3 1 0 , 1
x x
x
x x x
Hence, 1 3x , 1x . (ii)
2
2 2
2
2
2
2
2
3 1 2 3 d d
1 1
2 1 4 d
1
1 4 d
1
41 d
1
4
1
x x x xx x
x x
x xx
x
xx
x
xx
x Cx
(iii)
2
2
3
2 2
2 3
3
2 3
2
3 1 7 d
61
3 1 3 1 7 d d
61 1
4 4 7
1 1 6
4 4 4 4 73 2 3
3 1 2 1 1 3 1 6
4 75 6 5
1 6
4 31
1 6
6 4 31
p
p
p
x xx
x
x x x xx x
x x
x xx x
pp
pp
pp
p p p
2
31
6 37 55 0
3 11 2 5 0
11 5 or (N.A.)
3 2
p p
p p
p
−1 1 3
+ + −
18(i)
4(ii)
2 2
2
2
2
2
2 31
2 1
2( 3) 1
2
23 1
2
y x
yx
yx
y
x
y
x
y = 2
222 2
2
0
2 22
0
2 2
22
0
2 23 1 3 1 d
2 2
2 29 6 1 1
2 2
2 29 6 1 1 d
2 2
212 1 d (shown)
2
y yV y
y y
y yy
yy
d2 2sec , 2sec tan
d
yy
22
0
23
0
3
0
3
0
212 1 d
2
12 sec 1 2sec tan d
=24 tan sec tan d
24 2 3 sec d from (*)
124 2 3 3 ln 2 3
2
12 2 3 ln 2 3
yV y
19 (i)
2
2 2
2 2
2 2
2
1 2 1 1 4 4
2 241
2
4 4
4 4
2 4
xx x
x x
x x
x x
x
(ii)
2 2 1
0 0
2 1
1 1 14 d 4 4sin
2 2 2 2
1 4 4sin
4 2 4
kkx
x x x x
k kk k a
(iii) 2 24 4y x
22
21
2
xy
2
1
k
R
2
0
14 d
2
k
R x x
(iv)
1
Required area 4 with 1
1 3 4sin
2
3 46
2 3
3
R k
20
ln dx x x
= 2 2 1
ln d2 2
x xx x
x
= 2 2
ln where is an arbitrary constant.2 4
x xx c c
22 xu d
2 1d
uu
x
When x = 2 , u = 2. When x = 7, u = 3.
7
2
2d
11
xx
x
=
3
22
(2 ) d9
uu u
u
=
3 2
22
2 d9
uu
u
=
3
22
92 1 d
9u
u
=
3
1
2
2 3tan3
uu
= 1 22 3 3 2 3tan
4 3
= 13 22 6 tan
2 3
21 (a) Then
d
d
x xuu e e u
x .
d 1
d
x
u u
1 1 1 d d
22x xx u
e e uu
u
2
1 d
2u
u
-1
22
1 1 d tan
2 22
uu c
u
12
tan2 2
xec
(b)
2 2 2
1
0 0 1
4 3 d 4 3 d 4 3 da a
x x x x x x x x x
13 3
2 2
0 1
2 3 2 33 3
a
x xx x x x
321 1
2 3 2 3 2 33 3 3
aa a
322
2 2 33 3
aa a
22
2
1 d
1 ( 2)y
y = 1 1 ( 2)
ln2 1 ( 2)
yc
y
=
1 1ln
2 3
yc
y
222
x
xy
22
2
2 2 2 2
22
2
2
( 2)2
2( 2) ( 2)
2( 2)
1 ( 2)
2 2 where 2 and 2
1 ( 2)
xy
x
y x y x
yx
y
a by
When x = 2
1 ,
3
7y
Volume required
=
73
2
22
1 7 22 d
2 3 1 ( 2)y
y
=
73
2
2
7 22 d
12 1 ( 2)y
y
= 7
12
72 2
3
73
22
12 d
1 ( 2)y
y
= 5
4
73
2
1ln
3
y
y
= 5
4
3
4ln
3
2ln
= 5
ln 24
23
The intersection points are (2,1/3) and (3,1/2).
1/22 2 2
1/3
21/2
1/3
3
1 1Volume required= 3 2
2 3
9 1 4
2 1 3
191.0159
6
6.76 units
x dy
ydy
y
y
x
y=1
1x x=2 x=3
1/3 1/2
When , ,x t y 1
3 13
When , ,x t y 1 1
22 2
dx
dt t
2
1
dy x1
2
1
3
d dt
d
x
tt
2
23
1
4
dttt
2
223
1 1
4
a 3 b= 2
Using t = 2sin
When ,t
33
When ,t
24
1 1
2
1
2
1
3
x
y
cosdt
d
2
dttt
2
223
1 1
4
cos dsinsin
4
3
2 22
1 12
24 4
cos dec
4
3
21
4
cot
4
3
1
4
1 11
4 3
1 11
4 3
unit2
(=0.382)
24
25
At point P, t =1, x = 3, y = 4.
At point Q, t = 1 3 13
, ,2 4 8
x y
Area of the region
= 4
13
8
1 3 133 4 d
2 4 8x y
= 1
2 21
2
675(3 )(3 3) d
64t t t
= 5.32 units2
y = ex – 2,
When the graph crosses the x-axis, x = ln2.
2
0
12 d d
ln
kex
ee x x
x x
ln 2 2
0 ln 2
1
( 2) d ( 2) d dln
kex x
e
xe x e x xx
P
Q
C
2 0
ln 2 2
0 ln 22 2 ln ln
kex x
ee x e x x
2 1(2 2ln 2 1) ( 4 2 2ln 2) ln ln
2e k k
lnk = 2(e2 + 4ln2 – 7)
26
13
321 12
0 0
0
15
2
0
(2 1) 1(b) 2 1 d (2 1) d
3 3
1 1 (2 1)3 3 0
3 3 5
1 2 3 13 9 3 1
15 5 15
xx x x x x x
x
DIFFERENTIAL EQUATIONS
1. DHS 12/P1/Q11
(a)
2 de
d
x yz
x
2 22 2 2
2 2
d d d d de 2e e 2
d d d d d
x x xz y y y y
x x x x x
Given 2
1 4
2
d d2 e
d d
xy y
x x
2 1 4de e
d
x xz
x
1 4 2 1 2d d
e .e e (shown)d d
x x xz z
x x
Hence, 1 2d e dxz x
1 21
e2
xz C
2 1 2
1 4 2
1 4 2
d 1e e
d 2
d 1e e
d 2
1 1e e where and are arbitrary constants
8 2
x x
x x
x x
yC
x
yC
x
y C D C D
2. HCI 12/P1/Q8
(i) 2 2 36x y (ii) d
d
yky
t
Method 1
Implicit Differentiation w.r.t. time
d d2 2 0
d d
x yx y
t t
2 2
d d
d d
= ( )
(36 ) (shown)
x y y
t x t
yky
x
y k xk
x x
Method 2
Implicit Differentiation w.r.t x then using chain rule (rate of change equation)
d2 2 0
d
yx y
x
2 2
d d d d d d or
d d d d d d
d (36 ) = (shown)
d
y y x x x y
t x t t y t
x ky y k xk
xt x x
y
(iii) 2
2
2
2
2
2 4
2 4 4
d 2(36 ) (given)
d
d 2 d36
1 2 d 2 d
2 36
1ln 36 2
2
ln 36 4 '
36 e
36 e 36 e ( 0)
t
t t
x x
t x
xx t
x
xx t
x
x t C
x t C
x A
x A x A x
4
Using initial conditions, when 0, 4
4 3620
36 20e t
t x
AA
x
4
4
For to be 3,
36 9 27
27 36 20e9
e20
1 9ln 0.2 s
4 20
t
t
OY
OX
t
(iv)
3. NJC 12/P1/Q4
(i)
2
2
22 2 2 2
4 2 4
2
d d2
d d
d2 2 0
d
d
d
d shown
d
x ut
x utu t
t t
ut tu t t ut ut
t
ut u t
t
uu
t
(ii) 2
2
2
1 d 1dt
1'
'
where '
uu
t Cu
tt C
x
tx C C
t C
Since there was 0.2 milligrams of bacteria
after 15 minutes, then
2
0.25 10.2 0.05 0.2 0.0625
0.25 16C C
C
216
16 1
tx
t
Jesse’s model is not appropriate.
Based on Jesse’s model, the rod
would never fall flat on the
ground.
When t = 4,
216 4 256
or 3.9416 4 1 65
x
(iii) As
216, .
16 1
tt
t
The particular solution of the DE suggests that the amount of bacteria in the Petri
dish will grow indefinitely as time passes. Hence the model is not a realistic one.
4. RJC 12/P1/Q9
2d2 ( 1)e
d
xyy x
x
--- (1)
Let 2e xz y 2 2d d
e 2ed d
x xz yy
x x
Sub. into (1), 2d
2 ( 1)ed
xyy x
x
2 2de 2 e 1
d
x xyy x
x
d1
d
zx
x
( 1) dz x x
2
2
xz x c ,
22e
2
x xy x c
(i)
When 0, 1, 1.x y c
22The particular solution is e 1 .
2
x xy x
(ii)
2
22 2 2
2
3 22 2 2
3 2
d2 ( 1)e
d
d d2 e 2( 1)e (2 1)e
d d
d d2 2e 2(2 1)e 4 e
d d
x
x x x
x x x
yy x
x
y yx x
x x
y yx x
x x
2 3
2 3
2 3
2 3
d d dWhen 0, 1, 1, 1, 2
d d d
1 2Using Maclaurin's expansion, 1 ...
2! 3!
1 1 1 ...
2 3
y y yx y
x x x
y x x x
x x x
(iii) Replace x by -2x in the standard series expansion of ex and perform an expansion for
22e 1
2
x xx
up to and including the term in x3. Compare the coefficients of this
series with that of the expansion in (ii) to verify the correctness.
5. RJC 12/P2/Q2
(i)
d
, 0d
xkx k
t
1d dx k t
x
ln , 0x kt c x
e ktx A When 0t , 80x , thus 80A .
When 3t , 20x ,
320 80e k
1 1 1
ln ln 43 4 3
k .
Thus ln 4
380et
x
(ii)
2ln 4 1
When 6, 80e (80)16
t x
Just before the (n+1)th injection, the amount of drug present in the blood stream =
1
16nu
1
Immediately after the ( +1)th injection,
1amount of drug present, 80
16n n
n
u u
In the long run, the amount present approaches the value 85.3 (3 s.f.).
6. RVHS 12/P2/Q1
3 2
3 2
2 2 2
d(1 ) 0 (1)
d
d dLet
d d
Subs. into (1):
d0
d
d( 1) (shown)
d
yx x y x
x
u yu xy x y
x x
uy y x y x
x
uux x u x
x
3
3
3
2
3
3
333
1d d
1
ln | 1|3
1 e where e
e 1 1e 1 (shown) and f
3
xC
xx
u x xu
xu C
u A A
Au A y x x
x
(i)
(ii) 1A 1A
7. TJC 12/P1/Q11(b)
(i) Let (15 )dx kx x
dt .
Given that when t = 0, x = 5, 0.1dxdt
. 0.1 = k(5)(15 5) k = 1500
Hence, (15 )
500
x xdxdt
.
(ii) (15 )
500
x xdxdt
1500 1
(15 )dx dt
x x
500 1 115 15
dx t cx x
100 ln ln 153
x x t c
100 ln3 15
x t cx
as 0 15x
When t = 0, x = 5 100 100ln 0.5 ln 23 3
c
Hence, 100 2ln3 15
xtx
is the solution of the DE.
(iii) When x = 0.95(15) = 14.25 m2, 121.25t hours. (2dp)
It required a minimum of 122 hours for 95% of the soil area to become infected.