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8/11/2019 RI H2 Maths 2013 Prelim P1 Solutions
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8/11/2019 RI H2 Maths 2013 Prelim P1 Solutions
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H2 MA 9740/ P1/2013 RI Year 6 Preliminary Examination - Page 2 of 13
Hence,
2 3 sin 2 3 sin
6 6
2 3 sin sin6 6
2 3 sin cos cos sin sin cos cos sin6 6 6 6
1 1OR 2 3 2cos sin
2 6 6 2 6 6
PQ PR
4 3 cos sin6
34 3 sin
2
6sin
6 (since is small sin ) (shown).
2 3 and 6.a b
No. Solution
3(i) By GC,
2
3
4
14
21
30
u
u
u
3(ii)Conjecture:
2
1 5.nu n
LetnP be the statement
21 5nu n for .n
LHS =1 9u =
21 1 5 RHS
Hence1P is true.
AssumekP is true for some k
, i.e. 2
1 5k
u k .
To prove:1kP is true, i.e. to prove
2
1 ( 1) 1 5ku k
1
2
2
2
2 3
( 1) 5 2 3( 1) 2( 1) 1 5
1 1 5 RHS
k ku u k
k k
k k
k
Hence1kP is also true.
SincekP is true 1kP is true, and 1P is also true, by the
principle of Mathematical Induction,nP is true for all ,n
i.e. the conjecture is true.
No. Solution
4(i)2
4 4 2
4 4 2
ln 64
2ln ln 4 6
yy x x
y y x x
8/11/2019 RI H2 Maths 2013 Prelim P1 Solutions
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H2 MA 9740/ P1/2013 RI Year 6 Preliminary Examination - Page 3 of 13
Differentiate with respect tox:
3 3
43
2 2
44
d 2 d 4 4 12
d d
d 4 24 12
d
d 4 ( 3) 2 ( 3)
d 2 12 2 1
y yy x x
x y x
y yx x
x y
y xy x xy x
x yy
4(ii) When 2,y
4 2
2 2
2 2 2
16 6
8 2 0
8 or 2 (reject as 0)
2 2.
x x
x x
x x x
x
Hence,
4 2 2 (8 3) 4 2 2 (8 3)d or
d 2(16) 1 2(16) 1
40 2 40 2 or .
31 31
y
x
No. Solution
5(i)
OC OB BC k b a
5(ii) By the Ratio Theorem,
1 1 1 1
.2 2 2 2
kOY OC OA k
b a a b a
1 1 1 1
2 2 2 2
k kXY XO OY OA
b + b a
XY is parallel to OA.
1 1 1
2 2
XY XY k k
CB kOA k k
Hence the required ratio is 2 21 : 4k k
OA
BC
X Y
D
8/11/2019 RI H2 Maths 2013 Prelim P1 Solutions
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H2 MA 9740/ P1/2013 RI Year 6 Preliminary Examination - Page 4 of 13
No. Solution
6(i) 2 1 2
1 1 1 7
3 1 1
1
4 2 1 3 72
Position vector of the point of intersectionA is
OA
=2 1 31 11 1 3
2 23 1 5
6(ii) 1 2
1 1
1 1 2 2sin
33 6 3 2
.
6(iii)
BA OA OB
=3 2 11 13 1 1 .
2 25 3 1
Shortest distance
11 2 3 2 1
sin 1 .2 3 2 3 6
1
BA
OrShortest distance
1 221
1 11 12 21 11 12 .2 6 6 61
1
BA
8/11/2019 RI H2 Maths 2013 Prelim P1 Solutions
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H2 MA 9740/ P1/2013 RI Year 6 Preliminary Examination - Page 5 of 13
No. Solution
7 2
2
d
d
1d 1 d
1d 1 d
1ln ln 1 , where is an arbitrary constant
ln1
e , where e1
e 1 1 (Shown).
11 e e
e
t C
t
t t
t
NN kN
t
N tN kN
kN t
N kN
N kN t C C
Nt C
kN
NB B
kN
BN
Bk k Ak
B
Alternatively:
2
2
1d 1 d
1
d 1 d1
1ln , where is an arbitrary
1ln
1e , where
1 1e (Shown).
e
t C
t
t
N tN kN
N N t
kN
k t C C N
k t CN
k A A eN
A k NN k A
1When 0, 250 : 250 .
As , 10000 :
1 10000 0.0001 and so 0.0039.
t NA k
t N
k Ak
Therefore1
.0.0001 0.0039e t
N
Now, whenN= 750, 1
0.00011 750
750 e0.0001 0.0039e 0.0039
t
t
8/11/2019 RI H2 Maths 2013 Prelim P1 Solutions
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8/11/2019 RI H2 Maths 2013 Prelim P1 Solutions
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H2 MA 9740/ P1/2013 RI Year 6 Preliminary Examination - Page 7 of 13
2 2 1
10
0
2 0
0
1 1sin (2 ) sin (2 )
4 4
f ( ) f ( 1)
f (0) f (1)
f (1) f (2)
f (2) f (3)
f ( 1) f ( )
f ( ) f ( 1)
1 1 f (0) f ( 1) sin (2 )
4 4
nr r
n r rr
n
r
n
S x x
r r
n n
n n
n x
2 1
1
2 2 1
1
sin (2 )
1 sin ( ) sin (2 ) (proved)
4
n
n
n
x
x x
8(iii) Since 2 10 sin 2 1n x for all 0n and
11 0
4n as ,n
therefore 2sinnS x as n . Hence, 2sin .S x
No. Solution
9(i) Let $kbe the amount needed for the fund to award the scholarship for 2014.
(1.025) 2000k and thus2000
1951.22 19511.025
k (correct to the nearest dollar)
For additional amount needed :
Mtd 1 :
Additional amount needed
=2
2000$ $1903.63 $1904
(1.025)
Mtd 2 :
Let x be the additional amount needed.
1.025(1.025( ) 2000) 2000
3854.1951
3854.1951 1951.22 1904 (nearest $).
k x
k x
x
Hence 1904x (nearest $).
9(ii) Method 1
Amount needed
=2 10
2000 2000 2000....
1.025 (1.025) (1.025)
=
10
11
20001.02511.025
11.025
= 17504.13
= 17504 dollars (to nearest dollars)
8/11/2019 RI H2 Maths 2013 Prelim P1 Solutions
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H2 MA 9740/ P1/2013 RI Year 6 Preliminary Examination - Page 8 of 13
Method 2Either
10
1010
10
10
0
1.025 11.025 2000 0
1.025 1
1.025 12000
1.025 1 17504 (nearest $)1.025
a
x
x
Or
9
910
9
10
2000
1.025 11.025 2000(1.025) 2000
1.025 1
1.025 12000 2000(1.025)
1.025 117504 (nearest $)
1.025
b
x
x
9
(iii)
Method 1
Amount needed
2 3
2000 2000 2000...
1.025 (1.025) (1.025)
=2000 1
11.0251
1.025
= 80 000 dollars
Method 2
Either
0
1.025 11.025 2000 0
1.025 1
1.025 12000
1.025 1 180000 1 80000
1.025 1.025
n
nn
n
n n
a
x
x
(as1
01.025n
as n )
Or
8/11/2019 RI H2 Maths 2013 Prelim P1 Solutions
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H2 MA 9740/ P1/2013 RI Year 6 Preliminary Examination - Page 9 of 13
1
1
1
2000
1.025 11.025 2000(1.025) 2000
1.025 1
1.025 12000 2000(1.025)
1.025 1
1.025
2000 82000(1.025 1)
1.025
82000 8000080000
1.025 1.025
n
nn
n
n
n
n
n
b
x
x
(as1
01.025n
as n )
Method 3
For scholarship to continue indefinitely, yearly interest earned should be sufficient for thescholarship awarded for the following year. Hence
Minimum amount =2000
800000.025
No. Solution
10
2
2 2
2
2 5 2f
3 2 3 3
3 5 2 3 2 2 3 6 39 45 8 24 18
2 15 27 0
2 3 9 0
3 3 9 or Since .
2 2
xx
x
x x xx x x x
x x
x x
x x x
10(i)
3
9 2
2 3f ln ln 9 or ln
3 2
0 e or e .
x x x
x x
10(ii) 1 2 3f 0.5 9 (no solution) or 0.5
2 3 2
0.5 1.5 or 0.5 1.5
1 or 2.
x x x
x x
x x
8/11/2019 RI H2 Maths 2013 Prelim P1 Solutions
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H2 MA 9740/ P1/2013 RI Year 6 Preliminary Examination - Page 10 of 13
No. Solution
11(i) 2 k
2
2
2
2 2
16 16 0
2
13 5
2
yx x
yx
2 2
3 11
25 50
x y
Ellipse:
11(ii)
2
2
2 2
16 9 25
3 11
25 25
yx x
k
x y
k
Hyperbola : centre 3,1
Oblique Asymptotes : 3 1 y k x k and
1 3 y k x k
55
x
y 3 1 y k x k 1 3 y k x k
( 3,1)
2 2
3 11
25 25
x y
k
O
y
2 2
3 11
25 50
x y
x
5
5 2
3,1
0
8/11/2019 RI H2 Maths 2013 Prelim P1 Solutions
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H2 MA 9740/ P1/2013 RI Year 6 Preliminary Examination - Page 11 of 13
11(iii) Equation of a line of symmetry: 3 or 1x y .
Translation by 3 units in the positive x - direction
Translation by 1 unit in the negative y - direction
No. Solution
12(a) 22
1
23 32
11
23 3
1
ln d
ln 1d
3 3
ln
3 9
8ln 2 8 1
3 9 9
8ln 2 7
3 9
x x x
x x xx
x
x x x
12(b)
2 2
1
9 9d d
3 2 4 ( 1)
1 9sin
2
x xx x x
xc
221
20 0
9 1d 9sin 9 3
2 6 63 2
xx
x x
1 12 2
1122
2 1 1 1 11 2 d2 4 2 4
x x x x
12
1 12 2
2
20
9 1Since d 3 , so .
23 2
1 2 d 2 1 d 3a
x ax x
x x x x
OR
12
12
11 2 d (2)(1)
2x x
ax
12
y
2 2x
8/11/2019 RI H2 Maths 2013 Prelim P1 Solutions
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H2 MA 9740/ P1/2013 RI Year 6 Preliminary Examination - Page 12 of 13
12
1 12 2
12
2
20
2
2
2
74
9 1Since d 3 , so .
23 2
1 2 d 2 1 d 3
3
1 1( ) 2
4 2
70
4
1 1 4( ) 1 8
2 2
1Since ,
2
1 2 22
a
a
x ax x
x x x x
x x
a a
a a
a
a
a
No. Solution
13(i)
d d2 ; 4 10 2 2 5
d d
x yt t t
t t
Thus, 2 2 5d 2 5
.d 2
ty t
x t t
For min. pt.,d 5
0
d 2
yt
x
Hence coordinates of the min. pt. is 14.25,3.5
Coordinates ofAis (8,16) .
13(ii)
13(iii) The equation of tangent at Pis
O
14.25,3.5 x
y
C
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2 2
3 2 2
3 2 3 2
2
2 5 2 10 16 8
2 10 16 (2 5) 8
2 10 16 2 2 16 5 5 40
2 5 5 40
py p p x p
p
py p p p p x p
py p p p px p p x p
py px x p
2
2 5 5 40.py p x p 13(iv) Since the tangent passes through 0, 0 ,
25 40 0
2 2
p
p
Hence the possible coordinates of Pare
16,32 20 2
1 132 20 2 32 20 2 tan arg tan
16 16
0.228 arg 1.311
z
z
END OF PAPER