RAFFLES INSTITUTION 2013 Year 6 Preliminary Examination Paper 1 H2 Mathematics 9740 Solutions
No. 1
Solution
1
1 x2
1
1 2 2 x
12 32 1 2 1 x x2 2! 2
2
x 2 3x 4 2 8 d 1 Since sin 1 x , so dx 1 x2 1 sin 1 x dx 1 x2 1
x 2 3x 4 . 1 dx 2 8 x3 3x5 x c, where c is an arbitrary constant. 6 40 When x 0 , c sin 1 0 0 . Hence sin 1 x x where a
x3 3x5 6 40
3 . 40
No
Solution
2
P 2 3
6
Q
6
3
R
2 QPR = . 6 6 3 By sine rule, PQ PR 3 3 6 2 3 sin 2 3 3 sin sin 3 6 6 2 PQ 2 3 sin and PR 2 3 sin . 6 6
Hence,
PQ PR 2 3 sin 2 3 sin 6 6 2 3 sin sin 6 6 2 3 sin cos cos sin sin cos cos sin 6 6 6 6 1 1 OR 2 3 2 cos sin 2 6 2 6 6 6
sin 6 3 sin 2
4 3 cos
4 3 6sin 6 (since is small sin ) (shown).
a 2 3 and b 6. No. 3(i)
Solution By GC,
u2 14 u3 21 u4 30
3(ii) Conjecture: u n 12 5. n Let Pn be the statement “ un n 1 5 ” for n . 2
LHS = u1 9 = 1 1 5 RHS Hence P1 is true. 2
Assume Pk is true for some k , i.e. uk k 1 5 . 2
To prove: Pk 1 is true, i.e. to prove uk 1 (k 1) 1 5 uk 1 uk 2k 3 2
(k 1) 2 5 2k 3 (k 1) 2 2(k 1) 1 5 k 1 1 5 RHS 2
No. 4(i)
Hence Pk 1 is also true. Since Pk is true Pk 1 is true, and P1 is also true, by the principle of Mathematical Induction, Pn is true for all n , i.e. the conjecture is true. Solution
y2 y 4 ln x 4 6 x 2 4 y 4 2ln y ln 4 x 4 6 x 2 H2 MA 9740/ P1/2013 RI Year 6 Preliminary Examination - Page 2 of 13
Differentiate with respect to x:
4y 3
dy 2 dy 4 x3 12 x dx y dx
dy 4 y 4 2 3 4 x 12 x dx y
dy 4 xy ( x 2 3) 2 xy ( x 2 3) dx 2 2 y 4 1 2 y4 1 4(ii) When y 2, 16 x 4 6 x 2 x 2 8 x 2 2 0 x2 8
or x 2 2 (reject as x 2 0)
x 2 2.
Hence, 4 2 2 (8 3) dy 4 2 2 (8 3) or dx 2(16) 1 2(16) 1
40 2 40 2 or . 31 31
No. 5(i)
Solution C
B D X
Y A
O
5(ii)
OC OB BC b ka By the Ratio Theorem, 1 1 1 1 k a. OY OC OA b ka a b 2 2 2 2
1 1 k 1 k 1 XY XO OY b + b a OA 2 2 2 2
XY is parallel to OA. XY XY 1 1 k 1 k CB kOA k 2 2k Hence the required ratio is 1 k : 4k 2 2
H2 MA 9740/ P1/2013 RI Year 6 Preliminary Examination - Page 3 of 13
No. 6(i)
6(ii)
6(iii)
Solution 2 1 2 1 4 2 1 3 7 1 1 1 7 2 1 1 3 Position vector of the point of intersection A is 2 1 3 1 1 OA = 1 1 3 3 2 1 2 5
1 2 1 1 1 1 2 2 . sin 3 3 6 3 2
3 2 1 1 1 BA OA OB = 3 1 1 . 2 2 5 3 1 1 1 2 3 2 1 Shortest distance BA sin 1 . 2 3 2 3 6 1
Or Shortest distance 2 1 1 2 1 1 BA 1 2 1 2 1 1 1 1 2 . 2 6 6 6 1 1
H2 MA 9740/ P1/2013 RI Year 6 Preliminary Examination - Page 4 of 13
No. 7
Solution
dN N kN 2 dt 1 dN 1 dt N kN 2 k 1 dN 1 dt N 1 kN ln N ln 1 kN t C , where C is an arbitrary constant N t C 1 kN N Bet , where B eC 1 kN Bet 1 1 (Shown). N t 1 Bke k 1 k Ae t Bet Alternatively: 1 N kN 2 dN 1 dt 1 2 N dN 1 dt 1 k N 1 ln k t C , where C is an arbitrary N ln
ln
1 k t C N
1 k Aet , where A e C N 1 1 (Shown). Aet k N N k Ae t 1 When t 0, N 250 : 250 . A k As t , N 10000 : 1 10000 k 0.0001 and so A 0.0039. k
Therefore N
1 . 0.0001 0.0039e t
Now, when N = 750,
1 0.0001 1 750 750 e t t 0.0001 0.0039e 0.0039
H2 MA 9740/ P1/2013 RI Year 6 Preliminary Examination - Page 5 of 13
1 750 0.0001 ln 37 t ln 0.0039 117 1.1513 weeks 8 days (nearest day). Hence, the number of days required is 8 days. No. 8(i)
sin A sin A 1 cos A 4
2
Solution
2
sin 2 A sin 2 A cos 2 A 1 2 2sin A cos A 4 1 sin 2 A sin 2 2 A (shown). 4 1 1 1 S n sin 4 x sin 4 2 x 2 sin 4 22 x n sin 4 2n x 4 4 4 1 2 2 sin x 4 sin 2 x 1 sin 2 2 x 1 sin 2 22 x 4 42 1 2 sin 2 22 x 13 sin 2 23 x 4 4 1 1 2 2 n 1 n 4n 1 sin 2 x 4n sin 2 x 1 sin 2 2n x 1 sin 2 2n 1 x 4n 4n 1 1 sin 2 x n 1 sin 2 2n 1 x (proved). 4 sin 2 A
8(ii)
Alternatively, Let f(r )
1 sin 2 (2r x) 4r
H2 MA 9740/ P1/2013 RI Year 6 Preliminary Examination - Page 6 of 13
n 1 1 S n r sin 2 (2r x) r 1 sin 2 (2r 1 x) 4 r 0 4 n
f (r ) f (r 1) r 0
f (0) f (1) f (1) f (2) f (2) f (3) f (n 1) f (n) f (n) f (n 1) 1 1 sin 2 (20 x) n 1 sin 2 (2n 1 x) 0 4 4 1 sin 2 ( x) n 1 sin 2 (2n 1 x) (proved) 4
f (0) f (n 1)
8(iii)
Since 0 sin 2 2n1 x 1 for all n 0 and 1 0 as n , 4 n1 therefore S n sin 2 x as n . Hence, S sin 2 x .
No.
Solution
9(i)
Let $k be the amount needed for the fund to award the scholarship for 2014.
(1.025)k 2000 and thus k
2000 1951.22 1951 (correct to the nearest dollar) 1.025
For additional amount needed : Mtd 1 : Additional amount needed
2000 $1903.63 $1904 = $ 2 (1.025) Mtd 2 : Let x be the additional amount needed. 1.025(1.025(k x) 2000) 2000 k x 3854.1951 x 3854.1951 1951.22 1904 (nearest $).
Hence x 1904 (nearest $). 9(ii)
Method 1 Amount needed 2000 2000 2000 .... = 2 1.025 (1.025) (1.025)10 1 1 2000 1.02510 = = 17504.13 1.025 1 1 1.025 = 17504 dollars (to nearest dollars) H2 MA 9740/ P1/2013 RI Year 6 Preliminary Examination - Page 7 of 13
Method 2 Either a10 0 1.02510 1 1.02510 x 2000 0 1.025 1 1.02510 1 2000 1.025 1 17504 (nearest $) x 1.02510
Or b9 2000 1.0259 1 1.02510 x 2000(1.025) 2000 1.025 1 1.0259 1 2000 2000(1.025) 1.025 1 17504 (nearest $) x 1.02510
9 (iii)
Method 1 Amount needed
2000 2000 2000 ... 2 1.025 (1.025) (1.025)3 2000 1 = = 80 000 dollars 1.025 1 1 1.025 Method 2 Either an 0 1.025n 1 1.025n x 2000 0 1.025 1 1.025n 1 2000 1.025 1 80000 1 1 x n 1.025n 1.025
(as
80000
1 0 as n ) 1.025n
Or
H2 MA 9740/ P1/2013 RI Year 6 Preliminary Examination - Page 8 of 13
bn 2000 1.025n 1 1 1.025 x 2000(1.025) 2000 1.025 1 n
1.025n 1 1 2000 2000(1.025) 1.025 1 1.025n 2000 82000(1.025n 1 1) 1.025n 82000 80000 80000 1.025 1.025n
x
(as
1 0 as n ) 1.025n
Method 3 For scholarship to continue indefinitely, yearly interest earned should be sufficient for the scholarship awarded for the following year. Hence Minimum amount =
No. 10
f x
2000 80000 0.025
Solution
2 x5 2 3 2x 3 3 3 x 5 2 x 3 2 2 x 3
2
6 x 2 39 x 45 8 x 2 24 x 18 2 x 2 15 x 27 0 2 x 3 x 9 0 x 9 or x
10(i)
f ln x
2 ln x 9 or 3 9
10(ii)
3 2
3 Since x . 2 3 ln x 2 3 2
or x e . 0 xe 1 2 3 f x x 0.5 9 (no solution) or x 0.5 2 3 2 x 0.5 1.5 or x 0.5 1.5 x 1 or x 2.
H2 MA 9740/ P1/2013 RI Year 6 Preliminary Examination - Page 9 of 13
No. 11(i)
Solution
k 2
x
2
6 x 16
x 3
2
x 3
y 1
25
0
2
y 1
2
2
2
2
y 1
52
2
50
1
Ellipse:
y x 3 25
2
y 1
2
1
50
3,1 5 x
0 5 2
11(ii)
y k x 1 3 k
y
y k x 3 k 1
x 3 25
2
y 1 25k
2
1
( 3,1) 5 5 O
x
2
6x 9
x 3 25
2
y 1
2
k
y 1 25k
x
25
2
1
Hyperbola : centre 3,1
Oblique Asymptotes : y k x 3 k 1 and
y k x 1 3 k
H2 MA 9740/ P1/2013 RI Year 6 Preliminary Examination - Page 10 of 13
11(iii) Equation of a line of symmetry: x 3 or y 1 . Translation by 3 units in the positive x - direction Translation by 1 unit in the negative y - direction No. 12(a)
Solution
2
x 2 ln x dx
1 2
x3 ln x 3 1
2
1
x3 1 dx 3 x
2
x3 ln x x3 9 1 3
12(b)
8ln 2 8 1 9 9 3 8ln 2 7 3 9 9 dx 2 3 2x x
9 4 ( x 1)
2
dx
x 1 9sin 1 c 2
2
x 1 dx 9sin 1 9 2 2 0 6 3 2x x
2
9
0 1 2
1
1
1 1
1
1 2 x dx x x 2 2 1 2 12 2 4 2 4
Since
OR
1 2
12
2 0
9 3 2x x
1 2 x dx
1 dx 3 , so a . 2
2
a 1 2
2 x 1 dx 3
y
2x
2
a
1 2
3 6
1 2
12
1
x
1 2 x dx (2)(1) 2
H2 MA 9740/ P1/2013 RI Year 6 Preliminary Examination - Page 11 of 13
Since
1 2
12
2 0
9 3 2 x x2
1 2 x dx
1 dx 3 , so a . 2 a 1 2
2 x 1 dx 3
a
x 2 x 1 3 2
1 1 (a 2 a) 2 4 2 7 a2 a 0 4 a
1 1 4( 74 )
2 1 Since a , 2 1 2 2 a 2 No. 13(i)
1 8 2
Solution dx dy 2t ; 4t 10 2 2t 5 dt dt d y 2 2t 5 2 t 5 Thus, . dx 2t t dy 5 For min. pt., 0t dx 2 Hence coordinates of the min. pt. is 14.25, 3.5
Coordinates of A is (8,16) . 13(ii)
y
C
14.25,3.5
x
O
13(iii) The equation of tangent at P is
H2 MA 9740/ P1/2013 RI Year 6 Preliminary Examination - Page 12 of 13
y 2 p 2 10 p 16
2p 5 x p 2 8 p
py 2 p3 10 p 2 16 p (2 p 5) x p 2 8 py 2 p3 10 p 2 16 p 2 px 2 p3 16 p 5 x 5 p 2 40 py 2 px 5x 5 p 2 40 py 2 p 5 x 5 p 2 40. 13(iv) Since the tangent passes through 0, 0 ,
5 p 2 40 0 p 2 2 Hence the possible coordinates of P are
16,32 20 2
32 20 2 1 32 20 2 tan 1 arg z tan 16 16 0.228 arg z 1.311
END OF PAPER
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