Ri h2 Math p2 Solutions

Raffles Institution 2011 Year 6 Preliminary Examination Paper 2 H2 Mathematics 9740 SOLUTIONS Section A: Pure Mathematics [40 marks] 1

Since one of the root of the equation is 1  2i  1  2i   a 1  2i   10 1  2i   25  0 4

[6m]

3

  7  24i   a  11  2i   10  20i   25  0  11a  2ai=  22  4i  a(11  2i)  2(11  2i) OR: Compare real or imaginary part: 11a  22 or  2a  4 a=2

(since a is real)

Clearly, the other root is 1  2i z 4  2 z 3  10 z  25  0  ( z  1  2i)( z  1  2i)( z 2  bz  c )  ( z 2  2 z  5)( z 2  bz  5)

 2b  5  5  0  b0

(by inspection, c  5 ) (compare coefficient of z 2 )

z 4  2 z 3  10 z  25  ( z 2  2 z  5)( z 2  5)  0  z  1  2i, 1  2i, 5,  5

Let z =  (w  1 ) Therefore w = 2i,  2i, 1  5, 1  5

2(i) [1m]

Since a42 , a43 , a44 are in AP, a44  a43  a43  a42 4  3 2  a44  2a43  a42  2     .  81  81 81

(ii) [2m]

Since a24 , a34 , a44 are in GP, 4 4 1 1 a44  a24 r 2   r 2  r 2   r   81 9 9 3 1 Since the array contains positive real numbers, r > 0 so r  . 3

2(iii) © RI 2011

We have 2011 RI Year 6 Preliminary Examinations/H2 Mathematics 9740 – Page 1 of 11

[3m]

a24  a14 r  (a11  3d )r...(1) a42  a12 r 3  (a11  d )r 3 ...(2) Hence 4 1 4  a24  (a11  3d )   a11  3d 9 3 3 2 1 2  a42  (a11  d )   a11  d 81 27 3 Solving both equations (via GC) or simply subtracting the 2nd equation from the 1st, we 1 get a11  d  . 3 Alternatively, 1 1 a42  a41   a11r 3  . Solve simultaneously with (1). 81 81

(iv) [2m]

Method 2 akk  ak 1  (k  1)dr k 1

Method 1 akk  a1k r k 1

 (a11  (k  1)d )r k 1 1  1  1    (k  1)    3  3  3 1 1  (1  (k  1))   3  3 1 k  3

(v) [3m]

 a11r k 1  (k  1)dr k 1 k 1

k 1

k

11    33

k 1

11  (k  1)   3 3

k

1    (1  k  1)  3 1 k  3

k

1 1 2 3 n Sn  Sn   2  3    n 3 3 3 3 3 1 2 n 1 n  2  2    n  n 1 3 3 3 3 1 1 1 1 n   2  3    n  n 1 3 3 3 3 3 

1 3

(1   13  ) n

1

1 3



n 3n 1

1 n n  (1   13  )  n 1 . 2 3

Hence 2 1 n 3 1 n n S n  (1   13  )  n 1  S n  1  n   . 3 2 3 4  3  2(3n )

© RI 2011

2011 RI Year 6 Preliminary Examinations/H2 Mathematics 9740 – Page 2 of 11

k 1

3(i)

y

[1m]

1

x  a sin 4  , y  b cos 4 

2

O

3(ii)

At (a sin 4  , b cos 4  ),

[3m]

dx  4a sin 3  cos  d so

3(iii) [3m]

x

dy dy  dx d

and

dy  4b cos3  sin  , d

b cos 2  b dx 4b cos3  sin       cot 2  (Shown). 3 2 d 4a sin  cos  a sin  a

The equation of the tangent at (a sin 4  , b cos 4  ) is

b y  b cos 4    cot 2   x  a sin 4   . a Setting y  0 and solving for x gives x  a sin 2   cos 2   sin 2    a sin 2  .

 Coordinates of P are (a sin 2  , 0). Similarly, setting x  0 and solving for y gives y  b cos 2  .

 Coordinates of Q are (0, b cos 2  ) . 3(iv) [3m]

From (iii), OP  a sin 2  and OQ  b cos 2  , so

1 ab ab A  (OP)(OQ)  sin 2  cos 2   sin 2 2 . 2 2 8 A is maximum when sin 2 = 1, i.e.  

Maximum value of A 

© RI 2011



  for    0,  , 4  2

ab 2    ab sin  2    . 8 4 8 

2011 RI Year 6 Preliminary Examinations/H2 Mathematics 9740 – Page 3 of 11

Alternatively,

From (iii), OP  a sin 2  and OQ  b cos 2  , so

1 ab A  (OP)(OQ)  sin 2  cos 2  . 2 2







dA ab  2 sin  cos3   2sin 3  cos   ab sin  cos  cos 2   sin 2  d 2   For    0,  ,  2 dA   0, cos 2   sin 2   tan 2   1    when d 4



4

ab 2    ab    ab  1  Maximum value of A   sin   cos 2    .   2 8 4 4 2  2

4(i) [3m]

4(ii) [3m]

2 0 1  p       For l1 and l2 to intersect, 0   1 = 3    0  for some  ,   . 1 5 4  2         p  2    0     3    3,   6, p  0 5  2  3  Hence coordinates of A  (6, 3,16) q 2 l3 has equation r   0     1 ,   1 5        q    2   OA   0     1 1  5        Shortest distance from A to l3  2  1 5   

1 22  (1) 2  52

6 q  2   3    1  15   5     

1 0  5q  30  q    q 52  12 (since q  0)  30 



© RI 2011

13 q 15

2011 RI Year 6 Preliminary Examinations/H2 Mathematics 9740 – Page 4 of 11

4(iii) [4m]

q 2      OB   0     1 for some   1 5     15 13 Now AB  , so q  15 by (i). 15    2  Since AB  l1 , AB   1  0 5    9   2   2     3     1    1  0  5   5    15         (18  3  75)   (22  (1) 2  52 )  0 60 2 30  15   2   19  So OB   0   2  1   2   1   5   11         

4(iv) [3m]

Note that (0, 0,1) and (15,0,1) lie on l1 and l3 respectively, so they lie on 1 , and hence

15   0  15  1 a vector parallel to 1 is  0    0    0   15  0  .  1  1  0  0         1  2   0  0       So a normal vector to 1 is 0  1  5  (1)  5  .  0   5   1  1         0 0 0 So equation of 1 is r   5    0    5  1 1 1       0  r  5  1 1  

Section B: Statistics [60 marks] 5(i) [2m]

Number of ways  215  15C0  15C1

 32768 1 15  32752 OR 15

  15Cr  15C2  15C3  15C4    15C15  32752 r 2

© RI 2011

2011 RI Year 6 Preliminary Examinations/H2 Mathematics 9740 – Page 5 of 11

5(iia) [1m]

Number of ways  9!  362880

OR

 9C3  6C3  3C3   3!  362880 3

5(iib) [2m]

Number of ways  2!  6C1  7 !  60480

OR

 2!  6C1  7C1  6C3  3C3   3!  60480 2

5(iii) [2m]

Number of ways = (6  1)!  6  5  4 = 14400 OR

Number of ways   9  1 !   7  1 ! 3!  6  1 ! 3 P2  6 P2  14400

6(i)

Probability

[2m]

 2  4   3  5  23           5  10   5  10  50

6(ii) [5m]

Probability  P(transferring from A to B and drawing a white ball from B) + P(transferring from B to A and drawing a white ball from A)

1  23  3  5  3   4  4             4  50  4  9  6   9  6  491  900 P (ball transferred from A to B │final ball chosen is white) P  transfer from A to B and final ball chosen is white   P  final ball chosen is white   1   23   23       4 50 200  207        491  491  982   900  900 

7(i)

x (micrograms per litre)

[1m]

t (hours)

7(ii) [1m] © RI 2011

Product moment correlation coefficient =  0.920 (3 s.f.) 2011 RI Year 6 Preliminary Examinations/H2 Mathematics 9740 – Page 6 of 11

7(iii) [2m]

Equation of the regression line of x on t is x = 120.743 – 12.6179t i.e. x = 121 − 12.6t (to 3 s.f.) The gradient 12.6 means that the estimated decrease in the concentration of drug in the bloodstream is 12.6 micrograms per litre for every 1 hour increase in time.

7(iv) [2m]

Using the equation of the line in (iii), when x = 50, t = 5.61,  Estimated time is 5.61 hours. The estimate is not reliable as the scatter diagram in (i) shows that a curve is a better fit for the data points, so a linear model is not suitable.

8(i) [1m]

The probability of concluding that the population mean GPA score is more than 2.9, when it is actually 2.9, is 0.1.

8(ii)

Note: The t-test should be used, because the sample size is small, and the population variance is unknown.

[6m] To test H0 : μ = 2.9 vs H1 : μ > 2.9 Perform a 1 tail t-test at 10% level of significance. Under H0 , T 

X  0

~ t (n  1) S/ n where 0 = 2.9, x = 3.12, s = 0.72821, n = 15

Using a t-test, p-value = 0.131 (3 s.f.) Since p-value = 0.131 > 0.1, we do not reject H0 and conclude that there is insufficient evidence at the 10% level of significance that the population mean GPA score is more than 2.9. Assumption is that the GPA scores of the student population follow a normal distribution.

9(i) [2m]

Let X be the number of students, out of 16, who wore brown shoes to school. X ~ B (16, 0.17). P ( X  3)  0.244. (3 s.f.)

9(ii) [5m]

Let Y be the number of students, out of 60, who wore white shoes to school. Y ~ B  60, 0.58  . Since n = 60 > 50 is sufficiently large and p = 0.58 are such that np  34.8  5 and n(1  p )  25.2  5, Therefore Y ~ N  np, np (1  p)  approximately i.e. Y ~ N  34.8, 14.616  approximately.

© RI 2011

2011 RI Year 6 Preliminary Examinations/H2 Mathematics 9740 – Page 7 of 11

P Y  38   P Y  38.5   by continuity correction   0.167.  3 s.f.

10(i) [2m]

Let X be the number of defects in a box.  2.5  X ~ Po   3.2   40  i.e. X ~ Po (0.2) P  X  1  0.98248  0.982 (3 s.f.)

10(ii) [2m]

P(a package is accepted)  P  box has at most one defect and its barcode is not defective   0.98248  0.96  0.94318  0.943 (3 s.f.)

10(iii) [4m]

Let Y be the number of packages, out of 70, that are rejected. Y ~ B  70,1  0.94318 

Since n = 70 is large, p = 0.05682 is small such that np  3.9774  5 , therefore Y ~ Po  3.9774  approximately. P (more than 67 packages are accepted)  P Y  2   0.241 (3 s.f.)

11(i) [2m]

Let X be mass of an abalone in grams. X ~ N 180,  2





P  X  200   0.08

200  180    PZ    0.08    20    P  Z    0.92    From G.C., 20  1.4051

   14.234  14.2 (3 s.f.) (shown)

11(ii) [1m] 11(iii) [3m]

© RI 2011

P  X  165   0.14541  0.145 (3 s.f.)

 14.22  X ~ N 180,  15  

2011 RI Year 6 Preliminary Examinations/H2 Mathematics 9740 – Page 8 of 11



 



P X  180  5  P X  180  5  P ( X  180  5)

   2 P  X  185 

 or 2P( X  180  5)   

 2 P X  180  5

  or 2P( X  175) 

 0.173 (3 s.f.)

Alternative 1,

 14.22  X  180 ~ N  0,  15  



 



P X  180  5  P X  180  5  P ( X  180  5)



 2 P X  180  5



 0.173 (3 s.f.)

Alternative 2,

 X ~ N  2700, 3024.6 





P X  180  5  P   X  2700  75   P ( X  2700  75)  P(  X  2700  75)  2 P   X  2775   0.173 (3 s.f.)

11(iv) [3m]

Let C be cost of one abalone in dollars. 450 2 C X ~ N 81,  6.39  1000







T  C1  C2  C3  C4  C5 ~ N 5  81, 5   6.39 

2



 T ~ N  405, 204.1605  P T  420   0.147 (3 s.f.)

Alternatively,



Let W  X 1  X 2  X 3  X 4  X 5 ~ N 5 180, 5  14.2 

2



 W ~ N  900, 1008.2 

Probability required  P  450 W  420   1000   P W  933.33   0.147 (3 s.f.)

12(i) © RI 2011

Unbiased estimate of population mean 2011 RI Year 6 Preliminary Examinations/H2 Mathematics 9740 – Page 9 of 11

=

 ( x  50)  50  100  50 n

[2m]

120 1  49 or 49.167 (5 s.f )  49.2 (3 s.f ) 6

Unbiased estimate of population variance 2 ( x  50)    1   2    ( x  50)  n 1  n    2 1 (100)  (1158  )  9.0308  9.03 (3 s.f ) 119 120

12(ii) [4m]

Since n = 120 is large, by Central Limit Theorem, 9.0308 X N (49.167, ) approximately. n P ( X  49.9)  0.01 Method 1 : Using Standardization P ( X  49.9)  0.99 49.9  49.167 )  0.99 9.0308 n 49.9  49.167 From GC ,  2.3263 9.0308 n n  90.960 P(Z 

 Least value of n  91

Method 2 : Using table of values Key in Y1 = normalcdf(49.9, E99, 49.167, n 90 91 92

9.0308 / X )

P( X  49.9) 0.01033 > 0.01 0.00999 < 0.01 0.00965 < 0.01

 Least value of n = 91 The population is first divided into 4 strata according to 4 levels – Sec 1 to 4. Random samples, of sample sizes proportional to the relative sizes of the strata, are then drawn separately and independently (by using simple random sampling) from each stratum (level). OR (divide into 2 strata according to gender) © RI 2011

2011 RI Year 6 Preliminary Examinations/H2 Mathematics 9740 – Page 10 of 11

(iii)

© RI 2011

Stratified sampling is a better sampling method compared to simple random sampling because it is likely to give good representative samples of the population, i.e. students from each stratum (level or gender) are well represented as in the population.

2011 RI Year 6 Preliminary Examinations/H2 Mathematics 9740 – Page 11 of 11