Ri h2 Math p2 Solutions

(1)

Raffles Institution

2011 Year 6 Preliminary Examination Paper 2 H2 Mathematics 9740

SOLUTIONS

Section A: Pure Mathematics [40 marks]

1

[6m]

Since one of the root of the equation is 1 2i 



1 2i



4a



1 2i



310 1 2i







25 0 



 7 24i

 

a  11 2i

 

 10 20i



25 0  11 a2 i= 22 4ia  

a( 11 2 ) 2( 11 2 )  i    i

OR: Compare real or imaginary part: 11 a 22 or 2 a  4

 a = 2 (since a is real)

Clearly, the other root is 1 2i

4 ₂ 3 ₁₀ _{25 0} z  z  z  2 (z 1 2i)(z 1 2i)(z bz c)        2 2 (z 2z 5)(z bz 5)      (by inspection, c  ) 5  2    (compare coefficient ofb 5 5 0 z ) 2  b 0 4 ₂ 3 ₁₀ ₂₅ z  z  z (z22z5)(z25)  0 1 2i, 1 2i, 5, 5 z      Let z =  (w  1 )

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

[1m]

Since a₄₂,a₄₃,a are in AP, ₄₄

44 43 43 42 44 43 42 3 2 4 2 2 . 81 81 81 a a a a a a a          _ _    (ii) [2m] Since a₂₄,a₃₄,a are in GP, ₄₄ 2 2 2 44 24 4 4 1 81 9 9 a a r   r r   1 3 r 

Since the array contains positive real numbers, r > 0 so 1 3

r .

(2)

[3m] 24 14 11 3 3 42 12 11 ( 3 ) ...(1) ( ) ...(2) a a r a d r a a r a d r       Hence 24 11 11 42 11 11 4 1 4 ( 3 ) 3 9 3 3 2 1 2 ( ) 81 27 3 a a d a d a a d a d            

Solving both equations (via GC) or simply subtracting the 2nd equation from the 1st, we get a₁₁ d  1 3. Alternatively, 3 42 41 11 1 1 81 81

a a  a r  . Solve simultaneously with (1).

(iv) [2m] Method 1 1 1 1 11 1 1 ( ( 1) ) 1 1 1 ( 1) 3 3 3 1 1 (1 ( 1)) 3 3 1 3 k kk k k k k k a a r a k d r k k k            _   _{ }         _{ }       _{ } Method 2 1 1 1 1 11 1 1 ( 1) ( 1) 1 1 1 1 ( 1) 3 3 3 3 1 (1 1) 3 1 3 k kk k k k k k k k a a k dr a r k dr k k k                 _{ }   _{ }       _{ }         _{ } (v) [3m]

 

2 3 2 2 1 2 3 1 1 1 3 3 1 1 3 1 3 1 1 1 2 3 3 3 3 3 3 1 2 1 3 3 3 3 1 1 1 1 3 3 3 3 3 (1 ) 1 3 1 (1 ) . 2 3 n n n n n n n n n n n n S S n n n n n                                 Hence

 

1 3 1 2 1 3 1 (1 ) 1 . 3 2 3 4 3 2(3 ) n n n n n n n n S    _ S  _  _  

(3)

3(i) [1m] 3(ii) [3m] At ₍ _sin4 _{, cos}4 ₎_, b a   3 d 3

sin cos and 4 cos sin , d d 4 d y b x a          so 3 2 2 3 2

cos sin cos

cot (Shown). sin co d d d 4 d d d 4 s sin y y x b b b x a a a    _             3(iii) [3m]

The equation of the tangent at ₍ _sin4 _{, cos}4 ₎ b a   is





4 2 4 cos bcot sin . y b x a a       

Setting y and solving for 0 x gives





2 2 2 2

sin cos sin sin .

xa    a 

 Coordinates of P are _{( sin}2 _{, 0).}

a 

Similarly, setting x and solving for y gives 0 ybcos .2

 Coordinates of Q are (0, _cos2 ₎

b  .

3(iv) [3m]

From (iii), _sin2

OPa  and OQbcos2, so 2 2 2 sin co 1 ( )( ) 2 2 s 8 sin 2 . ab A OP OQ    ab 

A is maximum when sin 2 = 1, i.e.

4    for 0, , 2  _ _  

Maximum value of _{sin 2}2 _.

8 4 8 ab b A _  _a   x O y 2 1 4 4 , sin cos y a b x    

(4)

Alternatively,

From (iii), _sin2

OPa  and OQbcos2, so 2 2 1 ( )( ) sin cos . 2 2 ab A OP OQ   



3 3





2 2



2 2 2

sin cos 2sin cos cos sin

For 0, , 2 when d 2 sin cos d 2 d 0, cos sin t d an 1 4 A ab ab A                                  Maximum value of 4 2 2 1 sin cos . 2 4 4 2 2 8 ab ab ab A  _{ }  _{ }  _ _        4(i) [3m]

For l and₁ l to intersect, ₂

2 0 1 0 1 = 3 0 for some , . 1 5 4 2 p                                        2 0 3 3, 6, 0 5 2 3 p p                _       _{ } Hence coordinates of A(6, 3,16) 4(ii) [3m] l has equation 3 2 0 1 , 1 5 q                       r 3 2 2 2 2 2 2 0 1 1 5

Shortest distance from to

2 1 5 2 6 1 3 1 2 ( 1) 5 15 5 0 1 5 30 5 1 (since 0) 30 13 15 q OA A l q q q q q q                 _{ } _{ }                                _ _{  }              

(5)

4(iii) [4m] 2 0 1 for some 1 5 q OB         _{ } _{ }             15 13 Now , so 15 by (i). 15 AB q 1 2 2 2 2 Since , 1 0 5 9 2 2 3 1 1 0 15 5 5 (18 3 75) (2 ( 1) 5 ) 0 60 2 30 15 2 19 So 0 2 1 2 1 5 11 AB l AB OB                                        __ _ _{   }                                              4(iv) [3m]

Note that (0, 0,1) and (15,0,1) lie on l and1 l respectively, so they lie on3  , and hence 1

a vector parallel to  is ₁ 150 00 150 15 01 1 1 0 0                                   . So a normal vector to is ₁ 10 21 05 ( 1) 50 0 5 1 1                         _            . So equation of  is ₁ 05 00 05 1 1 1                            r 0 5 1 1 r           

Section B: Statistics [60 marks]

5(i) [2m] Number of ways 15 15 15 0 1 2 C C    32768 1 15 32752     OR 15 15 15 15 15 15 2 3 4 15 2 32752 r r C C C C C  



    _ 

(6)

5(iia) [1m] Number of ways  9! 362880 OR

 

3 9 6 3 3 3 3 3! 362880 C C C      5(iib) [2m] Number of ways 6 1 2! C 7! 60480     OR

 

2 6 7 6 3 1 1 3 3 2! C C C C 3! 60480        5(iii) [2m] Number of ways = (6  1)!  6  5  4 = 14400 OR Number of ways 



9 1 !

 

     7 1 ! 3! 6 1







! 3P₂6P₂14 004 6(i) [2m] Probability 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)









transfer from to and final ballchosen is white final ballchosen is white

P A B P  1 23 23 207 4 50 200 491 491 982 900 900                         7(i) [1m] 7(ii) [1m]

Product moment correlation coefficient =  0.920 (3 s.f.)

t (hours) x (micrograms per litre)

(7)

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)

[6m]

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

To test H0 : μ = 2.9 vs H1 : μ > 2.9

Perform a 1 tail t-test at 10% level of significance. Under H0 , 0 ~ ( 1) / X T t n 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. ~ (16, 0.17). X B ( 3) 0.244. (3 s.f.) P X   9(ii) [5m]

Let Y be the number of students, out of 60, who wore white shoes to school.





~ 60, 0.58 .

Y B

Since n = 60 > 50 is sufficiently large and p = 0.58 are such that 34.8 5

np  and (1n  p) 25.2 5, 

Therefore Y ~N np np



, (1p)



approximately i.e. Y ~N



34.8,14.616



approximately.

(8)







 







38 38.5 by continuity correction 0.167. 3 s.f. P Y  P Y   10(i) [2m]

Let X be the number of defects in a box.

2.5 ~ Po 3.2 40 X _  _  _ i.e. ~ Po (0.2)X P X



1



0.98248 0.982 (3 s.f.) 10(ii) [2m]





(a package is accepted)

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





~ 3.9774

Y Po approximately.





(more than 67 packages are accepted) 2 0.241 (3 s.f.) P P Y    11(i) [2m]

Let X be mass of an abalone in grams. _~



_180, 2



X N  P X



200



0.08 200 180 0.08 P Z      _  _   20 0.92 P Z     _  _   From G.C., 20 1.4051 14.234 14.2 (3 s.f.) (shown)      11(ii) [1m]



165



0.14541 0.145 (3 s.f.) P X    11(iii) [3m] 2 14.2 ~ 180, 15 X N_ _  _

(9)

















180 5 180 5 ( 180 5) 2 180 5 or 2 ( 180 5) 2 185 or 2 ( 175) 0.173 (3 s.f.) P X P X P X P X P X P X P X               _    _     _  _  Alternative 1, 2 14.2 180 ~ 0, 15 X  N_ _  













180 5 180 5 ( 180 5) 2 180 5 0.173 (3 s.f.) P X P X P X P X              Alternative 2,





~ 2700, 3024.6 X N 













180 5 2700 75 ( 2700 75) ( 2700 75) 2 2775 0.173 (3 s.f.) P X P X P X P X P X                    11(iv) [3m]

Let C be cost of one abalone in dollars.





2 450 ~ 81, 6.39 1000 C X N







2



1 2 3 4 5~ 5 81, 5 6.39 TC C C C C N  





~ 405, 204.1605 T N 



420



0.147 (3 s.f.) P T   Alternatively, Let W X₁X₂X₃X₄X₅~N



5 180, 5 14.2 





2







~ 900, 1008.2 W N  Probability required





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

(10)

[2m] 12(ii) [4m] = ( 50) 50 100 50 120 1 49 or 49.167 (5 s.f ) 49.2 (3 s.f ) 6 x n  _ _  _  



Unbiased estimate of population variance





2 2 2 ( 50) 1 ( 50) 1 1 ( 100) (1158 ) 9.0308 9.03 (3 s.f ) 119 120 x x n n  _        _ _     



Since n = 120 is large, by Central Limit Theorem, X N(49.167,9.0308) approximately. n ( 49.9) 0.01 ( 49.9) 0.99 49.9 49.167 ( ) 0.99 9.0308 49.9 49.167 From , 2.3263 9.0308 90.960 Least value of 91

Method 1 : Using Standardization

P X P X P Z n GC n n n         _   

Method 2 : Using table of values

Key in Y1 = normalcdf(49.9, E99, 49.167, 9.0308 / X )

n P X( 49.9)

90 0.01033 > 0.01 91 0.00999 < 0.01 92 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).

(11)

(iii)

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.