RI H2 Maths 2013 Prelim P2 Solutions

RAFFLES INSTITUTION 2013 Year 6 Preliminary Examination Paper 2 H2 Mathematics 9740

No. 1(i)

Solution

 2 7  2     1 : r   1   5 and l AC : r  2    1  , where         2 8  2        7  2    C is on l AC  OC   2    for some     8  2     2  7  2   2     C is on 1  OC   1   5 , i.e.  2      1   5  8  2   2   2        (14  4 )  (2   )  (16  4 )  5

   3 7  2  1          Hence OC   2   3  1  =  1 , i.e. OC  i  j  2k 8  2  2       

(ii)

 1   7   6   2           AB   5    2    3  //  1  //  2 , since  2 contains l AB 8  8  0   0           2  2  1   2 // normal of 1 //  1   2  

 2   2   2   1  Hence normal of 2 //  1    1    4  //  2   0   2    4   2           1  7  1  Equation of  2 is r   2    2    2  , i.e. r  (i  2 j  2k )  5  2   8   2       

(iii)

 2   2   2   1      Since  2    1   4  2  2  0 and  2    2   2  4  2  0 ,  1   2   1  2        

2i  2 j  k is  to both normal of 1 and normal of  2 . Hence 2i  2 j  k is parallel to l (verified). From (i) and (ii), we can deduce that C is on both 1 and 2 . Hence the cartesian equation of l is x  1 y  (1) z  2 1 x y 1 , i.e.     z2 2 2 1 2 2

No. 2(a) (i)

Solution

y

x O

From graph of y 

1 2x 1   shown above, R f   1  k ,  k  . x  2x  3 2   2

For gf to be a function, R f  Dg . 

1 1  k  2k i.e. k  2 2

(ii)

 11  R f   4,   2 R gf   0, 1  

(b)

Since y  h( x ) passes through (1,1) and (2, 2) , a (1)3  b(1) 2  c(1)  d  1 −−− (1) a (2)3  b(2) 2  c(2)  d  2 −−− (2) Now h ( x)  3ax 2  2bx  c Since (2, 2) is a maximum point, 3a(2) 2  2b(2)  c  0  12a  4b  c  0 −−− (3) Since (4,14) lies on graph of y  h  x  ,

either (4,14) or (4, 14) lies on graph of y  h( x ) , H2 MA 9740/P2/2013 RI Year 6 Preliminary Examination - Page 2 of 10

i.e. h(4)  14 or  14 . As h( x ) is strictly decreasing for x  2 , h(4)  h(2)  2 . Hence h(4)  14 and a (4)3  b(4) 2  c(4)  d  14  64a  16b  4c  d  14 −−− (4) Solving (1) to (4) gives

No. 3(i)

a  1 , b  4 , c  4 , d  2

Solution

dy  8sin  cos  d

y  2sin   y  4sin 2  

1    4 2

When y  2, sin 

When y  4, sin  1   



4 2

4 y dy  y 

 

8 8



2

4  4sin 2  8sin  cos   d 4sin 2 

2



4



cos  8sin  cos   d sin 

2



4









2



cos 2  d

4



2



4

1  cos 2 d 2 

sin 2  2   4   2    4     1   4       2  4 2   2

(ii)

When x  0, y  2

y

16 16 4 y  x  2 4  22 2 ( x  2)  4 y y

Since x  2 , x  2  2 Required area 



4

4 y y

x dy

2

H2 MA 9740/P2/2013 RI Year 6 Preliminary Examination - Page 3 of 10

2



4 2

 4 y   1   dy y  

4  4  2   y 2  2   2[2  (  2)]



 4 y dy  y 

 2(4   )

(iii)

2

Required volume   (42 )(2)    y 2 dx 0

 32   

2 0

2

  16  ( x  2) 2  4  dx  

 35.9 (3 s.f.)

No. 4(a)

Solution

Since z  i is a root, i3  2i  k  0  k  i

Hence the equation becomes z 3  2 z  i  0 . z 3  2 z  i  ( z  i)( z 2  az  1) Comparing coefficient of z 2 , a  i . For z 2  iz  1  0 , z 

i  i 2  4(1)(1) i  5 i  i 5   2(1) 2 2

Hence the other 2 roots are

i+i 5 i  i 5 . and 2 2

(b) (i)

1 1 1  [cos( )  i sin( )]  (cos   i sin  ) w r r

(ii)

500  3 w  40i w 500 (cos   i sin  )  3r  40i r

Comparing real and imaginary parts, H2 MA 9740/P2/2013 RI Year 6 Preliminary Examination - Page 4 of 10

500 500 cos   3r ,  sin   40 r r 3r 2 40r cos   , sin    500 500 2

2

 3r 2   40r      1  500   500  9r 4  1600r 2  250000  0 ( r 2  100)(9r 2  2500)  0 ( r  10)(r  10)(9r 2  2500)  0

Since r    , r  10 (iii) 3 4 Subst r  10 , cos   , sin    5 5 3 4   w  10   i   6  8i 5 5 

Alternative Solution 1: 500 From consider the modulus of both sides to get  3 w  40i, w 500  3r  40i  (3r ) 2  402 . This leads directly to the same equation for r as above, and r 500  6  8i. is solved similarly to get r  10. Then w  3(10)  40i Alternative Solution 2: 3r 2 40r From cos   , sin    , eliminate r to get 500 500 2

3  500 16  16 cos   sin    sin 2   1  cos 2   .  500  40 15  15 3 Solve this quadratic to get cos   , rejecting the other root, and proceed to get 5 500 40 4 Thus Finally r  10 and sin    cos   100. r2  r . 3 500 5 w  r cos   i sin  6  8i.

H2 MA 9740/P2/2013 RI Year 6 Preliminary Examination - Page 5 of 10

Section B: Statistics [60 marks]

No. (i)

Solution Obtain the name list of all patients and number them from 1 to N , N being the number of patients in the clinic’s records. Then use a random number generator to obtain 200 numbers from 1 to N , and the patients with the corresponding numbers will be selected as respondents.

(ii)

Use simple sampling to number of each age shown in the the sample size

Age (x) Sample size (y)

x  20

20  x  35 35  x  50 x  50

20 40 25 15  200  200  200  200 100 100 100 100  40  80  50  30 y proportional to the number of patients within each age group.

random choose y patients from group as table, with

(iii)

Simple random sampling may not give a fair representation of patients from different age groups.

No. (i)

Solution Number of arrangements  12!  6!  2  478955520

(ii)

Number of arrangements  6C5  (5  1)!  25  5  4  92160

6

Or: 6C1  (5  1)!  25  5 P2 ;

6  (6  1)!  25  4 ; 6   (7  1)!  25  (6  1)!  26 

No. (i)

Solution Let Rn denote the event that a red ball is removed on the n th draw.

Let Wm denote the event that a white ball is removed on the m th draw. r r 1 (or using tree diagram) P( R1  R2 )   r  w r  w 1 P( R2 )  P( R1  R2 )  P(W1  R2 ) r r 1 w r    r  w r  w 1 r  w r  w 1 r  rw P( R1  R2 ) r 1 P( R1 | R2 )   P( R2 ) r  w 1 

H2 MA 9740/P2/2013 RI Year 6 Preliminary Examination - Page 6 of 10

(ii)

P(W1  W2 )  1  P( R1  R2 ) r r 1  r  w r  w 1 2 wr  w2  w  (r  w)(r  w  1)

 1

(iii)

If exactly one white ball remains in the box when the removal stops, ( w  1) white balls must be removed in the first ( r  w  2) draws. Required probability  r  w  2 (r  w  2)!   w  1  ( w  1)!(r  1)! wr    (r  w)! (r  w)(r  w  1)  r  w   w! r !  w 

No. (i)

Solution To test H 0 :   150 vs H1 :   150 , where  denotes the population mean lifetime of batteries.

Since p  value = 0.0975 > 0.08, we do not reject H 0 and conclude that there is insufficient evidence, at 8% level of significance, that the population mean lifetime of batteries is less than 150 hours. (ii)

t 

 t  10317 n

n

 2  Under H 0 , T ~ N  0 ,  approximately by Central Limit Theorem, where n   0  150,   16.877 . p  value  P(T  t )  0.0975   10317  150    PZ  n   0.0975 16.877     n   10317  150 n   1.29593 16.877 n 10317 21.871    150  0 (shown) n n

Solve using GC, n  70 (iii)

It is NOT necessary to assume that T is normally distributed, since the sample size n is large, so by Central Limit Theorem, T is approximately normally distributed.

H2 MA 9740/P2/2013 RI Year 6 Preliminary Examination - Page 7 of 10

No. (i)

Solution

From GC, r  0.916 (3 s.f.)

(ii)

y 19.8

6 1 (iii)

11

x

A linear model predicts the average height would increase continuously with no upper bound. OR: The scatter diagram does not show the points lie close to a straight line. OR: The scatter diagram shows that as x increases, y increases at a decreasing rate. A quadratic model predicts that the average height reaches a maximum, then decreases with time and eventually takes on negative values.

(iv)

For the suggested model, the least square regression line is y  6.11ln x  6.10 . The product moment correlation coefficient is 0.993 (3 s.f.). When x  5.5, y  16.5 Hence the average height is 16.5 feet. Since x  5.5 is within range of value of x , and the product moment correlation coefficient is close to 1, suggesting a strong positive linear correlation between the 2 variables, the estimate is reliable.

No. (i)

Solution

X  Y ~ N(100  90, 25  36), i.e. X  Y ~ N(10, 61)

P  X  Y  5   P( X  Y  5)  P( X  Y  5)  0.73897  0.027394  0.766 (3 s.f.)

OR: P  X  Y  5   1  P  X  Y  5   1  P( 5  X  Y  5)  1  0.23363  0.766 (3 s.f.) (ii)

X 1  X 2  3Y ~ N(100  100  3(90), 25+25+32 (36)) i.e. X 1  X 2  3Y ~ N(470,374) P( X 1  X 2  3Y  495)  0.0981 (3 s.f.) Let T be the number of observations, out of 150, for which Y  90 . Then T ~ B(150, 0.5) . H2 MA 9740/P2/2013 RI Year 6 Preliminary Examination - Page 8 of 10

Since n  150 is large such that np  75  5 and n(1  p)  75  5 , T ~ N(75,37.5) approximately.

P(T  68)  P(T  67.5) (by continuity correction)  0.88966 (5 d.p.)

No. (i)

Solution

Let M be the number of substandard “medium” eggs, out of 10. m   Then M ~ B 10, .  100  2

8

m   m   P( M  2)  0.3  C2   1    0.3  100   100  From graphic calculator, m  18.6 (3 s.f.) or 21.5 (3 s.f.) (rejected since m  20 ) 10

(ii)

Let X be the number of “small” eggs, out of 30. Let Y be the number of “large” eggs, out of 10. Then X ~ B(30, 0.18) and Y ~ B(10, 0.08) . Required probability  P( X  5)  P(6  X  7)P(Y  1)  P( X  5)  [P( X  7)  P( X  5)][1  P(Y  0)]  0.53946  (0.84179  0.53946)(1  0.43439)  0.710 (3 s.f.)

(iii)

Let W be the number of “large” eggs, out of 60. Then W ~ B(60, 0.08) . Since n  60  50 is large, p  0.08  0.1 is small such that np  4.8  5 , W ~ Po(4.8) approximately. Required probability  P(W  4) = 0.476 (3 s.f.)

(iv)

Assume that the daily demand of eggs in the morning and that in the afternoon are independent.

(v)

Let T be the total daily demand of eggs, in trays of 30. Then T ~ Po(5 19  5  9), i.e. T ~ Po(140) . Let k be the number of trays of eggs the supermarket stocks up each day. We need H2 MA 9740/P2/2013 RI Year 6 Preliminary Examination - Page 9 of 10

P(T  k )  0.9 . From graphic calculator, when k  154 , P(T  k )  0.88859  0.9 when k  155 , P(T  k )  0.90332  0.9 when k  156 , P(T  k )  0.91655  0.9 Hence, least value of k  155 .

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