RI 2016 P2 A-Level H2 Math Solution

1 | P a g e Section A: Pure Mathematics Question 1 No. Suggested Solution Remarks for Student d 0.1d V t  1tan tan where depth of water2 r r h h hh      2 2 31 1 1 1 3 3 2 12V r h h h h          2 1 33 2 3 d 1 d d 4 d 1 36When 3, 12 1 36 d0.1 4 d d 0.0251d V h ht t V V h h h t h t                     Rate of increase of depth of water is 0.0251 m per minute. Raffles Institution H2 Mathematics Solution for 2016 A-Level Paper 2

2 | P a g e Question 2 No. Suggested Solution Rem arks for Stud ent (a)(i) 2 2 2 2 2 2 3 dcos d , cos d 1 2 d 1sin sin d 2 , sin d 1 2 1 1 dsin cos cos d , sin d 1 2 2 d 1sin cos sin 1, cos d vx nx x u x nx x ux nx x nx x x v nxn n x n vx nx x nx nx x u x nxn n n n x ux nx x nx nx c v nxn n n x n                         (ii)       22 2 2 2 3 2 2 2 2 2 2 2 2 2 2 2 1 2 2cos d sin cos sin 2 22 cos 2 cos 4 cos 2 2 cos 2If is even, cos d 4 2 2 6If is odd, cos d 4 2 6 x nx x x nx x nx nxn n n n nn n n nn n x nx x a n n n x nx x a n n                                       

3 | P a g e (b)    2 2 2 0 2 2 20 22 20 2 5 29 9 25 9 5 d d 1Volume d 9 2 d d 2 d 0 99 d 2 59 1 9 d2 1 9 1 d2 1 9 ln2 1 9 1 ln 9 ln 52 5 1 4 5ln2 5 9 u xy x u x x x u x x x x x ux x x x x ux u uu uu u uu                                                           Question 3 No. Suggested Solution Remarks for Student cos , 1 cos , for 0 2x t t y t t       (i)

4 | P a g e     0 1 cos 0 0 or 2 0 cos 0 1 or 2 cos 2 2 1 1,0 and 2 1,0 y t t x x                        cos , 1 cos , for 0 2 d d1 sin , sind d d sin 0 sin 0d 1 sin 0, , 2 From above points where meets the -axis, max point is when cos 1, 1 cos 2 1, 2 x t t y t t x y t tt t y t tx t t D x t x y                                   (ii)     cos 1 0 0 0 0 Area d 1 cos 1 sin d 1 cos sin sin cos d 11 cos sin sin 2 d2 1sin cos cos 24 1 3sin cos cos 24 4 a a a a a a y x t t t t t t t t t t t t t t t t a a a a                                 (iii) 1 2 d 1 d 2t y x       2 1 1 , 12 2 Thus, equation of normal is 1 2 2 2 1 1 is ,02 is 0, 1 1 1 1area of triangle 1 12 2 4 t x y y x y x E F OEF                                   

5 | P a g e Question 4 No. Suggested Solution Remark s for Student (a)  3 i 1 3 i 1z z       2arg where tan 5z    (a)(i) (ii) Equation of circle:     2 23 1 1 ...(1)x y    3 1 1 Im Re 0 2 2 Z1 Z2  3 1 1 Im Re 0 2 2 Z1 Z2 

6 | P a g e Equation of line: 5 2y x c  Sub (3, 1) into line, we have 15 171 2 2c c    5 17 ...(2)2 2y x  Solving (1) and (2), the two values are 3.37 + 0.0715i, 2.62 + 1.93i (b)(i) i 42 2i 8ew       1 3 8 13 3i i 2 i3 4 4 4 2 2 8 11 i 122 3 7i i i4 12 12 8e 2 e 2 e 2 e , 1,0,1 2e , 2e , 2e kk k z w z w z k z z z                                                          (ii)         1 1 21 arg * 2 arg * 2 arg * arg 2 arg since * is a positive real number2 1 arg 2 3 71 or or ...4 2 2 2 7 15or or ...4 4 4 4 Required 7 n n n n w w w ww w w w w w w w n w n n n                                     

7 | P a g e Section B: Statistics Question 5 No. Suggested Solution Remarks for Student (i)      1 1 2 1 4 1 11P P P + 7 2 7 3 7 6 42R B Y                                 (ii)     2 1 P 47 3P | 11P 11 42 BB W W           (iii)      1 1 2 1 4 1 4P P P 3! 3! 7 2 7 3 7 6 1029R B Y                         

8 | P a g e Question 6 No. Suggested Solution Remarks for Student P D A F Total Male 2345 1013 237 344 3939 Female 867 679 591 523 2660 Total 3212 1692 828 867 6599 (i)(a) 3939 59.69 606599  (b) 679 10.29 106599  (ii) Part (i) only takes into account gender and dept NOT age. Age group may not be evenly spread across all dept and gender. (iii) Null hypothesis, 0H : 37 Alternative hypothesis, 1H : 37 Perform an one-tailed test at 5% significance level. Under H0, 140~ N 37, approximately by Central Limit Theorem since 80 is large80X n      Managing director’s belief should be accepted 0H is rejected -value P 0.05 0 34.824 p X x x       Set of values of x is 0,34.8 (iv) Now, perform an one-tailed test at % significance level. Managing director’s belief should not be accepted 0H is not rejected

9 | P a g e  -value P 35.2 100 0.086809100 0 8.6809 0 8.68 p X                Set of values of  is 0,8.68 Question 7 No. Suggested Solution Remarks for Student (i) Number of ways is4 3 24P (ii) Number of ways is 10 6 4 3 3 3 720 120 24 576P P P      Applying complement principle (iii)   8 1 !3!1Required probability 10 1 ! 12   Grouping method (iv)   7 37 1 ! 5Required probability 10 1 ! 12 P   Slotting method

10 | P a g e Question 8 No. Suggested Solution Remarks for Student (a)(i) (3, 87.4) should be excluded. Read question carefully. Question requires students to use GRAPH paper, with given scale. Is (1,72.5) acceptable? (ii) Scatter diagram does not display linear relationship, since as x increases, y increases at a decreasing rate. Thus it should not be modelled as an equation of the form y ax b  (iii) cy dx  All x and y values are positive, thus d > 0 to account for large values of x when 0.c x Values of y increases as x increase, so c < 0 (iv) (1,72.5) (50,92.4) x y