RI 9740/02 2013

1 | Page   A x x y y Section A: Pure Mathematics Question 1 No. Suggested Solution Remarks for Student (i) We have f {1}\D   and g .R   Since gf ,RD the composite function fg does not exist. (ii) We have gf ( ) 1 2f ( ) 212 1 312 1 1 63. 1 xx x x x x          If 1(gf ) (5),x  then 5g f ( ) ,x so 5. 63 1 4x x   Therefore 1(gf ) (5) 4.  It is also fine to write this as 3( 1) .1 x x   The form given in the answer makes the next part easier. Question 2 No. Suggested Solution Remarks for Student (i) Consider one of the kite shapes: We have tan .6 3x y yx  The base of the prism is an equilateral triangle with side 22 3 .ay ax  The area of this base, S, is    2 (base) (height) 1 23 1 23 s i n23 3 23 . 2 4 ax ax a S x        Raffles Institution H2 Mathematics Solution for 2013 A-Level Paper 2

2 | Page   Therefore   21 32 . 34 aVS x x x  (ii) We have     2 23 23d3 43d4 3 63 .24 3 ax axV xx axax       At stationary values,   3 23 63 04 or (reject). 63 3 d 2 0d ax ax aaxx V x      We have      2 2 d3 23 6 3 63 2 34d 32 6 3 . V ax axx ax       At , 63 ax 2 2 d 30 ,d V ax   so this gives a maximum value of 2 31 32 3 .45 463 63 aa aVa         Product rule. 23 ax is rejected as it is too large, giving a “base” with zero length. For “real life” problems, you should use the second derivative test. Question 3 No. Suggested Solution Remarks for Student (i) Let f( ) l n( 1 2s i n ) ,yx x so e1 2 s i n .y x Differentiating implicitly gives 22 2 332 32 de2 c o sd ddee 2 s i n dd dd dde3 e e 2 c o sdddd y yy yy y y xx yy xxx yy yy xxxxx         When 0,x 0e, 1y y and 23 23 dd d 2, 4, 14.d dd yy y x x x   Therefore (0) 2, f (f0 ) 4(0) 0, f     and (0) 4.f1  A direct differentiation approach gets messy real fast. For such problems, use implicit differentiation as far as possible.

3 | Page   Maclaurin’s Theorem gives 2 3 3 2 (0) (ff ff( ) f( 0 ) 1! 2 0) (0) 7 ! 2 3! 2 3 xx x xx x x            You can use the standard series from MF15 to verify this answer if you have time. A faster, but less reliable way to check is to pick a small value of x, say 0.1, and verify that the value from this expression agrees with the value of f(0.1) from the GC. (ii) We have 23 3 23 23 () () ()es i n 1 2! 3! 3! 26 , ax ax ax nxnx ax nx an nnx anx x                     so comparing with 23 7f( ) 2 2 3xx x x    gives 2n and 12. aan   The next term in the series of es i nax nx is 23 33(1 )2 2 1 .26 3 x x    Thankfully this is non-zero, or we will have to find the next term. Question 4 No. Suggested Solution Remarks for Student (i) The acute angle between 1p and 2p is 11 1 26 23 12 12 6 2 16cos cos cos 40.4 . 2126 94 9 23 12                     (ii) We have 1 :2 2 1px y z  and 2 :6 3 2 1 .px y   Solving simultaneously with GC gives 17 66 25 .33 x z yz     The line of intersection can be obtained by solving the equations of the planes in Cartesian form simultaneously.

4 | Page   Therefore a vector equation for l is 7/6 5/3 , 1/6 2/3 0 . 1 x y z            r  This can be simplifies to 7 1 :1 , . 1 10 6 l        r  For this question there is no real need to simplify this equation. However, if the equation is used again later, simplifying helps. Picking 1  gives the point 1 1 1    on l, and of course the direction can be changed to a nicer multiple. (iii) Distance from A to 1p is 41 2 31 2 311 11 21 .332 2 ꞏ 2 2 1 1 1 c c c                        Similarly, distance from A to 2p is 41 6 31 3 312 12 27 .776 3 ꞏ 2 6 1 3 2 c c c                 Therefore   22 7( 1) 6( 7) 01 ( 112 1 4 13 35)( 49) 0 35 or 49.13 37 cc c cc c cc             We used the point 1 1 1    that is on both planes. Square to remove modulus on both sides.

5 | Page   Section B: Statistics Question 5 No. Suggested Solution Remarks for Student (i) Using a list of all the employees in alphabetical order, assign each employee a unique number from 1 to 100 000. We then use a random number generator to generate 90 random numbers from 1 to 100 000, and select the 90 employees assigned these numbers to form the required sample. Since the selection process is random, it might happen by chance that some countries might not have any employee in the sample. Hence the sample may not be representative. Be concise and to the point. The time spent for each part should be proportional to the marks allocated, so don’t spend too much time here. (ii) A more appropriate method to a get a good representation is stratified sampling. Group the employees into different strata according to the country they are based. Then, using simple random sampling, select the number of employees from each strata based on the percentage of employees in each country. For example, if r% of employees are from Singapore, select 9 10 r employees from Singapore. Either quota or stratified sampling will work to ensure that each country is represented. Question 6 No. Suggested Solution Remarks for Student We are given 2~N ( , ) .Y   2P( 2 ) 0.95 P 0.95 2 1.644853 1.644853 2 ----(1) aYa Z a a              P( ) 0.25 P 0.25 0.674490 0.674490 ---- (2) aYa Z a a               (1) (2) gives 0.43112.319 4 57 .33 aa  Therefore 0.674490 1.29aa   (3 s.f.), giving k = 1.29. Since both mean and variance is unknown, we have to standardize. Use invNorm from GC. invNorm again. With unknown a, we have to solve (1) and (2) by hand. Note more s.f. for intermediate values to maintain accuracy. Or store the values in GC.

6 | Page   Question 7 No. Suggested Solution Remarks for Student (i) The assumptions are :  Whether a packet contains a free gift is independent of whether other packets contain a free gift.  The probability that a packet contains a free gift is the same for all packets. The usual conditions to use the binomial distribution, phrased in context. Don’t state the obvious, like “a packet either contains a free gift or it does not”. (ii) When n = 20, we have 1~B 2 0 , ,20F   so  P 1 0.377354 0.377 (3 s.f.)F   Use binompdf. (iii) When n = 60, 1~B 6 0 , .20F   Since 60n is large, 1 0.0520p is small, and 35 ,np 3~P oF approximately. Therefore    P 5 1 P 4 0.18474 0.185 (3s.f.).FF    Be sure to check the conditions and state the approximate distribution used. Question 8 No. Suggested Solution Remarks for Student (i) P( ')P( | ') P( ') P( | ')P( ')P( ') (0.8)(1 0.7) 0.24. BABA B A BA A A      Recall definition of conditional probability. (ii) We have P( ) P( ') P( ) 0.24 0.7 0.94,AB BA A     so P( ' ') 1 P( ) 1 0.94 0.06.AB A B     Use a Venn diagram to see this. (iii) P( )P( | ) 0.88 P( ) 0.88P( )P( ) ABA BA B BB       Therefore   0.94 0.88P( P( ) P( ) 0.88 1 P( ) )0 . 5 .0.12B A BB B      This gives P( ) P( ) P( ) P( ) 0.7 0.5 0.94 0.26. A BA B A B      Same trick as (ii).

7 | Page   Question 9 No. Suggested Solution Remarks for Student (i) The unbiased estimates for the p