RI 2017 P2 A-Level H2 Math Solution

1 | P a g e Section A: Pure Mathematics Question 1 No. Suggested Solution Remarks for Student (i) Curve given by parametric equations 3, 2x y tt  To find points on the curve that lie on the line y = 2x, we have for some t, 2y x 32 2t t       2 3 3 3 3, 2 3 3 3, 2 3 t t t x y t x y              3,2 3 and 3, 2 3A B   Alternatively, express the curve in Cartesian form, 3 3 6 , 2x t y tt x x     And solve the simultaneous equations 6 and 2y y xx  .      2 2 62 3 3,2 3 and 3, 2 3 Length 2 3 4 3 60 2 15 x xx A B AB          (ii) 2 2 3 6, 2 d 3 d d 2, 2d d d 3 x y tt x x y y tt t t x        Equation of tangent at point P: 22 32 3y p p x p        2 2 22 2 3 2 43 y p p x p y p x p      Raffles Institution H2 Mathematics (9758) Solution for 2017 A-Level Paper 2

2 | P a g e   2 When 0 : 4 2 6When 0 : 43 6Thus, is ,0 and is 0,4 x y p y p x p x p D E pp           Mid-point F is 3, 2pp     which is the point P. 3 6, 2x y pp x   . That is, xy = 6. Question 2 No. Suggested Solution Remarks for Student (i)    1 13 133, 2 3 13 1 1562 3 2 u S d d          (ii)    13 1 13 13 13 13 3 13, 156 1 3 1 156 1 1 52 52 52 51 0...(1) ru S r r r r r r r                 Note that r = 1 satisfies (1). But if r = 1, then S13 = 3(13) = 39 ≠ 156. So common difference cannot be 1. Using GC, r = 1.210024 = 1.21 (3 s.f.) or –1.451067 = –1.45 (3 s.f.) (iii)    1 33 1.2100 100 3 12 Use GC to get inequality or table of values, Smallest 42 n n n        

3 | P a g e Question 3 No. Suggested Solution Remarks for Student (a)(i) f ( )y x f (2 )y x Scaling parallel to the x-axis by factor of 1 2.  Curve f(2 )y x cuts the axes at  1 ,0 , 0,2a b    (ii) f ( )y x f ( 1)y x  Translation of 1 unit in the positive x-direction.  Curve f ( 1)y x  cuts the x-axis at 1, 0a (iii) f ( )y x f ( 1)y x  f (2 1)y x  Translation of 1 unit in the positive x-direction followed by scaling parallel to the x-axis by factor of 1 2.  Curve f(2 1)y x  cuts the x-axis at 1, 02 a    (iv) We reflect the graph of f ( )y x about the line y x to obtain the graph of 1f ( )y x .  Curve 1f ( )y x cuts the axes at 0, , ,0a b (b)(i) a = 1 g is undefined when x = 1. (ii) 2g g g 1g 1 1 g 1 11 1 1 11 1 1 x x x x x x x x x                     Note that if g has an inverse and h gx x then 1g h  . Here, g gx x So immediately we actually know 1g g  Replace x with 2x Replace x with x - 1 Replace x with x - 1 Replace x with 2x

4 | P a g e   1 1 1Let 1 11 1 11 1 11 1 1g 1 g 1 y y x x x y x y x x x                 (iii)      2 1 2 1g g 1 1 1 1 0 or 2 b b b b b b          Question 4 No. Suggested Solution Remarks for Student (a) Area =    5.5 2 1 1 1 6 5 d 15.18752 x x x x         Note that we just use GC as question did not state “exact value” or “without using calculator”, etc. The value of 15.1875 is exact. (b)(i) Volume    2 1 20 21 2 0 1 2 0 d d 1 2 1 1 2 1 2 1 y ya y y a y y a y a a a a                            

5 | P a g e (ii)            2 2 2 2 2 2 2 2 2 2 42 1 2 1 1 4 1 1 4 4 4 4 0 4 16 4 4 4 4 1 8 8 Note that 1, thus 0 1 1. 1 1 1 1Thus, >1 or 12 2 b b a a b b a a b b a a b b a a a a a ab a a a a a a a a ab b                                          Given that the container is formed the same way, 1b , 21 1 2 a ab   

6 | P a g e Section B: Statistics Question 5 No. Suggested Solution Remarks for Student (i) 6 Rs, 3 Ys      6 5 5P 2 9 8 12 6 3 5 5P 3 2! 9 8 7 14 6 3 2 5 3! 5P 4 9 8 7 6 2! 28 6 3 2 1 4! 1P 5 9 8 7 6 3! 21 T T T T                                                    (ii)        2 2 2 2 2 22 5 5 5 1 20E 2 3 4 512 14 28 21 7 5 5 5 1 125E 2 3 4 512 14 28 21 14 75Var E E 98 T T T T T                                                        (iii) Let X denote no. of games, out of 15, where Lee takes at least 4 counters out of the bag.      19~ B 15, P 4 84 P 5 1 P 4 0.238 (3 s.f.) X T X X            Question 6 No. Suggested Solution Remarks for Student (i) Each of the 5 families forms a unit. These 5 family units can be arranged in 5! ways. For a given family, the 4 family members can be arranged among themselves, in 4!ways, and as there are 5 families, in 4! 4! 4! 4! 4!    ways. No. of required arrangements =   55! 4! 955514880

7 | P a g e (ii) Fathers are together with Red father (RF ) and Blue father (BF ) at the ends 1 2 3R BF F F F F OR 1 2 3B RF F F F F which can be arranged in 2 3!ways. Remaining Red and Blue family (M,D,S) can be arranged in 3! 3! ways. The fathers, Red and Blue family form a unit. Together with the remaining 9 people, they can be arranged in 10!ways. No. of required arrangements =     310! 3! 2 1567641600 (iii) Excluding the fathers, we have 15 people to arrange in a circle which can be done in 15 1 ! ways. There are 15 “slots” to include the fathers, which can be performed in 15 5P ways. The number of ways to arrange 20 individuals in a circle is 20 1 ! Required probability =    15 515 1 ! 1001or 0.25820 1 ! 3876 P  Question 7 No. Suggested Solution Remarks for Student (i) Every biscuit bar has equal chance of being selected, and chance of selection of one biscuit bar is not affected or influenced by the selection of another biscuit bar. (ii) Unbiased estimate of population mean is 7.7 32 31.8075 31.840x    Unbiased estimate of population variance is 2 0.2453269 0.245s   (iii) Null hypothesis, 0H : 32 Alternative hypothesis, 1H : 32 where  is the population mean mass of biscuit bars. Perform a two-tailed test at 1% significance level. Under H0, 0.2453269~ N 32, approximately by Centre Limit 40 Theorem since 40 is large X n      -value 2P 31.8075 0.0139699p X    Question is about claim is 32 grams.

8 | P a g e Since -value 0.0139699p  > 0.01, we do not reject 0H : 32 , and conclude that there is no significant evidence at 1% level to claim that the mean mass of biscuit bars is not 32 grams. (iv) Since the sample size is large and the sample is random, Central Limit Theorem can be applied such that the distribution of the sample mean is approximately normal. Question 8 No. Suggested Solution Remarks for Student (a) L1 and L2 for (i), L3 and L4 for (ii), L1 and L5 for (iii) (i) Note that perfect fit is required as question wants product moment correlation coefficient to be –1 NOT approximately –1. (ii) 8 points that form shapes like square or circle would suffice.

9 | P a g e (iii) (b)(i) (D) y a x b  Note that curvature-wise, it seems that log is also suitable. However, we noted x = 0 is defined for this scatter plot, so we have to reject (C) (ii) 4.18 74.0y x  Product moment correlation coefficient = 0.981 (iii) r = 0.981 is near 1 which suggest good fit of the model. Besides, the value 189 is within the given range of values of x, so we did not extrapolate the information.

10 | P a g e Question 9 No. Sug