2014 H2 Maths Paper 2 solution vetted Mar 2018

2014 A-Level H2 Mathematics Paper 2 Section A: Pure Mathematics 1 (i) 23 , 6x t y t dd 6 , 6dd xy ttt d d d 1 d d d y y x x t t t   When d1 0.4d y xt , 2.5t  . (ii) Tangent at P:   2163 py p x p   This line meets the y-axis at D. 0,Dx   216 0 3D py p p   3 6 3Dy p p p     23 , 6P p p and  0, 3Dp Let   239 22 ,ppM be the midpoint of PD. 23 2 (1)M px   , 9 2 29 (2)MM py p y    Sub (2) into (1):   2 23 22 2 9 27M M Mx y y  The locus of the midpoint has equation 22 27xy . 2 Let    2 22 9 13 2 5 92 5 9 x x A Bx C xxxx           229 13 9 2 5x x A x Bx C x       Comparing coefficients of 2 0 : 2 9 (1) : 5 2 1 (2) : 9 5 13 (3) x A B x B C x A C                          222 2200 2 2 2 220 0 0 2 22 21 0 00 11 1 33 22 33 1 2 2 3 3 3 3 8 1 2 2 3 3 38 2 23 9 13 3 3 8 d d 2 5 92 5 9 2 2 8 d d d2 5 9 9 ln 2 5 ln 9 8 tan ln1 ln 5 ln13 ln 9 tan tan 0 ln 5 ln13 ln 9 tan x x x x xx xxxx xx x xx x x xx                                         1 3 3 13 8 2 2 45 3 3ln tan  Aim: Eliminate p. Solving simultaneously, 3, 3, 8A B C  

3 (i) The distances of the stages: 8, 16, 24, … form an A.P. with first term 8 and common difference 8. (a) Total distance after 10 stages,  10 10 2 2(8) 9(8) 440 m S   (b) Total distance after n stages,   2 2 2(8) ( 1)(8) 4 4 metres n nSn nn     To run at least 5 km (5000 m), 5000nS  (i.e. 24 4 5000nn ) By GC, when 34n , 4760 5000nS  35n , 5040 5000nS   The least number of stages is 35. (ii) The distances of the stages: 8, 16, 32, … form a G.P. with first term 8 and common ratio 2. Total distance after n stages,     8 2 1 21 8 2 1 n n n s     By GC, when 10n , 8184 10000ns  11n , 16376 10000ns   After running 10 km, the athlete has completed 10 stages and 10000 8184 1816 m of the 11th stage.   10 11 1 2 8 2 4096OA     The athlete has not reached 11A . He is 1816m away from O and running towards 11A . 4(a) Out of syllabus O 4096 m 1816 m

4(b) (i) 3 1 2r   , 1 1 63tan      6 i i i 62e 64e 64e w w       (ii)    i i 6 6 2e * 2e nn n w w         1 1 i i 66 1 6 2e 2e n n n n        For * nw w to be real, 1 6 ,n kk  61nk  The three smallest positive whole number values of n are 5, 11 and 17. Alternative       1 1 i1 6 11 66 2e* 2 cos isin n n n n nn w w             For * nw w to be real, Im * nw w   = 0     1 6 1 6 1 6 sin 0 sin 0 , 61 n n n kk nk           6 (i) Number of teams 3 8 5 6 1 4 2 4 31500C C C C     (ii) Case 1: brother as midfielder No. of teams 3 8 4 5 1 4 1 4 4200C C C C     Case 2: brother as attacker No. of teams 3 8 4 5 1 4 2 3 12600C C C C     Total number of possible teams 4200 12600 16800   (iii) Case 1: The particular midfielder is in the team as midfielder. No. of teams 3 8 3 5 1 4 1 4 3150C C C C     Case 2: The particular midfielder is in the team as defender. No. of teams 3 8 3 5 1 3 2 4 2520C C C C     Case 3: The particular midfielder is not in the team. No. of teams 3 8 3 5 1 4 2 4 3150C C C C     Re Im x By the same reasoning, we only need to choose 10 players in addition to the particular midfielder in Cases 1 & 2. Notice the subscripts add up to 10 in each case. That’s because we only need to choose 10 players in addition to the brother.

Total number of possible teams 3150 2520 3150 8820    7(i) Let X be the number of rolls, out of 10, where the fair die shows a 6. Then 1~ B 10, 6X    P 3 0.15506 0.155 (3 s.f.) X   (ii) (iii) Out of syllabus 8(a) (i) (ii) (b)(i) (A) 1 0.9470r  (B) 2 0.9749r  (b)(ii) lnP c m d is the better model as the price of the used cars decreases at a decreasing rate according to the table. Thus, the decrease in price is not linear, which is what P am b suggests. Furthermore, the value of 21rr  , so there is a stronger linear correlation between ln m and P. From GC, 33700ln 196000Pm  (3 s.f.) (b)(iii) When 50m , 64000P The price of a car that is 50 months old is estimated to be $64000. 9 Out of syllabus 10 (i) (a)   1 2 1 2 1P 10 10 10 1000 500      (b)    P at least 1 1 P no     9891 10 10 10 44 125      (c)        P , , in any order P P P         3 1 2 3 3 4 4 1 4 10 10 10 10 10 10 10 10 10 29 500                          x y x x x x x x O x y x x x x x x O

(ii)                         P , , in any order | exactly 1 P , , in any order P exactly 1 PPPPPP P _ _ P _ _ P _ _ (where "_"means symbol is not "*") 4 4 1 4 2 3 2 3 1 2 2 2 1 10 10 10 10 10 10 10 10 10 10 10 10 10                                     3 3 1 4 2 10 10 10 10 10 1 8 9 9 2 9 9 8 1 10 10 10 10 10 10 10 10 10 71 1000 306 1000 71 306                               11 Out of syllabus