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