RVHS_H2_MATHS_P1_ANS

RVHS 2015 Y6 H2 MA Prelim Paper 1 (Solutions) Question 1 [5 Marks] (a)       2 2 2 74 223 2 3 2 023 2 1 2 013 x xx xx xx xx xx       13 or 1 2 2xx     (b) Sub yx ,                2 2 1 2 0 23 2 1 2 013 2 1 2 013 xx xx xx xx yy yy        Therefore, from part (a) 13 or 1 2 2 13 or 1 2 2 (No solution) 2 1 or 1 2 yy xx xx               

2 Question 2 [5 Marks] Using cosine rule,                2 2 2 1 2 2 2 1 2 2 1 22 1 2 2 222 24 2 ˆ2 cos 111 2 1 cos 222 2111 4 2! 1 2 (shown)4 1 184 11 11 221 8 82 2 2! 1 1 4 82 1 22 AC AB BC AB BC ABC AC                                                   44

3 Question 3 [8 Marks] (a) nnSn 32 2  )1(3)1(2 2 1  nnSn = )33()242( 2  nnn = 12 2  nn nu = 1 nn SS = )12()32( 22  nnnn = 41n 1nu = 4( 1) 1n = 43n 1 (4 1) (4 3)nnu u n n     = 4 Since  1nn uu constant, the given sequence is an arithmetic progression. (shown) 2015nS 201532 2  nn 0201532 2  nn Using GC, 5.32n or 31n Since n is a positive integer, least value of n = 32. (b) (i) Total height of an n-layered cake = 12 100 510...100 510100 51010               n kkk =                      12 20...2020110 n kkk =                           201 2011 10 k k n

4 =              n k k 20120 200 (shown) (b) (ii) When k = 19, Total height =              n 20 191200 19200 1 12020 n   Using GC, 17.86375281n Maximum number of layers is 17. (b) (iii) It is assumed that th e thickness of whipped cream used to join the different layers together is negligible. Question 4 [8 Marks] (a)       21 2 2 21 tan1 2 tan tan e d1 e d (shown) 1 e2 1 e2 x u u x x xx uu C C          (b) (i) 2 2 2 sec 1 tan 1 dx udu u x    x y 0 2π 32π 8ye     21tan1 2 tan 1 x xe y x     R 1

5 (b) (ii)       21 2 2 21 22 22 2 2 tan1π 132 20 1π2 tan16 0 π2 16 16 ππ 316 16 Volume generated tan eπ (1) d81 π1 e82 π e182 4 4 units8 x x x ex x e e ee                             Question 5 [9 Marks] (a) Let nP denote the statement 1 for 0, ! n nu n n n    . When 0n , 0LHS 1 (given by question) 01RHS 1 LHS0! u    Therefore, 0P is true. Assume that kP is true for some , 0kk . i.e. 1 ! k ku k  . We need to show that 1kP  is also true. i.e.   1 2 1! k ku k    .             2 1 2 2 2 22 1 1! 11 ! 1 ! 11 1! 2 1 1 1! 2 1! kk kkuu k k k k kk k k k k k k k k k k k      