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                     Therefore 1kP  is also true once kP is true. Since 0P is true, and kP is true implies that 1kP  is also true, by Mathematical Induction, nP is true for , 0nn .

6 (b)     22 1 1 1 21 2 1 1 2 23 34 45 1 1 12 22 ( 1) 1 (replacing by 1)( 2)! ( 1)! 1 ( 1)! 33 2 2 ! NN rj N j N jj j NN NN NN N r r j j rjrj jj j uu uu uu uu uu uu uu uu N N                                 Question 6 [9 Marks] (i) ab is the perpendicular distance of point A (with position vector a) to line OB. (ii) (a) 33  OBOC b 3 12 OAOCOD  = 3 1 (2(3b) + a) = 3 1 a + 2b A O B C D a b M 2 1 1 2

7 (b) Let ABAM  , ( ) (1 ) ---(1)OM OA AB          a b a a b Let ODOM  = 2 - - - (2)3  ab Since a and b are non-zero and non-parallel, (1) = (2) 1 3    ,  2 Solve simultaneously, 7 6 , 7 3 ba 7 6 7 1 OM (iii) Area of triangle OAC = OCOA2 1 12 = ba 32 1   8ba Shortest distance from A to OC = ab = b ba = 82 1 = 4 units A O B C D a b M 2 1 μ 1 – μ 1 – λ λ

8 Question 7 [10 Marks] (i) Method 1: Let iz x y where ,xy  2 22 8i 2 i 8i z x y xy     Comparing real and imaginary parts, 22 0xy xy   and 28 4 xy xy   When 2, 4 no solutionx y y   When 2, 4 2, 2x y y y x      2 2i or 2 2iz     Method 2: 2 i22 2 i 4 3ii 44 8i 8e where 2 2e for 0, 1 2 2e or 2 2e 2 2i or 2 2i k k z zk zk z z                          (ii) 4 2 2 2 64 8i 8i or 8i w w w w      The roots of the second equation are the conjugates of those of the first. Since two of the roots are those found in (i), representing them on argand diagram should give:

9 (iii)   2 2 2i 4i 0zz             2 2 2i 2 2i 4 1 4i 2 11 i 8i 2 11 i 2 2i2 2i or 2 z                Question 8 [10 Marks]  )1ln(sin 1   xy --- (1) )1ln(sin  xy Differentiate implicitly with respect to x: 1 1 d dcos  xx yy (shown) --- (2) Differentiate implicitly with respect to x: 22 2 )1( 1 d d)sin(d d d dcos  xx yyx y x yy 2 2 2 2 )1( 1 d dsind dcos      xx yyx yy (shown) --- (3) (i) Differentiate implicitly with respect to x: x yyx y x yy d d)sin(d d d dcos 2 2 3 3  3 2 2 2 )1( 2 d dcosd d d d d d)2)((sin                        xx yyx y x y x yy 3 3 2 2 3 3 )1( 2 d dcosd d d dsin3d dcos                xx yyx y x yyx yy --- (4) When x = 0, (1): 00sin 1  y (2): 1d d1d d)0cos(  x y x y (3):   1d d11)0sin(d d)0cos( 2 2 2 2 2  x y x y (4):

10      3d d21)0cos(11)0sin(3d d)0cos( 3 3 3 3 3  x y x y So, ...!3 3 !2 )1(10 32  xxxy ...2 1 2 1 32  xxxy (ii) 1.02 1 2 1 32       xxxy   1.02 1 2 1)1ln(sin1.0 321        xxxx Sketch          321 2 1 2 1)1ln(sin xxxxy , 1.0y and 1.0y . Solution set:  : 0.521 0.783, x x x    (iii) Consider     1 1 )1ln(1 1)1ln(sind d 2 1   xx xx = 2))1(ln(1)1( 1  xx 2))1(ln(1)1( 1  xx =  )1ln(sind d 1  xx =       ...2 1 2 1 d d 32 xxxx 2 2 31 xx (up to and including term in 2x ) x y y = 0.1 y = –0.1