AJC 2007 P1 ANS

2007 H2 Mathematics Prelim Paper 1 solutions Solution 1. 1, ( ) 3 2 4  x−   = 4 (2-3 ) ( 1 - 2  x)-3  = ( )( ) 2 3411 3 .... 2 2 2! 2  x x⎛ ⎞− −⎛ ⎞ ⎛ ⎞− − + − +⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠⎝ ⎠   = 213 3 ... 24 4  x x+ + + for ⏐x ⏐ < 2 Coefficient of xn = ⎟ ⎠  ⎞⎜⎝  ⎛  2 1 ( )( ) [ ]( )3 4 ... 2 1 ! 2 n n n ⎡ ⎤− − − +  ⎛ ⎞−⎢ ⎥⎜ ⎟⎝ ⎠⎢ ⎥⎣ ⎦   = ⎟ ⎠  ⎞⎜⎝  ⎛  2 1(-1)2n ( )( ) ( )( ) n nn ⎟ ⎠  ⎞⎜⎝  ⎛ ++ 2 1 21 21   = ( )( 2 1 2 12 ++⎟ ⎠  ⎞⎜⎝  ⎛  + nn n )  2.  y =(cos –1 x)2  dx dy = 2 cos-1  x  ⎟ ⎟  ⎠  ⎞ ⎜ ⎜ ⎝  ⎛  − − 21 1  x    x dx dy x 12 cos2 1  −−=⎟ ⎠  ⎞⎜⎝  ⎛ −   ( ) ( ) y x dx dy x 4cos41 21 2 2 ==⎟ ⎠  ⎞⎜⎝  ⎛ −   −  (proved) (i) Differentiating wrt x , ( ) dx dy dx dy x dx  yd  dx dy x 4221 2 2 2 2 =⎟ ⎠  ⎞⎜⎝  ⎛ −⎟⎟  ⎠  ⎞ ⎜⎜ ⎝  ⎛  ⎟ ⎠  ⎞⎜⎝  ⎛ −   ( ) 21 2 2 2 =⎟ ⎠  ⎞⎜⎝  ⎛ −⎟⎟  ⎠  ⎞ ⎜⎜ ⎝  ⎛ − dx dy x dx  yd  x   When x = 0 , y = (cos –10 )2 = 4 2π    ; dx dy = - π ; 2 2 dx  yd   = 2 ; By Maclaurin’s Theorem , y = 4 2π    - π  x + x2 + …. (ii) At x = 0, equation of tangent to the curve is y = 4 2π    -π  x 3a) a) 0 2 2 0 4 4 sin sin sin(sin) sin(sin) x xdx x xdx x xdx π π   π   π  − − = − +∫ ∫ ∫   1

2  Solution = 0 2 0 4 1cos2 1cos2 2 2  x xdx dx π   π   − − −− +∫ ∫   = 0 2 0 4 1 1 1 1sin2 sin2 2 2 2 2  x x x x π   π   − ⎡ ⎤ ⎡ ⎤− − + −⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦= 1 84 π   = +  b) 2 1 dx x u du = ⇒ = − 1 u   22 2 1 1 d ( 1xx2 2 u ) x du u u = − − − ∫ ∫   2 1 1 2 du u −= −∫   = 11 1sin 2 sin () 2 2 uc c 1 2  x −− + = −   −    Alternately: 11 2cos ( ) 2 c  x − +  4. (i) (ii)  Direction vector of is1l 0 2 5 ⎛ ⎞ ⎜ ⎟−⎜ ⎟⎜ ⎟⎝ ⎠ Equation of is ,1l 2 0 1 2 3 5 r  ⎛ ⎞ ⎛  ⎜ ⎟ ⎜= − +λ −⎜ ⎟ ⎜⎜ ⎟ ⎜⎝ ⎠ ⎝  %  ⎞ ⎟ ⎟⎟ ⎠ λ∈    4 13 3 OB ⎛ ⎞ ⎜=⎜⎜ ⎟−⎝ ⎠ uuur ⎟ ⎟ ⎟ ⎟ 0 5 ⎟ ⎟ ⎟ ⎟  ⎞ ⎟ ⎟⎟ ⎠  & 2 12 3 5 ON  ⎛ ⎞ ⎜ ⎟= − − λ⎜ ⎟⎜ ⎟+ λ⎝ ⎠ u u u r Then 2 14 2 65  BNONOB −⎛ ⎞ ⎜= − = − − λ⎜⎜ ⎟ + λ⎝ ⎠ uuur uuur uuur 0 2 2 0 1 42 2 5 6 5  BN  −⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟ ⎜• − = ⇒ − − λ • −⎜ ⎟ ⎜ ⎟ ⎜⎜ ⎟ ⎜ ⎟ ⎜ ⎟ + λ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ u u u r   2843025 0 29 58 2 ⇒ + λ + + λ = ⇒ λ = − ⇒ λ = − 2 10 4  BN  −⎛ ⎞ ⎜= −⎜⎜ ⎟−⎝ ⎠ u u u r   Equation of line BN : 4 1 13 5 3 2 r  ⎛ ⎞ ⎛  ⎜ ⎟ ⎜= +λ⎜ ⎟ ⎜⎜ ⎟ ⎜−⎝ ⎠ ⎝  %

3  Solution 5(i) (ii) cos 1 y x= −  1 1 cos 1 1 cos  yx  x y − − = − = ±   Since 1 1 xπ  − ≤ ≤, 11cos x y−= −   1 1: 1cos , , 1 1 f x xx− −  x− ∈ − ≤ ≤a    [ ) [ ] 0,  1 ,1 g  f   R  D  π   = ∞ ⊄ = −   ⇒  fg does not exist. For fg to exist, [ ] [ ]0,1 maximal 1, 0g g R D= ⇒ = −   6(i) (ii) 1 1 0 0 2 2 2 2  x xdx dx  x x − −= −− −∫ ∫   = 1 1 1 2 2 0 [(2)2(2)] x x dx − − − + −∫   = 13 1 2 0 2(2)4(2) 3  x x ⎡ ⎤ − − −⎢ ⎥⎣ ⎦ 1-1 2   =2(4 2 5) 3 −  1 2 1 1 1... 1 22 2 2 n n nS  n n n n n n = ⋅ + ⋅ + + ⋅ − − n n −   1 2 1 .... 2 1 2 2 2 1 1 1 2 ..... 21 2 2 n n n n n n n n n n n n nn n n n ⎡ ⎤ ⎢ ⎥ ⎢ ⎥= + + +− − −⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎡ ⎤= + + +⎢ ⎥− −⎣ ⎦   1 0 lim 2  xS dxn  x =→∞   −∫  =2(425) 3 −

4  Solution Alternative solution for (i) – by parts ( ) ( ) 11 0 0 13 2 0 13 2 0 22 2 2 2 2 2 2 2 2 3 4  22 2 3 2 ( 425) 3  xdx x x xdx  x  x x x  x x x ⎡ ⎤⎡ ⎤= − − − − −⎢ ⎥⎢ ⎥⎣ ⎦⎣ ⎦− ⎡ ⎤⎡ ⎤= − − + − −⎢ ⎥⎢ ⎥⎣ ⎦⎣ ⎦ ⎡ ⎤⎡ ⎤= − − − −⎢ ⎥⎢ ⎥⎣ ⎦⎣ ⎦ = − ∫ ∫   7a)Area A = 2 2 1 ln 1 ln21 2 4  xdx  x − ⋅ ⋅∫   = 2 1 1 1 1()ln 8  x dx  x xx ⎡ ⎤− + ⋅ −⎢ ⎥⎣ ⎦∫ ln2  = 2 1 ln 1 ln2 8  x  x x ⎡ ⎤− − −⎢ ⎥⎣ ⎦   = 1 5ln2 28 −   7b)Points of intersection of curves are (−5, 9) and (0, 4). Volume = 29 4 9 2 2 2 0 0 4 16(2 ) ( 2 ) 13  y ydy ydy dyπ π π   ⎛ ⎞−− − − − + −  ⎜ ⎟⎝ ⎠∫ ∫ ∫   = 466.52653− 8.3775593− 107.66306 = 350.4859107≈ 350 8i) Aftern leaps of the cheetah, the deer would have leaped n n 3 52 6 5 =× . Therefore the deer is at a distance ⎟ ⎠  ⎞⎜⎝  ⎛  +n 3 5 5 221 from the cheetah’s starting point. ii) Distance leaped by cheetah: 10 1,4  −== d a   Aftern leaps, the distance leaped by the cheetah = ( )⎥⎦ ⎤ ⎢⎣ ⎡ −−= 1 10 18 2 nnS c   To catch the deer, nS c 3 5 5 221  +≥   ( ) nnn 3 5 5 221 1 10 18 2 +≥⎥⎦ ⎤ ⎢⎣ ⎡ −−   ( ) nnnn 3 5 5 1071 20 4   +≥− −   nn n n 10012843 3 2402 +≥+−   0 1284 14332 ≤+− nn   ( )   ⇒ ( )0 12 1073   ≤−− nn 3 235 12 ≤ ≤n   Least number of leaps =12

5  Solution iii) Let k be the initial distance between the deer and the cheetah. For the deer to survive the chase, for all n values, nk S c 3 5+<   ( ) nk nn 3 51 10 18 2 +<⎥⎦ ⎤ ⎢⎣ ⎡ −−   nk nnn 100 6033 2402 +<+−   0 60 14332 >+− k nn   ⇒ Discriminant < 0 ⇒ 0 720 1432 <− k  ⇒  m401 . 28>k  least distance = 28.5 m 9a)let z = x + yi ( ) (*2( ) xyii ixyii+ + = + +  ( ) (1 2 2 xy i yix− + = − + + )1 ∴  ( ) 2 ----- (1) 121 ------(2)  x y  y x = − − + = +   Sub (1) into (2): ( )122 1 y y− − = − +   2 3  y∴ =  and 4 3  x = −  9b. i) 2 5 5 2 2 4 4 5 5 5 5 1 0 , , , , k i i i i i  z z e  ze e e e π   π π π π     − − − = ⇒ = ⇒ = 1 0   ii)( )  ( )5 5 5 5 z z+ − − =   5 5 1 5  z  z +⎛ ⎞⇒ =⎜ ⎟ −⎝ ⎠   2 55 5 k i ze  z π   +⇒ , k = 0,± ±  (from (i))=− 1, 2   ( ) ( ) 2 55 5 k i  z ze π  ⎛ ⎞⎜ ⎟⎝ ⎠⇒ +  = −   2 5 2 5 5 1 1 k i k i e  z e π   π   ⎛ ⎞⎜ ⎟⎝ ⎠ ⎛ ⎞⎜ ⎟⎝ ⎠ ⎛ ⎞ −⎜ ⎟⎜ ⎟⎝ ⎠⇒ =   ⎛ ⎞+⎜ ⎟⎜ ⎟⎝ ⎠   5 5 5 5 5 5 5 k k k i i i k k k i i i e e e e e e π π π   π π π   −⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠ −⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎛ ⎞ −⎜ ⎟⎜ ⎟⎝ ⎠= ⎛ ⎞+⎜ ⎟⎜ ⎟⎝ ⎠   Accept ei0

6  Solution = 5.2 sin 5 2cos 5 k i k  π   π   ⎛ ⎞⎜ ⎟⎝  ⎛ ⎞⎜ ⎟⎝ ⎠  ⎠= 5tan 5 k i  π   (proved) 10i) 2 9 33 3 3  x pxq qp y x p  x x + − − −= = + − ++ +   Asymptotes: 3, 3 yxp x⇒ = + − = −  ii) ( )2 9 31 3 dy q p dx  x − −= − +   ( )2 For 0, 3 9 3 39 3 dy dx  x qp  x q p = + = − − = − ± − −   For 2 turning points,9 3  0qp− − > 9 3 (shown)q p⇒ < −   iii) When p = 2,q = 1, 22 1 3  x x y  x + −= +   ( ) ( ) 4 3 2 22 2 2 6 0 2 1 2 3  x xx x  xx x x + − − − = + − = +  2 2 212 3  x x  x  x + −=+  ----- (1) 2 intersection points betweenC  & 2 2 y  x = ⇒ 2 real roots (shown) 11a)(i) 2 1r r u u+ += +  x  y 3 - p  p - 3  - 3  y = x + p – 3 2 2 y  x = r u   2 1 1 1r r  r r  u u u u + + + = +   1 11r  r  v   v+ = + k (ii) As ,v  andr  →∞r  → 1r v k + → k  k  11+ =∴

7  Solution 1 2 +=k k   01 2 =−−k k    2 5 1±=k    Since for all0>r u 1≥r    01 > =⇒   + r  r  r  u uv  for all 1≥r    2 51+=⇒k   (ans) 11b) i) me that the result is true i.e. (i) 11 =u   12 =u 2213   =+= uu u 43214   =++= uu u u   843215   =+++ = u uuuu   16543216   =++++= uuuuu u   (i 2,22 ≥= − nu n n   (iii) Let 2=n , LHS = 12 =u   RHS = 120 =  Therefore the result is true for n = 2. Assu for n = k, 2 ≥k    2 ,22 1 1 ≥==   − − = ∑ k uu k  k  i ik  For , 1 = + 1 1 i ik +=k n   ∑∑  − = +== 1 1 k  i k i k  uuuu   22 22  −−+= k k  2 2 . 2 −k   =   ( )11 k  2 2 2k    + −=  −= Therefore the result is true for 1+=k n . ence by induction, the result is true for allH 2,  ≥∈ n Z n . 12a)a) ( )  x x y x dx 1 [ ++dy 2 22 sincos]  =−  ---------------------- (1) g v = x – yUsin , dxdx −=1 ⇒ dydv dx dv dx dy −=1 --------------------- (2) 1 - Substitute (2) into (1) : dx dv + [ 1 + ] cos 2 x =2v x2sin dx dv = [ 1 + ] [cos2 x ] + 1 -2v x2sin   dx dv = cos2 x[ 2 + ]2v

8  Solution 2 1 2 dv v+∫  = ( dx x∫ + 2 cos 1 2 1 )   1 2 1 tan 2 v−  = ⎟ ⎠  ⎞⎜⎝  ⎛ +  x x 2sin 2 1 2 1  + C 1 2 1 tan 2  xy− − = ⎟ ⎠  ⎞⎜⎝  ⎛  +  x x 2sin 2 1 2 1  + C  y = x –   2 12 tan sin2 2 2 2  x x C  ⎛ ⎞⎛ ⎞+ +⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠   12b)dx R kx dt  = −  ,kis a positive constant At x= 1.5 R , 30 0 2 dx  R Rk  dt  = ⇒ − =  ⇒ k = 3 2  Thus , 2 3 dx R x dt  = −  ( shown ) i) 2 3 1 1dx dt   R x =−∫ ∫  2 3 1 2ln 3  R x t − − = c+  2 2 2ln 3 3 3  R x t − = − − c   2 3 2 3 t   R xAe− − =   ( 2 3 3 2 t  ) x RAe− = −   Att = 0 , x = 0 , 0 =( ) R A−  ie A = R  ⇒  2 331 2 t   R x −⎛ = − ⎜⎝ ⎠ e  ⎞ ⎟  ii) Ast  → ∞ , t  e3 2−  → 0 x →  3 2  R ⇒  α  = 3 2  R  ie regardless of time, the amount of drug in the patient’s body will never exceed3 2  R. 13a) 2 2 4 2 (18) (18)  L y x  x x = + − = + −   2 4 2 3 (18) 2 4 2(18)(1  Lx x dL L x x dx = + − = + − −)

9  Solution At min pt, 0dL dx =  3 3 4 36 2 2 18 0  x x  xx ∴ = − + − =   From GC, x =2 is the only solution. Therefore the point is (2, 4) 2 , 0 2 , 0 Minpoint. dL x dx dL x dx + − = > = < ∴   13b)i) 2 6 46 2 6 2 6 3  x  x  x hr  r h = ⇒ =+ = ⇒ = ++   2 21 1(6) (2)6 3 3 V rhπ π  = + −   2 2 3 1 2 ( 6 ) 3 3 1 1 ( 6