JJC H2 MATHS P1 Solution

1 2014 H2 Mathematics Prelim P1 Worked Solutions Qns Solution Notes 1 Let the amount of money for a blue voucher, a yellow voucher and a red voucher be x, y and z respectively. Then 3 4 7 27.40 (1) 5 2 4 20.80 (2) 2 8 5 45.00 (3) xyz xyz xyz       From G.C., 2, 5, 0.20xyz 4 2 43.40 43.40 4(2) 2(0.2) 75 x py z p   2 (i) Let nP be the statement  n nu 2813 1  for all 0n  . LHS of P0 = 3 0 u (Given) RHS of P0 =  32813 1 0   P0 is true. Assume that kP is true for some 0k  , ie  k ku 2813 1  . We want to prove 1kP is true, ie   1 1 2813 1    k ku . 1LHS of kP = kk uu 211  (Given) = 1 –  k 2813 2  =  k 23 16 3 1  =  k 2)2(3 8 3 1  =   1 2813 1   k 1kP is true Since P0 is true and kP is true 1kP is true, by Mathematical Induction, nP is true for all 0n  . (ii) The sequence is divergent as n  ,   n 2 does not converge to a finite number

2 3 when r = 0, 3 = 3b r = 1, 1 = (4  a)b Alternative Method (r + 1)4 + (r + 1)2 + 1  (r2 + ar + 3)(r2 + r + b) (r4 + 4r3 + 6r2 + 4r + 1) + (r2 + 2r + 1) + 1 (r2 + ar + 3)(r2 + r + b) Comparing coefficients: r0: 3 = 3b r3: 4 = 1 + a a = 3, b = 1 22 22 2 2 42 11 ( 3 3 ) ( 1 ) 13 3 ( 1 ) ( 3 3 ) 2( 1) (1 )(1 )1 rr r r r r rr r r rr r rr            42 0 22 0 1 (1 )(1 )1 11 1 21 3 3 N r N r r rr rr r r                 22 22 11 13 11 37 1 2 11 11 1 1 3 1 3 11 13 3 NN N N NN N N                     = 2 11123 3 NN   42 2 1 N r r rr  = 1 42 1 1 (1 )(1 )1 N r r rr       2 2 11 1 2 3 (1 ) 3 (1 ) 3 11 1 23 1 NN NN        

3 4 (i) Area of region R 3 2 0 253 3 d 4 1.37 (to 3 s.f.) x x        (ii) Equation of new curve 2 2 2 35 4 2 2 4 y x y x      Volume of revolution 23 2 0 3 2 2 0 3 2 2 0 3 1 0 2 2 d 4 48 4 d4 4 124 1 d (Shown)4 4 124 l n 2 s i n 2(2) 2 2 12 3 24 l n 3 43 23 x x xx x xx x xx xx                               5 222 22 4 2 2 24 2 24 1 3 3 rl k kl r lrh kk rhr rr krVr r rk rV              24 4 24 d2 d3 3 Vk r r r kr   

4 24 4 24 24 4 42 2 4 dAt stationary point, 0d 2 3 3 2 3 3 V r kr r kr kr r rk kr         6 (a) (i) 2 d1ln d vux x x  d1 1 d u vx xx  21 11 1 ln d 11 1ln d n n n xx x x xxx x           1 ln 1 ln 1 1 ln 1 + 1 n n nx n nn n nn          (a) (ii) 21 1l n 1ln d + 1lim n nxx nnx     = 1 (b) dsec sec tand xxa a    When x = a, sec 1 cos 1 0   . When x = 2a, 1sec 2 cos 23    .

5   2 22 22 23 0 3 2 0 3 2 0 3 0 d sec sec tan dsec tan d sec 1 d tan 3 3 a a xa xx aa aa a a a a                       7 Let An denote the distance ran on the nth training session and Sn denote the total distance ran for the n training sessions. (i) An = 7.5 + 0.8(n  1) = 6.7 + 0.8n (ii) Sn  475 2(7.5) 0.8( 1) 4752 n n  7.1n + 0.4n2  475 From GC, n = 26.7 Least n = 27 (iii) For the modified training session, let Bn denote the distance ran on the nth training session and and Gn denote the total distance ran for the n training sessions. B6 = 14.93 x(1.2)5 = 14.93 x = 6 (nearest integer)

6 (iv) Gn =   1 16 5 6 5 6   n = 30  15 6  n 1 N n n G   =    N n n 1 5 6 130 =     N n N n n 11 5 6 3030 =   N N 301 130 5 6 5 6 5 6   =    N N 30 130 6 5 6  = 6GN – 30N 8 (i) 2 45 2 xxy x   2(2 ) 45y xx x     2 (4 )( 5 2 ) 0xyx y   For real values of x, discriminant 0 . 2(4 ) 4 ( 5 2 ) 0yy    2 4y 2 or 2yy   (ii) 2 22 45 (2 )22 45 (2 ) x xB Axxx x xA x B          Comparing coefficients of 2x , 1A Comparing constants, 54 1 AB B     2 45 1 (2 )22 xxyx xx     1211 1 22 22 TTyx yx y xxx x           The graph of 1yx x can be transformed to the graph of 2 45 2 xxy x   using the following transformations in succession. T1: Translation of 2 units in the positive x direction. T2: Reflection in the x-axis.

7 (iii) 9(i) least a = 4 (ii) when x > 4, f( x) = y = 2 82   x x xy – 2y = 2x – 8 x(y – 2) = 2y – 8 x = 2 82   y y f –1 : x  2 82   x x , 0 < x < 2 (iii) g(x) = x2 – 6x + 7 = (x – 3)2 – 2, x < 3 Dg = (–, 3) Df = (4, ) Rg = (–2, ) Rf = (0, 2) fg does not exist since Rg = (–2, –) (4, ) = Df gf exists since Rf = (0, 2)  (–, 3) = Dg y O (1, 2) (3, 2) 2x 2yx 2 2 5 2(0, ) (1 , 2 ) (3 , 2 ) 2x 2yx - 2 x

8 gf(x) = 28g 2 x x    = 72 8262 82 2              x x x x , x > 4 From GC: Rgf = (–1,7) Alternative method (Mapping method):    fg4, 0, 2 1, 7    Rgf = (–1,7) 10 (i) Given A, B and C are collinear, ACk A B     k ca ba 1kk cb a (shown) (ii)    1 () 1 kk kk        ac a b a ab aa ˆsin 90 1kk  ab n 0 9 k It is the area of a parallelogram with sides OA and OC. (iii) Area of triangle OAC = 3 x area of triangle OAB 13 22  ac ab 93s i n 9 0 27 k   ab 3 3 k k   (iv) Length of projection of OC onto OA = 12 12 12 36     ca a ca a When 3k  , 3- 2cb a   2 3- 2 32 3(0) 2(3) 18     ca b aa ba aa

When 3k  , 34 cb a   2 34 34 3(0) 4(3) 36       ca b aa ba aa

34  cb a

9 11 (a)  1i1izt t   11 itt tt     Let 1 x t t ------(1) 1 yt t ------(2) (1) + (2): 2x yt (1) (2): 2xy t 2 2 xy x y   22 4xy 11 (b) If  arg argp q , arg arg argp p qP O Qq   1 22t a n 3 a a  1 22 tan or 1.183 rad 23 6p qa a  P Q 5a 3a Im Re a

10 12 (i) 23 2 ,4 dd 2, 3dd d3 d2 xt yt xy tttt yt x       32 Tangent at 34 2 P py px p  323 8yp x p  (ii) Since the tangent passes through the origin, subst. x = 0 and y = 0 into the equation of tangent in part (i). 3 80 2 p p     4, 12 4, 12 x y P    (iii) 0, 4, 0x yt   (iv)       4 12 0 22 2 05 2 1Area 4 12 dy2 24 3 dt 324 5 24 5 x tt t            x y ‒4 O