SAJC 9758 2024 Promo Solutions

2024 JC1 H2 Maths Final Exam Marking Scheme Page 1 of 22 No Solution 1 (i) 55 24 ( 4)! n nS n   As n ,  5 04! n n   55 5 24 ( 4)! 24 n nS n     , which is unique and finite. Hence, the series converges. Hence, 5 24S  . (ii) 2 1 r r u     3 r r u    (replace r with 2r ) 2SS 55 7 24 24 6!    7 6! 7 720 (iii) For 2n , 1nn nuSS  55 54 24 ( 4)! 24 ( 3)! nn nn      45 ( 3)! ( 4)! nn nn   21 (4 ) 5(4 ) ! nnn     21 81 6 5(4 ) ! nn nn     2 71 1 (4 ) ! nn n   11 561 9 24 5! 120uS

2024 JC1 H2 Maths Final Exam Marking Scheme Page 2 of 22 For 1,n 21 7 11 19 (1 4)! 120RHS   2 71 1 (4 ) ! n nnu n   where n   2 (i) 2, 1 1 d1 d 1 , dd 21 xt y t xy tt tt       d1 d2 1 yt xt   As 1t , d d y x  . The tangent to the curve approaches a vertical line. 2 (ii) (iii) At point (2 , 1 1 )Tt t  , Equation of the tangent at T x y (0,2) (2,1)

2024 JC1 H2 Maths Final Exam Marking Scheme Page 3 of 22 (1 1 ) ( 2 ) 21 21 21 ( 1 1 ) 2 21 21 2 ( 1 ) 2 21 2 21 tyt x t t ty t t tx t ty t t tx t ty tx t                     (iv) When 2, 22 1 2 1 2 x t t t     When 1 2t  , equation of tangent: 11 122 222 2 1 212 yx yx       When tangent meets x-axis, 02 ( 1 2 )yx  P(2(1 2),0) When tangent meets y-axis, 01 2xy Q(0, 1 2) 2 1 (2)(1Area of 2)(1 triangl 2)e OPQ 2 3 2 2 units    3(i)  22 1 22 22 2 2 :4 5 2 0 154 1 [Vertical Hyperbola with centre at origin (0,0)]25 Cy x yx yx    Asymptotes

2024 JC1 H2 Maths Final Exam Marking Scheme Page 4 of 22 22 2 As , 54 5 4 55 Asymptotes: and 22 x yx xy yx y x       Axial Intercepts of C1 (From GC)   (0, 5) & (0, 5) or 0, 2.24 & 0, 2.24 (ii) 2 :2 3Cy x  (0, 0) 5 2 yx (0, 5) 1 22 : 1, 0 54 y C x x  (0, 5) 5 2 yx

2024 JC1 H2 Maths Final Exam Marking Scheme Page 5 of 22 Points of Intersection between C1 and C2 (From GC) (0.364, 2.27) & (4,5) (iii)  22 2 11 2 21 Note that :1 : 5 154 4 2 3 5 1 Upper Half of 4 yx xCC y xxC C          From the graphs in (ii), 0 0.36364(5sf) or 4 0 0.364(3sf) or 4 xx xx     4(a) (i) (1.5, 0) (4,5) (0.364, 2.27) 2 :2 3Cy x  (0, 0) (0, 5) (0, 5) 5 2 yx 5 2 yx (0, 3) 1 22 : 1, 0 54 y C x x 

2024 JC1 H2 Maths Final Exam Marking Scheme Page 6 of 22 (ii) (b) 22 2 11 44 ( 2 )y ax x a x a   2 22 2 2 11 44 xx a y y yy ax x a        Sequence of transformations: (order is not important) 10, 2   2y O O

2024 JC1 H2 Maths Final Exam Marking Scheme Page 7 of 22 1. Translation of 2a units in the negative x-direction. 2. Reflection about the x-axis. Alternative Method: 2 2 22 11 4( 4 ) 11 (2 ) ( 2 ) 4 xx a y y y ax x x a x yy x aa x x a            Sequence of transformations: (order is not important) 1. Translation of 2a units in the negative x-direction. 2. Reflection about the x-axis. 5(a) For sum to infinity to exist, 2sin 1 (*) 12 s i n 1 11 sin22 (1)66                 1 2( # )12 s i n 10 1 2sin since 2sin 12 1sin 0.253 (2)4              Since 66   , therefore 0.253 0.526 (final answer)

2024 JC1 H2 Maths Final Exam Marking Scheme Page 8 of 22 Alternatively, students can graph 1 12 s i ny   From the graph, 1 212 s i n   0.253 0.526 (final answer) (b) Let a denote the first term and r be the common ratio of the geometric progression respectively. Likewise, let b and d denote the first term and common difference of the arithmetic progression.  ar4 = b + 6d …(1) a r 8 = b + 24d …(2) a r 10 = b + 49d …(3) (2) – (1): ar8 – ar4 = 18d …(4) (3) – (2): ar10 – ar8 = 25d …(5) Eq(5)/Eq(4):   82 44 1 25 181 ar r d dar r        42 22 4 2 42 42 1 25 1811 25 181 18 25 25 18 25 25 0 (@) rr rr r r rr rr             2 2 25 25 4(18)( 25) 25 2425 2(18) 36r     y x 0 6x 

2024 JC1 H2 Maths Final Exam Marking Scheme Page 9 of 22 25 2425 36 1.436 or 1.436 r   Since for both values of r, |r|= 1.436 > 1, the geometric progression is not convergent. Alternatively, students can use GC to solve 4218 25 25 0rr  The 4 roots are 1.436, 1.436, 0.821i, 0.821i However we only need to consider real roots. Hence 1.436 or 1.436r  . 6(i) Since the line 1y cuts the graph  fyx at two points, f is not a one-to-one function. Therefore f does not have an inverse. (ii) Maximum 5 2k  . Let fyx . For f 5D, 2   , 1 25y x  y x O ൬0, 1 5൰ x = y = 0 1y

2024 JC1 H2 Maths Final Exam Marking Scheme Page 10 of 22 1 1 125 125 51 22 51f( ) 22 Replacing by , 51f( ) 22 x y x y x y y y yx x x         1 ffDR0 ,  1 51f: , w h e r e , 0 .22xx x x    (iii) fR0 , gD[ 2 , )  Since fgRD , gf exists. (iv)   2 2 2 2 1gf g 25 1 3125 1 3152 13 52 152 61 4 1,52 x x x x x x x x               where 6, 5ab .