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 M