ACJC 9758 2022 Promo Solutions

1 2022 ACJC H2 Math Promo Marking Scheme Qn Solution Remarks 1(i)     3 3 1 2 eln cos ln e ln cos 3ln cos d 1 sin 3d 2 cos 1 3tan 2 x y x xx xx yx xxx x x        Setter: YXF M1 apply laws of logarithm A1 (ii)     1 ln 2 2 1 ln ln ln ln ln 1 d 1 ln 2 lnd d ln (ln 2)d xxyx y x xx y x x y x x xy x x y y x xx       Alternative: 1 ln 2 1 ln ln ln 1 1 d 1 1 ln 2 lnd 1 d 1 ln 2lnd d1 2ln lnd xxyx y x xx y yxx y x x x y yxy x x y y x yxx       B1 apply ln on both sides and apply laws of logarithm M1 apply implicit differentiation A1 OR B1 apply ln on both sides and apply laws of logarithm M1 apply implicit differentiation A1

2 2(i)        2 2 2 8851 10 , 22 51 2 88 10 2 02 51 14 10 20 02 10 31 14 02 10 31 14 02 2 7 5 2 02 272 or 52 xxx x x x x x x x x xx x xx x xx x xx                         Setter: NSH M1 move to LHS and combine M1 factorising the numerator A1 (ii) 88 10 51 or 012 88 10 51 or 01 2 1 2 1 72 or or 052 22No solution. 77 22 77 x xxx xx x xxx x x                   M1 replace x with 1 x A1 3(i) 2 31 ax bx cy x   Substitute ( 1, 4) and ( 3, 2) into equation, 8 ----- (1) 9 3 16 ----- (2) a b c a b c              2 2 2 3 1 3d d 31 ax b x ax bx cy x x        Substitute d3 and 0d yx x  ,     0 6 8 3 9 3 21 3 0 ----- (3) a b a b c a b c           From GC, 1, 0 and 7a b c   . Setter: NSH M1 substituting in ( 1, 4) or ( 3, 2) . M1 d 0d y x  A1

3 (ii)   2 7 31 1 64 3 9 9 3 1 xy x x x       2 2 1 39 3 1 0 7 3 7 3 1 39 1 7 9 x x x x xx x x        2 7 31 xy x   B1 shape B1 vertical asymptote and y-intercept [ECF allowed based on value of c found in part (i).] B1 equation of oblique asymptote 4(i) Setter: YKX B1 equations of asymptotes B1 shape of graph B1 coordinates of axes intercepts (Do not penalise if not in coordinate form.) –6 –4 –2 2 4 6 5 10 x y –6 –4 –2 2 4 6 5 10 x y (0, 7) y x O C x y

4 (ii) B1 equations of asymptotes B1 shape of graph B1 coordinates of axes intercept and turning point (Do not penalise if not in coordinate form.) 5(i) 62 1 3 6 1 32 6 2 10 1 3 2 2 6 1 0 1 6 4 6 0 2 2 2 1 ON OA AN AN ON                                                         Setter: LCH B1: AN M1: 2 10 1 AN   A1 (ii) '2 2 3 1 2 2 6 2 1 1 3 OA ON OA                               M1 ratio theorem or addition of vectors A1 x y

5 (iii) Let D denote the point(s) that are 33 units away from A on line L. Then 62 1 OD       and 32 6AD         . (can be taken from (i)).          2 2 2 2 2 32 6 3 3 3 2 6 27 6 24 18 0 4 3 0 3 1 0 1 or 3                                    4,1,0 or 0,3, 2DD M1 distance formula to find  A1 both coordinates 6(i)              f 2 1 f 2 cos 2 2 cos 2 2sin 2 1 sin 2sin 2 1 sin r r r r r r             2k  Setter: YKX B1 usage of factor formula B1 2k  (ii)            11 1sin 2 1 f 2 1 f 2 2sin f 0 f 2 1 2sin nn rr r r r             f2  f4  f4  f6    ... ... f 2 3 n   f 2 2 n   f 2 2 n   f 2 1 n   f 2 1 n         2 2 f2 1 cos 0 cos 22sin 1 1 cos 22sin 1 2sin2sin sin sin n n n n n                     √B1 using (i) M1 method of difference A1 cosine double angle formula leading to given answer

6 (iii)          1 1 1 1 1 2 2 sin 2 1 sin 2 1 1 sin 2 1 sin 1 sinsin n r n rr n r rr r n                     M1 changing running index A1 final answer 7(a)       1 22 33 2 163 OE OA AB OE OC CD                      a b a ab a 6b a ab Comparing the coefficients of a and b,    211 3 6 61,16 16 6 6 51 16 16 8OE             3a b a+ b 8 Setter: LCH B1 OE (at least 1 correct) or equivalently equation of any one of the two lines. M1 comparing coefficients A1 (b)(i) Let the foot of perpendicular from point D to line OE be N, with position vector n. Then   de en ee  2 32 4 3 2     de efd ee ed ed B1 obtaining foot of perpendicular using vector projection M1 applying ratio theorem or addition of vectors A1 both answers (ii)   1 1 3 3 2 2 2 4 d e d d eOD OF       OR   1 1 322 2 2 2 2 4 d e d e deOE DE              M1 area of triangle formula A1 value of k

7 8(i) 2 2 2 dd anddd d d d d d d dWhen , , ln and d Equation of tangent: ln ln Equation of normal: 1ln 1 ln x a y a t t t t y y t a t tx t x t a ayt p x y a p ppx ay a p p x p y px a a p ay a p x pp ay x a ppp                           Setter: YXF M1 d d y x M1 – finding equation using formula or substituting to find c. A1 equation of tangent and normal (ii) When 0, lnx y a a p   When 20, ln ax y a p p   Area of APB 2 2 22 3 1 ln ln2 1 2 ( 1) 2 aaa a p a p pp aaa pp ap p                    M1 – subtracting the y- intercepts or finding lengths using pythagoras theorem A1 (iii) B1 correct shape with x- intercept labelled (iv) Equation of tangent when 1p : ln1y x a a xa      Note that both the tangent and the line y mx a pass through the point (0, a). Range of m: 10 m   . √B1 equation of tangent B1 correct range of m O x y (a, 0)

8 9(i)        2 f 1 5 3 4x x x x       fR 4,0 Setter: NSH B1 9(ii) The line yk where k cuts the graph of  gyx at most once. Hence g is one-one and 1g exists. B1 (iii)     1 2 2 1 g 34 34 34 3 4 3 g ( ) 3 4, D ( 4,0) yx xy xy yx xx                     M1 making x the subject A1 for 1g ( )x A1 for 1gD  (iv) 11h h( 2)2 1h h(3)2 1ln 42       M1 Getting 1h 2   or h(3) . A1 (v) B1 one cycle drawn B1 2nd cycle drawn B1 end points and asymptotes labelled 10(i) Equation of planes are: 6 4 2 4 10 56 56 3 2 1 2 10 14 14 r r       Setter: LCH M1 dividing by magnitude of normal to both sides, or any equivalent method A1 equations of both planes x y O x y (–4, 0) (1, 0) (5, 0) (6, 0) x = 5 (0, 0)

9 (ii) 11 61 41 22 14cos cos 40.261 56 6 41 22                           M1 formula to find angle between 2 planes A1 (iii) 6 1 6 3 4 1 10 2 5 2 2 2 1                                                 Let 0, 6 4 4 (1) (2) Subst. (2) into (1) 6 4 4 2 4 6 2 3 2 3 2 2 z xy x y k x k y k y y y k y k x k k x k                              1 2 2 3 : 2 3 5 01 k Lk                      r ,  M1 cross produc