ACJC 9758 2024 Promo Solutions

ANGLO-CHINESE JUNIOR COLLEGE 2024 H2 MATHEMATICS 9758/01 2024 ACJC H2 Math Promo Marking Scheme Qn Solution 1          2 32 4 32 4 012 12 32 24 1 012 3 012 xx xx xx xx x xx x xx          21 x 32 4 12 x xx   Replace x by x ,  21 1 since 1 always true 11 x xx x         2(i)     2 2 2 22 22 2 2 22 2 2 1t a n Differentiating w.r.t : d2s e cd d21 t a nd d2 1 1 (shown)d Differentiating w.r.t : dd d22 2 1 2dd d dd d 2 1 (shown)dd d yx x yyx x yyx x yyy x x yy yyy yxx x yy yyy yxx x               2(ii) When 0: 1xy   2dd 121 1 1 1dd 2 yy xx     222 22 d1 d 110d2 d 4 yy xx     The Maclaurin expansion for y is: 2 211 111 ... 1 ...24 2 28 xyx x x       0–2 1 ++ – –

ANGLO-CHINESE JUNIOR COLLEGE 2024 H2 MATHEMATICS 9758/01 3(i) 3(ii) f( ) f( ) f( ) f( ( ) ) f( 2 ) A B C yx yx k yx k yx k kx k         1, 2ab k  3(iii) From (i), we see that the curve f( 2 )yx k is a reflection of the curve f( )yx about the line xk . Hence if g( ) g(4 )xx for all x, then the curve g( )yx is the same as when it is reflected about the line 2x . Hence line of symmetry is 2x . 4(i) 4(ii) 0x  , a minimum point 6x  , a (stationary) point of inflexion/inflection 4(iii) From the graph,  f2 5  , so the tangent has equation 5y x . Thus, the tangent passes through the point  2,10 . Hence, the equation of the normal is: 11 5 210 2 or 55 5yx y x      y x O

ANGLO-CHINESE JUNIOR COLLEGE 2024 H2 MATHEMATICS 9758/01 5(i) 5(ii)  22g( ) 3 12 13 3 2 1xx x x  . Hence  g 1,R  f 2 ,3D  . Therefore gfRD , thus fg exists. Put gR as the domain of f. therefore  fg 0, 0.695R  5(iii)  2 g( ) 3 2 1xx Hence turning point is at 2x , therefore largest k is 2.  2 32 1 , 2 1122 33 yx x yyxx     Since 2x , 12 3 yx  1 1h: 2 , 1 3 xxx  6(i)   1 3 1 3 1 2 12 aa b ab ba b ab OC OA AN OC OB BM                              Comparing the coefficients of a and b, 11 22 13 33            y x O

ANGLO-CHINESE JUNIOR COLLEGE 2024 H2 MATHEMATICS 9758/01 13 32 5 22 34,55 31 2 11 55 5 5ca b a bOC                6(ii) ac  is the length of projection of vector a onto the line with direction vector c. 6(iii) Since OA is the diameter of the circle, 2 πOCA . 0OC AC      2 2 2 22 2 0 21 55 21 55 21 1 55 2 1 2 cca ca c ca ab aa b =a a a          221 2: 2 : 12 aac a c c  Alternatively, Since OA is the diameter of the circle, 2 πOCA . ac c 22 22 2 12 1 55 21 55 21 1 55 2       aa bcc aa b c aa c 221 2 2 :2 : 1    ac a c ac A O C

ANGLO-CHINESE JUNIOR COLLEGE 2024 H2 MATHEMATICS 9758/01 Alternatively, 21 55ca b 2 22 2 21 55 21 55 21 1 55 2 1 2 ca a b a ca a ab ca a a ca a           21cos , 2ca a   where AOC  . cos 2 a c  cos c a  , from the right-angle triangle OCA. 2 ac ca 22 2 2 :2 : 1 ac a c ac    6(iv) Triangle OCA is a right-angle triangle with :2 : 1ac  . By Pythagoras’ Theorem, :: 2 : 1 : 1ac CA   , and hence triangle OCA is an isosceles triangle. Area of Triangle OCA 2 2211 1 1 22 4 2 ca a    Alternatively, Area of Triangle OCA 1 2 ac 1 sin2 ac  , where AOC  . Since 1cos 2 c a  from (iii), π 4  . Area of Triangle OCA 211 1 422 22 2 aac a a     7(i) d π2s i nd6 x att    and d cosd y att 

ANGLO-CHINESE JUNIOR COLLEGE 2024 H2 MATHEMATICS 9758/01 dd d dd d cos π2s i n 6 cos 312s i nc o s22 cos cos 3 sin yy x xtt at at t tt t tt           Gradient of normal when t  : cos 3 sin 1 3 tan (shown)cos     7(ii) To find Q, let 0x  : ππ π 3ππ 4πcos 0 or or 66 2 2 3 3tt t      π 4π 33sin or sin (rej) or 33 2 2ya a a a         Thus, at 30, 2Qa  , 4π 3t  . (Shown) 7(iii) Using parts (i) and (ii): Gradient of normal at Q is  24π13 t a n 1 3 43    Equation of normal at 30, 2Qa  : 3 402ya x   34 2yx a  7(iv) When the normal intersects the x-axis, sub 0y : 3304 28x ax a   Thus, 3 ,08Ta  . To find P, let 0y : sin 0 0 or πtt ππ2c o s o r 2c o s π 3 o r 3 ( r e j )66xa a a a          Thus,  3, 0P a .

ANGLO-CHINESE JUNIOR COLLEGE 2024 H2 MATHEMATICS 9758/01 3 38 3 8 3 1 (independent of , shown)8 OT aaOP a    8(a)(i) 2 1 1 41 2 1 11 22 ( 2 1 ) n r n rn n      As n , 1 02(2 1)n  . 2 1 11 241r r     . Alternative: 2 1 11 1 lim lim 141 2 1 2 2 nnr n rn n       Since 1 2 is a constant, the series converges. 8(a)(ii) Replace r with 1r :    1 61 6 1 7 16 22 11 11 2 1 (2 3) 2( 1) 1 (2( 1) 3) 1 21 ( 21 ) 11 41 41 16 2( 1) 1 12 1 16 23 1 3 5 13(2 3) Nr N rr N r N rr rr r r rr rr N N N N N N                          8(b) 2 111 22u   3 111 1 2 u    4 112 1u   Alternative: Use GC  50 1 121 7 1 7 1 1 62 26.5 r r u       

ANGLO-CHINESE JUNIOR COLLEGE 2024 H2 MATHEMATICS 9758/01 9(i) 13.05 1.75 1.3tan tanAPQ APQ x x      13.45 1.75 1.7tan tanBPQ BPQ x x      Hence, 11 1.7 1.3tan tanBPQ APQ x x         9(ii) 22 22 22 22 1.7 1.3 d d 1.7 1.311 1.7 1.3 1.7 1.3            xx x x x x x At stationary point, d 0dx   : 22 22 1.7 1.3 01.7 1.3  xx   22 221.7 1.3 1.3 1.7xx  22 2 1.7 1.3 1.3 1.7 2.211.7 1.3x    1.487x  (as x is positive) 9(iii) Method 1: First Derivative Test 1.48x  1.487x  1.49x  d dx  0.000398 0 0 0.000202 0 Graph By the First Derivative Test, this gives a maximum point. Method 2: Second Derivative Test   2 222 22 22 d3 . 4 2 . 6 d 1.7 1.3 xx x xx    When x = 1.487, 2 2 d 0.0597 0d   x By the Second Derivative Test, this gives a maximum point. It gives the distance which maximises the angle APB, so it makes for easier throwing / more chance to throw object through the hole / gives the widest leeway / better accuracy. 9(iv) By chain rule: 22 22 dd d 1 . 7 1 . 3 0.1dd d 1 . 7 1 . 3       x tx t x x When x = 1: d 0.04625 0.1 0.00463 rad/sdt    10(i) 2 22 2 ax ax ay x   

ANGLO-CHINESE JUNIOR COLLEGE 2024 H2 MATHEMATICS 9758/01         2 2 22 2 2 2 22 2 2 2d d 2 26 4 2 2 2 43 2 2 x ax a ax ax ay x x ax ax a ax ax a x ax ax a x                 Alternatively: 2 22 2 22 ax ax a aya x xx       2 d2 d 2 ya ax x   10(ii) For C to have no stationary points, d 0d y x  has no solutions.  2 2 2 43 2 04 3 2 2 ax ax a ax ax a x        No solutions, therefore 0D        2 44 3 2 0 44320 42 0 02 aa a aaa aa a       Alternatively:    2 2 2 d2 0d 2 22 2 2 ya ax x aaax ax         For no solutions, 2 0a a   Hence 02 a . 10(iii) y x O  ky x

ANGLO-CHINESE JUNIOR COLLEGE 2024 H2 MATHEMATICS 9758/01 10(iv)  2 32 2122 2 x xkxx x k x x x      Sketch ,0kyk x . From diagram, there are 2 positive roots and 1 negative root. 11(i) 2l : 1 32 y z  , 5x  50 12 31              r 1l : 0 11 0 a b        r Since 1l is perpendicular to 2l , 00 210 1 b               20b  2b (shown) Since 1l intersects 1p at  ,1 ,0a ,      21 40 4a   24a  2a (shown) 11(i