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

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