2022 TKSS AMath Prelims P2 Ans

2022 Sec 4 Preliminary Exam Add Math Paper 2 Mark Scheme Qn Working Mark allocated 1 3 4 54x y  , 4 3 1 0x y   . 3 4 54 54 3 ...(1)4 4 3 1 4 3 ...(2) x y xy x y y x xy        sub (1) into (2) 54 3 54 34 3 4 4 x x x x             2 2 2 216 12 12 54 3 3 54 216 0 18 72 0 x x x x x x x x           ( 12)( 6) 0 12 or 6 x x x x      53 3(12)sub 12 into (1), 4 1 4 2 53 3(6)sub 6 into (1), 4 9 x y x y       112, 4 2x y  or 6, 9x y  M1 substitution leading to one variable only M1 solve equation A1 for x values A1 for y values 4 m 2(a) 3 sin 3cos 3cos 3 sin Let 3cos 3 sin cos( ) cos cos sin sin R R R                      3tan 3 6       22 2 2 3 3 12 2 3 R R R     3sin 3cos 2 3cos( ) 6       B1 for α or oe Do not award if tan α is -ve M1 for R A1 oe.

2022 Sec 4 Preliminary Exam Add Math Paper 2 Mark Scheme (b) 1 1 3 sin 3cos 1 2 3 cos( ) 1 6          Minimum value of y 1 is min when cos 0 6 6 2 3 y             Hence, and its corresponding value of  is 3  . B1 answer M1 find angle using their minimum A1 oe. 6 m 3(a) 2 2 4 2 2 2 2 2 ( 3) 28 3 28 0 ( 4)( 7) 0 7 or 4 (NA) 2 ln7 1 ln7 0.9732 x x x x x x x x e e e e e e e e x x              B1 quadratic form and RHS=0 B1 factorised form B1 exact or oe (b)        4 16 4 16 4 4 4 4 4 4 4 4 4 4 4 2 2 4 4 log log log log log loglog log log 16 log 16 1 1log log log log2 2 Let log 1 1 2 2 1 1 2 2 0 ( 1) 0 0 or 1 log 0 or log x x x x x xx x x x x x y x y y y y y y y y y y y y x x                               0 1 1 4 or 4 1 or 4 x x x      M1 change base M1 attempt to express into quadratic equation M1 factorisation M1 change from log to exp form A1 8 m

2022 Sec 4 Preliminary Exam Add Math Paper 2 Mark Scheme 4(a) (alternate angles) (tange r nt chord theorem) ) = Triangle is simila to Tr (common an iangle . A gl (A e A ) ACD CAS ABC CAS ACD ABC C A AB DAC ADC CB           B1 for statement with property B1 for statement with property (-1m for missing concluding statement; name of test not required) (b) 2 ( as is a parallelogram) AC AD AB AC AC AB AD CS AB AD SC SADC       B1 ratio o.e B1 AD SC with reason and leading to answer AG (c) Method 1: (tangent chord theorem) 180 (Angles sum of triangle) 180 BCA ABT CBA CAD BCA CBA CAD ABT             Method 2: (tangent chord theorem) (tangents from external point) 180 (adjacent angles on a straight line) 1 CAS CBA ABT DAT CAS CAD DAT CBA CAD ABT              80 B1 statement and property B1 statement and property AG B1 statement and property B1 statement and property AG 6m 5(a) 2 2 6 2 4 0 2 3 0 (2 3)( 1) 0 11 1 2 x x x x x x x             B1 form inequality B1 factorise (-1m if factors are not in integral form) B1 (b) Since 2( 8) 6 0m x x m    , therefore 042  acb . 2 2 2 2 6 4( 8) 0 36 4 32 0 4 32 36 0 8 9 0 ( 1)( 9) 0 1 or 9 m m m m m m m m m m m m                  Also, B1 discriminant B1 0 M1 factorisation A1

2022 Sec 4 Preliminary Exam Add Math Paper 2 Mark Scheme 8 0 8 m m    Hence, 9m . The smallest integer value of m = 9. A1 only if 8 0m  is accounted for 8 m 6(a)     9 2 8 72 18 2 2 9 9 ...1 2 a a ax x x xx x x                                18 15 2 129 36 ...x ax a x    B1 18 15 9x ax B1 2 1236a x 6(b)   9 92 2 1 9For , T = r r r a ax x rx x                             92 18 2 18 3 9= 9 9 r r r r r r r axr x a x xr a xr                       3For , 18 3 3 5 x r r      5 5 5 9 1265 126 126 1 1 a a a a          M1 skill leading to 3x M1 their coefficient A1 (c)   9 2 6 For term independent of in , 18 3 0 6 9Term independent of 1 84 6 ax x x r r x                 9 2 3 3 3 2 1 21 1 .... 126 84 ... Term independent of 1(84) ( 2)( 126) 336 x xx x x x                         M1 finding term independent of x with their ‘a’ M1 correct pairing A1 8 m

2022 Sec 4 Preliminary Exam Add Math Paper 2 Mark Scheme 7(a) 2 ( at the centre 2 at circumference) 2 BOD BCD x        Area of sector BOD 2 2 1 (10) (2 )2 100 cm x x   2 1Area of (10)(10)sin2 50sin( ) 50sin( 2 ) 50sin 2 cm COD COD BOD x x           Area of sector Area of 100 50sin 2 A BOD COD x x      B1 sub into formula B1 sub into formula B1 ( 2 ) x  seen leading to answer AG 7(b) d 100 50(2 cos 2 )d 100 100cos 2 A xx x     Sub. 6x  , d 100 100cos 2d 6 150 A x           Let the rate of x be d d x t and rate of change of A be t A d d . 1d 0.2 radian s (Given)d x t  2 1 d d d d d d 150 0.2 30 cm s A A x t x t       Hence, the rate of change of A is 2 130 cm s  . M1 attempt to find d d A x M1 chain rule with substitution (their d d A x ) A1

2022 Sec 4 Preliminary Exam Add Math Paper 2 Mark Scheme (c) d d d50 d d d 50 100 50(2cos 2 ) 1 2 2cos 2 1cos 2 2 basic angle = 3 2 42 , 3 3 2, (NA)3 3 3 x A x t x t x x x x x x                  M1 knowing how to formulate relation M1 solve equation by making cos2x the subject A1 9 m 8(a)       2 2 1 2 1 2 1 2 2 1 2 2 1 2 1 2 1 2 d 1 2 2 1d 2 2 2 2 1 x x x x x x x x e xe e xx xe x e e e x x                     B1 product rule B1 derivative of 2 1xe  B1 derivative of 2 1x  AG 8(b)       2 1 2 2 1 2 2 1 2 2 1 2 1 2 2 1 2 d 2 1 d 2 d 1 x x x x x x e x x x e x x x e x x e e c x e c                          M1 converse statement M1 expand or split ( ) ( ) d ( ) d ( ) d f x g x x f x x g x x       A1 2 1 2 12 d x xe x e c    M1 simplify to answer AG (c)   1 12 1 2 2 1 11 22 2 1 0 2 1 d 1 4 1 2.47 ( 3s.f)4 x xxe x x x e e e e                   The area between the graph and the x-axis in the interval 1 1 2 x  lies below x-axis. Alternative: The graph lies below x-axis in the interval 1 1 2 x  . B1 oe. B1 key words in bold and interval 1 1 2 x  9 m

2022 Sec 4 Preliminary Exam Add Math Paper 2 Mark Scheme 9(a) 1Equation of : 1 2 Gradient of 2 Equation of : 7 2( 3) 2 13 QS y x PR PR y x y x         B1 soi M1 sub in coordinates A1 9(b) At , 0 2 (2,0) 1 1 2 132 5 142 35 5 415 3 45 ,15 5 S y x S x x x x y R                  B1 M1 equate A1 x or y-coordinate A1 o.e 9(c) 1 Horizontal distance between and Horizontal distance between an 8 = 6 8 2 6d 14 S Q Q Q T T S Q       Q is the midpoint of S and T. Or Q is equidistant from S and T. M1 compare distance soi