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