2022 SGSS AMath Prelims P1 Ans

1 2022 AM 4E5N Prelim Paper 1 Marking Scheme Solutions: 1 3 33(1 ) 2 3(1 ) 2d d1 1 1 x x x xx x x             2 23(1 ) d 1x x x        33(1 ) 2ln(1 )(3)( 1) x x c    3(1 ) 2ln(1 )x x c     2 3 2 2 1 22 1 2 1 2 p q       3 3 2 2 2 2(2) 1 2     7 5 2 1   7 5 2  7, 5p q   3 34 π3V r 2d 4πd V rr  d d d d d d V V r t r t  2 d20 4 4 π d rr t   2 d 4 d π r t r When r = 11, 2 d 4 d π(11) r t  ≈ 0.0105 cm/s (correct to 3 s.f.) 4 2 22 5 48 0x xy y    --- (1)

2 Solutions: 8 2y x  --- (2) From (2): 2 8y x  --- (3) Put (3) into (1): 2 22 5 (2 8) (2 8) 48 0x x x x      2 2 22 10 40 (4 32 64) 48 0x x x x x       2 2 22 10 40 4 32 64 48 0x x x x x       28 8 16 0x x   2 2 0x x   ( 1)( 2) 0x x   1 or 2x x  Put 1x  into (3): 2( 1) 8y   10 Put 2x  into (3): 2(2) 8y  4  Coordinates of the points of intersection are (–1, –10) and (2, –4). 5 Let 2 2 2 8 3 1 (2 1) 2 1 (2 1) x x A B C x x x x x        Multiplying throughout by the denominator, 2 28 3 1 (2 1) (2 1)x x A x Bx x Cx       When 0x  , 1A When 1 2x  , 2 1 1 18 3 12 2 2 C              3 1 2 2 C 3C  When 1x  , 2 28(1) 3(1) 1 (1)(3) 3 3(1) B     12 9 3 3 B   2B

3 Solutions: 2 2 2 8 3 1 1 2 3 (2 1) 2 1 (2 1) x x x x x x x        6(a) (b) 1 24 3 2x x   2 2(2 ) 3 2 (2 )4 x x  2(2 ) 16(2 ) 12 0x x    Let 2xu . 2 16 12 0u u   2( 16) ( 16) 4(1)( 12) 2(1)u       16 304 2u  0.7177978871 or 16.71779789u u  2 0.7177978871 or 2 16.71779789x x  (no solutions) * ln16.71779789 ln 2x  ≈ 4.06 ** (correct to 2 d.p.) 7(a) (b) 2C( ) 40 480 1442x x x   240( 12 ) 1442x x   2 2 2 12 1240 12 1442 2 2x x                     240 ( 6) 36 1442x      240( 6) 2x   Minimum value of C(x) = 2 Corresponding value of x = 6

4 Solutions: (c) Minimum average cost per chair = $2 Number of chairs produced = 6000 8(a) (b) (c) (Alternate Segment Theorem) ( s in the same segment) ABC CAF DEF     Since ABC DEF 