2022 AMath Prelims P1 w Ans

2022 4E5N AM Prelim P1 Mark Scheme 1 Qn Solutions Remarks 1a 2 2 4 3 4 3 0 nx x n nx x n        For no real roots, D < 0      2 2 2 2 4 4( )( 3) 0 16 4 12 0 4 12 16 0 3 4 0 4 1 0 n n n n n n n n n n                 1n or 4n Since 0n , 1n  1b (2 1) 3x x   and 2 4x     22 3 0 2 3 1 0 x x x x          2 4 0 2 2 0 x x x      31 or 2x x  2 2 x   The final solution range are 2 1 x   or 1.5 2 x  . 0 -1 -2 2 1.5 -1 4

2022 4E5N AM Prelim P1 Mark Scheme 2 Qn Solutions Remarks 2 x ey x3        3 3 2 3 2 3 1d d 3 1 x x x x e ey x x e x x    For increasing functions, d 0d y x    013 2 3  x xe x Since 03 xe and 02 x , 3 1 0 1 3 x x    - 3 y x 27 19  yx 32 33  yx 32  ------- Equation 1 216)2(8  xy 2 1 423 2222  x y 2 14 2 3 22   xy 2 1423  xy ------- Equation 2 Substitute Eqn 1 into Eqn 2:

2022 4E5N AM Prelim P1 Mark Scheme 3 Qn Solutions Remarks 1( 2 ) 4 2 2 5 9 2 2 9 5 xx x x       93 2 5 6 5 y y         4i LHS sin sin(2 ) x x x   sin sin 2 cos cos 2 sin x x x x x   2sin 2sin cos (cos ) sin (2cos 1) x x x x x x    2 2 2 sin 2sin cos 2sin cos sin 4sin cos RHS x x x x x x x x       4ii xx sin23sin  2 2 2 2 4sin cos sin 2sin 4sin cos sin 2sin 0 4sin cos sin 0 sin (4cos 1) 0 x x x x x x x x x x x x x          sin 0x or 2 1cos 4x (NA) 0,x   Alternative: 2sin (4sin 5) 0x x   sin 0x or 2 5sin 4x (rejected)

2022 4E5N AM Prelim P1 Mark Scheme 4 Qn Solutions Remarks 5i   d ln(2 cos )d 1 2 sin2 cos 2 sin 2 cos x xx xx x x x x          5ii 2 6 2 6 4 2sin d2 cos 2 sin2 d2 cos x xx x x xx x           2 6 2 ln(2 cos ) 2 ln( cos ) ln cos2 3 6 x x                              32 ln ln 3 2 2 ln 3 3 2 62 ln 2 3 3                                  3a and 3b -

2022 4E5N AM Prelim P1 Mark Scheme 5 Qn Solutions Remarks 6i 2 3 2 3 2 102 5 23 2 10 5 23 2 5 23 10 0 x x x x x x x x x           Let f 3 2( ) 2 5 23 10x x x x    . f       3 2 ( 2) 2 2 5 2 23 2 10 0         By Factor Theorem, ( 2)x is a factor. -