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. - 2 3 2( 2)( ) 2 5 23 10x ax bx c x x x       By comparison, 2a 2 10 5 c c   2 5 4 5 9 a b b b      3 2 2 2 5 23 10 0 ( 2)(2 9 5) 0 ( 2)(2 1)( 5) 0 x x x x x x x x x             12, ,52x  6ii     3 2 2 1 5 1 23 13 0y y y          3 2 2 1 5 1 23( 1) 10 0y y y       Since 1y x  , 11, ,62y

2022 4E5N AM Prelim P1 Mark Scheme 6 Qn Solutions Remarks 7ai Maximum value of 4y  Minimum value of 2y   4 ( 2) 12c    7aii 3a 0.5b 7b 7c There are 5 solutions

2022 4E5N AM Prelim P1 Mark Scheme 7 Qn Solutions Remarks 8i 2 8 1 6 16 8 8 = ( ) 8 ( ) r r r r r pT x r x p xr                   For term independent of x , 16 8 0 2 r r    3 2 =7 8 ( ) =7 2 T p     228 7p  0.5p   0.5p  (ve rejected) 8ii For constant term, there is a 8x term in the expansion of 8 2 6 px x      . 16 8 8 3 r r    3 8 8 4 8 ( 0.5) = 73T x x             8 8 2 8 16 8 8 6 11 1 4 7 7 ... 2x x x x x x x            The constant term in   8 8 2 6 11 2x x x           8 81 7 7 x x    0  There is no constant term. 9i

2022 4E5N AM Prelim P1 Mark Scheme 8 Qn Solutions Remarks v 1 2 3 4 5 lg F 1.74 1.78 1.82 1.86 1.91 See graph behind 9ii Gradient 1 2 1 2 y y x x   = 0.0374 to 0.0453  lg lg lg vF km F m v k    lg 0.0374 to 0.0453 1.09 to 1.11 m m   lg 1.69 to 1.71 48.9 to 51.3 k k     48.9 1.09 v F  9iii Since lg100 2 , 7.1v m/s from the graph 10i Equation of AC is  30 7 4 7 2 y x y x     

2022 4E5N AM Prelim P1 Mark Scheme 9 Qn Solutions Remarks 10ii From equation of AC,  0, 2A and  0,27D Diagonals of kite are perpendicular. 1 7 BDm  Equation of BD is 1 277y x  Solve AC and BD simultaneously: 17 2 27 7 49 14 189 50 175 13 2 x x x x x x         When 13 2x , 1 17 3 2 262 2y       1 13 ,262 2M     10iii M is the midpoint of BD. Let  ,B x y 1 1 0 273 ,26 ,2 2 2 2 x y            13 2 72x   and 126 2 27 262y         7, 26B 10iv Because both triangles share the same height CM, 1 4 BD BE 

2022 4E5N AM Prelim P1 Mark Scheme 10 Qn Solutions Remarks 7 5 35Ex     27 1 4 22Ey      35, 22E 11i Midpoint of chord joining the two given points   2 2 4 2,2 2 2, 1           As the perpendicular bisector of a chord passes through centre of a circle, the y-coordinate of its centre 1 Radius of the circle =  4 1 5   Let centre of circle be  , 1x  . B D E 1 unit 4 units 1 7 4 28 radius