AISS Prelim AMath Paper 2 Marking Scheme

3 AISS PRELIM/4E/4049/01/2024 [Turn over 1 A curve has the equation sin 2 2c o s 2 xy x  . (a) Show that the gradient function can be expressed in the form  2 cos 2 2 2c o s 2 kx x   , where k is a constant. [3] sin 2 2c o s 2 xy x           2 22 2 22 2 2 2c o s 2 2 c o s 2 s i n 2 2 s i n 2d d 2c o s 2 4c os2 2cos 2 2sin 2 2c o s 2 4c os2 2 c os 2 sin 2 2c o s 2 4c os2 2 2c o s 2 x xx xy x x xx x x xx x x x x          (b) Find the acute angle between the tangent to the curve at 12x  and the line 0y . [ 3 ] 2 4cos 26gradient of tangent 1.1386 2c o s 6       Angle required =   1tan 1.1386 48.7  (1 dp) M1: correct quotient/pdt rule M1: differentiate sin 2x and cos 2x correctly A1: use of identity to reach answer M1: correct gradient value M1, A1 (accepts 0.850 rad)

4 AISS PRELIM/4E/4049/01/2024 2 (a) Factorise 33 27x k as a product of a linear and a quadratic factor. [2]    33 2 227 3 3 9x kx k x k x k    (b) Hence solve    3 27 3 10xx x  , expressing non-integer roots in surd form. [3]    321: 27 3 3 9kx x x x           233 9 3 1 0xx x xx     230 o r 3 9 1 0xx x x   23 or 4 1 0xx x    41 6 4 1 2 25 x    (c) Find the value of k given that 33 27x k leaves a remainder of 351 when divided by 2x . [ 2 ] Let 33f( ) 2 7x xk 33f (2) 2 27 351k  3 351 8 7 27 3k  B1: identify x=-3 as a root from Hence M1: apply quad formula correctly A1 M1: applying remainder thm correctly A1 M1: (x+3k) A1

5 AISS PRELIM/4E/4049/01/2024 [Turn over 3 The diagram shows a L-shaped rod ABC where AB and BC have length 18 m and 8 m respectively and angle ABC is 90°. The rod is hinged to a wall at A so as to rotate in a vertical plane. The rod AB makes an acute angle θ with the vertical wall surface OA. (a) Given that G is a point directly below C, show that cos sinOG p q   , where p and q are constants to be found. [2] Using triangle BCD: Angle DBC = θ cos 8 8cos BD BD     Using triangle ABE: Angle BAE = θ sin 18 18sin BE BE     OG = BD + BE = 8cos 18sin , 8, 18pq (b) Express OG in the form  cosR   where 0R and 09 0   . [3] D E B1: either correctly establishing BD or BE B1: showing the other component and clear indication that OG is a sum of the 2 values G

6 AISS PRELIM/4E/4049/01/2024  22 1 1 8cos 18sin 1881 8 c o s t a n 8 9388 cos tan 4 19.7 cos 66.0 OG                (c) Find the length of OG and the corresponding value of θ if G is at maximum displacement from O. [ 3 ] max 388 19.7 m OG   1 9when cos tan 1 4   1 1 9tan 0 4 9tan 66.04      