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        M1: 2281 8 M1: 1 18tan 8  A1: correct evaluation and form, accept √388 A1: 388 or 19.7m M1: 1 9tan 0 4  A1: 66.0o

7 AISS PRELIM/4E/4049/01/2024 [Turn over 4 A circle 1C has equation 22 641 2xy xy . (a) Find the radius and the coordinates of the centre of 1C . [3]     22 22 32 22 2 64 1 2 32 1 2 3 2 32 5 xx yy xy xy        Radius = 5 units Centre = (3, -2) (b) Find the equation of the tangent to the circle at the point P (7, 5) . [3] Gradient of normal = 2( 5 ) 3 37 4   Gradient of tangent = 4 3 Eqn of tangent: 457 3yx  44 3 33yx (c) Another circle 2C has centre (8 , 4 ) and radius 7 cm. Find the shortest distance between the 2 circles. [2] Distance between centres of circle =    22 3 8 2 4 157   Shortest distance = 157 7 5 0.530 cm (3 s.f.) M1: completing the square or using formula A1 B1 M1 A1, or 3y = 4x – 43 M1award for correct pts and their -1/m used M1 A1

8 AISS PRELIM/4E/4049/01/2024 5 (a) Prove the identity    3 3 sin cos 1 sin cos tan 1cos AA A A AA   . [4]      3 22 3 22 3 33 3 33 3 33 33 3 sin cos 1 sin cos cos sin cos sin cos sin cos cos sin cos 1 cos cos sin 1 sin cos sin cos cos cos sin sin cos sin cos cos sin cos cos cos tan 1 AA A ALHS A AA A A A A A A AA A A A A AAA A A A A AA A AA AA A              OR      3 22 3 32 22 32 3 33 3 33 33 3 sin cos 1 sin cos cos sin cos sin cos sin cos cos sin sin cos sin cos sin cos cos sin cos cos sin cos cos sin cos cos cos tan 1 AA A ALHS A AA A A A A A A A A AA AA A A A A AA A AA AA A          M1: correct expansion of terms M1: expand & simplify M1: applying identity A1: manipulation to RHS M1: applying identity M1: correct expansion of terms M1: simplify A1: manipulation to RHS

9 AISS PRELIM/4E/4049/01/2024 [Turn over (b) Hence solve    3sin cos 1 sin cos 2cos 0AA A A A  exactly, for A  radians. [4]       3 3 3 3 sin cos 1 sin cos 2cos sin cos 1 sin cos 2cos tan 1 2 tan 1 tan 1 A AA A A AA A A A A A A          Basic angle = 4  Quad: Q2, Q4 3 ,44A   M1: simplification to single trigo equation M1: correct basic angle – must be acute A1, A1

10 AISS PRELIM/4E/4049/01/2024 6 The diagram shows the graph of 2 4yx a x and cbxxy  2 . The graph of 2 4yx a x touches the x-axis at P. Points M and O are the -interceptsx of the graph of cbxxy  2 . The origin O is the mid-point of MP. (a) Find the values of a, b and c. [4] Since origin (0, 0) is on cbxxy  2 , c = 0 Discriminant of 2 4yx a x = 0 since curve intersects x-axis once only: 2 4(1)(4) 0 4 a a   Since P is on positive x-axis, a < 0: 4a OR 2 2 2 44 24 aayx a x x     . Since curve intersects x-axis once only: 2 40 4 4 a a   Since P is on positive x-axis, a < 0: 4a Coor of P = ,0 (2 ,0)2 a Coor of M =  ,0 ( 2 ,0)b  2b B1 M1: using discriminant or the equation must be a perfect square A1 B1 M1: using discriminant or the equation must be a perfect square A1

11 AISS PRELIM/4E/4049/01/2024 [Turn over (b) The graph of 2 4yx a x and cbxxy  2 intersects at N. Find the coordinates of N. [2] 22 2 44 2 64 2 3 22 1 6233 9 x xx x x x y         Coordinates of N = 21 6,39   (c) The graph 2yp x q x r  has its turning point at N and passes through point P. Find the values of p, q and r, where r > 0. [3] Graph with turning point at N and passes P (downward opening): 2 21 6 , 039yp x p     At P (2,0) , 2 21 602 39p   1p 2 2 2 21 6 39 44 1 6 399 44 33 yx xx xx           41, 3pq r   OR (longer method): form 3 equations with coor of N, P and either 2nd x-intercept or derivative and solve 2yp x q x r  𝑑𝑦 𝑑𝑥 ൌ 2𝑝𝑥 ൅ 𝑞 Turning point at 21 6,39   : 2𝑝 ቀ ଶ ଷቁ ൅𝑞ൌ 0 43p q ---- (1) Graph passes through P:   2 02 2pq r  42 0pq r  A1 M1: equating and solving correct x (allows ecf) M1: correct completed square form A1: correct p A1: correct q & r M1: establishing 3 equations correctly

12 AISS PRELIM/4E/4049/01/2024 S u b s t ( 1 ) : qr ----- (2) Graph passes through N: 2 16 2 2 93 3 pq r     46 9 1 6pqr Subst (1) and (2): 3691 6 12 16 4 3 qqq q q    7 A particle P, travels in a straight line, so that its displacement, s m, from O at time t seconds, is modelled by 321 533stt  . (a) Find the value of t when particle P return to its initial position. [2] Initial position when t = 0, s = -3 32 32