SCSS 2024 AM Prelim P2 MS

2024 4G3 Additional Mathematics Preliminary Examinations Marking Scheme Section A (41 marks) 1 A man bought a new car. The value of the car depreciated with time so that its value, $,P after t months’ use is given by 175 000 , ktPe −= where k is a constant. (a) Find the value of the car, $,P when the man bought it. [1] When 0,t= ( )0 175000 , k Pe − = 175000P= [B1] The value of the car is expected to be $162 000 after eight months’ use. (b) Show that 0.01.k = [2] When 8,t= 8175000 162000te− = [M1] 8 162000 175000 ke− = 1 162000ln8 175000k =− 0.0096487k = 0.01k = [A1] (c) Use the result from part (b) to determine the age of the car correct to the nearest month, when its value reached half of the original value when the man bought it. [2] 0.01 175000175000 2 te− = [M1] 0.01 1 2 te− = 10.01 ln 2t−= 11 ln0.01 2t = − 69.31t = 70t = months [A1]

2 2 A calculator must not be used in this question. (a) Show that cot15 3 2.= + [4] LHS cot15= ( ) 1 tan 60 45= −  tan 60 tan 451 1 tan 60 tan 45 − = +   [M1 – application of additional formula] ( )( ) 311 1 3 1  −= + [M1 – exact values of trigonometric functions for special angles] 13 31 += − 1 3 3 1 3 1 3 1 ++= −+ [M1 – multiplication by conjugate] ( ) ( ) 2 2 1 2 3 3 31 ++= − 2 3 4 2 += [A1] 32=+ RHS= (shown) OR cot15 cos15 sin15 cos60cos 45 sin 60sin 45 [ 1]sin 60cos 45 cos60sin 45 1 1 3 1 22 22 [ 1] 3 1 1 1 22 22 13 22 31 22 1 3 1 3 [ 1] 3 1 1 3 3 2 3 1 31 2 3[ 1] M M M A = += −  +  =  +  + = − ++= −+ ++= − =+

3 OR cot15 1 tan(45 30) 1 tan 45tan 30 [ 1]tan 45 tan 30 111 3 [ 1]11 3 1 3 1 3 [ 1] 3 1 1 3 3 2 3 1 31 2 3[ 1] M M M A = − += − + = − ++= −+ ++= − =+ OR cot15 cos15 sin15 cos 45cos30 sin 45sin 30 [ 1]sin 45cos30 cos 45sin 30 1 3 1 1 2222 [ 1] 1 3 1 1 2222 13 22 31 22 1 3 1 3 [ 1] 3 1 1 3 3 2 3 1 31 2 3[ 1] M M M A = += −  +  =  +  + = − ++= −+ ++= − =+

4 (b) Use the result from part (a) to find an expression for 2cosec 15 , in the form 3pq+ where p and q are integers. [2] 2cosec 15 21 cot 15= +  [M1 – application of special identities] ( ) 2 1 3 2= + + 1 3 4 3 4= + + + 8 4 3=+ [A1] ( )4 2 3=+ OR ( ) ( ) ( ) ( ) 2 2 cosec 15 1 sin 15 1cot15 cos15sin15 132 sin15 cos15 132 sin(45-30) cos(45 30) 13 2 [ 1 app of correctly formed special iden tities] 2 3 2 2 3 2 44 32 62 10 1632 4 4 3 8[ 1] M A  =  =     = +   = +   −  = +  − −+  += − = +  =+ 3 Given that 22 0 532 d , 24 m xxe e x − −= where m is a positive constant, (a) show that 424 12 5 0.mmee − + = [3] 22 0 532 d 24 m xxe e x − −= 2 2 0 53 44 mx x ee − += [M1 – integration] 2 2 5 5 3 14 4 4 m m ee − + − + =  [M1 – substitution of limits] 2 2 5 304 m me e+ − = [A1] 424 12 5 0mmee − + = (shown)

5 (b) Use the result from part (a) and a suitable substitution to find the value of .m [4] 424 12 5 0mmee − + = Let 2 ,mue= 24 12 5 0uu− + = [M1 – substitution] ( )( )2 5 2 1 0uu− − = [M1 – factorisation of quadratic equation] 5 2u= or 1 2u= 2 5 2 me = or 2 1 2 me = 52 ln 2m= or 12 ln 2m= [M1 – application of ln] 15ln22m= or 11ln22m= 0.458m= or 0.347m=− (reject) Therefore, 0.458.m= [A1 – includes rejecting 0.347m=− ] 4 The point A lies on the curve ln .y x x= The tangent to the curve at A is parallel to the line 2 3.yx=+ (a) Find the exact coordinates of .A [4] 2Am = d 2d y x = [M1] Given 2 ln ,y x x= d1 lnd y xxxx =+  [M1 – differentiation using product rule] d 1 lnd y xx =+ 1 ln 2x+= ln 1x= xe= [M1] When ,xe= lny e e= ye= ( ),A e e [A1]

6 The normal to the curve lny x x= at A meets the line 23yx=+ at the point .B (b) Show that the -coordinatex of B is ( )2,ke − where k is a constant to be found. [3] Let equation of normal at A be .y mx c=+ 1 ,2m=− ( ),A e e ( )1 2y e x e− =− − [M1] 11 22y x e e=− + + 13 22y x e=− + - eq (1) 23yx=+ - eq (2) ( ) ( )12= 13 2322x e x− + = + [M1] 53 322xe=− 36 55xe=− ( )3 25xe=− (shown) [A1] 5 (a) Prove the identity 1 cos 2 sin 2 tan .1 cos 2 sin 2 xx xxx −+ =++ [3] LHS 1 cos 2 sin 2 1 cos 2 sin 2 xx xx −+= ++ ( ) ( ) 2 2 1 1 2sin 2sin cos 1 2cos 1 2sin cos x x x x x x − − + = + − + [M1 – application of trigonometric identity where 2x is removed and this step allows for factorization next] 2 2 1 1 2sin 2sin cos 1 2cos 1 2sin cos x x x x x x − + += + − + 2 2 2sin 2sin cos 2cos 2sin cos x x x x x x += + ( ) ( ) 2sin sin cos 2cos sin cos x x x x x x += + [M1 – factorisation and simplification] sin cos x x= [A1] tan x= OR

7 LHS 1 cos 2 sin 2 1 cos 2 sin 2 xx xx −+= ++ ( ) ( ) 22 22 1 cos sin 2sin cos 1 cos sin 2sin cos x x x x x x x x − − + = + − + 2 2 2 2 2 2 2 2 cos sin cos sin 2sin cos cos sin cos sin 2sin cos x x x x x x x x x x x x + − − += + + − + [M1 – application of trigonometric identity where 2x is removed and this step allows for factorization next] 2 2 2sin 2sin cos 2cos 2sin cos x x x x x x += + ( ) ( ) 2sin sin cos 2cos sin cos x x x x x x += + [M1 – factorisation and simplification] sin cos x x= [A1] tan x= (b) Hence solve the equation 1 cos 2 sin 2 3cot 21 cos 2 sin 2 xx xxx −+ =++ for 0 180 .x   [4] 1 cos 2 sin 2 3cot 21 cos 2 sin 2 xx xxx −+ =++ tan 3cot 2xx= 3tan tan 2x x= 2 3tan 2 tan 1 tan x x x = − [M1 – application of double angle formula] ( ) 23 1 tan tan 2 tan x x x − = 222 tan 3 3tanxx=− 25tan 3x= 2 3tan 5x= 3tan 5x= [M1 – forming trigonometric equation] Range: 0 180 .x   x is in all quadrants 1 and 2. Reference angle 1 3tan 5 −= [M1 – reference angle] 37.76124= 37.8x= and 180 37.76124x= −  142.2x= [A1 – for both answers]

8 6 The diagram shows two rods AB and BC rigidly hinged at B so that angle 90 .ABC= The lengths of AB and BC are 2 cm and 8 cm respectively. The point C is fixed on horizontal ground and the rod BC rotates in a vertical plane with the rod BC inclined at an angle  to the ground. (a) Show that the height, cm,h of A above the ground is given by sin cos ,h a b =− where a and b are integers to be found. [2] 2cosBD = 8sinBE = [M1 – both BD and BE ] 8sin 2cosh =− [A1] 2 cm

9 (b) Using the values of a and b found in part (a), express h in the form ( )sin ,R − where 0R and 0. 2  [4] 8sin 2cos− ( )sinR =− sin cos cos sinRR    =− cos 8R  = - eq (1) sin 2R  = - eq (2) [M1 – application of addition formula and forming 2 equations] 2282R=+ [M1 – R ] 68R= 2 17 / 8.25R= 2tan 8 = [M1 – tan ] 0.24498 = rad Therefore 8sin 2cos− ( )2 17 sin 0.245=− [A1] (c) Hence, state the maximum value of h and find the corresponding value of . [3] ( )2 17 sin 0.24498h =− max 2 17 / 8.25 cmh = [B1 – maxh ] At maximum, ( )sin 0.24498 1 −= [M1 – ( )sin 0.24498 1 −= ] ( )0.24498 2 −= 1.82 = rad [A1]

10 Name: ( ) Class: ________________________________ Section B (49 marks) 7 The cubic polynomial ( )f x is such that the coefficient of 3x is 1 and the roots of ( )f0 x = are ,m 2m and ( )1, m− where 0.m It is given that ( )f x has a remainder of 30 when divided by 1. x− (a) Show that 322 3 30 0.m m m− + − = [3] Given roots, ( ) ( )( )( )f 2 1x x m x m x m= − − − + [M1]