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]