2023 CCHM Prelim AMath P1 - MS

2023 Preliminary Exam/CCHMS/Secondary 4/Additional Mathematics/4049/01 Name: Class: Class Register Number: PRELIMINARY EXAMINATION 2023 SECONDARY 4 ADDITIONAL MATHEMATICS 4049/01 Paper 1 Thursday 24 August 2023 2 hours 15 minutes MARKS SCHEME This document consists of 19 printed pages and 1 blank page. [Turn over

2 2023 Preliminary Exam/CCHMS/Secondary 4/Additional Mathematics/4049/01 Mathematical Formulae 1. ALGEBRA Quadratic Equation For the equation 2 0ax bx c+ + = , 2 4 2 b b acx a −  −= Binomial expansion nrrnnnnn bbar nbanbanaba ++     ++     +     +=+ −−− ......21)( 221 , where n is a positive integer and ! ( 1) ... ( 1) !( )! ! n n n n n r r r n r r  − − +== − . 2. TRIGONOMETRY Identities 22sin cos 1AA+= 22sec 1 tanAA=+ 22cosec 1 cotAA=+ sin( ) sin cos cos sinA B A B A B =  cos( ) cos cos sin sinA B A B A B= tan tantan( ) 1 tan tan ABAB AB = sin 2 2sin cosA A A= 2 2 2 2cos 2 cos sin 2cos 1 1 2sinA A A A A= − = − = − 2 2 tantan 2 1 tan AA A= − Formulae for ABC sin sin sin a b c A B C== 2 2 2 2 cosa b c bc A= + − 1 sin2 ab C=

3 2023 Preliminary Exam/CCHMS/Secondary 4/Additional Mathematics/404/01 [Turn over 1 The points R and S have coordinates ( )3, 2 3 and ( )5, 4 5 respectively. Show that the gradient of RS can be expressed in the form 15ab+ , where a and b are integers to be found. [4] ( )( ) ( ) ( ) ( ) 22 4 5 2 3 5 3Gradient of 5 3 5 3 4 5 2 3 5 3 = 53 20 4 15 2 15 6 2 14 6 15 2 2 7 3 15 2 7 3 15 7, RS a −+= −+ −+ − −−−= −= − = =− = 3b=− B1 – 4 5 2 3 53 − − M1 – √ 53 53 + + M1 – either numerator or denominator expanded correctly A1

4 2023 Preliminary Exam/CCHMS/Secondary 4/Additional Mathematics/4049/01 2 Given that cos p = and that  is acute, express in terms of p, (a) sin , [1] (b) ( )tan 90 , − [2] (c) cos 2 . [2] 21 p− 2 2 1 tan 1 1 1 p p p p = = − = − ( ) ( ) 2 2 2 2 2 2 cos 2 1 2sin 1 2 1 1 2 1 1 2 2 2 1 p p p p =− = − − = − − = − + =− B1 M1 – 1 their tan A1 M1 – uses any cos 2 formula correctly A1

5 2023 Preliminary Exam/CCHMS/Secondary 4/Additional Mathematics/404/01 [Turn over 3 Express ( )( ) 2 2 12 32 31 2 1 2 xx xx ++ −+ in partial fractions. [5] ( )( ) ( ) ( ) ( )( ) ( ) ( )( ) ( ) 2 22 22 2 22 12 32 31 2 1 22 1 2 2 12 32 31 2 2 1 2 2 1 1let , 2 2550 4 8 let 2, 15 5 3 let 0, 31 4 2 31 32 2 3 24 2 8, 2, 3 12 32 31 8 2 3 2 1 22 1 2 2 x x A B C xxx x x x x A x B x x C x x A A x C C x A B C B B B A B C xx xxx x x ++ = + +−+− + + + + = + + − + + − = = = =− =− =− = = − − = − + = = = = =− ++ = + −−+− + + M1 – realising the form of partial fractions M1 – realising the need to eliminate the denominator A1, A1, A1 – do not award last A1 if not expressed in partial fractions

6 2023 Preliminary Exam/CCHMS/Secondary 4/Additional Mathematics/4049/01 4 The expression 32ax bx b++ leaves a remainder of R when divided by ( )1x+ and a remainder of 52R− when divided by ( )2x+ . (a) Show that 23 5 ab −= . [4] (b) Given further that 8ab=− and ab , find the value of a and of b. [3] ( ) ( ) ( ) ( ) ( ) 32 32 when 1, 11 2 --- (1) when 2, 2 2 5 2 8 5 5 2 --- (2) sub (1) into (2), 8 5 5 2 2 8 5 5 10 2 10 5 8 5 2 5 3 2 23 (shown)5 x a b b R a b R x a b b R a b R a b a b a b a b b b a a ba ab =− − + − + = − + = =− − + − + = − − + = − − + = − + − − + =− + − − =− + + =− + −= ( )( ) 2 2 8 8 8 3 2 5 40 3 2 3 2 40 0 3 10 4 0 10 or 43 2.4 2 (reject) 4, 2 ab b a a a aa aa aa aa bb ab =− −= − − += − =− + − − = + − = =− = = =−  = =− M1 – realises ( )f1 R−= M1 – realises ( )f 2 5 2 R− = − DM1 – award only if both M1 above is achieved A1 AG M1 – allow slips only for LHS A1 √M1 – finding ‘b’

7 2023 Preliminary Exam/CCHMS/Secondary 4/Additional Mathematics/404/01 [Turn over (c) Using the values of a and b found in part (b), explain why the equation 32 0ax bx b+ + = has only one real root and state its value. [4] ( ) ( ) ( )( ) ( )( ) ( )( ) 32 32 3 2 2 2 2 2 2 2 4 2 2 0 2 1 0 when 1, 2 1 1 1 0 By factor theorem, 1 is a factor. 1 2 1 0 by comparing terms, 21 1 1 2 1 0 For 2 1 0, discriminant 1 4 2 1 7 0 2 1 0 has no real roots. xx xx x x x x cx x c c x x x xx xx − − = − − = = − − = − − + + = − + =− = − + + = + + = = − =−   + + = 324 2 2 0 has only 1 real root of 1.x x x− − = = M1 – correct method to find quadratic (division or inspection) B1 - factor M1 – realising the need to find discriminant / solve quadratic equation A1 – no real roots + x = 1

8 2023 Preliminary Exam/CCHMS/Secondary 4/Additional Mathematics/4049/01 5 Find the coordinates of the stationary points of the curve ( ) 2 3xy x −= and determine the nature of each stationary point. [7] ( ) ( ) ( ) 2 2 1 2 2 2 2 2 2 3 2 3 2 23 3 69 9 6 6 9 d 19d 9 1 dFor stationary points, 0d 910 9 1 9 3 or 3 0 or 12 d 18d 18 At 3,0 , d 18 2 0d 3 3 3,0 is a minimum poi xy x xx x x x xx y xx x y x x x x x y y xx x y x − − − −= −+= = − + = − + =− =− = −= = = =− =− = = = =   ( ) ( ) ( ) 2 32 nt At 3, 12 , d 18 2 0d3 3 3, 12 is a maximum point y x −− = =−  −  − − B1 M1 – sets d d y x to 0 A1 – for (3,0) A1 M1 – their 2nd derivative or use of 1st derivative test A1 A1 – for (-3, -12)

9 2023 Preliminary Exam/CCHMS/Secondary 4/Additional Mathematics/404/01 [Turn over 6 The height above ground, h metres, of a ball, released by a machine can be modelled by the equation 20.2 6 3h x x=− + + where x is the horizontal distance travelled by the ball in metres. (a) State the height above ground at which the ball is released. [1] (b) Express 20.2 6 3h x x=− + + in the form ( ) 2 a x b c−+ , where a, b and c are constants to be found. [2] (c) Using your result from (b), explain why the height of the ball can never be more than 48 metres. [2] (d) Hence, explain if this machine is safe for use in an indoor stadium with a ceiling height of 45 metres. [1] 3 metres ( ) ( ) ( ) ( ) 2 2 2 2 2 2 0.2 30 3 0.2 30 15 15 3 0.2 15 225 3 0.2 15 48 h x x xx x x =− − + =− − + − + =− − − + =− − + ( ) ( ) ( ) ( ) 2 2 2 2 For 0.2 15 48, 15 0, 0.2 15 0 0.2 15 48 48 Since the ball reaches a maximum height of 48 m, it will never reach a height of more than 48 m. x x x x − − + − − −  − − +  B2, 1 : –1 for each error √ M1 – their square term A1 Since the ball can reach a maximum height of 48m which exceeds the ceiling height of 45 m, this machine is not safe for use in the indoor stadium. B1 – to have comparison of maximum height with ceiling height.

10 2023 Preliminary Exam/CCHMS/Secondary 4/Additional Mathematics/4049/01 7 (a) Solve the equation ( )3 18 3 40xx += . [3] (b) Solve the equation ( )22log 3 log 6yy= + + . [5] ( ) ( ) ( ) 2 3 18 3 40 3 18 3 40 0 xx xx += + − = Let y = 3x. ( )( ) 2 18 40 0 20 2 0 20 or 2 yy yy yy + − = + − = =− = 32 ln 3 ln 2 3 20 (rej. 3 0) or ln 3 ln 2 ln 2 ln 3 0.631 (3 sig. fig.) x x xx x x = = =−  = = = ( ) ( ) ( ) ( ) ( ) 22 22 2 2 2 2 2 22 2 2 log 3 log 6 log log 6 3 log log 6 3 log 2 log log 6 31 2 2log log 6 3 log 3 6 yy yy y y y y yy y y = + + − + = − + = − + = − + =  = + Comparing, ( )( ) 2 2 2 36 3 18 3 18 0 6 3 0 6 or 3 (rej. 0) y y yy yy yy yy =+ =+ − − = − + = = −   M1 – taking ln A1 M1 – solving of quadratic equation B1 – change of base law M1 – power law M1 – quotient law M1 – removal of log/comparing A1