2023_CCHM_Prelim_AMath P1 - MS
Uploaded by hima · 8 October 2023
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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 deno
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