presbyterian high school 2023 A math paper 2 answer key

Name: Index No.: Class: PRESBYTERIAN HIGH SCHOOL ADDITIONAL MATHEMATICS 4049/02 Paper 2 21 August 2023 Monday 2 hrs 15 min MARKING SCHEME Setter: Tan Chee Wee Vetter: Tan Lip Sing This question paper consists of 18 printed pages and 0 blank page. Mathematical Formulae

2 1. ALGEBRA Quadratic Equation For the equation Binomial expansion 1 2 2( ) ... ... , 12 n n n n n r r nn n na b a a b a b a b b r − − −     + = + + + + + +           where n is a positive integer and 2. TRIGONOMETRY Identities Formulae for 1 sin2 ab C= 2 0,ax bx c+ + = 2 4 2 b b acx a −  −= ! ( 1)...( 1) ( )! ! ! n n n n n r r n r r r  − − +== − 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 2cos2 cos sin 2cos 1 1 2sinA A A A A= − = − = − 2 2tantan 2 1 tan AA A = − ABC sin sin sin a b c A B C== 2 2 2 2 cosa b c bc A= + −

3 1 An object is heated in an oven until it reaches a temperature of X C. It is then allowed to cool. Its temperature,  C, when it has cooled for time t minutes, is given by 630 100(0.8) t  =+ . (a) Find the value of X. [1] 0 630 100(0.8)X =+ 130X = B1 (b) Find the value of  when t = 8. [1] 8 630 100(0.8) =+ 104 = B1 (c) Find the value of t when 95= . [3] 695 30 100(0.8) t =+ 665 100(0.8) t = 6(0.8) 0.65 t = M1 6lg(0.8) lg 0.65 t = lg(0.8) lg 0.656 t = M1 6lg 0.65 lg 0.8t = 11.6t= A1 (d) A sketch of the graph of  against t is given below. State the value of p. [1] 30p= B1 C t mins p 0

4 A calculator must not be used in this question. 2 (a) In the diagram, triangle ABC has an area of ( )8 2 4+ cm2, angle 4BAC = radian and AB = ( )2 2 2+ cm. Find the length of AC, leaving your answer in the form ( )2pq + cm, where p and q are integers. [5] Area = BACACAB  sin2 1 ( )( )128 2 4 2 2 222 AC + = +  M1 ( )( )18 2 4 2 2 2 AC+ = + ( )( )16 2 8 2 2 AC+ = + 16 2 8 22 AC += + M1 16 2 8 2 2 2 2 2 2 +−= +− M1 32 2 32 16 8 2 42 − + −= − M1 24 2 16 2 −= 12 2 8=− A1 (b) Find