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 cos75 , giving your answer in the form 4 ab− , where a and b are integers. [3] cos75 cos(30 45 )= +  cos75 cos30 cos45 sin30 sin 45=  −   M1 3 2 1 2cos 75 2 2 2 2=  −  M1 62cos 75 4 −= A1 A B C

5 3 (a) Prove that cosec2 cot 2 tanx x x−= . [3] 1 cos 2cos 2 cot 2 sin 2 sin 2 xec x x xx− = − 1 cos 2cos 2 cot 2 sin 2 xec x x x −−= M1 22sincos 2 cot 2 2sin cos xec x x xx−= M1 for either formula sincos 2 cot 2 cos xec x x x−= cos 2 cot 2 tanec x x x−= AG1 (a) Hence solve 2cosec2 cot 2 2sec 3x x x− = − for 0 360x   . [5] 2cos 2 cot 2 2sec 3ec x x x− = − 2tan 2sec 3xx=− ( ) 2tan 2 1 tan 3xx= + − M1 2tan 2 2 tan 3xx= + − 22 tan tan 1 0xx− − = M1 (2 tan 1)(tan 1) 0xx+ − = tan 0.5x=− or tan 1x= M1 Basic angle 26.6= or 45= 180 26.6 , 360 26.6x= −  −  45 ,180 45x=  +  45 ,153.4 ,225 ,333.4x=     A1, A1

6 4 (a) Solve ( ) 19 5 2 3xx ++= . [5] ( ) 23 5 2 3 3xx+ =  Let 3xu= 2 56uu+= M1 2 6 5 0uu− + = ( 1)( 5) 0uu− − = 1u= or 5u= M1 31x = or 35x = 0x= lg 5 lg 3x= M1 0x= 1.46x= A1, A1

7 (b) Solve 2 4 100 1002log log ( 9) log 1xx + − =− . [5] 2 4 100 1002log log ( 9) log 1xx + − =− 2 4 100 100 1log log ( 9) log 2xx + − =− 1 2 2 100 100log ( 9) log 4xx −  + − = M1 2 100 91log 2 x x + = M1 quotient law 12 29 100x x + = 2 9 10x x + = M1 2 9 10xx+= 2 10 9 0xx− + = M1 ( 1)( 9) 0xx− − = 1x= or 9x= A1

8 5 The diagram shows a quadrilateral ABCDE where triangle ABC is similar to triangle DEC. AB = 15 cm, DE = 9 cm, angle ACD = 90 o and angle ABC is a variable angle , where 0 o <  < 90 o. (a) Show that the perimeter, P cm, of the quadrilateral is given by 24 24sin 6cosP = + + . [4] In ABC , cos 15 BC = M1 either 15cosBC = sin 15 AC = 15sinAC = In DCE , cos 9 EC = M1either 9cosEC = sin 9 DC = 9sinDC = Therefore 15 9P AE DB= + + + 24 15sin 9cos 9sin 15cosP    = + − + + M1 24 24sin 6cosP = + + (shown) a.g. A1 B D E 9 cm 15 cm  C A

9 (b) Express P in the form ( ) kR ++sin . [4] 24sin 6cos sin( ) R   + = + 226 24 612 6 17R or= + = M1 6tan 24 = M1 14.0 = M1 612 sin( 14.0 ) 24P = +  + A1 (c) Find the value of  when the perimeter is 38 cm. [2] 24 612 sin( 14.03 ) 38+ +  = 612 sin( 14.03 ) 14+  = M1 14sin( 14.03 ) 612  +  = 14.03 34.46+ =  20.4 = (1 d.p.) A1

10 6 A piece of wire 60 cm long is bent to form the shape shown in the figure. This shape consists of a semi-circular arc, radius, r cm, and an equilateral triangle on the opposite ends of a rectangle of length 4x cm. (a) Express x in term of r. [2] (b) Hence show that the area enclosed, A cm2, is given by 260 ( 3 4 ) 2A r r = + − − [3] 2 2 12460sin222 1 rrxrrA ++= M1 for 2 areas, M2 for all 3 areas 22 3 60 4 128 2 8 2 rrA r r r  −−=  +  + 2 2 2 2 13 60 4 2A r r r r r = + − − + 2 2 2 160 3 4 2A r r r r = + − − 260 ( 3 4 ) 2A r r = + − − (shown) AG1 2(2 ) 2(4 ) 60r x r + + = M1 4 8 60r r x+ + = 8 60 4x r r = − − 60 4 8 rrx −−= A1 4x r