ahmad ibrahim 2023 A math paper 2 answer key

AHMAD IBRAHIM SECONDARY SCHOOL GCE O-LEVEL PRELIMINARY EXAMINATION 2023 SECONDARY 4 EXPRESS Name: Class: Register No.: MARKING SCHEME ADDITIONAL MATHEMATICS Paper 2 Candidates answer on the Question Paper. 4049/02 11 August 2023 2 hours 15 minutes READ THESE INSTRUCTIONS FIRST Write your name, class and index number on all the work you hand in. Write in dark blue or black pen. You may use an HB pencil for any diagrams or graphs. Do not use staples, paper clips, glue or correction fluid. Answer all questions. Give non-exact numerical answers to 3 significant figures, or 1 decimal place in the case of angles in degrees, unless a different level of accuracy is specified in the question. The use of an approved scientific calculator is expected, where appropriate. You are reminded of the need for clear presentation in your answers. At the end of the examination, fasten all your work securely together. The number of marks is given in brackets [ ] at the end of each question or part question. The total number of marks for this paper is 90. _________________________________________________________________________ This document consists of 20 printed pages. For Examiner’s Use /90

2 AISS PRELIIM/4E/4049/P2/2023 Mathematical Formulae 1. ALGEBRA Quadratic Equation For the equation ,02 =++ cbxax a acbbx 2 42 −−= Binomial expansion 1 2 2() 12 n n n n n r r n n n na b a a b a b a b b r − − −     + = + + + + + +           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 bc A=

3 AISS PRELIIM/4E/4049/P2/2023 [Turn over 1 (a) Find the range of values of x for which the expression 232 x− is negative. [2] (b) Find the set of values of the constant k for which the curve 2yx= lies entirely above the line ( )1.y k x=+ [3] ( ) 2 2 1 0 x k x x kx k + − −  For the quadratic expression to be always positive, ( ) 2 2 40 40 40 40 b ac kk kk k − + + −   2 2 3 2 0 2 3 0 33 or OR22 66 or 22 x x xx xx − − −  −  M1 A1 M1: finding roots or factorisng A1 M1

4 AISS PRELIIM/4E/4049/P2/2023 2 (a) Find the range of values of k such that the line 3xy+= intersects the curve 22 22x x y k− + = . [4] ( ) ( ) ( )( ) 22 22 22 2 2 2 3 3 --------(1) 2 2 3 2 2(9 6 ) 2 18 12 2 0 3 14 18 0 Since the line and curve intersects, 40 14 4 3 18 0 196 216 12 0 20 12 0 5 3 xy yx x x x k x x x x k x x x x k x x k b ac k k k k += =− − + − = − + − + = − + − + − = − + − = − − − −  −+ − +   (b) State a possible value of k if if there is no intersection between the line and the curve. [1] Any value that is < 5 3 . M1 A1 M1 M1 manipulation to get quadratic equation in 1 unknown B1

5 AISS PRELIIM/4E/4049/P2/2023 [Turn over 3 A polynomial, P, is ( ) 22 1nx k x k− + + where n and k are positive integers. (a) Explain why 1x− is a factor of P for all values of k. [2] ( ) ( ) ( ) ( ) 22let f 1 f 1 1 1 =0 since remainder =0, 1 is a factor. nx x k x k kk x = − + + = − − + − (b) Given that k = 4, find the value of n for which 2x− is a factor of P. Hence factorise P completely. [4] ( ) ( ) ( ) ( )( )( ) ( )( )( )( ) 2 22 2 2 42 f 5 4 f 2 2 16 Since 2 is a factor, 2 16 0 2 f 5 4 = 1 2 3 2 = 1 2 1 2 n n n x x x x n x x x x x x x x x x x = − + =− − −= = = − + − − + + − − + + M1 A1: must mention remainder = 0, or by factor theorem M1, must write factors ( )( )12xx−− first since it’s a hence question A1 M1 A1

6 AISS PRELIIM/4E/4049/P2/2023 4 (i) By expressing the function in the form ( ) 2 h a x m n= − + , where a, m and n are constants, explain whether the projectile can reach a height of 3 metres. [2] ( ) ( ) ( ) ( ) 2 2 2 2 2 2 3 1.5 = 2 1.5 0.75 = 2[ 0.75 0.75 0.75] = 2 0.75 2.625 maximum point is 0.75, 2.625 Therefore the projectile cannot reach a height of 3m since the maximum height is 2.625m. h x x xx x x =− + + − − − − − − − − − − (ii) Given that the defence structure is 1.4 metres horizontally from the catapult and 0.8 metres above the ground, justify if the projectile will hit the structure. [2] A projectile was launched from a catapult to hit a defence structure on a fort. The height, h metres, of the projectile above ground is given by the equation 22 3 1.5h x x=− + + , where x metres is the horizontal distance from the catapult. M1 for completing the square A1 for comparing 2.625 and 3 M1 for determining if the point lies on the equation A1 ( ) ( ) ( ) ( ) 2 2 2 Subs 1.4,0.8 into 2 3 1.5 2 1.4 3 1.4 1.5 = 1.78 0.8 Since the point 1.4,0.8 does not lie on the curve 2 3 1.5, therefore the projectile will not hit the structure. h x x h h x x =− + + =− + +  =− + +

7 AISS PRELIIM/4E/4049/P2/2023 [Turn over (iii) Sketch the curve of 22 3 1.5.h x x=− + + [2] O B! correct shape with turning point B1 correct y-intercept h h t

8 AISS PRELIIM/4E/4049/P2/2023 5 (a) (b) Differentiate ln ( )sin x with respect to x. [2] The diagram shows part of the curve cot ,yx=− cutting the x-axis at ,02   . The line 3y=− intersects the curve at P. (i) State the value of px , the x-coordinate of P. [1] 13 tan 1tan 3 6 x x x  = = = B1

9 AISS PRELIIM/4E/4049/P2/2023 [Turn over (ii) Explain why the expression 2 cot dx Px x  − does not give the area of the shaded region. [1] The shaded area is below the x-axis. If we 2 cot dx xQ x  − , we will get a negative value for the area. Thus 2 cot dx xQ x  − does not give area of the shaded region. (iii) Find the exact area of the shaded region. [3] ( ) 2 6 2 6 22 1 tan when 3 cot dx [ln sin ] 1ln1 ln 2 1ln 2 units or ln units2 y x y x x     =− =− − − = =− =−  B1 M1 M1 A1

10 AISS PRELIIM/4E/4049/P2/2023 6 (a) Without using a calculator, show that ( ) 71cos 2 612 4  =− . [3] ( ) 7cos cos12 4 3 cos cos sin sin4 3 4 3 2 1 2 3 2 2 2 2 1 264           =+              =−                =−    =− (b) Evaluate 2212 0 3cos sin dx x x  − exactly. [4] ( )   12 12 2 2 2 2 00 12 0 12 0 12 0 31 3cos sin d (2cos 1 1) 1 2sin 1 d22 31 cos 2 cos 2 1 d22 2cos 2 1 d sin 2 sin 6 12 6 12 x x x x x x x x x xx xx       −