XMSS 2024 AM Prelim (P1) Mark Scheme

This document consists of 18 printed pages. [Turn over XINMIN SECONDARY SCHOOL SEKOLAH MENENGAH XINMIN Preliminary Examination 2024 CANDIDATE NAME Mark Scheme [Draft 3] CLASS INDEX NUMBER ADDITIONAL MATHEMATICS Paper 1 Secondary 4 Express Setter : Mr Johnson Chua Vetter : Ms Low Yan Jin Moderator : Ms Pang Hui Chin Candidates answer on the Question Paper. No Additional Materials are required. 4049/01 23 August 2024 2 hour 15 minutes READ THESE INSTRUCTIONS FIRST Write your name, register number and class in the spaces at the top of this page. 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 the questions. Give non-exact numerical answers correct 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. 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. Errors Qn No. Errors Qn No. Accuracy Simplification Brackets Units Geometry Marks Awarded Presentation Marks Penalised For Examiner’s Use 90 Parent’s/Guardian’s Signature:

2 Mathematical Formulae 1. ALGEBRA Quadratic Equation For the quadratic equation ax 2 + bx + c = 0, 2 4 2 b b acx a −  −= 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 ( ) ! ( 1)...( 1) ! ! ! n n n n n r r n r r r  − − +== − 2. TRIGONOMETRY Identities sin 2 A + cos 2 A = 1 sec 2 A = 1 + tan 2 A cosec 2 A = 1 + cot 2 A sin (A ± B) = sin A cos B ± cos A sin B cos (A ± B) = cos A cos B  sin A sin B tan ( A ± B ) = tan tan 1 tan tan AB AB  sin 2A = 2 sin A cos A cos 2A = cos2 A – sin2 A = 2cos2 A – 1 = 1 – 2 sin2 A tan 2A = 2 2 tan 1 tan A A− Formulae for  ABC sin sin sin a b c A B C== a 2 = b 2 + c 2 − 2bc cos A  = 2 1 bc sin A

3 [Turn over 1 A triangle has a base of length ( )6 2 7+ cm and an area of ( )17 7 7+ cm2. Find, without using a calculator, the perpendicular height to the base of the triangle, in cm, in the form ( )7ab+ , where a and b are integers. [3] 2 Solve the equation 51 xx− + = . [4] ( ) ( ) ( ) ( )( ) ( )( ) 2 1 6 2 7 17 7 7 --- [M1]2 2 17 7 7 6 2 7 17 7 7 3 7 --- [M1] 3 7 3 7 51 17 7 21 7 49 37 2 4 7 2 1 2 7 -- h h h h h h +  = + + = + +− = +− − + −= − += =+ -[A1] Alternative ( )( ) ( )( ) 34 14 7 6 2 7 --- [M1] 6 2 7 6 2 7 204 68 7 84 7 196 36 4(7) 8 16 7 8 1 2 7 ---[A1] h h h h +− = +− −+−= − += =+ ( ) ( )( ) 2 2 2 5 1 --- [M1] 5 1 5 1 25 10 1 11 24 0 3 8 0 3 or 8 (rej) xx xx xx x x x xx xx xx − + = − = + − = + − + = + − + = − − = == [A1] [A1: no A1 if students do not reject] Students must show the relevant method in solving quad. equation (either factorisation or quad. formula). Failure to do so would result in the loss of M1; A1 would still be awarded accordingly.

4 3 The equation of a curve is 42 1 xy x += + , where x > 1− . (a) Find d d y x , leaving your answer in the form ( )1 n ax b x + + . [2] Method 1: Quotient Rule ( ) ( ) ( ) ( ) 1 2 3 2 3 14 1 4 2 1d 2 --- [M1]d1 2141 1 = 1 4( 1) (2 1) = 1 23 = ---[A1] 1 x x xy xx xx x x xx x x x −+ − + + = + ++− + + + − + + + + Method 2: Product Rule ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 31 22 3 3 3 3 d1 4 2 1 1 1 (4) --- [M1]d2 2 1 4 = 11 2 1 4( 1) = 1 2 1 4 4 = 1 23 = --- [A1] 1 y x x xx x xx xx x xx x x x −−= + − + + + +−+ ++ − − + + + − − + + + + + (b) Explain why the curve is an increasing function. [2] ( ) ( ) 3 3 Since 1, 1 0, and 2 3 0 23 0, for 1 0 1 Since 0, is an increasing function. x xx x dy x dxx dy ydx − +  +  +  − →  +  M1 A1

5 [Turn over 4 The line 2 3 12xy+= intersects the curve 2 48yx=− at points A and B. Find the value of p and q for which the length of AB can be expressed as pq . [6] Alternative 2 2 2 2 2 3 12 --- (1) 4 8 --- (2) From (1): 2 12 3 ---(3) Subst (3) into (2): 2(12 3 ) 8 --- [M1] 24 6 8 6 16 0 --- [M1] ( 2)( 8) 0 2 or 8 when 2, 2 12 3(2) 3 (3, 2) --- [A1] when 8, 21 xy yx xy yy yy yy yy yy y x x A y x += =− =− = − − = − − + − = − + = = =− = =− =  =− = 22 2 3( 8) 18 (18, 8) --- [A1] Length of : (18 3) ( 8 2) --- [M1, allow ecf] = 325 =5 13 [ 5, 13 ]--- [A1] x B AB pq −− = − − + − −  = = ( ) 2 2 2 2 2 2 2 3 12 --- (1) 4 8 --- (2) From (1): 12 2 ---(3)3 Subst (3) into (2): 12 2 4 8 --- [M1]3 12 2 489 144 48 4 36 72 4 84 216 0 ---[M1] 21 54 0 ( 3)( 18) 0 3 or 18 when 3, 12 xy yx xy x x x x x x x xx xx xx xx x y += =− −= − =− − =− − + = − − + = − + = − − = == = = 22 2(3) 3 2 (3, 2) --- [A1] when 18, 12 2(18) 3 8 (18, 8) --- [A1] Length of : (18 3) ( 8 2) --- [M1] = 325 =5 13 [ 5, 13 ]--- [A1] y A x y y B AB pq − =  = −= =− − − + − −  = = No M1 if students do not show working for factorisation/ quadratic formula. Award the remaining marks accordingly.

6 5 The height, h m, of a baseball above ground t seconds after it has been hit is given by 224 4h c t t= + − , where c is a constant. (a) If c = 1.65, express h in the form 2()h p q t r= + + where p, q and r are constants to be determined. Hence, state the maximum height attained by the baseball and the time at which this occurs. [4] (b) Find the range of values of c if the baseball did not reach a height of 40 m. [2] 2 2 2 2 2 2 1.65 24 4 =1.65 4( 6 ) =1.65 4( 6 3 ) 4(3 ) --- [M1 for completing the square. Awd if wrong val subst.] =37.65 4( 3) --- [A1] max. height: 37.65 ---[A1] time: 3 seconds or 3 ---[A1 h t t tt t t c t t = + − −− − − + + −− = ] 2 2 2 2 4 24 40 4 24 40 0 Since ball did not reach 40m, there is no solution to the equation. 4 0 4, 24, 40 24 4( 4)( 40) 0 --- [M1] 576 640 16 0 16 64 4 --- [A1] t t c t t c b ac a b c c c c c c − + + = − + + − =  −  =− = = − − − −  − +    Alternatively 2 2 22 2 2 24 4 4 6 4 4[( 3) (3) ] 4 4[( 3) 9 ] 4 4( 3) 36 --- [M1] 36 40 4 --- [A1] h c t t ctt ct ct tc c c = + − =− − − =− − − − =− − − − =− − + + +  Note: M1 is not awarded if students wrote (-3)2 Awarded even if completed square form is partially correct.

7 [Turn over 6 A curve is such that 2 63 2 d 12 15d xxy eex −=+ . The point ( )0, 2P − lies on the curve and the normal to the curve at P is parallel to the y-axis. Find the equation of the curve. [6] 63 63 63 tangent at P d 12 15 dd 12 15 = --- [M1]63 = 2 5 ddat 0, 2 5 --- [M1 for subst. 0 into ]dd = 3 m : 0 3 0 --- [M1] 3 d d xx xx xx y e e xx ee c e e c yyx c x xx c c c y − − − =+ ++ − −+ = = − + = −  − = =   632 5 3xxeex −= − + 63 63 63 63 2 5 3 d 25 = 3 --- [M1, allow ecf]63 5 = 333 at 0, 2, 15 2 ---[M1, allow ecf]33 2 = 2 4 5 3 4 ---[A1]33 xx xx xx xx y e e x ee xd ee xd xy d d d eeyx − − − − = − + − + +− + + + = =− + + =− +− =−  = + + − 

8 7 The equation of a curve is 2y kx kx p= + + , where p and k are constants. (a) Show that 4 kp for which the curve lies completely above the x-axis. [3] (b) In the case where k = 2 and p = 4, find the values of m for which the line 4y mx=− is a tangent to the curve. [4]