YTSS A Math 2024 Prelim P1 MS

NAME: Solution ( ) CLASS: YISHUN TOWN SECONDARY SCHOOL PRELIMINARY EXAMINATION 2024 SECONDARY 4 EXPRESS / 5 NORMAL ACADEMIC ADDITIONAL MATHEMATICS PAPER 1 (4049/01) DATE : 21 August 2024 DAY : Wednesday DURATION : 2 h 15 min MARKS : 90 READ THESE INSTRUCTIONS FIRST Do not turn over the cover page until you are told to do so. Write your name, class and class index number in the spaces at the top of this page. Write in dark blue or black pen. You may use a 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 marks for this paper is 90. MARKS OBTAINED FULL 1 4 2 4 3 5 4 6 5 6 6 6 7 8 8 7 9 8 10 9 11 10 12 8 13 9 TOTAL 90 This question paper consists of 21 printed pages and 1 blank page. O

2 2024 YTSS Preliminary Examination Secondary 4E5N Additional Mathematics EXP (4049/01) Mathematical Formulae 1. ALGEBRA Quadratic Equation For the equation , Binomial Expansion where is a positive integer and 2. TRIGONOMETRY Identities Formulae for 02 =++ cbxax a acbbx 2 42 −−= ( ) nrrnnnnn b...bar n...banbanaba ++     ++     +     +=+ −−− 221 21 n ( ) ( ) ( ) ! 1...1 ! ! ! r rnnn rnr n r n +−−=−=      AA AA AA 22 22 22 cot1eccos tan1sec 1cossin += += =+ ( ) ( ) ( ) A AA AAAAA AAA BA BABA BABABA BABABA 2 2222 tan1 tan22tan sin211cos2sincos2cos cossin22sin tantan1 tantantan sinsincoscoscos sincoscossinsin − = −=−=−= = = = =   ABC Cab Abccba C c B b A a sin2 1 cos2 sinsinsin 222 = −+= ==

3 2024 YTSS Preliminary Examination Secondary 4E5N Additional Mathematics EXP (4049/01) [Turn over Answer all questions. 1 (a) Write down the turning point of the curve 2 1yx=− + , stating clearly whether it is a maximum or minimum point. [2] (b) In a clearly labelled diagram, s ketch the graph of 2 1yx=− + , indicating clearly the coordinates of the turning point and axial intercepts. [2] Maximum point (0, 1) B1 maximum B1 (0, 1) x O y (0, 1) 1 B1 shape of graph B1 label

4 2024 YTSS Preliminary Examination Secondary 4E5N Additional Mathematics EXP (4049/01) 2 A cylinder has radius (2 2)− cm and volume (7 8 18) − cm3. Find, without using a calculator, the height of the cylinder, in cm, in the form ( 2)ab+ , where a and b are integers. [4] Volume, 2(2 2) (7 8 18)h − = − ( ) ( ) 2 7 8 18 22 7 8 18 4 4 2 2 14 2 18 6 4 2 6 4 2 6 4 2 84 2 112 108 72 2 36 32 12 2 4 4 1 3 2 h   − = − −= −+ −+= −+ + − −= − += =+ The height of the cylinder is (1 3 2)+ cm. B1 B1 for expansion of 2(2 2)− M1 for correct numerator or denominator A1

5 2024 YTSS Preliminary Examination Secondary 4E5N Additional Mathematics EXP (4049/01) [Turn over 3 (a) Explain why the principal value of 1 2cos 2 −  cannot be 4 − . [1] For any cosy = , 10 cos y − , Hence principal value cannot be 4 − . B1 10 cos y − *Accept since 1 2cos 24 − = , 1 2cos 24 − − . (b) It is given that cos p =− , where 0p , and  is an obtuse angle. Without using a calculator and leaving your answers in terms of p, find the value of (i) cos 2   − , [2] 2 cos sin2 1 p   −= =− (ii) 2cot  . [2] 2 2 2 2 1cot tan 1 p p  = = − M1 2 1 tan  , A1 OR B2 2 21 p p− M1 sin A1 OR B2 21 p−

6 2024 YTSS Preliminary Examination Secondary 4E5N Additional Mathematics EXP (4049/01) 4 (a) Find the remainder when 324 3 11 3x x x+ + − is divided by 2x+ . [2] Let 32P( ) 4 3 11 3x x x x= + + − P ( 2) 4( 8) 3(4) 11( 2) 3 45 − = − + + − − =− The remainder is 45− . (b) When a polynomial g( )x is divided by 1x− and 21x− , the remainders are 1 and 2 respectively. Find the remainder when g( )x is divided by ( 1)(2 1)xx−− . [4] Degree of remainder is less than degree of divider. Let the remainder be ax b+ . Then g( ) ( 1)(2 1) Q( )x x x x ax b= − − + + g(1) 1 1 (1) 1g2 2 1 2 (2)2 1(1) (2) : 1 2 2 Sub. into (1): 3 ab ab a a b = +=  = += − =− =− = Therefore the remainder is 23x−+ . M1 sub ( 2)− A1 M1 M1 1g(1) 1 and g 2 2 ==  M1 solve for a and b A1

7 2024 YTSS Preliminary Examination Secondary 4E5N Additional Mathematics EXP (4049/01) [Turn over 5 (a) Prove the identity 2 2 2 2cos cos sin sin sin ( ) sin ( ) cos 2A B A B A B A B A− − + − = . [4] Proof 1 ( )( ) ( ) ( ) 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 22 LHS cos cos sin sin sin ( ) sin ( ) cos cos sin sin sin cos cos sin sin cos cos sin cos cos sin sin sin cos cos sin cos cos sin sin sin cos cos sin cos 2 R A B A B A B A B A B A B A B A B A B A B A B A B A B A B A B B A B B AA A = − − + − =− − + − = − − + = + − + =− = = HS (proven) Proof 2 ( ) ( ) ( )( ) 22 LHS cos cos sin sin sin ( ) sin ( ) cos cos sin sin cos cos sin sin sin ( ) sin ( ) cos ( ) cos ( ) sin ( ) sin ( ) cos ( ) cos 2 RHS (proven) A B A B A B A B A B A B A B A B A B A B A B A B A B A B A B A B A = − − + − = + − − + − = − + − + − = + + − = = (b) Hence find the value of A for 02 A−   such that 2 2 2 2cos cos sin sin sin ( ) sin ( ) 0.241A B A B A B A B− − + − = − . [2] 2 2 2 2cos cos sin sin sin ( ) sin ( ) 0.241 cos 2 0.241 A B A B A B A B A − − + − = − =− Reference angle 1.3274= 2 ( 1.3274) 0.907 (3 s.f.) A A =− − =− M1 factories ( ) 22sin cosBB+ M1 addition formula sin ( )AB M1 22( )( )a b a b a b+ − = − A1 A1 B1 B1 M1 cos ( )cos ( )A B A B−+ M1 22 ( )( )a b a b a b− = + − M1 ()A B A B++−

8 2024 YTSS Preliminary Examination Secondary 4E5N Additional Mathematics EXP (4049/01) 6 (a) The curve 1x yy+= and the line 2 6 0xy− + = intersect at the points A and B. Find the x-coordinate of A and of B. [3] 1 (1)x yy+= 2 6 0 2 6 (2) xy xy − + = =− Sub. (2) into (1), 2 2 26 1 26 60 ( 2)( 3) 0 2 or 3 2 12 y yy y y y yy yy yy xx − += − + = + − = − + = = =− =− =− M1 for substitution. M1 for ( 2)( 3) 0yy− + = A1 for both x-coordinates

9 2024 YTSS Preliminary Examination Secondary 4E5N Additional Mathematics EXP (4049/01) [Turn over (b) Hence solve for the value(s) of x in the simultaneous equations 2 lg lg 1,lg lg 6. x yy x y += =− [3] 2 lg lg 1 (1)lg lg 6 lg 2lg 6 0 (2) x yy x y xy += =− − + = From (a), 2 lg 2 or 10 x x − =− = 12 lg 12 10 x x − =− = B1 Quotient Law B1 substitution A1 (Accept 2 110 , , 0.01100 − )

10 2024 YTSS Preliminary Examination Secondary 4E5N Additional Mathematics EXP (4049/01) 7 The equation of a curve is 2 2 ln xy x= , where 0x . (a) Show that the gradient of the curve can be written as ln r p q x x + , where p, q and r are integers. [2] 2 2ln xy x= 2 4 4 3 2 ( ) 4 ln