JWSS 4E-5N AM P1 MS (Internal Use)

Class Index Number Name: Setter: Ms Sharron Chiam Jurong West Secondary School Preliminary Examinations 2020 80 ADDITIONAL MATHEMATICS 4047/01 Secondary Four Express/ Five Normal Academic 27 August 2020 Paper 1 1000 - 1200 2 hours Candidates answer on the Question Paper. No Additional Materials are required. 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 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. 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 of the marks for this paper is 80. After checking of answer script Checked by Signature Date Student This document consists of 17 printed pages.

2 JWSS Preliminary Examinations 2020 Additional Mathematics (4047/01) Secondary 4E/ 5NA Mathematical Formulae 1. ALGEBRA Quadratic Equation For the equation ax2 + bx + c = 0, . Binomial Expansion ( ) nrrnnnnn bbar nbanbanaba ++     ++     +     +=+ −−− 221 21 , where n is a positive integer and ( ) ( ) ( ) ! 1...1 !! ! r rnnn rnr n r n +−−=−=      . 2. TRIGONOMETRY Identities 1cossin 22 =+ AA AA 22 tan1sec += AAec 22 cot1cos += BABABA sincoscossin)sin( = BABABA sinsincoscos)cos( = BA BABA tantan1 tantan)tan(  = AAA cossin22sin = AAAAA 2222 sin211cos2sincos2cos −=−=−= A AA 2tan1 tan22tan −= Formulae for ABC C c B b A a sinsinsin == Abccba cos2222 −+= Cabsin2 1= 2 4 2 b b acx a −  −=

3 JWSS Preliminary Examinations 2020 Additional Mathematics (4047/01) Secondary 4E/ 5NA 1 A cone with base radius ( )5 2 3+ cm and a slant height l cm has a curved surface area of ( )51 3 3− cm2. Without using a calculator, obtain an expression for l in the form of ( )3ab+ , where a and b are integers. [4] Solution: Curved surface area of cone rl= ( ) ( )5 2 3 51 3 3l + = −  ( ) ( ) 51 3 3 5 2 3 l − = + [M1] – Correct use of formula 51 3 3 5 2 3 5 2 3 5 2 3 −−= +− [M1] – Correct use of conjugate surds ( ) ( ) 255 102 3 15 3 6 3 25 4 3 − − += − [M1] 273 117 3 13 −= ( )21 9 3=− cm [A1] 2 The acute angles A and B are such that ( )tan 8AB+= and 1tan 5A= . Without using a calculator, find the exact value of cosB . [5] Solution: ( )tan 8AB+= tan tan 81 tan tan AB AB + =− 1 tan5 811 tan5 B B + = − [M1] – Correct use of double angle formula 11 tan 8 1 tan55 BB + = −  [M1] 18 tan 8 tan55 BB+ = − 13 39tan55 B= tan 3B= [M1] By Pythagoras’ Theorem, 2213r=+ [M1] 3 1 B r

4 JWSS Preliminary Examinations 2020 Additional Mathematics (4047/01) Secondary 4E/ 5NA 10r = 1 10cos 10 10 B= 10cos 10B= [A1] 3 (i) Find the range of values of m for which the curve 2(4 ) 4 1y m x x m= + − + + has a maximum point. [1] Solution: For curve to have a maximum point, 40 m+ 4m− [B1] (ii) Find the range of values of m for which the curve 2(4 ) 4 1y m x x m= + − + + is always negative for all real values of x. [3] Solution: For curve to be always negative, 2 40b ac− ( ) ( )( ) 2 4 4 4 1 0 mm− − + +  [M1] – Either seen for correct use of discriminant ( ) 216 4 5 4 0mm− + +  216 4 20 16 0mm− − −  24 20 0mm− −  [M1] 2 50mm+ ( )50mm + 5m− or 0m [A1] (iii) Hence state the range of values of m for which the curve has a maximum point and is always negative for all values of x. [1] Solution: 5m− [B1] x 0

5 JWSS Preliminary Examinations 2020 Additional Mathematics (4047/01) Secondary 4E/ 5NA 4 (a) It is given that ( ) 22f d 3 lnx x x x kx c= + + where k and c are constants. (i) If ( ) 2 1 f d 2 e x x e =+ , show that 2k=− . [3] Solution: ( ) 2 1 f d 2 e x x e =+ 2 2 2 1 3 ln 2 e x x kx c e + + = + [M1] ( ) 2 2 23 ln 3ln1 2e e ke c k c e + + − + + = + ( ) 2 2 232e ke c k c e+ + − + = + 2 2 232e ke k e+ − = + [M1] 2 2 232e e k ke− − = − 2222e k ke− = − ( ) ( ) 222 1 1e k e− =− − 2k =− [A1] (ii) Find ( )f x . [2] Solution: ( ) 22f d 3 ln 2x x x x x c= − + ( ) 2 1f 6 ln 3 4x x x x x x = + −  [M1] – Correct use of product rule 6 lnx x x=− [A1] – Either seen ( )6ln 1xx=−

6 JWSS Preliminary Examinations 2020 Additional Mathematics (4047/01) Secondary 4E/ 5NA (b) The figure shows part of the curve ( )fyx= . ( )10, 2− and ( )4, 9− are two points on the curve. Given that 4 10 d 30yx − − = , find 9 2 d.xy [2] Solution: Area of A + B 30= Area of B ( )62= 12= Area of A 30 12=− [M1] 18= Area of C ( )74= 28= 9 2 dxy = Area of A + C 28 18=+ 46= [A1] y x y x A B C

7 JWSS Preliminary Examinations 2020 Additional Mathematics (4047/01) Secondary 4E/ 5NA 5 The diagram shows a triangle PQR in which the point P is ( )2, 8− , the point R lies on the x-axis and angle PQR is 90 . The equation of QR is 2 5 64yx+= . (i) Find the coordinates of Q. [5] Solution: 2 5 64yx+= 2 5 64yx=− + 5 322yx=− + Gradient of QR 5 2=− Gradient of PQ 2 5= [B1] Equation of PQ: ( )282 5 c= − + [M1] – Attempt to find c ( )282 5 c= − + 44 5c=  Equation of PQ is 2 44 55yx=+ . [A1] 5 322yx=− + ---- (1) 2 44 55yx=+ ---- (2) (1) = (2) 5 2 44322 5 5xx− + = + [M1] 29 116 10 5x −−= 8x=

8 JWSS Preliminary Examinations 2020 Additional Mathematics (4047/01) Secondary 4E/ 5NA Subst. 8x= into (1) ( )5 8 322y=− + 12=  ( )8,12Q [A1] (ii) Given that M is the midpoint of PR and that PQRS is a rectangle, find the coordinates of M and of S. [3] Solution: When 0y= , ( )2 0 5 64x+= 12.8x= [M1]  ( )12.8, 0R Coordinates of M 2 12.8 8 0,22 − + += ( )5.4, 4= [A1] Let ( ),S x y ( )8 12, 5.4, 422 xy++ = 8 5.42 x+ = and 12 42 y+ = 8 10.8x+= 12 8y+= 2.8x= 4y=−  ( )2.8, 4S − [A1]

9 JWSS Preliminary Examinations 2020 Additional Mathematics (4047/01) Secondary 4E/ 5NA 6 (a) Solve the equation ( ) ( )lg 20 5 lg 10 1xx+ − − = . [3] Solution: ( ) ( )lg 20 5 lg 10 1xx+ − − = 20 5lg 1 10 x x + = − [M1] 20 5 1010 x x + =− [M1] 20 5 100 10xx+ = − 15 80x= 15 3x= [A1] (b) Given that loga px= and loga qy= , express in terms of x and y, (i) log pq a , [2] Solution: loglog log a pq a aa pq= [M1] 1 1 logaaog p q= + 1 xy= + [A1] (ii) log p aq . [3] Solution: log log logp p paq a q=+ [M1] log log log log aa aa aq pp=+ [M1] 1 y x += [A1] Or loglog log a p a aqaq p= [M1] log log log aa a aq p += [M1] 1 y x += [A1]

10 JWSS Preliminary Examinations 2020 Additional Mathematics (4047/01) Secondary 4E/ 5NA