MGS Paper 1 MS

Class Index Number Name : __________________________________________ This question paper consists of 23 printed pages and 1 blank page. METHODIST GIRLS’ SCHOOL Founded in 1887 PRELIMINARY EXAMINATION 2022 Secondary 4 Monday ADDITIONAL MATHEMATICS 4049/01 15 August 2022 Paper 1 2 h 15 min Candidates answer on the Question Paper. No Additional Materials are required. READ THESE INSTRUCTIONS FIRST Write your class, index number and name 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 figure, 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. 90

Page 2 of 24 Mathematical Formulae 1. ALGEBRA Quadratic Equation Bino mial Expansion 2. TRIGONOMETRY Identities Formulae for ∆ABC sin sin sin abc ABC= = a2 = b2 + c2 − 2bc cos A ∆ = 2 1 bc sin A 02 = + +c bx ax a ac b bx 2 42 − ± −= ( ) nr r nnnnn b b ar nb anb ana b a + +     + +     +     + = + −−− 2 21 2 1 ( ) ! ( 1)...( 1) !! ! n n nn n r r rnr r  − −+= = − 1cos sin22 = +A A AA 22 tan1sec + = B A B A B Asin cos cos sin)sin( ±= ± B A B A B Asin sin cos cos)cos( = ± B A B AB A tan tan1 tan tan)tan(  ±= ± A A Acos sin2 2sin = A AA 2tan1 tan22tan −=

Page 3 of 24 1 A triangle with a base of ( )32+ cm has an area of ( )12 7 3+ cm2. Find the perpendicular height of the triangle, leaving your answer in the form ( )3ab+ cm, where a and b are integers. [4] ( ) 1 3 2 12 7 32 h+ ×= + 1 12 7 3 3 2 2 32 32 h +−= × +− 1 12 3 24 7(3) 14 3 2 34h −+ −= − 1 233 21h −−= − 1 3232 h= + 643h= + M1 M1 M1 A1

Page 4 of 24 2 (a) The equation of a curve is 2 4 16y px x p= −+ . Find the range of values of p given that the curve lies completely above the x-axis. Discriminant = 2( 4) 4( )(16 )pp−− = 216 64 p− 216 64 0p−< 216 64 p< 2 1 4p > 1 2p<− , 1 2p> (reject as p > 0) [3] (b) Find the value of h for which the line 2y xh= + is a tangent to the curve 22 65yx x= −+ . 2y xh= + --- (1) 22 65yx x= −+ --- (2) [4] Sub (1) in (2): 22 2 65xh x x+= − + 22 625 0x xx h− − +−= 22 85 0xx h− +−= Discriminant = 2( 8) 4(2)(5 ) 0 h−− −= 64 40 8 0 h−+= 3h=− M1 A1 A1

Page 5 of 24 Given that 5tan 12A= and 3sin 5B=− , and that A and B are in the same quadrant, without using a calculator, calculate the values of (a) cos ec ( )A− , )sin( 1)(eccos AA −=− - M1 Asin 1 −= 1 5 13 = −− 13 5= - A1 [2] (b) sin 2B , sin 2 2sin cosB BB= 3 4 24sin2 2 5 5 25B   =− −=     - A1 [1] (c) tan( )AB+ . BA BABA tantan1 tantan)tan( − +=+ 53 12 4 531 12 4  +  =  −×   - M1 56 33= - A1 [2] 3 13 y −12 x A −5 5 y −4 x B −3

Page 6 of 24 4 (a) Solve the equation ( ) 0212172 12 =+−+ xx . ( ) 0212172 12 =+−+ xx ( ) ( ) 2 12 2 17 2 21 0xx ×− += Let 2xu= , 22 17 21 0uu− += (2 3)( 7) 0uu− −= 3 2u= , 7u= 32 2 x = , 27x = 3lg2 lg 2x =  , lg2 lg7x = 3lg 2 lg2x = , lg7 lg2x= 0.585x= , 2.81x= [4] M1 M1 M1 A1

Page 7 of 24 Explain why the equation ( ) 212 17 2 0xx p+ − += has no solution if 1368p> . Discriminant = 2( 17) 4(2)( ) p−− = 289 8 p− Since 1368p> , 8 289p− <− 289 8 0p−< ∴Discriminant < 0 Hence, the equation has no solution if 1368p> . (b) [2] Let 2xu= , 22 17 0u up− += D < 0, ( ) 2 17 4(2)( ) 0 p−− < 289 8 0p−< 1368p> Hence, the equation has no solution if 1368p> . M1 A1 A1 M1

Page 8 of 24 5 The mass, m grams, of a radioactive substance remaining, t days after being measured is given by 0.0110 0.2 tme −= + . (a) Find the initial mass. Initial mass = 0.01(0)10 0.2e− + = 10.2 g [1] (b) Sketch the graph 0.0110 0.2 tme −= + for t ≥ 0. [2] A1 m t 0 0.2 10.2 0.0110 0.2 tme −= + B1 – Correct shape B1 – Correct intercept

Page 9 of 24 (c) Find the least number of days it takes before the amount of substance is reduced to 5% of its initial mass. 0.0110 0.2 0.05(10.2)te− +< 0.0110 0.2 0.51te− +< 0.0110 0.31te− < 0.01 0.031te− < 0.01 ln(0.031)t−< 347.37t > Least number of days = 348 [3] (d) Explain why the mass of the radioactive substance can never be less than 0.2 g. [2] Since 0.01 0te− > 0.0110 0 te− > 0.0110 0.2 0.2te− +> Therefore, the mass of the radioactive substance can never be less than 0.2g. M1 M1 A1 A1 A1

Page 10 of 24 6 The polynomial baxaxx +−− 232 23 has a factor 12 −x and leaves a remainder of –8 when divided by 1x− . (a) Find the values of a and of b. 32 11 1 123 2 022 2 2f a ab   = − − +=       13 044 a ab− −+= 741ab−= -- (1) (1) 2 3 2 8f a ab=− − += − 5 10ab−= 5 10ba= − -- (2) Sub (2) in (1): 7 4(5 10) 1aa− −= 7 20 40 1aa− += 13 39 0a− += 3a= 5(3) 10b= − = 5 [4] M1 M1 M1 A1