2022 TKSS AMath Prelims P1 Ans

2 Marking Scheme Additional Mathematics Paper 1 CANDIDATE NAME ADDITIONAL MATHEMATICS 4049/01 Paper 1 Thursday 11 August 2022 2 hours 15 minutes Candidates answer on the Question Paper. No additional materials are required. READ THESE INSTRUCTIONS FIRST Write your name, class and index number on 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, highlighters, glue or correction fluid. Answer all the questions. Give non-exact numerical answers correct to 3 significant figures, or 1 decimal in the case of angles in degree, unless a different level of accuracy is specified in the question. The use of a 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. TANJONG KATONG SECONDARY SCHOOL Preliminary Examination 2022 Secondary 4 MARKING SCHEME CLASS INDEX NUMBER

2 4049/1/Sec4Prelim2022 Mathematical Formulae 1. ALGEBRA Quadratic Equation For the equation ax2 + bx + c = 0, x = a acbb 2 42  . Binomial Theorem (a + b)n = an +       1 n an  1 b +       2 n an  2 b2 + . . . +       r n an  r br + . . . + bn, where n is a positive integer and       r n = !)!( ! rrn n  = ! )1).......(1( r rnnn  2. TRIGONOMETRY Identities sin2 A + cos2 A = 1 sec2 A = 1 + tan2 A cosec2 A = 1 + cot2 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 A B A B   sin 2A = 2 sin A cos A cos 2A = cos2 A  sin2 A = 2 cos2 A  1 = 1  2 sin2 A tan 2A = A A 2tan1 tan2  Formulae for ABC C c B b A a sin sin sin  a2 = b2 + c2  2bc cos A  = 2 1 bc sin A

3 4049/1/Sec4Prelim2022 2(i) 3 2 2d 3 2d y x ax bx c y x ax bx        B1 (ii) For decreasing function, 2 2 d 0d 3 2 0 From 1 3, 3( 1)( 3) 0 3 6 9 0 y x x ax b x x x x x              By comparison, a= −3 and b = −9 B1 M1 B1 AG Their (i) < 0 Form eqn with factors (x+1)(x-9) ignore 3 Both correct Total: 4m 1 2 2 2 2 2 2 2 2 2 2 2 3 (2 ) (2 ) 1 0 6 3 4 4 1 0 5 10 4 1 0 0 ( 10 ) 4(5)(4 1) 0 20( 1) 0 20( 1)( 1) 0 x x m x m x x mx x m mx x x mx m D m m m m m                            m < −1 or m > 1 M1 B1 M1 A1 B1 Remove / substitute to 1 variable factorisation Total:5m -1 1

4 4049/1/Sec4Prelim2022 3a 2 3 2 32 2 2 2 3 2 2 2 3 51 2 2 7 1 2 2 100 5 56 2 2 5 2 52 3 5 2 3 2 52 5 5 2 3 5 2 3 5                                          a = -1, b = 2, c = -3.5 M1 B1, B1 Attempt to reduce to base 2, 3, 5 3 correct B1, B1 2 correct B1 3b 12 550000 1 100           = $89792.82 M1 A1 Form exponential function 2 dp Total: 5m 4 2 2 2 2 2 1 2 3 52 ( 2)( 1) 3 5 ( 2)( 1) ( 2) ( 1) 3 5 ( 1) ( 2) 11 2,3 3 2 1 11 2 22 3( 2) 3( 1) x x x x x x x x A B x x x x x A x B x A B x x x x x x                              M1 B1 M1 A1 B1 Long division seen Correct partial fractions Use comparing coefficient or substitution of x √ From correct PF Total: 5m

5 4049/1/Sec4Prelim2022 5(i) 1 sin 3( ) 1 p q q    Since y max = 3, p = 3-1 = 2 B1 B1 Substitution seen 2 – 1 seem (ii) G1 P1 Correct shape, 1.5 cycles seen Max point at y =3, min at y = -1 (iii) Between 0 2 x   , there will be 3 cycles of the graph sin 3y p x q  . For line 2 1y x   , when 2 , 3x y   For graph sin 3y p x q  , when 2 , 1x y   The line will intersect the graph at 5 points within the domain. L1 B1 Straight line passing through (0, -1) and ( π, 1) o.e. Accept sketch (iv) 0 B1 Total: 7m

6 4049/1/Sec4Prelim2022 6 2 2 ( 2)( 1) (1)( 1) ( 2)2( 1) ( 1)[( 1) 2 4] ( 1)(3 3) 3( 1)( 1) y x x dy x x xdx x x x x x x x                    2 2 0 3( 1)( 1) 0 1, 1 0, 4 (1)( 1) ( 1)(1) 2 dy dx x x x y d y x x xdx             At (1, 0) 2 2 2(1) 0d y dx   (1,0) is a minimum point At (-1, 4) 2 2 2( 1) 0d y dx    (-1, 4) is a maximum point M1 A1 M1 M1 B1 B1 Product rule seen o.e. Solve their derivative = 0 Use of 1st / 2nd derivative Total: 6m

7 4049/1/Sec4Prelim2022 7(i) 2 2 2 2 2 2 2 2 2 2 cot tan cot tan 1 tantan 1 tantan 1 tan 1 tan sin1 cos sec sin1 cos 1 cos cos sin cos 2 LHS x x x x xx xx x x x x x x x x x x x RHS                2 2 2 2 2 2 cot tan cot tan sin sin cos cos sin sin cos cos cos sin cos sin cos sin cos 2 LHS x x x x x x x x x x x x x x x x x x x RHS            B1 B1 B1 1 sincot tan cos xx or x x Rewrite in terms of sin x or cos x / simplify fraction Double angle formula to arrive at AG (ii) 2 cos 2 cos 2 cos 1 cos 0 (2 cos 1)(cos 1) 0 1cos , cos 12 0 ,120 , 240 ,360 x x x x x x x x x               B1 M1 A1, A1 Use identity For both answers in each pair 0, 360 and 120, 240 Total: 4m

8 4049/1/Sec4Prelim2022 8(i) 2 3 2 3 2 4 5 ( 1)( 5) if ( 1) is a factor, 2(1) 9(1) 3(1) 0 4 if ( 5) is a factor, 2( 5) 9( 5) 3( 5) 0 490 x x x x x m m x m m                     M1 B1 B1 Substitute the value of 1 or 5 to form eqn =0 (ii) When m = 4, (x – 1) is a factor (By Factor Theorem) 3 2 2 3 2 2 2 9 3 4 ( 1)(2 4) 3 4 7 2 9 3 4 0 ( 1)(2 7 4) 0 ( 1)(2 1)( 4) 0 1,4, 0.5 x x x x x ax compare coeff x a a x x x x x x x x x x                         B1 M1 M1 B1 (x-1) as a factor soi Find quadratic factor Factorise completely = 0 (iii) 2 4 6 2 2 9 3 4 0 1 11, 2 y y y x y y        M1 A1 soi √ Total: 9m

9 4049/1/Sec4Prelim2022 9(i) At , 2 1 (1, 2) 2 8 4 By symmetry (4, 0) At , 0 3sin 04 4 A y x A Period B OR B y x x                  B1 B1 (ii) Area X   4 1 4 1 3sin 4 43 cos 4 12 cos cos 4 12 2 1 2 12 21 2 6 2 2 ( ) x dx x shown                                     B1 M1 B1 AG Correct integral Seen (ignore limits) Substitution of limits Evaluate special angles leading to answer (iii)     4 1 4 1 2 2 ln 2[ln 4 ln1] 2 ln 4 2.773 6 2 2 2.352 ln 4 AreaY dxx x AreaX AreaY             B1 B1 B1 o.e. B1 Negative integral Total: 9m

10 4049/1/Sec4Prelim2022 10(i) 20 3(7) (7) 7 20 k k      M1 A1 substitution (ii) 23 7 6 20 4 4 v t kt a t when t a        M1 A1 dva dt seen (iii) 2 3 2 3 2 2 2 3 20 7 10 7 , 0, 0, 0 0, 10 7 0 ( 10 7) 0 0, ( 10) ( 10) 4(1)( 7) 10.662 v t t s t t t c At A t S c When S t t t t t t t t s                          B1 M1 A1, A0 Correct integral with +c -1m if c=0 is not seen DM awarded only when quadratic formula or completed square is seen Minus 1 mark if t = 0 not shown / t is divided (iv) 3 2 3 2 0, (3 1)( 7) 0 7 , 7 [(7 10(7) 7(7))] 196 8, (8 10(8) 7(8) 184 when v t t t At B t m At t                 Total distanc