2020 MARKING SCHEME 4E AMATHS PRELIM P1 (Hougang)

Hougang Secondary School Mathematics Department MARKSCHEME Mathematics Syllabus Express/Normal Academic/Normal Technical Subject : Additional Mathematics Examination : Prelim 2 Level : 4E/5N/4AO Paper : 1 Qn Working Marks Remarks 1(i) 304 4 4xy 3yx ---------------Eq1   1 443 3 3y x 4 4y x --------------Eq2 Sub (1) into (2): 434x x      2 2 3 4 4 3 4 4 0 2 3 2 0 xx xx xx          2x or 2 3x 6y or 2y M1 M1 M1 A1 A1 Indices law Indices law Perform substitution/elimination Answer for both x 1(ii) B1 B1 Shape, pass through origin 2(i)   9 1 2 9 r r r kTx r x          939 rr kxr  9 3 0r 3r   39 6723 2 k k     M1 M1 M1 A1 General term Term indep of x Form equation

Qn Working Marks Remarks 2(ii)   12 term + constant+...x x   9 3 1 10 ()3 r r rej      2 ... + constant+...x  constant term = 2 (672) = 1344 M1 M1 A1 Recognising the need to check for 1/x term 3 Base area =     21 4 5 22    2 1 16 8 5 5 (2)2 21 8 5 cm     50 5 101 21 8 5 50 5 101 21 8 5 21 8 5 21 8 5 242 5 121 121 1 2 5 h cm       M1 M1 M1 A1  16 8 5 5 Forming equation of h using prism volume rationalising 4(i) 0.004(1000)500 9.1578 9.16 (3 ) Re R R g sf    M1 A1 Sub t = 1000 4(ii) 10 500 50100 gg 0.004 0.004 0.004 50 500 1 10 1ln ln10 ln 0.1 0.004 575.6 t t t e e e t t years          Year 2595. M1 M1 A1 10percent of original Attempting to solve for t Do not round up

Qn Working Marks Remarks 4(iii) B1 Shape, include R- intercept 500 No need to penalise the negative t portion of the graph 5(i) 1sec cos 5sec 4 A A A   M1 A1 Knowing reciprocal identity 5(ii) sin( ) sin cos cos sinA B A B A B   = 3 15 4 8 5 17 5 17 77 85                  M1 A1 Correct use of formulae 6 22 4 5 10 6x kx k k    22 4 10 0x kx k    for all values of x      2 2 4 4(2)(10 ) 0 16 80 8 0 2 2 5 0 2.5 2 kk kk kk k             M1 M1 M1 M1 A1 Forming quad ineq (Allow slips) Correct a,b,c Using D < 0 Factorisation 7(i) 6 10 60 2 3 BC ADmm     3 2 DCm  Eqn of DC is  366 2 1.5 3 yx yx     At C, 0 1.5 3 2 (2,0) x x C    M1 M1 M1 M1 Gradient Perpendicular gradient Eqn DC Alternatively, accept using gradient of DC. Find point C R t

Qn Working Marks Remarks 10 2 8 0,22 ( 4, 4) BCM     Eqn of perpendicular bisector is  344 2 3 102 yx yx     M1 A1 Midpoint of BC