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 Eqn 7(ii) Area ABD   2 0 10 6 01 10 8 6 102 1 0 60 60 100 48 02 26units         M1 A1 Correct use of shoelace method 7(iii) 6 0 10 60 , 1044 3 ,112 E E     B1 8(i) 2 2 3 1 3 1 32 1 ()12 Vol r h hVh V h shown        M1 M1 r = h/2 correct volume formulae 8(ii) 2 2 1 4 12.5 (3)4 10 / min9 dV hdh dV dV dh dt dh dt dh dt dh mdt        M1 M1 A1 Able to form chain rule and use h = 3 Accept 0.354m/s E A(0,10) D(6,6)

Qn Working Marks Remarks 8(iii) 2 2 21 24 Ar hAh      1 2 dA hdh  2 1 10(3)29 21 / min3 dA dA dh dt dh dt m      M1 M1 M1 A1 Forming eq of area Differentiate A Forming chain rule (rates) 9(i) Least value of g(x) = 7 Greatest value of g(x) = 3 B1 B1 9(ii) Period of f(x)= 360 B1 Accept 2 9(iii) Period of g(x)= 360 B1 Accept 2 9(iv) Sine graph B1 B1 tan graph B1 B1 Shape, (90,3),(270,2) (0,2), (180, 2), (360,2) shape asymptote at x = 180 9(v) 3 solutions B1 No ecf 10(i)   2ln 3yx 2 2 3 dy x dx x              22 22 2 22 22 2 22 3 2 2 2 3 2 6 4 3 26 3 x x xdy dx x xx x x x        B1 M1 A1 differentiation quotient law (270,7) (180,2) (90,3) (360,2)

Qn Working Marks Remarks 10(ii) 2 2 03 0 dy x dx x x       2 ln 3 0 ln 3 (0,ln 3) y y   When x = 0, 2 22 0 6 2 (0 3) 3 dy dx   Since 2 2 0dy dx  , (0,ln3) is a minimum point. M1 A1 M1 A1 0dy dx  Stationary point Using second derivative test Min pt 11i(a) 25 40s t t 10 40vt When t =1,v = 30 m/s 10a m/s2 M1 A1 A1 dsv dt 11i(b) When v = 0, 10 40 0 4 t t   P changed its direction when t = 4. M1 A1 v = 0 11i(c) When t = 4, 25(4) 40(4) 80 s sm   When t = 10, 25(10) 40(10) 100 s sm   Total distance travelled = 80+180m Average speed = 26m/s M1 M1 M1 A1 ECF up to 3 method marks 11ii(a) 0.5(0)9 3( ) 31 2 ek k k    M1 A1 11ii(b)     0.5 0.5 0.5 32 ' 3(0.5) ' 1.5 t t t de de de    Since 0.5 0,te  for all values of t, 0.51.5 te > 0, d is increasing. M1 M1 Differentiate d Reasonable explanation