[FFM] [2023] 4E AM 4049 Prelims P1 MS

2023 Sec 4 Express Additional Mathematics Paper 1 Preliminary Examinations Marking Scheme No. Solution Marks AO 1(i) cosec 1 sin 1 7 4 4 7  = =    = B1 AO 1 1(ii) cos30 (tan 45 sin 60 ) 33 122 3 3 2 3 3 or 2 4 4  +  =+  +=+ B1 (special angle for cos 30 and sin 60) B1 AO 1 2 Sub. 3y ax=− into 227yx=+ 2 2 2 7 3 2 10 0 x ax x ax + = − −+= Let 2 4 0,b ac− ( )( ) ( )( ) 2 2 ( ) 4(2)(10) 0 80 0 4 5 4 5 0 or 80 80 0 4 5 4 5 or 80 80 a a a a a a aa − −  − − +  − +  −   −   M1 (Form quadratic equation) M1 (Discriminant is negative) M1 (Factorise using surds) A1 AO 1

3 11 22 11 22 11 22 1 1 1 1 1 1 2 2 2 2 2 2 1 1 1 1 2 2 2 2 11 22 3 252 3 1 1 252 2 2 3 3 1 1 252 2 2 2 xx xx xx x x x x x x x x x x y Ae Be dy Ae Bedx dy y e edx Ae Be e e Ae Be Ae Be A e B e − − − − − − −− =+ =− + = −  + = − − +     + = − + − +       By comparing coefficients, 31 222 22 1 31 522 5 AA A A BB B =− = = =− + =− B1 (Differentiate y correctly) M1 M1 (Compare coeff. for A or B correctly. FT from previous M1) A1 A1 AO 2 4 ( ) ( ) 2 2 2 3 82 Sub. 13.5, 313.5 8 2 3 24 27 0 (3 27)( 1) 0 9 (reject) or 1 V h h V hh hh hh h =+ = =+ + − = + − = =− ( )3 282 = 3 12 dV hdh h =+ + Sub. 1, 3(1) 12 = 15 h dV dh = =+ 1 = 815 8 = cm/s or 0.533 cm/s (to 3 s.f.)15 dh dh dV dt dV dt= − −− M1 (Simplify to get quadratic equation) A1 B1 (Differentiate correctly) M1 (Substitute into Chain Rule. FT for dV/dh.) A1 AO 2

5(i) 0 (400) 000.841 400 ln 0.841 0.00043290 = 0.000433 (to 3 s.f.) (Shown) kt k A A e A A e k k = = = − − M1 (Sub. A correctly) AG1 AO 3 5(ii) 0.00043290 000.5 0.00043290 ln 0.5 ln 0.5 0.00043290 1601.2 = 1601 years (to nearest whole number) tA A e t t −= −= = − OR 0.000433 000.5 0.000433 ln 0.5 ln 0.5 0.000433 1600.8 = 1601 years (to nearest whole number) tA A e t t −= −= = − M1 (Sub. A correctly) A1 AO 2 5(iii) 0.00043290(3200)100 25.025 = 25.0 grams (to 3 s.f.) Ae −= OR 0.000433(3200)100 25.017 = 25.0 grams (to 3 s.f.) Ae −= M1 (Substitute correctly) A1 AO 2 6(i)   bisects angle ) (alternate segment theorem) ( (angles in the same segment) (p   en ) rov DAR ABD CBD ABD CBD CAD DAR A BSD ABC CD  =  =  =   = B1 (2 statements correct) B1 (3 statements correct) AG1 AO 3 6(ii) (common angle) (Alternate segment theorem) ARD ARC RAD RCA  =  = RAD is similar to RCA (AA similarity or 2 pairs of corresponding angles are equal) M1 (Both statements correct) AG1 (Similarity test must be stated) AO 3

6(iii) 2 (proven) RA RD RC RA RA RC RD =  =  Form proportional ratios and conclude AG1 AO 3 7(i) For 8 2 3 1      − x x , ( ) ( ) 83 1 2 24 5 8 1 8 = 1 r r r r r Tx r x xr − + −  =−   − For constant term, 24 5 0 4.8 (N.A.) r r −= = Hence, there is no constant term because r must be a positive integer/whole number. M1 (Form r +1 term) AG1 (Show that power of x is not 0 and conclude accordingly) AO 3 7(ii) For ( ) 8 35 2 1 1xx x  −+ , 24 5 6 30 5 = 6 r r − =− = and 24 5 5 6 35 5 = 7 r r − + =− = ( ) ( ) ( ) 67 6 11 5 6 6 88 1 (1) 167 (28 8) 20 x x x x x −− − −    − + −       =− = Hence the coefficient of 6x− is 20 (Shown) M1 M1 M1 AG1 AO 3

8(i) ( ) 2 2 2 2 2 2 2 4 9 2( 2 ) 9 2 2 1 1 9 2 ( 1) 1 9 2( 1) 2 9 2( 1) 7 xx xx xx x x x −+ = − + = − + − + = − − + = − − + = − + Stationary point is (1, 7). B1(completed sq) B1 AO 1 8(ii) Sub. 33yx=+ into 22 4 9y x x= − + : 2 2 22 2 4 9 3 3 2 7 6 0 (2 3)( 2) 0 1.5 or 2 7.5 9 (2 1.5) (9 7.5) 5 = or 2.52 5 or 2.52 x x x xx xx xx yy AB h − + = + − + = − − = == == = − + − = M1 (Equate and factorise) A1 (For both coordinates) M1 (Apply distance formula) A1 AO 2 9(i) 21 12 xyx x +=− − Since 1 2 0 1 2 x x −  y is not defined at 1 2x= . B1 AO 1

9(ii) 2 2 2 2 2 21 12 (1 2 )(2) (2 1)( 2)1 (1 2 ) 2 4 4 2 = 1 (1 2 ) 4 (2 1)(2 3) 4 2 2 = 1 or or 1(1 2 ) (1 2 ) (1 2 ) 1 2 xyx x dy x x dx x xx x x x x x x x x +=− − − − + −=− − − + +− − + − +− − +− − − − 2 3 2 33 8(1 2 ) ( 2) 16 16 = or (1 2 ) (2 1) dy xdx xx −= − − − −− [Quotient Rule M1 –correct] M2 – dy/dx fully correct] A1 B1 AO 1 9(iii) 2 22 For stationary points, 0. 410 (1 2 ) (1 2 ) 4 or 4 4 3 0 1 2 2 (2 3)(2 1) 0 31,22 71,22 dy dx x x x x x x x x y = −= − − = − − = − = − + = =− =− Coordinates of stationary points are 3 7 1 1, and ,2 2 2 2     −−       (or equivalent) M1 FT (Equate derivative to 0) A1, A1 AO 1 9(iv) 2 23 2 2 16= (1 2 ) 3Sub. , =2 02 dy dx x dyx dx − − = 2 2 1Sub. , = 2 02 dyx dx=− −  Hence, y has a minimum point at 3 2x= and a maximum point at 1 2x=− . Note: By 2nd derivative test only M1 FT (Find second derivative value for either point) A1 A1 AO 1

10(i) 3 2 5.13 03 = −− −−=BDm Equation of BD is 2( 3) ( ( 3))3 232 3 2 1 or 3 2 33 yx yx y x y x − − = − − + = + = − = − Equation of AD: 15 22yx=− + 2 1 51 =3 2 2 77 62 3 1 xx x x y − − + = = =  coordinates of D are (3, 1). B1 M1(FT for value of gradient) M1 (FT from equation of BD) A1 (no FT) AO 2 10 (ii) 3 2 ACm =− Equation of AC is 32 ( 1)2 37 or 2 3 722 yx y x y x − =− − =− + + = B1 (FT from gradient of BD) B1 (no FT) AO 2 10 (iii) Note CD is not parallel to the y-axis. Let E be the mid-point of AC. AC: 37 22yx=− + ----- (1) BD: 2 1 3yx=− ----- (2) 3 7 2(1) (2) : 1 02 2 3 13 9 62 27 13 xx x x − − + − − =  − =− = M1 (FT from eqn. of AC and BD) OE AO 2

2 27 13 13 5 = 13 y  = − 27 5 is , .13 13E   Let C be (x, y), 1 2 27 5,,2 2 13 13 1 27 2 5 and 2 13 2 13 41 16 13 13 41 16 is , 13 13 xy xy xy C ++    =       ++ == = =− −  2 411 3 3 11 13Hence, area of 162 2 3 1 2 13 1 48 41 123 483 6 ( 6 1)2 13 13 13 13 14 units ABCD − = −− = − + + + − − − − + = A1 (no FT) M1 (evaluate the ‘shoelace’. FT for coordinates of C and D) A1 (no FT) 11(a ) 1 2 9 3 8 3 389 xx x x − =− − =− Let 3x be y, 2 2 89 9 72 0 y y yy − =− −+= 2 2 2 ( 9) ( 9) 4(1)(72) OR 42(1) 9 207 = = ( 9) 4(1)(72)2 = 207 y b ac− −  − −=− − −− − OR M1 (or equivalent method) M1 (solve for y or discriminant) AO 3

2 2 1 809 231 239 or 42 9 9 yy y b ac − + = − = − =− Since the discriminant is negative, there is no solution for the equation. AG1 (mention “no solution”) 11 (b) (i) Since 90 (angle in a semi-circle),XYZ =  by Pythagoras’ Theorem, ( ) ( ) ( )( ) ( )( ) 222 50 2 28 2 50 2 5 2 2 2 28 2 2 7 2 2 102 4 14 (Shown) XZ = + + − = + + + − + =− B1 (state circle property for angle XYZ) M1 (apply Pythagoras’ Theorem) AG0 AO 3 11 (b) (ii) Gradient tan 28 2 50 2 2 7 2 5 2 2 5 2 2 5 2 2 10 14 2 14 10 2 (25)(2) 2 8 14 8 48 14 1 66 YXZ= −= + −−= +− − − += − −= =− B1 (tangent ratio) M1 (rationalise denominator) A1 AO 2

12(i) 2 2 Sub. 48 when 2, (2) (2) 48 2 4 48 2 24 ---------(1) v pt qt vt pq pq pq =− == −= −= −= 2 At max. speed, 0 when 2, 2 (2) 0 4 0 -----------(2) dv p q