[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 , ( )