Regent 4E5N AM Prelim Examination Paper 2 - Marking Scheme

Regent Secondary School Additional Mathematics Sec 4 Express Preliminar Examination 2020 Paper 2 (Setter: Ms Su RY) Marking Scheme Qn Solution Marks Total Marker’s Report 1i 22 4 3 0xx− + = Sum of roots ()+ 4 2 2 b a=− −=− = Product of roots () 3 2 c a= = 22+ 2 2 ( ) 2 3(2) 2( ) 2 1   = + − =− = (Shown) M1 M1 M1 A1 4 1ii Sum of new roots 22 22 33 6 16 7   = + + + = + + =+ = Product of new roots ( ) 22 2 2 2 2 2 22 2 ( 3)( 3) 3 3 9 3( ) 9 3 3(1) 92 57 4         = + + = + + + = + + + = + + = New quadratic equation M1 M1

2 2 5770 4 4 28 57 0 xx xx − + = − + = A1 3 2i General Term ( ) ( ) ( ) ( ) 7 7 1 72 7 7 7 r r rrr rr kxr x k x xr kxr − − − −  =  = = To find term with x3, 7 2 3 2 r r −= = Coefficient of x3 = 27 3 k  =21k2 To find term with x, 7 2 1 3 r r −= = Coefficient of x = 37 3 k  =35k3 23 32 2 21 35 35 21 0 7 (5 3) 0 30 (rej) or 5 kk kk kk kk = −= −= == M1 M1 M1 M1 M1 A1 6 2ii 7 2 7 0 r r −= = Coefficient of 7x = 07 0 k  = 1 To find term with x5, 7 2 5 1 r r −= = M1

Coefficient of 5x = 7 3 1 5     = 21 5 ( ) ( ) 7 2 2 7 5 77 7 15 211 5 ( ...) 5 21 20 kxx x x x x xx x −+  = − + + =− =− Coefficient of x7 = 20− M1 M1 A1 4 2ii Alternative Method: 2 7 6 2 7 5 (1 5 )[ 7( ) ...] 21(1 5 )[ ...] 5 kx x x x x x x − + +  = − + + Coefficient of x7 = 1 21− 20=− M1 M1 M1 A1 4 3i 0 0 0 lnln lnln mktm emm emm kt kt +−= = = − − t 2 4 6 8 10 mln 3.88 3.73 3.58 3.43 3.28 See graph at the end. B1 C2 3 3ii 0 4.03 0 ln 4.03 (acceptable range 4.02 - 4.03) 56.3mg (3sf) m me = = = M1 A1 2 3iii 3.28 3.88 10 2 0.075 0.075 ( 0.01) k k −−= − =− = M1 A1 2

3iv 4.03 Half original mass = 2 28.13046... ln(28.13046...) 3.33685... Time taken = 9.2 hr e = = M1 A1 √ based 3ii 2 4i 2 2 2 2 26 40 4(2)(6 ) 0 48 8 0 ( 12)( 4) 0 12 4 y x kx k b ac kk kk kk k = + + − − − −  − +  + −  −   M1 M1 A1 3 4ii When k = 2, 2 2 2 2 2 2 2 2 4 4 2 2 4 4 2 (2 ) 8 0 40 (2 ) 4(2)(8) 0 4 60 0 ( 10)( 6) 0 10 or 6 y x x y mx x x mx x m x b ac m mm mm mm = + + =− + + = − + − + = −= − − = − − = − + = = =− M1 M1 M1 A1 4 5i ( ) ( ) 2 4cos 8cos 2 3sin 6sin 3 3 4 4 6sin 8cos 14 6sin 8cos AF AB Perimeter     == == = + + + + + = + + M1 M1 A1