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 3 5ii ( ) ( ) 22 14 6sin 8cos 6sin 8cos sin 6 8 10 8tan 6 53.13 14 10sin 53.13 Perimeter R R P         = + + + = + = + = = = = + +  M1 M1 A1 3 5iii ( ) ( ) ( ) 23 14 10sin 53.13 10sin 53.13 9 sin 53.13 0.9 64.158 53.13 64.158 ,115.842 11.028 ,62.712 11.0 ,62.7 Basic angle      = + +  +  = +  = = + =   =   =   M1 M1 A1 3

6i 54 −== tdt dva Decelerating means 054 −= ta 4 5t seconds M1 M1 A1 3 6ii When instantaneously at rest, 0252 2 =+−= ttv 0)2)(12( =−− tt 2 1=t , 2=t M1 A1 2 6iii 2(2 5 2)s vdt t t dt= = − + = 3225 232 tt tc− + + When t =0, s =0 , so c =0 Hence, ttts 22 5 3 2 23 +−= M1 A1 2 6iv When 0=t , 0=s When, 2 1=t . 24 11)2 1(2)2 1(2 5)2 1(3 2 23 =+−=s When 2=t , 3 2)2(2)2(2 5)2(3 2 23 −=+−=s When3, 2 3)3(2)3(2 5)3(3 2 23 =+−=s Total distance travelled 11 11 2 3 2( ) ( )24 24 3 2 3d = + + + + 33 4= m M1 M1 M1 M1 A1 5

7i 62)( 23 +++= bxaxxxf Since ( 2)x+ s a factor, 06)2()2()2(2)2( 23 =+−+−+−=− baf 062416 =+−+− ba 1024 =− ba --------------- (1) 156)3()3()3(2)3( 23 =+++= baf 1563954 =+++ ba 4539 −=+ ba 2−=a and 9−=b M1 M1 A1, A1 4 7ii 06922)( 23 =+−−= xxxxf ( ) 0)32(2 2 =+++ kxxx 222 24 xxkx −=+ 6−=k ( ) 0)362(2 2 =+−+ xxx 2−=x or ( ) ( )22 )3)(2(46)6( 2 −−−−=x 2−=x or 2 33=x M1 M1 A1, A1 √ based on 7i 4 7iii 0618816 23 =+−− yyy ( ) ( ) ( ) 06292222 23 =+−− yyy Consider 2yx= 22 −=y or 2 332 =y 1−=y or 4 33=y M1 M1 A1, A1 4

8i 22 tan cot 2cos 2 tan cot sin cos cos sin sin cos sin cos 1 sin cos 1 1 (sin 2 )2 2 sin 2 2cos 2 A A ec A LHS AA AA AA AA AA AA A A ec A += + =+ += = = = = M1 M1 M1 A1 4 8ii tan cot 5 2cos 2 5 5cos 2 2 15 sin 2 2 2sin 2 5 AA ec A ec A A A += = = = = Basic angle = 1 2sin ( )5 − = 23.57818 2A is in 1st of 2nd quadrant 2A 23.57818 ,180 23.57818 ,360 23.57818 ,180 23.5 7818 360 =  −  +  − + 11.8 ,78.2 ,191.8 ,258.2A=     M1 M1 M1 B2 5 9i 3(cos 3cos )d xxdx − xxx sin3)sin(cos3 2 +−= xxx sin3)sin1(sin3 2 +−−= x3sin3= (Shown) M1 A1 2 9ii  − 1 5.0 3 )sin2sin3( dxxx  −= 1 5.0 1 5.0 3 sin2sin3 dxxxdx   1 5.0 1 5.0 3 ][cos2cos3cos xxx +−=   ]5.0cos1[cos25.0cos35.0cos1cos31cos 33 −++−−= = -0.181 (3sf) M1, M1 M1 A1 4

10i 2 1 2 When 0, 2 4 ( 4) 0 0 or 4 (4,0) 11 2( )2 11 When 4, 1 2 y xx xx xx xx A dy dx x x x dy dx − = = = −= ==  =− =− = = Gradient of AB = -2 Equation of AB 0 2( 4) 28 yx yx − =− − =− + M1 M1 M1 A1 4 10ii 110 1 2(1) 8 6 (1,6) x x y y B −= = =− + = M1 M1 A1 √ based on 10i 3 10iii 4 1 4 3 2 2 1 2 12 3 6 2 2 932 2 252 ( ) 936 510 units6 x x dx xx − +    = − +   = − − − + =  M1 M1 A1 √ based on 10i and 10ii 3 11i Centre ( ) 2 8 3 11,22 5,7 ++= = Radius B1

= 222 5   ( 3() 7)− + − = 5 units M1 A1 3 11ii (x − 5) 2 + (y − 7) 2 = 25 B1 1 11iii Consider x = 0, (0 − 5) 2 + (y − 7) 2 = 25 y 2 −14y = 49 = 0 Consider b2 − 4ac = 142 − 4(49) = 0 Since b2 − 4ac = 0 , y axis is tangent to the circle. M1 A1 2 or ( ) 2 70 7 y y −= = Since there is only 1 point of intersection, y-axis is tangent to the circle. M1 A1 2 11iv Gradient OQ = 7 11 58 − − = 4 3 Gradient of tangent at Q = 3 4− Equation of tangent at Q = 311 ( 8)4 3 174 yx yx − =− − =− + M1 M1 A1 3