SCSS 2024 EM Prelim P2 MS

1 2024 4E/5N EM Prelim Paper 2 Marking Scheme (90 marks) 1a ( ) ( ) 3 1 2 5 34 4 3 1 3 2 5 12 4 6 15 27 2 2 27 xx xx xx x x +− +  − +  −   M1 A1 1bi ( ) ( ) ( ) 2 0.2 3 1.5 0.2 2 1.5 7 / 1.754 a a −−= +− =− − B1 1bii ( ) ( ) 23 2 2 2 3 2 2 3 2 3 2 2 3 2 32 2 32 2 bca bc a b c b c ab ac b c ab b c ac b a c ac c acb a c acb a −= + + = − + = − − =− − − =− − −−= − += − M1 M1 Either A1 1c ( ) ( ) ( )( ) ( ) 22 22 2 2 52 13 2 3 5 2 3 2 3 13 2 3 10 15 2 6 2 3 6 9 10 17 6 2 9 9 8 8 3 0 8 ( 8) 4(8)( 3) 2(8) 1.2906, 0.29057 1.291, 0.291 (3 ) x xx x x x xx x x x x x x x x x x xx x x x dp −=−− − − − =−− − − + = − − + − + = − + − − = − −  − − −= =− =− M1 – common denominator M1 – either expansion M1 A1A1

2 2a y = 7 B1 2b 7 points – B2 4 points – B1 Curve – B1 2ci 3.25 1.2x−   B1 2cii 2 2 2 120 2 12 1.5 0 1.52 2 1 1.5 1.5 1.65 0.25 xx xx xx y x or x + + = + + − = − + − =− =− =− =− M1 A1A1 2ciii ( ) 2 22 2 21 22 122 12 xx x x +−    = + − −       = + − Therefore, the minimum point is ( )1, 2−− M1 - either x or y value seen A1 – in the form ( )2x p q++

3 3a ( )3 3 2 2 2 32 32 3 2 2Vol of hemisphere = 33 18 13Vol of cone = 32 19 34 3 4 318 3 4 918 4 18 4 9 8 ( ) x x xy xy xy x x y x x y xy x y x shown        =    =    = = = = = M1 M1 A1 3b ( ) 2 2 22 22 2 2 3 2 38 2 964 4 265 4 2658.1394 / 2 3CSA of cone = 2 3 8.13942 38.356 l y x xx xx x xx xl xx x   =+  =+  =+ = =   =  = ( ) ( ) 2 2 2 22 2 3CSA of cylinder =2 3 2 28.274 CSA of hemisphere =2 3 56.549 3rim area= 3 2 21.206 xx x x x xx x     =  = −  = M1 4 correct – M3 3 correct – M2 2 correct – M1

4 3b 2 2 2 2 2 2 Total SA = 450 38.356 28.274 56.549 21.206 450 144.385 450 3.1167 1.7654 x x x x x x x + + + = = = = ( ) Total ht, = 3 3 8 3 3 14 14 1.7654 24.7156 24.7 (3 ) h y x x h x x x hx h h h sf ++ = + + = = = = M1 M1 A1 4a 1 25 B1 4b 16 3 5n nT n −= M1 - 3n− M1 – 5n A1 - Tn 4c ( ) ( ) ( ) ( ) ( )( ) ( ) ( ) ( ) ( ) ( ) 1 22 22 16 3 1 16 3 5 1 5 16 3 3 16 3 5 1 5 13 3 16 3 1 51 13 3 16 16 3 3 51 13 3 16 16 3 3 51 16 ()51 nn n nTT nn nn nn n n n n nn n n n n n nn n n n n n nn shownnn + −+ −− = − + − − −=− + − − − += + − − + − − = + − − − + += + −= + M1 M1 A1 4d ( ) ( ) 1 For 0 5 1 0 16 051 0nn n nn nn TT+  + − + − B1

5 5a ( ) ( ) 24grad of line 8 ( 4) 1 2 1 2 14, 4 into 2 144 2 2 1 22 l y x c subst y x c c c yx −−= −− =− =− + − =− + =− − + = =