4E Northbrook AM P1 2026 Soln for student

NORTHBROOKS SECONDARY SCHOOL MATHEMATICS DEPARTMENT 4EXP ADDITIONAL MATHEMATICS Prelim Paper 1 MARKING SCHEME Qns Answer 1(a) Volume = 2rh ( ) ( ) 23 5 4 3 5 r− = + ( ) ( ) 2 3 5 4 35 r   − = + ( ) ( ) ( ) ( ) 2 3 5 4 3 5 3 5 3 5 r −− = +− 2 9 5 15 12 4 5 95r − − += − 2 27 13 5 4r −+= 1(b) 3xy−= --(1) and 22 3 19 0x xy y− + + = --(2) From (1) 3yx=− Sub 3yx=− into (2) 22 3 ( 3) ( 3) 19 0x x x x− − + − + = 2 2 23 9 6 9 19 0x x x x x− + + − + + = 2 3 28 0xx− + + = 2 3 28 0xx− − = ( 4)( 7) 0xx+ − = 4x=− or 7x= 7y=− or 4y= Coordinates are ( )4, 7−− and ( )7,4

2(a) 6 3 14 12 xy =−  dd 12dd yx tt= d 12d y x = ( ) 5 d 3 1 61d 4 12 12 yx x    =−        5 312 1 8 12 x=−  5 32 1 12 x=− 5 521 12 x=− 36x= 6 3 36 14 12y =−  48y= 2(b) 2 f ( ) 1 xex x= − ( ) ( ) 22 2 1 2 ( 1)f ( ) 1 xxx e ex x − − − = − ( ) ( ) 2 2 32f ( ) 1 xexx x − = − for 1x , change condition to x > 1.5 2 0xe  , ( ) 2 10 x− , ( )3 2 0x− ( ) ( ) 2 2 32 0 1 xex x − − f ( )x is a decreasing function

3(a) ( )( ) 2 22 7 30 3 5353 x x A Bx C xxxx − + + =+ −+−+ ( ) ( )( ) 227 30 3 3 5x x A x Bx C x− + = + + + − Let 5x= 28 28 A= 1A= Let 0x= ( )3 1 3 ( 5) C= + − 0C = Let 1x= ( )20 1 4 ( 4) B− = + − 6B= ( )( ) 2 22 7 30 3 1 6 5353 x x x xxxx −+ =+ −+−+ 3(b) ( ) 2 2 d2ln 3d3 xxxx += + 3(c) ( )( ) 2 2 7 30 3 d 53 xx x xx −+ −+ 2 16 d53 x xxx=+ −+ ( ) ( ) 2ln 5 3ln 3x x c= − + + + 4(a) 103 represents the height of the roller coaster car at the start 4(b) 23 24 103tt −+ = ( ) 23 8 103tt −+ = ( ) 2 3 4 16 103t − − + = ( ) 2 3 4 55t−+

4(c) The minimum height is 55 so the section of the roller coaster track has met the safety requirements 5(a) 8 2 12x x  − ( ) 8 1 2 8 12 r r rTx r x − +  =−  ( ) 8 8 2 1 8 21 rr r r rT x x r − − − + =− Power of x = 83 r− 8 3 0r−= 8 3r= Since r is not an integer, there is no independent term in the expansion. 5(b) 8 2 12x x  − = ( ) ( ) ( ) 2 8 7 6 22 88 112 2 2 ...12x x x xx        − + +              = 8 5 2256 1024 1792 ...x x x− + + 5(c) 8 32 11 52 xxx   +−     = ( ) 8 5 2 3 1 5 256 1024 1792x x xx  + − + + = ( ) ( )( ) 52 3 1... 1024 5 1792 ...xxx + − + + = 27936x Coefficient = 7936

6(a) 2 13y x x=− ( ) ( ) 22d d d 1 3 1 3d d d y x x x xx x x= − + − ( ) ( ) ( )( ) 1 2 2 d1 1 3 3 1 3 2d2 y x x x xx −= − − + − ( )( ) 2d3 1 3 2d 2 1 3 yx xxx x −= + − − ( ) 23 (2 ) 2 (1 3 )d d 2 1 3 x x xy x x − + −= − 2d 4 15 d 2 1 3 y x x x x −= − ( )4 15d d 2 1 3 xxy x x −= − 6(b) ( )0 1 2 4 15 d 2 1 3 xx x x− +− − = ( )0 1 4 152 d 2 1 3 2 1 3 xx x xx− −+ −− = ( )0 1 4 151 d 1 3 2 1 3 xx x xx− −+ −− = 0 1 2 02 1 1 (1 3 ) 1 3 1 ( 3)2 x xx − −   − +−  − = 2 ( 2)3+− = 4 3− 7(a) Amplitude = 3 Period = 180 or 

7(b) 7(c) 4k = or 2k =− 8(a) @ 0y= 3x= (3,0)A= 8(b) Grad of AD = 50 123 − =−−− Grad of DC = 1 11 − =− 5 3 5 13 ( 2) k k +− =−− 5 2 3 2kk− = + 24k = 2k = 8(c) C = (6, 13) Midpt of DC = 2 6 5 13,22 − + +  = ( )2,9 Gradient of perpendicular = 1− Equation of perpendicular bisector: ( )9 1 2yx− =− − 11yx=− +

8(d) Area = 3 8 6 2 31 0 3 13 5 02 − = ( )1 9 104 30 18 ( 26) 152 + + − − − − = 68 units2 9(a) 90BFA =  (given) 90ABC =  (angle in semicircle) ABC BFA = FAB BCA = (angle in alt. seg) triangle BFA is similar to triangle ABC 9(b) Since triangle BFA is similar to triangle ABC AB BF AC AB= 2AB AC BF =  9(c) TBA is a right angle triangle 2 2 2TA TB AB=+ TAC is a right angle triangle 2 2 2TA TC AC=− 2 2 2 2TB AB TC AC + = − ( )( ) 22TB AB TC AC TC AC + = + − 10(a) 2 32xy = 2 32y x= 222A x xy=+ 2 2 3222A x x x  =+  2 642Ax x=+

10(b) 2 d 64 4d A xxx =− 2 6404 x x=− 2 644 x x = 3 64 4x = 1.720508028x= 1.72x= 10(c) 2 23 d 128 4d A xx =+ @ 1.720508028x= 2 2 d 37.69911184 0d A x = The stationary value of A is a minimum value. Hence with this dimension, the cost of the production will be minimum 11(a) From B to FC sin 50 d = 50sind = From B to AE cos 30 d = 30cosd = 50sin 30cos .W =+

11(b) ( )50sin 30cos sin R   + = + 2250 30R=+ 3400R= 10 34R= 30tan 50 = 30.963756 = ( )50sin 30cos 10 34 sin 31.0  + = +  11(c) ( )10 34 sin 31.0 40+  = ( ) 40sin 30.963756 10 34  +  = 12.3 = 12(a) 12(b) 1 1xex− =− 1ln ln 1xex− =− ( )11 ln 12xx− = − ( ) ( )2 1 ln 1xx− = − ( )21yx=− y x 1 2 ( )ln 1yx=−

13 21 2 x dy edx − =− @ ( ) 24,e− 21 2 dy edx =− Equation of tangent: ( ) 22 1 ( 4)2y e e x− =− − − 2 2 21 22y e x e e=− − + 221 2y e x e=− − @ 0y= 2210 2 e x e=− − 1 12 x=− 2x=− ( 2,0)B = − Area = ( ) 3 22 4 1 d 2 2 x e x e − − −   3 22 4 2 x ee − − = − −  334 222 4 22e e e −−− −  = − − − −   ( ) 3 22222e e e − =− + − 3 2 22ee − =− 6.942795779= units 6.94= units