2026 GDLSS P2 MS

Marker’s Report (2026 4Exp Amath Prelim) Paper 2 Q Descriptor Remarks 1i ( ) 0.14 0.14 0 1 50 1 100 t kP e k e k − − = + = + = 0.14 0.14(9) 100 1 100 1 77.9026 77 tP e P e P − − = + = + = ≈ 1ii 0.14 100 1 tP e−= + When t becomes very large, 0.14te− will approach zero. Therefore P approaches 100. This means that the population of cuckoos will not exceed 100. 2 ( ) ( ) ( ) 2 2 2 2 9 2 52 3 9 5 2 230 4 10 4 10 x k x kx x kx x x kx x kx x x kx + + −> + − + + − − −+> − + +> + − +> ( ) ( )( ) 2 2 2 2 0 and 4 0 4 41 1 0 16 8 4 0 8 12 0 26 a b ac k kk kk k > −< −− < − + −< −+< << 3 ( ) ( ) ( ) ( ) ( )( ) 3 93 3 3 33 3 2 33 3 3 2 2 2 log 18 4log log 2 4 2 4 loglog 18 log 2 4 log 9log 9 log 18 log log 2 4 log 9 18 9 2 4 18 18 36 20 2 10 2 or 1 (rej) xx x x xx xx xx xx xx xx + = ++ + = ++ + = ++ = + = + −−= − += = =−

4a 2 21 2 21 21 3 3 3 21 d 4ed d 4ed d 4(e ) d2 d2sub 2, 4 de 22(e ) 4 e 4 d 2(e ) 4d x x x x y x y dxx y cx yx x c c y x −+ −+ −+ − −+ = = = + − = = −− − += − − =− ∴= − − ∫ 4b 21 21 21 e0 2(e ) 0 2(e ) 4 4 d 4d d 0, thus it has no stationary point.d x x x y x y x −+ −+ −+ > −< − − <− <− ∴≠ 4c 21 21 21 3 33 21 d 2(e ) 4d 2(e ) 4 2(e ) 42 when 2, e , e (e ) 8 8 e 48 x x x x y x y dx y xc xy c c yx −+ −+ −+ − −− −+ = −− = −− −= −+− = = ∴ = −+ = ∴= − + ∫

5a 23 33 2 12 1 2 3 2 23 2 2 132 323 2 1 9 3 2 6 2 23 3 3223 23 23 23 19,32ab − −− −+ ××= ×  × × = × = × ∴= = 5b ( ) ( ) 12 22(3 ) 4(3 ) 3 2(3)(3 ) 4(3 ) 3 (3 ) 6(3 ) 4(3 ) 9(3 ) 3649 3 11 n nn n n n nnn n n ++ −+ = −+ =−+ = −+ = Since n is a positive integer, 3n is a multiple of 3. There are 2 prime factors, 3 and 11. Thus, the largest prime factor of 122(3 ) 4(3 ) 3n nn++ −+ is 11. 6

6a 22 22 22 2 22 4 6 12 0 4 6 12 0 ( 2) 4 ( 3) 9 12 0 ( 2) ( 3) 5 ( 2, 3) 5 xy xy x xy y xy xy Ar +++−= +++−= + −+ + −− = ++ += −− = 6b (1,1)Q 6c 71 4 51 3 3 4 : 371 (1) 44 37 44 7 , 0, 3 7( ,0)3 PQ QR m m Eqn QR y mx c cc yx At R y x R −−= =−− ∴= − = + = − +⇒= = −+ = ∴= ∴ 6di Intersection of perpendicular bisector of AQ and Intersection of perpendicular bisector of QR is the centre B. (accept other methods e.g. distance between the points, midpoint of AR – reasoning must be shown)

1 Perpendicular bisector of : 1Midpoint = , 1 2 43, gradient of bisector = 34 3 1 111 () 42 8 3 11 48 Perpendicular bisector of : 51Midpoint = , 32 3 , gradient of4 AQ QR AQ AQ m cc yx QR AQ m −− = ∴⊥− ∴− =− − + ⇒ =− ∴= − −   = −∴ 2 12 4 bisector = 3 1 4 5 31()2 3 3 18 4 31 3 18 13Solve and ,62 cc yx y yB ⊥ ∴= +⇒= − ∴= − ⇒= − 6dii 2217 3( ) ( 0)63 2 125 18 25 5 29 5 5 5 10 632 r r r r = − +− − = ×= × = = 6e 221 3 125( )( )6 2 18xy− ++ = Also accept 2218 6 18 54 84 0xxy y−+ + −= 7a ( ) ( )( )22f6 x x x Ax Bx C= +− + + By observation, ( ) ( )( ) 22f 61x x x x Bx= +− + +

Given that ( )f 2 36−= − ( )( )36 4 2 5 925 42 2 B B B B −= −−+ = −+ =− =− ( ) ( )( ) 22 43 2 f 6 21 7 13 6 x xx x x xx x x = +− − + = −− + − 7b ( ) ( ) ( ) ( ) ( ) 32 32 3 7 60 f 2 2 32 72 6 0 2 is a factor of f xxx xx − − += −= − −− −−+ = ∴+ ( ) ( ) 32 2 3 7 60 20 xxx x ax bx c − − += + ++= By comparing coefficient, 1, 26 3 a c c = = = Sub 1x= ( ) ( ) ( )( ) ( ) 23 1 2 1 (1) 1 3 3 34 31 23 3 15 5 b b b b b −= + + + −= + −= + =− =− ( ) ( ) 22 530x xx+ −+= ( ) ( ) ( )( ) 2 2 5 30 5 5 41 3 2 5 13 2 xx x x − += −− ± − −= ±= 5 132, 2x ±∴= −

8a 36 3 12 , 36 , 0 3 12 6 12 2 2 + − = = = + − = − = t t v v t when c t t v t a Generally well done. Most students were able to integrate correctly and include +c. However, some did not recognise that v=36 when t=0, resulting in an incorrect velocity expression. 8b s m v t When tdtdvWhen tdtdv / 48 36 ) 4 ( 3 ) 2 ( 12 2 2 , 0 6 12 = + − = = = = − = Mixed performance. Some students did not recognize that maximum velocity occurs when the acceleration is zero, 0dvdt = . Others were able to find the maximum by completing the square. 8c 6 ), ( 2 0 ) 2 )( 6 ( 0 12 4 , 0 2 = − = = + − = − − = t NA t t t t t v When Generally well done. The majority of students recognized that at instantaneous rest, v=0. However, some students forgot to reject the negative value of t. 8d 23 23 23 23 6 36 0, 0, 0 6 36 0, 0 6, 6(6) (6) 36(6) 216 10, 6(10) (10) 36(10) 40 216 216 40 472 472 47.2 /10 s t t tc When s t c s tt t When t s When t s s When t s s Total dist m Average Speed m s = −+ + = = = = −+ = = = = − + ⇒= = = − + ⇒= − = + += = = Most students understood that average speed requires finding the total distance travelled and were able to integrate to obtain the displacement s. However, students still struggled to identify the key timings needed to determine the particle’s motion and hence the total distance travelled. In this question, they only needed to find the times when the particle was at instantaneous rest, t=6s and t=10s, and determine whether there was a change in direction at these points. This would allow them to calculate the total distance travelled correctly.

9a 44 2 22 2 2 22 2 2 sec tan (sec tan )(sec tan ) 1 sin(1)( )cos cos 1 sin (shown)c os LHS x x x xx x x xx x x = − = −+ = + += Generally well done. Most students were able to prove the result using either the LHS or RHS, with a variety of valid approaches. 9b 44 2 2 22 2 2 3 5 1 3 5 sec (2 70 ) tan (2 70 ) 4 1 sin (2 70 ) 4cos (2 70 ) 1 sin (2 70 ) 4(1 sin (2 70 )) 5sin (2 70 ) 3 sin (2 70 ) sin 50.768 0 180 70 2 70 290 (2 70 ) , 180 , 180 , 360 , (2 xx x x xx x x x x x x α α α α αα − −° − −° = + −° =−° + −° = − −° − °= − °= ± = = ° °≤ ≤ ° − °≤ − °≤ ° − ° = °− °+ °− − 70 ) 50.768 , 129.232, 230.768 , 309.232 (rej), 50.768 60.4 ,99.6 ,150.4 ,9.6x − °= ° ° °− ° = ° ° °° Generally ok. Most students were able to use the “hence” from the previous part and solve the equation correctly. However, some students forgot about the plus/minus when taking the square root, resulting in them not considering all four quadrants. Many also missed the first solution, 50.768° which lies in the fourth quadrant. 10ai ( )ln ln ln ln ln ln ln bt bt bt y ae y ae y ae y a bt = = = + = + Plot ln y against t t (years) 1 2 4 3 5 y (units) 950.6 480.6 30.8 99.5 9.8 ln y 6.86 6.18 3.43 4.60 2.28 Not very well done. Many students lost accuracy marks due to rounding of ln y. Students also need to clearly state what is being plotted against what, rather than simply presenting the equations without explanation. 10aii The incorrect reading is 480.6y= . ln 5.7 298.9 y y = = Students need to identify the incorrect reading clearly by writing and labelling it as the “incorrect reading”. Simply

circling the point on the graph is not sufficient. 10aiii gradient 6 2.8=1.75 4.5 1.16 b= − − =− Generally ok. 10b Method 1 2px qy x+= 2 1 qy px x yp xx qq = −+ = −+ Plot yx on the vertical axis and x on the horizontal axis. Vertical intercept = 1q thus we can read off the vertical incept from the graph and thus find the value of q. Gradient = pq− , we can calculate the value of the gradient and equate to pq− and thus find the value of p. Poorly done overall. Most students were unable to rearrange the given relationship correctly into the form for a straight-line graph. Students need to ensure that the constant term does not contain any variables such as x or y. Method 2 2 2 2 1 11 px qy x ypq xx yp x qx q += += = − Plot 2 y x on the vertical axis and 1x on the horizontal axis. Gradient = 1q , we can calculate the value of the gradient and find the value of q. Vertical intercept = pq− we can read off the vertical incept from the graph and equate the value of q thus find the value of p.

11a 2 2 2 2 2 2 sin (cos ) sin ( ) (cos sin ) ( sin cos ) (cos sin ) (2cos ) 22 2 cos 2 (cos sin ) 2 sin 2 cos 2 cos 2 sin 2 sin 0 (shown) x xx x xx x xx x xxx x ye x dy e x xedx dy exxdx dy e x xe x xdx dy exdx d y dy ydx dx e x