SAJC 2026 Prelim P2 Ans

1 St Andrew’s Junior College H2 Maths Prelim Exam P2 Solutions Section A 1 (a) 2 2 2 3 2 6 2 4 2 2 4 2 1 (1 sin 2 ) 1 1 sin 2 Differentiate with respect to d2 (1 sin 2 ) 2 cos 2 0d dsince 0, (1 sin 2 ) cos 2d d 1 cos 2d d cos 2 (shown)d d cos 2 (1 sin 2 )d (1 sin 2 ) ( y y x x x yy x y xx yy x y x x y y xx y y y xx y y x y y xx y x y                           4 1 4 4 2 2 1 sin 2 ) (1 sin 2 ) 1 1 1 2 (shown) x y x y y y y                 Alternative mtd:   1 2 3 2 31 32 2 23 6 2 6 2 2 2 1 (1 sin 2 ) 1 sin 2 d 1 (1 sin 2 ) ( 2cos 2 )d 2 (1 sin 2 ) cos 2 cos 2 (shown) d cos 2 cos 2 (1 sin 2 ) (1)d 1 1From 1 sin 2 1 sin 2 1 11 sin 2 sin 2 1 (2) Subs y x x y x xx x x y x y y x y x y xx y x yx x x y y                                       22 6 6 4 2 2 2 4 titute (2) into (1) d 1 2 1 1 1 1 1 2 (shown)dy y y y yx y y y                           

2 (b) 2 4 2d 2dy y yx      Differentiate with respect to x, 2 3 2 2 3 2 2 3 2 d d d d2 8 2d d d d d d d d 4d d d d d 0d d 4d y y y y y yx x x x y y y y y yx x x x y x y y yx                Differentiate with respect to x, 3 2 3 d d d 12d d dy y y yx x x  When 0x , 1 1 1 sin 0 y   3d (1) cos 2(0) 1dyx   2 3 2 d 4(1) 1 3d yx    3 2 3 d 12(1) (1) 1 11d yx    2 2 3 3 2 3 0 0 0 2 3 2 3 d d d1 ... d d 2! d 3! 3 111 1 ... 2! 3! 3 111 ... 2 6 x x x y y x y xy x x x x x x x x x x                   (c)   2 2 2 2 2 3 2 2 e e e 1 sin 2 1 sin 2 3 11e 1 ... 1 ... 2 2 6 e 1 2 3 ... x x x x xx x x x x x                         (d)   20.1 0 0.12 2 0 0.12 2 3 0 e d 1 sin 2 e 1 2 3 d e 0.8202(to 4 d.p) x x x x x x x x x             

3 2(a) Scheme A : k, k + 0.35k , k + 2(0.35k), … Total salary(in $) in the first n years         2 1 0.352 0.35 1.652 7 3340 n k n k nk n nk n        (b) Scheme B : 2, 1 , 1 ,...100 100r rk k k           Let 1 100rx      4kx  279841 ----- (1) 2 3 4 2 3 4 4 2 3 (2) 918 918781 (2) 27984 781Since 0, : (1) 1 1 1 918781 1 23From GC, . 20 231 100 1 279841 20 15 kx kx kx kx kx kx kx kxx kx x x x x r r                    (c) Under Scheme B, total salary(in $) in the first n years  1.15 1 1.15 1 nk    For Scheme B to exceed Scheme A,

4         1.15 1 1.15 1 1.15 1 Since 7 3340 7 33 40 0, 1.15 1 n n nk n n k k n             1 2 1.15 1 1.15 3340 1 7 n n Y Y n      n   1 1.15 1 1.15 1 n Y     2 7 3340Y n n 16 55.717 58 17 65.075 64.6 18 75.836 71.55 OR From GC, for 1 2Y Y , least n = 17. 3(a)   2 d d 2 y b ax x   At stationary points, d 0dyx  , i.e.,   2 02ba x  Solving, 2 bx a  .

5 Either 0a and 0b or 0a and 0b Or ab > 0 Or a and b have the same sign. Or 0ba  Alternative (Discriminant mtd)     2 2 2 0 2 2 0 4 4 0 ba x a x b ax ax a b           For two stationary points,     2 4 4 4 0 0 a a a b ab      (b) At x -intercept, 02bax x  2 2 0ax ax b   Since     2 discriminant 2 4 4 0a ab a a b      when 0b a  . Thus, C does not have any x -intercepts. (c) (d)  2 2P a , Substituting  2 2P a , into L ,  2 2 2 2RHS m a m a LHS     Thus,  2 2P a , lies on L .

6 Observe that the gradient of L varies for different values of m and when m a , L coincides with the oblique asymptote of C . Thus, L does not intersect C when m a . 4(a) (b) 2d d 3 cos sin 2 sin cosd dx y a t t a t tt t  2 d d 2 sin cos 2d dd 3 cos sin 3cos d y y a t tt xx a t t t t    (c) Gradient of tangent at point P 2 23 2 2 2 3cos 3 1cos 2 t t    Basic angle 1 1cos 42         4t  At point P, 3 3 1cos 4 2 2 2 ax a a              2 2 1sin 4 2 2 ay a a              2, ,2 4 22 2a a a aP P             O x y

7 Gradient of normal 1 3 2 2 2 2 3    Equation of normal at point P, 3 2 2 2 2 2 3 3 3 2 1 2 4 82 2 2 2 2 2 a ay x a ay x y x a                 (d) When 0y , 5 2 2 06 3a x  5 5 3 5 5 26 6 82 2 2 2 4 2 3 a x a a a         Coordinates of point Q is 5 5 2,0 ,0 84 2 a a            When 0y , 3 2 1 04 8x a  1 1 4 1 28 8 123 2 3 2 6 2 4 a x a a a          Coordinates of point R is 2 ,012 a     5 2 2 5 1 17 2 28 12 8 12 24QR a a a a                 Area of  PQR 2 2 1 17 2 2 2 24 17 2 units96 a a a             Alternative mtd: (exclude this in students’ copy)

8 Area of  PQR 2 2 2 5 2 2 2 1 4 8 12 4 2 0 02 2 1 2 5 2 2 12 2 8 2 17 2 units96 a a a a a a a a a a a                              Section B 5 Let  P B x Since , are independentB C ,  P 5xB C        1P ' P P 5 5 xC B C C B             P ' ' P P P 1 10 5 4 1 5 10 B A C B A B C B xx x             B A C

9           P ' P ' ' P ' ' ' 1 21 15 5 4 21 5 5 2 4 5 5 A B C B A B C x x x                   Alternative:   3 1 4 1 1 1 1P ' 5 10 5 10 5 5 5 2 4 5 5 A B x x x x              2nd part : 1 10 1 0 1 (1)5 5 10 1 0 5 (2)5 4 1 1 11 10 1 1 (3)5 10 8 8 8 2 4 10 1 0 (4)5 5 2 x x x x x x x x x                               Taking intersections of (1), (2), (3), (4), 1 1 8 2 x   Maximum value of  P B C = 1 1 15 2 10      Minimum value of  P B C = 1 1 15 8 40     Alternative Solution B A C

10 Let  P B C x  Since , are independentB C ,       1P P P 5 5B B C C x x            1P ' P P 5C B C C B x                P ' P ' ' P ' ' ' 1 21 1 55 5 4 21 55 5 2 45 A B C B A B C x x x                             P ' P ' ' P ' ' ' 1 21 15 5 4 21 5 5 2 4 5 5 A B C B A B C x x x                   6 (a) Number of arrangments = 7 6 (6 1)! 840C    (b) Step 1 : Choose 4 other people from 1 girl(excluding the 2 particular girls) and 4 boys to be seated Step 2 : Arrange the 4 people Step 3 : Slot in and arrange the 2 particular girls Number of arrangments = 5 4 4 2 4 other people 2particular girls (4 1)! 2! 360C C          Alternative: Number of arrangements = Number of ways where there are no restrictions – Number of ways where the 2 particular girls are seated together – Number of ways where only 1 of the 2 particular girls is seated = 840 – 5((5-1)!)(2!) – (2)((6-1)!) = 360