NYJC 2026 Prelim P2 Ans

2026 NYJC J2 H2 Mathematics Preliminary Examination 9758/2 solutions 1 of 17 Q1 Suggested Answers (a) 1 1 , 1 , 1nn ua u u n+ = = −  21 11u u a= − = − ( )32 1 1 1u u a a= − = − − = (b) The sequence is oscillating/alternating/periodic/cyclic/repeating/ (between a and 1 – a). (c) 101 1 49n n u = = ( ) ( )1 3 99 101 2 4 100... ... 49 51 50(1 ) 49 50 49 1 u u u u u u u aa a a + + + + + + + + = + − = += =−

2026 NYJC J2 H2 Mathematics Preliminary Examination 9758/2 solutions 2 of 17 Q2 Suggested Answers (a) Method 1: ( ) ( ) 2* * * 3i ii i3 i 3i 3 3 1 i3 z w z zz z zw a bz w a b = −= = =− = + = − + Method 2: ( )( ) ( ) ( )( ) ( ) 2*3i ii 3i i i i 1 ii3i i 3 1 i3 z w z a b a bw ab abw a b w a b = +−= − +− = = + −− = − + (b) O Im Q Re P

2026 NYJC J2 H2 Mathematics Preliminary Examination 9758/2 solutions 3 of 17 Q3 Suggested Answers (a) Given that 21 ,3V r h k== 2 3kh r = 2 2 2 2S rl S r l=  = Since 2 2 2 2 2 4 9and , kl r h h r = + = ( ) ( ) 2 2 2 2 2 2 4 2 2 2 2 4 2 4 2 24 2 9 9 (Shown) S r r h r r h krr r kr r     =+ =+  =+    =+  Alternative Method Since the slanted height 22 ,l r h=+ 2 2 6 2 42 91 9kl r r k rr = + = + 6 2 6 2 2 1 99S rl r r k r k rr  = = + = + ( ) 22 2 6 2 2 4 22 99 kS r k r rr   = + = +  (shown) (b) By implicit differentiation, 2 23 3 d 1824 dSkSr r r =−  2 23 3 d9 2 (*)dSkSr r r = − −−− At stationary points, d 0dSr = 2 23 3 920 kr r   −= since S > 0 62 1 2 6 29 9 2 rk kr = = Differentiating (*), 2 22 22 24 d d 27 6d dS S k Srr rr   + = +    At 1 2 69d , 0,2dkSr r== 2 2 2 2 24 d 27 60 d Sk rSrr  = +  since 2 0,r  4 0,r  2 0k  and S > 0

2026 NYJC J2 H2 Mathematics Preliminary Examination 9758/2 solutions 4 of 17 S is a minimum at 1 2 69 .2kr = (c) Minimum S = 99.8 cm2 r = 0 O r S

2026 NYJC J2 H2 Mathematics Preliminary Examination 9758/2 solutions 5 of 17 Q4 Suggested Answers (a)(i) (a)(ii) ( ) 1 fy x= (b)(i) ( )77g( ) g(5 ) g( 2 ) g 5 422 3 2 3g g( ) 3 cos22 1         + = − + −    = + = − +       =− x O Fig. 1.1 y x O Fig. 1.2 y

2026 NYJC J2 H2 Mathematics Preliminary Examination 9758/2 solutions 6 of 17 (b)(ii) ( )gyx= (b)(iii) ( )gyx=

2026 NYJC J2 H2 Mathematics Preliminary Examination 9758/2 solutions 7 of 17 Q5 Suggested Answers (a) Since every horizontal line yk= , k  intersects the graph of ( )fyx= at most once. f is one-to-one and 1f − exists. (b) Let 2 67y x x= − + ( ) 2 32yx= − − ( ) 2 32 32 xy xy − = + =  + Alternatively, 2 6 7 0x x y− + − = ( 6) 36 4(7 ) 322 yxy − −  − −= =  + Since 3, 3 2x x y = + + 1f ( ) 3 2xx− = + + (c) 3a=− For 27g: 3xx x ++  , g( 3)− is undefined. Thus 3.x− OR 3x− because it will make the denominator of 27g( )= 3xx x ++ to be zero. (d) 2 7 1g : 2 33xx xx+ =+++  1. Translation by 3 units.in the direction of the x-axis 2. Translation in the direction of the y-axis Order can also be 2 followed by 1. (e) ( )fR 2,= −   gD \ 3= −  or ( ) ( ), 3 3, −  −− Since −2 > −3, therefore fgRD , thus gf exists. y x (3, −2) y = f(x)

2026 NYJC J2 H2 Mathematics Preliminary Examination 9758/2 solutions 8 of 17 (f) Method 1: ( ) ( ) ( )( ) ( ) 2 g f 4 g 4 6 4 7 g1 = − + =− ( ) ( ) 2 1 7 5 1 3 2 −+== −+ Method 2: ( ) ( ) ( ) ( ) 2 2 2 g f g 6 7 2 6 7 7 6 7 3 x x x xx xx = − + − + + = − + + ( ) ( )( )( )( ) 2 2 2 4 6 4 7 7 5g f 4 24 6 4 7 3− + +== − + + Q6 Suggested Answers (a) Case 1: (4C, 1L, 1S) No of ways = 8 6 5 4 1 1 2100C C C  = Case 2: (3C, 2L, 1S) No of ways = 8 6 5 3 2 1 4200C C C= Case 3: (3C, 1L, 2S) No of ways = 8 6 5 3 1 2 3360C C C  = Total ways = 2100 4200 3360 9660+ + = (b) Total ways =  33 21 Arrange (unit of Slot in remaining RChoose and fix 2 Reds 3B) + 3Yat end 2! 4! 3! 2592CC    =   (c) ( ) ( ) 6 36 1 ! 3! 5Probability (0.35714)9 1 ! 14C−== −

2026 NYJC J2 H2 Mathematics Preliminary Examination 9758/2 solutions 9 of 17 Q7 Suggested Answers (a)(i) Since A and B are independent, P( ) P( ) P( )A B A B =  4 1 4 ( 1)P( ) P( ) 4 3 2 ( 4)( 3)( 2)r r r rA B WRR r r r r r r −− = =   = + + + + + + (a)(ii) 4P( ) 4A r= + 4 1 12 ( 1)P( ) 3 4 3 2 ( 4)( 3)( 2) WRR r r r rB r r r r r r −−=    =+ + + + + +  4 ( 1) 4 12 ( 1) ( 4)( 3)( 2) 4 ( 4)( 3)( 2) 121 4 4 12 8 r r r r r r r r r r r r r r −− =+ + + + + + + = + += = (b) 8 red and 4 white discs in the bag Possible values of X = 1,2,3,4,5 82P( 1) P( ) 12 3XR= = = = 4 8 8P( 2) P( ) 12 11 33X WR= = =  = 4 3 8 4P( 3) P( ) 12 11 10 55X WWR= = =   = 4 3 2 8 8P( 4) P( ) 12 11 10 9 495X WWWR= = =    = 4 3 2 1 8 1P( 5) P( ) 12 11 10 9 8 495X WWWWR= = =     = (c) P( 2 draws to get first red disc) P( 2) 2 8 10 3 33 11 X =  = + = P(exactly 1 person who takes 2 draws to get 1st red disc) 1 10 20 211 11 121  =   = Required probability P(Ben needs 3 draws | exactly 1 person takes 2 draws) P( 3 and exactly 1 person takes 2 draws) P(1 person takes 2 draws) P( 3)P(Aden takes 2 draws) 20 121 4 255 20 5 1 12 1 1 0 1 X X = ==  == ==

2026 NYJC J2 H2 Mathematics Preliminary Examination 9758/2 solutions 10 of 17 Q8 Suggested Answers (a) ( ) ( ) ( ) ( )E 0 0.4 1 4 0.6 1.8 2.4 3 1.8 0.2 Y p p p p = + + − = −= = (b) ( )~ B 10,0.6X ( ) ( )( ) ( ) ( ) ( ) P 3 7P 7 3 P3 P 7 P 2 1 P 2 0.82042 0.98771 0.83063 0.831 (3 s.f.) XXX X XX X   =   − = − =  = (c) ( )~ B 10,0.6X ( ) ( )E 10 0.6 6X == ( ) ( )( )Var 10 0.6 0.4 2.4X == ( ) ( )E 80 6 480T == ( ) ( )Var 80 2.4 192T == (d) Central Limit Theorem applied to sample total: Since X is not normally distributed, 1 80 ...T X X= + + is approximately normally distributed by Central Limit Theorem with large sample size (n = 80). Hence, ( )( )80 10 TW = will also be normally distributed. Alternatively (without reference to distribution of T) Central Limit Theorem applied to sample mean: Since sample size 800 is large, ( )( )80 10 TW = is approximately normally distributed by Central Limit Theorem. Here is it is implied that W is the average of 800 customers (over 80 days). (e) From (c) and (d), ( )~ N 480,192T approximately 800 TW = , ( ) 480E E 0.6 800 800TW = = = ( ) 2 192Var Var 0.0003 800 800TW = = = ( )~ N 0.6,0.0003W ( )P 0.63 0.041632 0.0416 (3 s.f.) W  =