2024 NYJC Prelim Exam H2MATH Paper 2 (Solutions)

2024 NYJC J2 H2 Mathematics Preliminary Exam 9758/2 1 of 14 Q1 Suggested Answers (a) Area of loop = ( )1 3 d2 x x x   −+ α = −3,  = 1 (b) Area of loop ( ) ( ) ( ) 1 3 2 2 0 2 24 0 2 35 0 2 2 1 3 d 2 4 2 d 4 4 d 414 35 256 units15 x x x u u u u u u u uu − = − + = −  =− =−  =    2 3 3 d21 d ux ux uu x =+ =+ = or 3 d1 d 23 d1 d2 ux u x x u xu =+ = + = When x = −3, u = 0 When x = 1, u = 2

2024 NYJC J2 H2 Mathematics Preliminary Exam 9758/2 2 of 14 Q2 Suggested Answers (a) Vector equation =r a where 0 1 The equation gives the position vector of points on the line segment OA. (not line OA) (b) Vector equation ( ) − =r a b 0  r is parallel to AB ( ), kk= −  r a b The equation gives the position vector of points on the line passing through O parallel to AB. (c) ( ) 0−=r r a 0OR AR =  OR AR⊥ Points on a sphere (or circle) with OA as a diameter. (d) ( ) ( )−  − =r a r b 0 ( ) ( ) is parallel to−−r a r b ( ) ( )− = −r a r b where   1 ( )1 − = −r a b ( ) 1 1 1 1        −−= = +    − − −    abr a b , where   1 since a  b Alternatively, ( ) ( )−  − =r a r b 0 AR is parallel to BR Since R is a common point, A, B and R are colinear i.e. R lies on the line AB ( )= + −r a b a , where  R O A r a r-a R O A Another possible R

2024 NYJC J2 H2 Mathematics Preliminary Exam 9758/2 3 of 14 Q3 Suggested Answers (a) 1 1 3iz =− + , 2 1iz =− − , 1i 12 3 2ez −= (b) 2i 3 1 2ez  = 3i 4 2 2ez −= 1i 12 3 2ez −= Method 1: Properties of modulus & argument ( ) ( )( ) 11 2 2 2 23 23 21 222 zz zz zz = = = ( ) 1 1 2 32 23 arg arg arg 2arg 2 3 1 23 4 12 19 12 z z z z zz      = − −    = − − − −       = ( ) 1 2 23 19 5arg 2 12 12 z zz      = − =− ( ) 5i 121 2 23 1 e 2 z zz −= (b) Method 2: Using exponential form ( ) 2i 3 1 22 31ii23 4 12 19i 12 2e 2e 2e 1 e 2 z zz         −−         =           = Im Re Show modulus (length) and argument (angle), if possible.

2024 NYJC J2 H2 Mathematics Preliminary Exam 9758/2 4 of 14 19i2121 e 2  −= 5i 121 e 2 −= (c) Let i 4 ezr = Method 1: Properties of modulus & argument ( ) ( ) 14 2 23 1 42 23 1 1 2 zz zz z z zz r = = = ( ) ( ) 1 4 1 422 2 3 2 3 arg arg argz z z z z z z z     =+          5 12=− + Since ( ) 14 2* 23 zz zz is purely real, 5 ,12 kk  − + =  5 12k  =+ ( )57 or 12 12    = − −   4 552 cos isin12 12z  = +  or 772 cos isin12 12    − + −        (c) Method 2: Using exponential form ( ) ( ) 5i i1214 2 23 5i 12 1 ee 2 e 2 zz r zz r    − −+  =  = 12 2 r r=  = Since ( ) 14 2* 23 zz zz is purely real, 5 ,12 kk  − + =  5 12k  =+ ( )57 or 12 12    = − −   4 552 cos isin12 12z  = +  or 772 cos isin12 12    − + −       

2024 NYJC J2 H2 Mathematics Preliminary Exam 9758/2 5 of 14 Q4 Suggested Answers (a) 2 1 4 8 3 2 3 2 1 AB r r r r =+− + − − ( ) ( )2 1 2 3 1A r B r− + − = When 11,22rB= =− When 31,22rA== 2 1 1 1 1 4 8 3 2 2 3 2 1r r r r =− − + − −  (b) 33 2 22 1 1 1 4 8 3 2 3 2 1 11 3 11 35 11 57 11 6 5 6 3 11 6 3 6 1 1 1 1 2 1 3 1 2 6 1 1 2 61 nn rr r r r r nn nn n nn ==     =−   − + − −     −  +−  +−=    +− −− +− −− −   = − =   −−     (c) 3 2 2 1 1 1As , 0 and so 6 1 4 8 3 2 n r n n r r = → → → − − +  2 2 11 4 8 3 2r rr  = =  −+ (d) ( )( ) ( )( ) 33 2 2 2 1 2 2 1 1 1 4 8 3 4 8 3 4 8 3 1 1 1 1112 6 1 2 2 1 1 1 1 2 2 1 6 1 4 2 2 1 6 1 2 2 1 6 1 nn r n r r n r r r r r r nn nn n nn n nn = + = =       =−     − + − + − +         = − − −   −−    =−  −− = −− = −−   

2024 NYJC J2 H2 Mathematics Preliminary Exam 9758/2 6 of 14 Q5 Suggested Answers (a) Volume of wok ( ) 2 16 256 d q yy − =− 3 16 256 3 q yy − =−  3 4096256 409633 qq  = − − − +  3 8192256 33 qq= − + 3 8192256 330033 qq − + = Using GC, q = −7.01245634 Depth of wok = −7.01245634 – (−16) = 9 cm (correct to nearest integer) (b) Method 1 (direct integration): Volume of flat frying pan ( ) 0 2256 d r yy=− 03 256 3 r yy =−  3 0 256 3 rr  = − −  3 2563 r r=−  3 125 43 66 4r r −= 3 256 14643 r r−= Using GC, r = −6 (b) Method 2 (using result from part (a)): Volume of flat frying pan = Volume of hemisphere − 3 8192256 33 rr −+ ( ) 3 3 3 2 819216 2563 3 3 2563 rr r r   = − − +  =−  3 125 43 66 4r r −= 3 256 14643 r r−= Using GC, r = −6

2024 NYJC J2 H2 Mathematics Preliminary Exam 9758/2 7 of 14 (c) Let time taken to fill the wok to full capacity be T seconds Method 1: d3 d 55 V tt = 3dd 55V t t= 23 110 tVC=+ When t = 0, V = 0, C = 0 When t = T, V = 3.3, 233.3 110 T= T = 11 Method 2: d3 d 55 V tt = 3.3 0 0 3dd 55 T V t t=   2 3.3 0 0 3 55 2 T tV =  233.3 55 2 T=  T = 11

2024 NYJC J2 H2 Mathematics Preliminary Exam 9758/2 8 of 14 Q6 Suggested Answers (a) (Treat as there is an ‘invisible’ car occupying a lot) Number of arrangements without restriction = 10! 3628800= (b) Number of arrangements with BB and RR together = 8! 2! 2! 161280  = (c) Method 1 Number of arrangements with B1B2 together = 9! 2! 725760= Similarly, number of arrangements with R1R2 together 9! 2! 725760= Number of arrangements with at least 2 adjacent cars = 725760 725760 161280 1290240+ − = By complement method, required number of arrangements is 3628800 1290240 2338560= − = (c) Method 2 Number of arrangements with B1B2 together and R1R2 not together = 8 27! 2! C 2! 564480   = Similarly, number of arrangements with R1R2 together and B1B2 not together 564480= By complement, required number of arrangements is 3628800 161280 564480 564480 2338560− − − = (c) Method 3 Number of arrangements with red cars separated (no restrictions on blue cars) = 9 28! C 2! 2903040  = Number of arrangements with red cars separated and blue cars together = 8 27! 2! C 2! 564480   = By complement, required number of arrangements is 2903040 564480 2338560−=

2024 NYJC J2 H2 Mathematics Preliminary Exam 9758/2 9 of 14 Q7 Suggested Answers (a) 18 12 18 12 10 1 9 30 30 10 10 18! 12! 18! 12! · ·!(18 0 )! ( 2)!(10 )! ( P( ) P( 1) ( 1)!(17 )! ( 3)!(9 )! 1)!(17 )!(9 )!( 3)! !(18 )!(10 )!( 2)! ( 1)( 3) (18 )(1 s s s s S s S s C C C C CC s s s s s s s ss s s s s s s s ss s − + − =  = + −  − + − + − + + − +  − − + + +  − − )s− 22 4 3 28 180 32 177 5.53 s s s s s s + +  − +   Thus s = 6. (b) Outcome Table Absolute Difference No. on Square 1 2 3 4 5 6 No. on Triangle 1 0 1 2 3 4 5 2 1 0 1 2 3 4 3 2 1 0 1 2 3 4 3 2 1 0 1 2 Probability Distribution x 0 1 2 3 4 5 P(X = x) 4 24 7 24 6 24 4 24 2 24 1 24 (c) 1 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3 32 P( 2 | 3) P( 2, 0, 1) 2! 3!P( 1, 1, 1) P( 1, 2, 0) 3! P( 3, 0, 0) 2! 6 4 7 2!24 24 24 7 7 6 4 4 4 3! 3!24 24 24 24 24 24 2! 336 or 0.218 (1543 X X X X X X X X X X X X X X X X = + + = = = = = = = = + = = =  + = = =        =       + ++             = to 3 s.f.)

2024 NYJC J2 H2 Mathematics Preliminary Exam 9758/2 10 of 14 Q8 Suggested Answers (a) Let L be the length of a randomly chosen rectangular cotton fabric. ( ) 2~ N 24, 1.5L ( )P 23.5 0.36944 (5 s.f.) 0.369 (3 s.f.) L= = (b) Let B be the breadth of a randomly chosen rectangular cotton fabric. ( ) 2~ N 20, 1.2B Perime