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  =+