2024 NYJC Prelim Exam H2MATH Paper 1 (Solutions)

2024 NYJC J2 H2 Mathematics Preliminary Exam 9758/1 1 of 16 Q1 Suggested Answers Make w the subject from i3wv+ + = : i3wv= − − Sub into 2 i 2 0vw− + = ( ) 2 i i 3 2 0vv− − − + = 2 i 1 3i 0vv− + + = Using quadratic formula, ( )i 1 4 1 3i 2v  − − += i 5 12i 2  − −= ( ) use GC to find 5 12ii 2 3i 2 − −−= ( )i 2 3i 2 +−= or ( )i 2 3i 2 −− 1i=− or 1 2i−+ 2 2iw=− − or 4i−+ Alternative (1) Sub i3vw= + + into 2 i 2 0vw− + = ( ) 2 i 3 i 2 0ww+ + − + = ( ) ( ) 22 2 i 3 i 3 i 2 0w w w+ + + + − + = Using quadratic formula, ( ) ( ) ( ) 2 6 i 6 i 4 10 6i 2w − +  + − += ( )6 i 5 12i 2 − +  − −= ( ) ( ) use GC to find 5 12i6 i 2 3i 2 − +  −−−= 2 2i, 4 iw=− − − + Using i3vw= + + We have 1 i, 1 2iv= − − + Alternative (2) From the equation i3wv+ + = Multiply by i , we have ( )i 1 3i i 1wv− + = ( ) 2 i 2 0 2vw− + = (1)+(2) : 2 i 1 3i 0vv− + + = Using quadratic formula,

2024 NYJC J2 H2 Mathematics Preliminary Exam 9758/1 2 of 16 ( )i 1 4 1 3i 2v  − − += i 5 12i 2  − −= ( )i 2 3i 2 −= use GC to find 5 12i−− ( )i 2 3i 2 +−= or ( )i 2 3i 2 −− 1i=− or 1 2i−+ 2 2iw=− − or 4i−+ Q2 Suggested Answers (a) dsin cos d xx a a a a   = −  = − dcos sin d yy a a a  = −  = ( ) 2 d sin d 1 cos 2sin cos22 1 1 2sin 2 cos 2 = cot 2sin 2 ya xa        = − =  −−  = (b) When 3  = , 31 , 3 2 2x a a y a= − = ( ) 1 3 d 1 1and cot 3d6 tan 6 y x  = = = = Equation of tangent is 13 32 3 2y a x a a − = − +  332 3y x a a  = − + When 3xa = , 33 2 233y a x a a a = − + = Therefore, the tangent passes through 1 , 23 aa  Or When 2,ya= 3 31 32 3 2 33 3 a a x a a x a  = − +  = = Therefore, the tangent passes through 1 , 23 aa 

2024 NYJC J2 H2 Mathematics Preliminary Exam 9758/1 3 of 16 Q3 Suggested Answers (a) ( ) 21sinyx −= Differentiate wrt x: ( ) 1 2 21 d1 2 sind 1 d1 2sind y xx x yxx x − − = − −= Differentiate wrt x: ( ) 2 2 2 22 2 2 2 d d 21 dd 11 dd12 dd y x yx xx xx yyxx xx − − = −− − − = Alternatively, 21d1 2sind yxx x −−= ( ) ( ) 2 221 2 2 d1 2sind d14 d yxx x yxy x − −= −=  Differentiate wrt x : ( ) ( ) ( ) 22 2 2 2 2 2 2 2 2 d d d d1 2 2 4 d d d d dd12 dd dd1 2 0 dd y y y yxx x x x x yyxx xx yyxx xx −  − =  − − = − − − = (b) Differentiate wrt x : ( ) 3 2 2 2 3 2 2 d d d d1 2 0 d d d d y y y yx x xx x x x − − − + =  ( ) 32 2 32 d d d1 3 0 d d d y y yxx x x x− − − = Differentiate wrt x : ( ) ( ) 4 3 3 2 2 2 4 3 3 2 2 4 3 2 2 4 3 2 d d d d d1 2 3 0d d d d d d d d1 5 4 0d d d y y y y yx x x x x x x x y y yxx x x x − − − + − =  − − − = When 2 3 4 2 3 4 d d d d0, 0, 0, 2, 0 and 8d d d d y y y yxy x x x x= = = = = =

2024 NYJC J2 H2 Mathematics P