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 Preliminary Exam 9758/1 4 of 16 24 24 2 8 ...2! 4! 1 ... 3 xxy y x x     = + +       = + + (c) ( ) 24 21 2 4 1 3 1sin 3 y x x x x x− + + When 1 2x= , 2 2 4 1 1 1 1 1sin 2 2 3 2 −     +           2 13 36 48   39 2  Percentage error of approximation 39 2 100%   − = 0.608% (3 s.f.)= Approximation is quite accurate as the value of x is close to zero and the series used terms up to x4.

2024 NYJC J2 H2 Mathematics Preliminary Exam 9758/1 5 of 16 Q4 Suggested Answers (a) ( )   f ( ) 3 cos sin cos = cos cos sin sin cos 3 --- (1) sin 1 --- (2) x x x R x R x x R R     = + = − + = = (1)2 + (2)2: ( ) 2 23 1 2R= + = ( ) 12 1: tan(1) 6 3  −== (b) 26 0 1 df ( ) xx       6 2 0 1 d 2cos 6 x x   =   −     6 2 0 1 sec d46 xx  =−   6 0 1 tan46 x   =−   1 tan 0 tan46  = − −   13 1243 == (c) 12 0 1 df (2 ) xx   12 0 1 d 2cos 2 6 x x  =  −    12 0 1 sec 2 d26 xx  =−   12 0 11 ln sec 2 tan 22 2 6 6 xx       = − + −            1 ln sec0 tan 0 ln sec tan4 6 6     = + − − + −        1 2 2ln1 ln4 33  = − + −  11ln4 3 =− 11ln 3 ln 348==

2024 NYJC J2 H2 Mathematics Preliminary Exam 9758/1 6 of 16 Q5 Suggested Answers (a) Method 1: ( ) ( ) ( ) ( ) 2 2 22 2 41 d44 2 2 4 9 d44 2 4 92 d d 44 2 92ln 4 4 2 94ln 2 2 x xxx x xxx x xxxx x x x C x xC x − ++ +−= ++ +=− ++ + = + + + + + = + + + +      Method 2: Use of partial fractions ( ) ( ) 222 4 1 4 1 4 4 2 22 x x A B x x x xx −− = = ++ + + ++ ( )4 1 2x A x B− = + + Solving, A = 4, B = −9 ( ) 22 4 1 4 9 dd4 4 2 2 x xxx x x x − =−+ + + +    ( ) 94ln 2 2xC x= + + + + (b) 1 2 0 41 d44 x xxx − ++  1 14 22 10 4 4 1 4 1 dd4 4 4 4 xx xxx x x x −−=− + + + + +  ( ) ( ) 1 14 10 4 994ln 2 4ln 2 22xx xx    =− + + + + +    ++    ( )9 9 94ln 4 4ln 2 4ln 3 3 4ln 44 2 4         =− + − + + + − +                 9 9 14ln 4ln 2 4ln 3 4ln4 4 2=− + + − − 2 3 14ln 99 2 44 32 14ln 27 2 =−   =−

2024 NYJC J2 H2 Mathematics Preliminary Exam 9758/1 7 of 16 Q6 Suggested Answers (a) ( ) ( ) 2 2 2 2 2 5 13 5 5 13 10 25 13 25 3 Since 0, 3 25 3 25 a d a d a a d a d a a d a ad d a ad d ad d a d ad ++ = + + = + + + = + = = = (b)(i) Sum to infinity = 21 0.5 b b=− (ii) ( ) ( ) 1 0.5 2 1 0.51 0.5 n n n b Sb − = = −− ( ) ( ) 22 2 2 0.004 since as all the terms are positiv e 2 1 0.5 2 1 0.5 0.004 0.5 0.5 0.002 since 0 n n n n nn nn S S b S S b b b b −   − − −  −   20.5 0.5 0.002 0nn− −  Using GC, n 20.5 0.5 0.002nn−− 8 0.00189 > 0 9 −5.0710−5 < 0 Smallest n = 9

2024 NYJC J2 H2 Mathematics Preliminary Exam 9758/1 8 of 16 Q7 Suggested Answers BM = sin and OM = cos Length of height, AM 1 cos =+ Length of base, BC 2sin= Area of triangle, ( )1 1 cos 2sin2S =  +  1sin sin 22=+ Alternatively, Area of triangle, ( ) ( ) ( ) 22112 1 sin 1 sin 222S   =  − +  1sin sin 22=+ d cos cos 2d S  =+ For maximum area, d 0d S  = cos cos 2 0+= ( )( ) 2cos 2cos 1 0 2cos 1 cos 1 0 1cos or cos 1 (rejected since is acute)2      + − = − + = = =− Therefore, 3  = 2 2 2 2 d sin 2sin 2d d 3 3When , 0 3 d 2 S S    =− − = =−  Thus S is maximum when 3  = Since 2 , 3BOC BAC  =  = = ( at centre = 2  at circumference) As the triangle is isosceles, all the angles in the triangle are 3  . Therefore, maximum area occurs when triangle ABC is equilateral, ie, when 3  = Maximum area 1 2 3sin sin 33 2 3 4 = + =   1 B C A M O 1

2024 NYJC J2 H2 Mathematics Preliminary Exam 9758/1 9 of 16 Q8 Suggested Answers (a) (b) (c) y-intercept 1 0b a− OR y-intercept 1 0b a− (give one of the diagrams will do) 1y ax b xa= + + − 121 2x x= + + − ( ) 222y r x a a b=+ − − + + ( ) 2 16 2 5x= − − + (upper semicircle only) 2a= and 1b= (and r = 4) Using GC, the x-coordinates of the points of intersection are 2.29 and 3.52

2024 NYJC J2 H2 Mathematics Preliminary Exam 9758/1 10 of 16 For ( ) 212 1 16 2 5 2xx x+ +  − − +− (upper semicircle only) 2 2.29x or 3.52 6x (the circle is only defined for [−2, 6]) Q9 Suggested Answers (a) Coordinates of A are (1,0) (b) 2 lny x x= ( ) 2d1 2 ln 2ln 1d y x x x x xxx = + = +  ( )2ln 1 0xx += x = 0 or 1ln 2x=− Since x > 0, 1 2ex − = Coordinates of B are 1 12 1e , e 2 − − −  (c) Method 1: Area required ( )( ) 1