NYJC 9758 2025 prelim P1 Solution

2025 NYJC J2 H2 Mathematics Preliminary Exam 9758/1 Marking Guide 1 of 16 Q1 Suggested Answers Since C has an oblique asymptote with gradient 1.5, 1.5p= C contains the point (3, 7): 2 2 1.5(3)7 31 1.5 1 (3) 0.5 3 0.5 3 qr xx q qq x r y ++ =− + − − = = − + Turning point at (3, 7): 2 2 2 2 1.5 1 5 d .5(3 (3 )( 1) ( 0.5 3 )(1) d ( 1) (3(3) )(3 1) ( (3) 0.5 3 )0 (3 1) (9 )(2) 14 0 2 0.5 3( 2 6 ) ). y x q x qx q xx q q q q q r x+ − − + + −= − + − − + + −= − + − =  =−  = − − =

2025 NYJC J2 H2 Mathematics Preliminary Exam 9758/1 Marking Guide 2 of 16 Q2 Suggested Answers Method 1: Let the radius of the sphere be r and the height of the cone be h. 2 2 2h r l h l r+ =  = − Total volume, 231 3 42 3 2 3V h l r =  + ( ) 23 4 3l l r r= − + 22d ( 1) 4d V lrr = − + 22d 04d V rlr =  = Since r > 0, 2 lr = 2 2 d 8 0 for d2 Vl rrr =  = , ie minimum volume when 2 lr = . Method 2: Let the radius of the sphere be r and the height of the cone be h. 2 2 2h r l h l r+ =  = − Total volume, ( ) 32 3 21 3 4 42 3 2 3 3V h l r l h l h   =  + = + − ( ) 22d 40d V l l hh = − − = ( ) 22 4l l h=− ( )2 22 2 l l h l l h lh =− =− = or ( ) ( ) 2 32 3 reject2 l h l lh lh l h =− = = ( ) 2 2 d 8d V lhh =− When 2 lh= , 2 2 d 40d V lh = , ie minimum volume when 2 lr = . 2l

2025 NYJC J2 H2 Mathematics Preliminary Exam 9758/1 Marking Guide 3 of 16 Q3 Suggested Answers (a) Method 1 22 AB BC AC AB BC AC+ =  + = ( ) ( )AB BC AB BC AC AC + + = ( 2 =v v v ) ( ) ( ) ( ) 2 2 2 2 2 2 2 2 2 2 2 since 0 AB AB AB BC BC BC AC AC AB BC AB BC AC AC AB BC AC AB BC AC AB BC  + + =  + + =  + = + = = So AB is perpendicular to BC and hence 90ABC =  . Method 2 ( ) ( ) 2 2 22 2 2 2 2 2 22 AB BC AC AB BC AC AB BC AB BC AB BC AB BC AB BC + =  + =  + = +  + = + + ( ) ( ) 22 2 2 2 2 2 2 0 AB BC AB AB BC BC AB BC AB BC AB BC AB BC AB BC  + = + +  + = + + = So AB is perpendicular to BC and hence 90ABC =  . (b) Method 1 By ratio theorem, ( ) ( )11d a cOD OA OC   = = − + = − + . Area of triangle ABD ( ) ( ) ( ) ( )( ) 1 2 1 2 1 12 AB BD  = = −  − = −  − + − b a d b b a a c b ( )( ) ( ) ( )1 12   = −  +  −  + b a b c a c a b ( )( ) ( ) ( )1 12   = −  +  −  + a b b c a c a b since b a a b =−  . ( ) ( ) ( )1 2   =  +  − a b b c a c 2 a b b c c a=  +  +  since 01 . So 2k = . A D C B

2025 NYJC J2 H2 Mathematics Preliminary Exam 9758/1 Marking Guide 4 of 16 Method 2 Since triangles ABD and ABC share the same height, Area of triangle ABD ( )( )Area of triangle 1 1 2 ABC AB AC    = +− =  ( ) ( )2 2 2 b a c a b c b a a c a b b c c a    = −  − =  −  −  =  +  +  So 2k = .

2025 NYJC J2 H2 Mathematics Preliminary Exam 9758/1 Marking Guide 5 of 16 Q4 Suggested Answers (a) (b)(i) (b)(ii) y = f(a – x) y x C’(a – c, – b) B’(0, b) E’(a – e, 0) D’(a – d, 0) A’(2a, 0) x = – n x P(0, p) y y = m y = 0 y = |g(x)| x P(0, p) y y = 0 y = g(|x|)

2025 NYJC J2 H2 Mathematics Preliminary Exam 9758/1 Marking Guide 6 of 16 Q5 Suggested Answers (a) ( ) ( ) ( ) 1 11 11 2 1 2 2 2 3 ( 3 ( 17 17) 3 17 ( 17 17 ( 1) 2 1) ( 1)(2 1)66 17 1) 2 ( 1)(2 1) ) ( 1) 1 (1 3 7 22 (2 ) 11 2 n rm nn r m r m n m n r r r m r rr r r r n nm nm r n n m m m nnm n n mm m mn =+ = + = + = = = + −+ + + − = + − + = − + − +   =  + ++ + − − + = + + − + + +  − − −−      (b) From GC, the term 45 can be found in both series. (c) The nth term of J, 1 1 1 13 2 (3 1) (3 1) (3 3 ) (3 )(3 1 2 2 ) 2 nn n nn n n bbj b b b − − − − = − − − =− =− = Given b is a positive odd integer, then nj is a positive odd integer since 13n− is odd. Since the arithmetic progression is 1, 3, 5, …, the set of all positive odd integers, each of nj must be a part of the arithmetic progression.

2025 NYJC J2 H2 Mathematics Preliminary Exam 9758/1 Marking Guide 7 of 16 Q6 Suggested Answers (a) ( ) ( ) ( ) If , forms a rhombus. 1arg 2 1 2 p q OPRQ r      = =+ − = + (b) 'POQ = − (b)(i) Method 1: ( ) ( )arg ' 2 ince 2 sq    =− −  Method 2: ( ) ( )arg ' 2q     = − − = − (b)(ii) ( ) ( ) ' Let be the foot of perpendicular from ' to the real axis. Real part of ' = cos 2 (Adjacent side of ' ) Imaginary part of ' = sin 2 (Opposite side of ' but negative) ' cos qq FQ qq OQ F qq OQ F qq    = −+   − +    = − ( ) ( )2 + i sin 2q  + − + Alternative: ( ) ( )' cos 2 i sin 2q q q    = − − −  –2   Q P R O Re Im Q’

2025 NYJC J2 H2 Mathematics Preliminary Exam 9758/1 Marking Guide 8 of 16 Q7 Suggested Answers (a) 2sin d 2sin cosd x x   = = When 0, 0x == . When 1 ,24x == ( ) ( ) ( )   1 2 24 200 2 4 20 24 0 24 0 4 0 4 0 16 16sin d 2sin cos d1 1 sin 16sin 2sin cos d cos 4sin 2sin cos d 0 , cos coscos 2 8sin d 4 1 cos 2 d 4 2sin 2 2 x xx                            =− − = =   =  = =− =− =−      (b) ( ) ( ) 2 1 122 111 2222 1 111 2222 1 1 d 1 12 11 d d 11 1122 11 1122 11 22 112 1 2 122 12 2 b a b a b a b a x x xx xx xx x x x x x x x x x x x x − −+ − − −=+ − + − +           − + − +         =−  +                           =− − + + − +                =−    1 1 1 1 2 2 2 2 22 11 2222 1 1 12 1 2 1 22 2 2 112 2 2 1 2 122 a a b b a a b b              − − + + − + −                         =− + − + + − +      

2025 NYJC J2 H2 Mathematics Preliminary Exam 9758/1 Marking Guide 9 of 16 Q8 Suggested Answers (a) Note that the sequence is an arithmetic progression with common difference 1.5 50 1 11 1 49(1. 3 5 7 .5 0.75 ) 99 u u u u u = =+ + = (b) Method 1 22 1 42kn n k uu uu = + ++= is a sum of n terms of an arithmetic sequence with first term 2 0.75 1.5 2.25u = + = and common difference 2(1.5) 3= ( ) ( ) 2 4 2 2(2.25) ( 1)(3)2 1.5 32 n nu u n n n u + ++ = +− =+ From GC, for ( )1.5 3 2025,2 n n+ least 37n= Method 2 ( )( ) 1 2 1 2.25 2 1 3 2025 n k n kk uk == = + −  Using GC, least 37n= (c)(i) 25 5 25 2 5 k k k u u u uu u u = =      2 1 0.75 ( 1)(1.5) 0.75 (5 1)(1.5) 0.75 (25 1)(1.5) 0.75 ( 1)(1. .5) 6.75 36 75 15.75 1 k k k + − + − + −= =−= = +  (c)(ii) Method 1 Common ratio 15.75 3 36.75 7== ( ) 13 13 3 7 3 7 ) 64.31144 61 36.75 1 ( .3 1 41 S = = − = − Method 2

2025 NYJC J2 H2 Mathematics Preliminary Exam 9758/1 Marking Guide 10 of 16 Using GC, 113 1 336.75 64.3117 r r − =   =   