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