NYJC_8865_2022_Prelim_Solutions_vetted

2022 NYJC J2 H2 Mathematics Prelim 8865/1 Marking Guide 1 of 9 Q1 Suggested Answers For ( ) 23 4 2 0k x x k− + −  , ( )( ) ( )( ) 2 2 2 3 4 0 and 2 4 3 4 0 3 and 4 12 16 04 3 and 4 3 1 04 3 and 4 1 1 04 31 and or 144 k k k k k k k k k k k k k k k −  − − −   + −   − −   + −   −  Combining solution on a number line, 1k  Q2 Suggested Answers (i) 32y ax bx cx d= + + + At ( )0,2 , d = 2 (ii) 2d 32d y ax bx cx = + + Since ( )0,2 is a maximum point, c = 0 1195, 3 −− is a minimum point, ( ) ( ) 2 3 5 2 5 0 75 10 0 (1) a b c ab − + − + = − = −−−−−−−−− ( ) ( ) 32119 5 5 23 125125 25 (2) 3 ab ab − = − + − + − = −−−−−− Using GC, 2 3a=− , 5b=− Hence 322 523y x x=− − + Q3 Suggested Answers (i) 3 eln ln e 3ln(1 2 ) 3ln(1 2 ) (1 2 ) x x x x x x  = − − = − − − ( )3 1 2 d e dln 3ln(1 2 )dd (1 2 ) 12 32 1 2 16 122 x xxxx x x x xx −  = − − − −=−  − =+ −

2022 NYJC J2 H2 Mathematics Prelim 8865/1 Marking Guide 2 of 9 (ii) ( ) 2e 1 e 1 e2 1 2 2 2 32d 94 12 d 2 12 9ln 2 12 9ln 2 12 9ln 1 2 12 9 14 2 12 5 xx x xx x x x x e e e ee ee − − −  − = − + = − + = − + − + + − = − + − = − −   Q4 Suggested Answers (i) 37 37 37 37 14 2e d 14 14ed 0 14 14e 1e ln1 3 7 3 7 x x x x yx y x x x − − − − =+ =− =− = =− = At 3 7x= , 337 7314 2e 6 2 87y − = + = + = Coordinates of turning point: 3 ,87   (ii) (iii) 44 d1 14 2e & 14 14e d yxy x −−=  = + = − Equation of the tangent: ( ) ( )( ) 4414 2e 14 14e 1yx −−− + = − − ( ) ( ) 4 4 4 44 14 14e 14 14e 14 2e 14 14e 16e yx yx − − − −− = − − + + + = − + (iv) Area required ( )( ) 1 3 7 4 4 0 114 2e d 1 16e 14 2e2 xxx − − −= + − + + ( ) 30,2e

2022 NYJC J2 H2 Mathematics Prelim 8865/1 Marking Guide 3 of 9 137 2 0 2e7 7.16487 5.57 (3 s.f.) x x −= − − = Alternative: Area required ( ) 11 3 7 4 4 00 14 2e d 14 14e 16e d 5.57 (3 s.f.) xx x x x− − −= + − − + =  Q5 Suggested Answers (i) Volume of container, For maximum volume, (rejected) or r E.g. 16.65 E.g. 16.67 10.4615 > 0 0 −2.0948 < 0 Slope / − \ Hence gives a maximum volume of 29088.82 cm3 Alternative method When ,  gives maximum volume of 29088.82 cm3 (ii) Amount of discarded metal Alternative method Amount of discarded metal = ( )( ) 22 2 40rr  − ( ) 22 80 rr =− 100 4hr=− 2V r h= ( ) ( ) 2 23 100 4 4 25 rr rr   =− =− ( ) 2d 4 50 3d V rrr =− d 0d V r = ( ) ( ) 24 50 3 0 4 50 3 0r r r r − =  − = 0r= 50 3r = 50 3 −   50 3 50 3 +   d d V r 50 3r = ( ) 2 2 d 4 50 6d V rr =− 50 3r = 2 2 d 200d V r =− 50 3r = ( ) 2100 40 2 40 rh= − − ( ) 24000 2 40 100 4rr= − − − 2160 2rr =− ( ) 22 80 rr =−

2022 NYJC J2 H2 Mathematics Prelim 8865/1 Marking Guide 4 of 9 (iii) ( ) 2e 1.05 2 80 e 1.05rrC D r r r r −−= + = − + From the sketch, the least cost to discard the waste metal from a container is $7.99.