TJC 9758 2025 Prelim P2 Solutions

Section A: Pure Mathematics Solutions 1(a) ( ) 31 ln 1 e2 xy=+ 23e 1 eyx=+ Diff. wrt x, 23 d2e 3ed yx y x = (Shown) -- (1) Alternative solution: 3ln 1 e xy=+ ( ) ( ) ( ) ( ) 1 33 2 3 1 3 3 2 33 23 3e 1 ed 3e d 2 1 e2 1 e d2 1 e 3e d d2e 3ed xx x x x xx yx y x y x y x − + == ++ += = ( ) ( ) 33 3 2 3 1ln 1 e ln 1 e 2 i.e. 2 ln 1 e e 1 e xx x y x y y  = + = +  = +  = + (b) Diff. wrt x, 23 2 2 2 2d2 d de 4 9ed e yyx y x y x +=    ---- (2) When 110, ln(1 1) ln 222xy= = + = , 2 ln2e e 2y == Sub into (1): d3 d4 y x = Sub into (2): ( ) 22 2 d32 2 8 9d4 y x += => 2 2 d9 d8 y x = 21 3 9ln 22 4 16y x x= + + + (c) ( ) ( ) ( ) ( ) 3 3 2 2 1eln ln 1 e ln 22 1 3 9ln 2 ln 22 4 16 1 3 9ln 22 4 16 x x xx xx − − −  + = + − =− + + + + =− − + + −

2(a) 2 2 2 (2 1)(2 1) n r rr= −+ Replace r by r +1, 12 12 21 1 21 2 2 (2( 1) 1)(2( 1) 1) 2 (2 1)(2 3) 22 (2 1)(2 3) (2 1)(2 3) 1 1 2 5 2(2 1) 3 15 11 3 4 1 rn r rn r n r rr rr rr n n += += =− = − = = + − + + = ++ =+ + + + + = − + −+ =− +    (b)(i) 1 4 5 for 1,2,3, .nnx x n+ = + = Since sequence converges to L, 1as , , . nnn x L x L +→ → → 2 2 4 5 4 5 --- (*) 5 4 0 5 41 2 5 41Since are positive numbers, . 2 n LL LL LL L xL =+ =+ − − = = += (ii) 22 1 (4 5 ) (5 4) (fr (*)) 5( ) nn n x L x L xL + − = + − + =− (iii) If ,nxL 22 1 11 1 1 1 1 5( ) 0 ( )( ) 0 0 since 0, 0, so 0 nn nn n n n n x L x L x L x L x L x L x L xL + ++ + + + + − = −  − +   −    +  

3(a) 1f : e 2 xx −+ , 0x and a is a positive constant. (b) 2g : 1 4 , 2x x x x+ −  Every horizontal line will cut the curve at most once. Hence g is 1-1. Therefore, 1g− exists. (c) ( ) ( ) 2 f 52 yx yx = = − − ( ) 2 25xy− = − 25xy=  − Since 2,x 1g ( ) 2 5xx− = − − ( 1 ggD R ,5− = = − (d) From graph of f in (i), Rf = 13,22    ( ,2 gD − =  gf exists. (e) Rgf = ( ) ( )( 22 31 225 2 ,5 2 − − − −  11 19,44 =  y = f-1f(x) x = a y = f(x) y = f-1(x) y x 3 2 • ( )2,5 x y 1 2 3 2 x y • 3 2 • O 19 4 11 4

4(a) Circumference of base = length of arc AB 2 2 ra ar   = = (b) (i) Let h = height of cone ( ) 2 2 2 2 22 2 2 22 2 4 44 h a r aa a   =− =− =− 21 3V r h = ( ) ( ) 2 2 4 2 42 2 2 2 42 6 4 2 2 4 1 9 1 49 16 4 4576 V r h aa a        = =  − =− (c) Method 1 ( ) 26 2 4 6 4 dd 4d 576 d Va     =− ( ) ( ) 6 2 3 5 4 6 3 2 2 4 16 6576 2 8 3576 a a       =− =  − 2d d V  = 0 when ( ) 3 2 28 3 0   −= 2 2 82 233   =  = ( ) 2 2 6 2 2 4 24 d 48 30d 576 Va    =− When 22 3= , 2 2 6 2 4 6 2 24 d 8 64 4 48 30 0d 576 3 9 27 V a a      = − =−         Therefore V is max when 22 3= Method 2

( ) 6 2 4 2 2 4 4576 aV   =− Differentiating w.r.t.  , ( ) 6 2 4 6 4 dd24 d 576 d VaV     =− ( ) ( ) 6 2 3 5 4 6 3 2 2 4 16 6576 2 8 3576 a a       =− =  − d d V  = 0 when ( ) 3 2 28 3 0   −= 2 2 )82 2(33 0   = = ( ) 226 2 2 4 42 dd2 2 48 30d d 576 V V aV    + = − When 22 3= , d 0d V  = 2 6 2 4 6 2 24 d 8 64 42 48 30 0d 576 3 9 27 V a aV      = − =−         Since V > 0, 2 2 d 0d V   Therefore V is max when 22 3= (d) Note that: 22 3= 2222 2 3 3 aara  = = =  2 2 2 2 2 2 2 1 1 33 3 h a r a a a h a = − = − =  = 23 223 r a r hha=  =  = ( ) 3221 1 2 23 3 3 ww yW r h y y = = = (e) d d d d d d W W y t y t=

2 2 d215 d d1 d 30 yy t y ty  − =  =− When y = 2, 1d1 cm sd 120 y t −=− Rate of change of y is 11 cm s120 −− .

Section B: Probability and Statistics [60 marks] 5 [Solution] (a) Required Probability = P(Dave hits 2 targets and Rafael hits 0 or 1)+ P(Dave hits 1 and Rafael 0) ( ) ( ) 2 7 8 2 2 2 7 3 2 2 2210 10 10 10 10 10 10 10 10              = + +                           0.1932= (b) P(Day was bad given Dave wins) ( )( ) ( )( ) ( )( ) 0.3 0.39 0.7 0.1932 0.3 0.39 0.464 (3s.f.) = + =

6 [Solution] (a)(i) Required number of ways = 12 5 51         =3960 (a)(ii) Required number of ways = 6 6 5 15002 3 1     =       Alternative method: Case 1: 2 not identical at 3rd storey 66 2! 60023    =       Case 2: 2 identical at 3rd storey 66 3! 90023 2!    =     (a)(iii) Case 1: 2 in 1st storey, 1 in 2nd storey, 2 in 3rd storey Number of ways = 3 3 6 5 6752 1 2 1      =         Case 2: 1 in 1st storey, 2 in 2nd storey, 2 in 3rd storey Number of ways = 3 3 6 5 6751 2 2 1      =         Case 3: 1 in 1st storey, 1 in 2nd storey, 3 in 3rd storey Number of ways = 3 3 6 5 9001 1 3 1      =         Total number of ways = 2250 (b) Number of ways to arrange the 10 participants = ( ) ( ) 5 5 1 ! 2! 768−= Number of ways to slot in the 2 game masters = 5 2!2   =20 Required number = 768 20 15360=

7 Solution (a) Let X be the number of faulty bowls in a box of 20 bowls. ~ B(20, 0.08)X P( 2) 1 P( 2) = 1 0.78795 = 0.21205 (5 s.f.) = 0.212 (3 s.f.) XX = −  − (b) Let W be the number of faulty bowls in a carton (of 12 boxes). B(12 20, 0.08), i.e. B(240, 0.08)WW  P( 15) P( 14) = 0.12933 (5 s.f.)WW =  Required Probability 32 2 (0.12933) (1 0.12933) 0.0437 (3 s.f.) C=− = OR Let A be the number of cartons out of 3 contains fewer than 15 faulty bowls each. B(3,0.12933)A Required Probability = P(A = 2) = 0.0437 (3 s.f.) (c) B(12, 0.21205)Y E( ) 12 0.21205 2.5446Y np= =  = = 2.54 (3 s.f.) Var( ) (1 ) 12 0.21205 (1 0.21205) 2.0050 (5 s.f.) = 2.01 (3s.f.) Y np p= − =   − = (d) B(12, 0.21205)Y Let 1 35...T Y Y= + + . Since sample size 35 is large, by Central Limit Theorem, N(35 2.5446, 35 2.0050)T  approximately i.e. N(89.061, 70.175)T approximately Required probability = P( 85) 0.314 (3 s.f.)T =

[Solutions] 8( a) Let W and L be the marks obtained in the written paper and lab -based practical respectively. 2~ N(62, )W  and 2~ N(56, 12 )L P( 85) 0.95W = 85 62P 0.95Z  − = Using GC, 85 62 1.64485 − = 13.983 = (5 s.f.) (b) 22~ N(62 56, 13.983 12 )WL− − + ~ N(6, 339.524)WL− ( ) ( )P 9 1 P 9 9 0.643W L W L−  = − −  −  = (3 s.f.) (c) ( ) 2 2 2 20.6 0.4 ~ N 0.6 62 0.4 56, 0.6 13.983 0.4 12T W L= +  +   +  ( )~ N 59.6, 93.42874T ( )P 60 0.517T = (d) 12 93.42874~ N 59.6, nT T TT nn + + + =  ( )P 58 0.95T  Method 1: From GC, n ( )P 58T  98 0.9494 < 0.95 99 0.9502 > 0.95 Therefore, the least n is 99. Method 2 (More tedious)