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   =− Differen