ASRJC 9758 2023 Prelim P1 Solution

[Turn over 1(a) Let x, y and z denote the number of wins, draws and losses by Lucy’s favorite team this season respectively. x + y + z = 38 …Eq(1) 3x + y + 0z = 54 …Eq(2) x + y – 2z = 8 …Eq(3) From GC, x = 13, y = 15 and z = 10 ∴ Lucy’s favorite team won 13 games this season. (b) Points scored by Mark’s favorite team = 3(13 – 2) + (15 + 5) = 53 < 54 Hence Lucy’s favorite team performed better this season. Alternatively, Since Mark’s favorite team won 2 games fewer and drew 5 games more, this team scored –3(2) + 5 (1) = –1 point more. Hence Lucy’s favorite team performed better this season. 2(a) d (tan )d xx d sin()d cos x xx= 2 2 cos sin ( sin ) cos xx x x −−= 22 2 cos sin cos xx x += 2 1 cos x = 2sec (Shown)x= (b) sin 2 cotxx cos2sin cos sin xxx x =  22cos (Shown)x=

2 © ASRJC 2023 (c) ( ) ( )15 30 cosec 10 tan 5 dx xx π π∫ ( ) ( ) 15 30 1 dsin 10 cot 5 xxx π π=∫ 15 2 30 1 d2cos 5 xx π π=∫ 215 30 1 sec 5 d2 xx π π= ∫ 15 30 11 tan 525 x π π =  1 tan tan10 3 6 ππ= −  11 310 3 = −  12 10 3 =  3 15= 3(a) sin cosec d PQ PQ d θ θ = = 2PQ QR RS a++= ( )2 2 cosecQR a d θ= − ( )2 cosecQR a d θ= − ( )1Area 2 d QR PS= + ( ) ( ) ( ) 1 2 cosec 2 cot 2 cosec2 1 4 4 cosec 2 cot2 d ad d ad da d d θθ θ θθ = − + +− = −+ ( ) 22 cot 2cosecad d θθ= +−

3 © ASRJC 2023 [Turn over (b) 22 2d cosec 2 cosec cotd A dd θ θθθ = −+ dWhen 0d A θ = , 22 2cosec 2 cosec cot 0dd θ θθ−+ = ( )cosec cosec 2cot 0 cosec 2cot 1cos 2 3 θθθ θθ θ πθ −= = = = 22 2d cosec 2 cosec cotd A dd θ θθθ = −+ θ 1.046 1.04723 =π 1.048 d d A θ 0.002769 2d 0 – 0.0018519 2d Slope / – \ Using the first derivative test, 3 πθ = gives a maximum value for A. Alternative method 22 2d cosec 2 cosec cotd A dd θ θθθ = −+ ( ) 2 22 2 3 2 2 d 2 cosec cot 2 cosec cosec cotd A dd θθ θ θ θθ = −+ At 3 πθ = , 2 2 2 d 2.309 0d A dθ = −> 3∴= πθ gives a maximum value for A. At 3 πθ = , 22 cot 2cosec 33A ad d ππ= +−  2 2 1222 33 23 ad d ad d  = +−   = −

4 © ASRJC 2023 4 (a) 1l : 12 43 xy z+−= = 1 14 : 2 3, 01 l λλ −   ⇒ = +− ∈   r  1 2 0 −   and 3 1 1  − are position vectors of 2 points on 1l , substituting into p2 , 67 33 7 α αβ −+= −+ = 4 3. 3 0 1 49 α β αβ      −=       += Solving, 1, 13αβ= −= (b) 1 03 3 1 1 1 3 31 12 3 1        = ×− = =       −−    n p1 : 1 10 1   =− r. 0xyz+−= 3327xyz−++= Solving using GC, line of intersection is 15 7 1 1 , 6 06 µµ −     = +∈       r  . Alternative method 3 37 2 −  = r. Since P1 is on P2, sub P1 into equation of P2

5 © ASRJC 2023 [Turn over 33 37 22 9332 4 7 85 7 78 5 t st st tst st ts ts −     −=    +   −+−++= −+ = += . Sub s into equation P1 03 78 115 12 0 03 78 11155 1 12 30 73 155 1 18 5 05 7 1 1 where 5 16 t t t t t µµ   +   = +−               = + +−               = +        = +∈    r r r r  (c) Distance = 0 101 342 1000   −     −−     . = 0 11 36 100   − . 6 10 =

6 © ASRJC 2023 5 (a) 2 22 341 2 12 12 3 3(2 1)(2 3) 4(2 1)(2 3) (2 1)(2 1) (2 1)(2 1)(2 3) 12 24 9 16 16 12 4 1 (2 1)(2 1)(2 3) rrr rr rr rr rrr r r rr r rrr −+−++ + +− − ++ + −= −++ + +− − + + −= −++ 8 20 (2 1)(2 1)(2 3) r rrr += −++ (b) 1 1 1 4 2 13 1 4 13 5 34 1 35 734 1 57 9 34 1 2523 2134 1 2321 21 341 21 5 (2 1)(2 1)(2 3) 1 8 20 4 (2 1)(2 1)(2 3) 1341 42 2 12 21 3 14 1 43143 1 1 2 22 3 n r n r n r nn n nn n nnn nn r rrr r rrr rrr = = = = −+ + −+ + −+ + −+ −− − + + −++ += − + −+ −−+ + −+ −+  = − −+  + ++ = + −  −+ + + ∑ ∑ ∑  1 12 3 21 1 3 3 42 3 2 1 n nn  + ++ = +−  ++ 2 2 1 3(2 3) 3 (2 1)(2 3) nn nn +− += − ++ 22 3 (2 1)(2 3) n nn += − ++

7 © ASRJC 2023 [Turn over (c) Sum to infinity 2 3= 22 2 0.0043 3 (2 1)(2 3) 2 0.004(2 1)(2 3) k kk k kk  +−− < ++ + <++ From GC, the smallest value of k is 63 Alternative solution: 2 0.004(2 1)(2 3) k kk + <++ 2 2 0.004(2 1)(2 3) 4 242 497 0 k kk kk +< + + − −> 1.9884 62.488k or k<− > Since and 1kk∈≥ , the smallest value of k is 63 k 2 (2 1)(2 3) k kk + ++ 62 0.0040315> 0.004 63 0.0039675 < 0.004 64 0.0039056 < 0.004

8 © ASRJC 2023 6(a) 2cot 3 dxx∫ 2cosec 3 1 dxx= −∫ cot3 3 x xc= − −+ (b) 24 22 43 d48 xx xxx −+ −+∫ 24 222 48 5 d48 48 xx xxx xx −+= −−+ −+∫ 4 22 51 d( 2) 4 xx= − −+∫ 4 1 2 52tan ( )22 xx − −= − 11554 tan (1) 2 tan (0)22 −−  =− −−     52 8 π= − (c) 2cotx θ= 2d 2cosecd x θθ =− When 23 23, 2cot = , tan = 3, 33 3x πθ θθ= = When 2, 2cot 2, tan = 1, 4x πθ θθ= = = 22 23 22 3 4 d(4 ) x xx − +∫ 2 24 22 3 4 4cot ( 2cosec ) d(4 4cot ) π π θ θθθ −= −+∫ 2 23 22 4 4(1 cot ) (2cosec ) d16(1 cot ) π π θ θθθ −= +∫ 2 23 22 4 1 (1 cot ) (2cosec ) d4 (cosec ) π π θ θθθ −= ∫ 2 3 2 4 1 (1 cot ) d2 (cosec θ) π π θ θ−= ∫ ( ) 223 4 1 cos sin d2 π π θ θθ= −−∫

9 © ASRJC 2023 [Turn over 3 4 1 cos 2 d2 π π θθ=− ∫ 3 4 1 sin 2 22 π π θ=−  12sin sin43 2 ππ= −− 13 142 = −−  13 48= − 7(a)(i) 22tan 1 secxx+= ( ) 2 2tan 1 secxx− += 22 22 123 yx+= 22 22 132 xy−= , 3, 0xy>< (ii) (b) When 1, 2sin 1 =1, sin = 0, 0x t tt= += When 12, 2sin 1 =2, sin = , 26x t tt π= += d2sin 1, 2cosd xxt t t=+= Required Area = Area of trapezium – 2 1 dyx∫ = ( )1 2 4 (1)2 +− 6 0 (2cos3 4sin )(2cos ) dt t tt π +∫

10 © ASRJC 2023 = 3− 6 0 (4cos3 cos 8sin cos ) dtt tt t π +∫ (shown) = 3− 6 0 (2cos4 2cos2 4sin2 ) dt t tt π ++∫ 6 0 sin43 sin2 2cos22 t tt π = − +− 2sin 33 sin 2cos ( 2)233 π ππ  = − + − −−   3331 4 = −+  332 4= − units2 8(a) ( ) 1 3 sin 2 3 sin 2 1 y x yx = + += Differentiate implicitly w.r.t. x: ( ) ( ) 2 d3 sin2 2cos2 0d 1d 2 cos2 0d d 2 cos2 0d yx yx x y yxyx y yxx + +=   += += Alternatively ( ) ( ) ( ) ( ) 1 2 2 3 sin 2 d 3 sin 2 2cos 2d d 2cos 2d yx y xxx y yxx − − = + = −+ =− Differentiate implicitly w.r.t. x: ( ) 2 2 2 2 2 2 dd 2 2sin2 2 2 cos2 0dd dd 4 sin2 4 cos2 0dd yy y xy xxx yy y xy xxx −+ = −+ =