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 −     = +∈     