ACJC 9758 2023 Prelim P1 Solution

1 2023 H2 Maths Prelim Paper 1 Marking Scheme Qn Solutions 1 ( ) 32f x x ax bx c=+ ++ 32 1 abc− =+++ ( ) 2f' 3 2x x ax b=++ 032 ab= ++ ( ) 1 fy x= has a vertical asymptote at 5x= implies that ( )fyx= has an x-intercept at 5x= . 0 125 25 5 a bc= + ++ 33 23 25 5 125 abc ab a bc ++= −  += −  + += − By GC, ( ) 32f 5 7 35xx x x=− +− . [ 5, 7, 35a bc= −== − ] 2 * 2 4i ----(1) 2 3i ----(2) wz zw + = −+ += From (1): 2 4i *wz= −+ − Substituting into (2): ( )2 3i 2 4i *zz+= −+ − 3i * 14 6izz+ = −− Let iza b= + , ( ) ( ) i 3i 3 14 6i 3 3 i 14 6i ab a b a b ab ++ += −− +++= − − Comparing real and imaginary parts: 3 14 ---(3) 3 6 ---(4) ab ab += − += − Solving simultaneously with (4) 3(3)− , we have 8 36 4.5bb− = ⇒= − Hence 0.5a=− , and 0.5 4.5iz= −− 2 4i *wz= −+ − : 2 4i 0.5 4.5i 1.5 0.5iw= −+ + − = − −

2 3(i) Method 1: Algebraic Manipulation 1 0 d 1 x x x+∫ 1 0 ( 1) 1 d 11 x x xx += − ++∫ 11 00 11d d 1 xx x x = +− +∫∫ ( ) ( ) 131 22 0 2 1213 xx= +− + ( ) ( ) ( ) ( ) 3 1 31 2 2 22 222 22 1 2133   =−−−     42 222 233   = − −−     42 233 2(2 2)3 = − = − Method 2: Integration by Parts Let ux= d1 d 1 v x x = + d 1d u x = 21vx= + 1 0 d 1 x x x+∫ 1 1 00 2 1 2 1dxx x x= +− + ∫ ( ) 13 21 0 0 212 12 3 xxx  += +−  ( ) ( ) 33 22 4220 2 1 3 = −− −   42 233= − ( ) 2 223= −

3 Method 3: Using Substitution (Change of Variable) Let 1ux= + . Then d 1d u x = . When 0, 1xu= = When 1, 2xu= = 1 0 d 1 x x x+∫ 2 1 1du u u −=∫ 2 1 1 du u uu = −∫ 112 22 1 duuu − = −∫ 231 22 1 2 23 uu= −  2 221 2 213   = −− −   42 233 2(2 2)3 = − = − 3(ii) Area of 1st rectangle: 2 11 11 1 1 1 1 1 1f 11 1 11 nn nn n n n n n nn n nn n      = = = =  +++  + Area of 2nd rectangle: 2 22 121 1 1 2 1 2f 22 2 21 nn nn n n n n n nn n nn n      = = = =  +++  + Area of nth rectangle: 2 11 1 1 1 21 nn n nnnnfnnn n n n nn nn n n n nn n      = = = =  ++  +

4 Total area of all the rectangles: 11 2 3fff f n n nnn n    = +++ +        11 2 3 123 2 n nn n n n n = + + ++ +++  All the rectangles above the curve enclosed the region bounded by the curve and the x-axis and include a small portion above the curve. Thus, the total area of all the rectangles is more than the area bounded by the curve and x-axis. Hence, ( ) 11 2 3 2 223123 2 n nn n n n n  + + ++ > − +++  . 4(i) ⋅pq is the length of projection of q onto p. OR ⋅pq is the length of projection of q onto the line with direction vector p. 4(ii) ( )OM = ⋅pqp   (vector projection, since M lies on OP.) 2OM∴= p   (shown) OR Since M lies on line OP, for some OM λλ= p   . Since M is the foot of the perpendicular from Q to the line OP, 0QM OP⋅=     ( ) ( ) 2 0 0 2 2(as