EJC 9758 2023 Prelim P2 Solution

2023 EJC Prelim Paper 2 Solutions Q1 4 5i 7 (1) (1 i) 8 30 (2) zw zw += −+= By (2), 4 16(1 i) 60(1 i) (3)zw++= + (3) (1) :− (16 11i) 53 60iw+= + 4iw= + By (1): 4 5i(4 i) 7z+ += 3 5iz= − Q2 (a) Consider the following cases: 2 7 3 13xx x− += − ( )( ) ( ) 2 2 6 10 0 6 6 4 1 10 21 6 76 6 2 19 22 3 19 xx x −−= ±− − = ±±= = = ± ( ) 2 7 3 13xx x− −+=− ( )( ) ( ) 2 2 8 16 0 8 8 4 1 16 21 4 xx x −+= ±− = = The roots are 3 19± and 4. (b) From the graph, the solution is 3 19 4 or 4 3 19xx− << <<+

Q3 (a) secyz x= dd sec tan secdd yz zxx xxx= + Sub in the DE, ( )dπ sec tan sec sec 3 π tan 0d dπ sec tan π sec 3 sec π sec tan 0d dπ sec 3 sec 0d dπ sec 3 secd d3 (shown)d π zzxx xzx x x zzxx xzxzxx x z xzxx z x zxx zz x  + + −= + +− = += =− −= 3 π 3 π 13dd π 3ln π e e where e xC x C zxz z xC z zA A − + − −= −= + =± = =± ∫∫ 3 π 3 π esec cos e x x y Ax y xA − − = = When π 3x= and 2y= , 3 π π3π2cos e3 e A A − = = 3 π 31 π cos e e 3cos e where 1 and π x x yx yx a b − −  ∴=  = = =−

(b) 31 πe cos x y x − = For vertical asymptotes, consider cos 0x= The asymptotes closest to y-axis are π 2x=− and π 2x= Q4 (a) 33 Ax y xy+−= Differentiating with respect to x, 22 0dd3 d3 d yyx xy xy x+− − = ( ) 223 d 3dyy yxx x =−− + (*) When d 0d y x = , 20 3 yx= −+ 23yx= (shown) (b) Sub 23yx= into 33 Ax y xy+−= , ( ) ( ) 332 2 33 Ax x xx+−= 3 6327 3xAxx+ −= 63 027 2 Ax x− − = ( ) ( ) ( ) ( ) 2 3 2 4(272 ) 2 27x A−−− ± − = − ( ) 3 22 2 27 1 27 Ax ±= + 3 71 27 12 Ax ±= + 33 11 127 1 27 13 27AxA= ±+ +± =

For more than one stationary point, Discriminant 1 027A+> So 1 27A>− (c) From (*), ( ) 223 d 3dyy yxx x =−− + Differentiate with respect to x , ( ) 2 2 2 13 d dd d ddd 6 6d y yy yxx xx xy x y =−  +−+−  When d 0d y x = , 23yx= , so we have ( ) 222 2 d 63 d 3 yxx x x − =−  ( ) 2 4 2 d27 6 d yxx x x − =− 2 24 3 d6 6 d 27 27 1 yx x xx x −= − −− = When 3 71 27 12 Ax += + , 2 23 d6 6 0 d 27 1 1 27 y xx A =− + = −< − so this is a maximum point. When 3 71 27 12 Ax −= + , 2 23 d66 0 d 27 1 1 27 y x Ax = =− + > − so this is a minimum point. Q5 (a) Since p and q are perpendicular, a direction vector parallel to q is 2 0 1   − .

Normal to q 22 2 1 0 2 4 22 11 4 2 −         = × −= −= −        −−     Equation of q: 1 51 r2 3 2 7 2 42 227xyz    = −=    ++=  (b) 2 10xz−= 227xyz++= Solve using GC: 15 2 51 4 xz yz zz = + = − = 52 115 404 x yz z      =+−       Let 1 4 zµ = , equation of m is 52 1 5 , 04 µµ      = +− ∈       r  (c) Let F be the foot of perpendicular from A to q. Method 1 21 : 1 2 , 62 AFl αα      = +∈    −   r  Since F lies on AFl , 2 12 62 OF α α α +