NYJC_VJC_TJC_9649_2025_Prelim P2 Solutions

1 2025 JC2 FM Prelim Paper 2 Solution 1 Suggested Solution (a) (i) ( ) ( ) ( )fg fg fg fg fg ff gg fg xy xx yy xy xy ∂∂∇+ = + + +∂∂ ∂∂ ∂∂=+ ++ ∂∂ ∂∂    ∂∂ ∂∂=+++  ∂∂ ∂∂   =∇ +∇ ij ij ij ij (ii) ( ) ( ) ( )fg fg fg gf gffg fg gg fffg f ggf xy xx yy xy xy ∂∂∇= + ∂∂ ∂∂ ∂∂ + ++  ∂∂ ∂∂    ∂∂ ∂∂= ++ +  ∂∂ ∂∂   =∇+∇ ij =i j ij ij (b) (i) ( ) 22 22 0 100 0 100 H xy xy = ⇒−+= ⇒+= The set of points of S which lie on the x-y plane ( ){ } 22 2, : 100xy x y= ∈ +=  which is the set of points on the circle with centre ( )0, 0 and radius 10. (ii) The highest point of S is ( )0, 0,100 . The directional derivative of H at the highest point of S in the direction of any unit vector u is ( ) ( ) ( ) ( ) ( ) ( )( ) 0, 0 , 0, 0 0, 0 20 20 0 DH H xy HH xy =∇ ∂∂= + ∂∂ = −− = u u i ju i ju   

2 2 Suggested Answers (a) d d yx t= Differentiating with respect to t , 2 2 dd dd xy tt= Using chain rule, d dd d d dd d x xy x xt yt y= = 2d d xx ax byy = −+ . (b) 2 dd 2dd ux ux xyy = = 1d 2d d 22d u au byy u au byy = −+ += Integrating Factor is 2d 2ee ay ay∫ = Multiplying throughout by 2e ay , ( ) 2 22 22 de 2e 2 ed d e 2ed ay ay ay ay ay u a u byy u byy += = Integrating throughout with respect to y , 2 22 22 2 e e ed ee 2 ay ay ay ay ay bbuy y aa bbyCaa = − = −+ ∫ 2 2 e2 aybbuy Caa −=−+ (c) 2 0.04 2 0.04 25 625 e d 25 625 ed y y xy C y yCt − − =−+  =−+ 4 4 0 2500 625 e 1875e C C −= −+ =− 4 0.04d 25 625 1875e ed yy yt −= −−

3 3 (a) We have 3 1, 1ωω= ≠ . Then 3 4 521,ω ω ωω ω= ⇒= = . 3 2 110 1 ωωω ω −++ = = − since 3 1, 1ωω= ≠ ( ) ( )( )( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 245 22 22 223 2 2 222 22 22 22 42 42 1 1 49 ω ωωω ωω ωω ωω ωω ω − −−− = −− −− = −− = −++ = −−+ = (b) ( ) ( ) ( ) ( ) ( ) ( ) 22 2 2 2 2 22 rcz rc z r c c z cz c rz rz r zz c z cz c zz rz rz r zcz c zrz r zc zr zc zr ∗∗ ∗∗ ∗ ∗∗ ∗ ∗ ∗ ∗∗ ∗ − +− =− ⇒− − + =− − + ⇒ − − + = −− + ⇒ −− = −− ⇒− =− ⇒−=− which is the perpendicular bisector of the line segment joining the points representing the complex number c and the real number r. Alternative Solution: Write izxy= + and let ic ab= + . ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )( ) ( ) ( )( ) ( ) 22 ii ii ii 22 rc rcz rc z ra b x y ra b x y r a x by bx r a y r a x by bx r a y r a x by ∗∗ − =− +− =−− + +−+ −     =− −+ +− +− −− +− =−− which is the line with axial intercepts ( ) 2222 ,0 , 0,22 rc cr ra b   −−    −   provided ,0r ab≠≠ OR a line with gradient ra b − and y-intercept 2 2 2 cr b − .

4 Qn 4 (a) Area of sector OAB, 2 0 1 d 2Tr α θ=∫ 2 0 2 0 2 2e d e (e 1) k k k k k θα αθ α θ= =  = − ∫ 11 d22e ed kk rr k θθ θ= ⇒= Length of arc AB, 1 2 11 22 2 2 0 22 20 2 2 20 2 0 22 0 d dd 44e e d