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 4 e ( 1) d 12( 1) e d 2( 1) e 2( 1) (e 1) kk k k kk rSr k kk k k kk α θθα θα θα αθα θθ θ θ θ = +  = + = + = + = + = +−  ∫ ∫ ∫ ∫ (b) Consider T S : 1 2 2 22 (e 1) (e 1) 12( 1) (e 1) k k k Tk k S kk α α α −= = + ++− As k →∞ , 2 1 1 k k → + and /e1 kα → , and so 1T S → TS∴≈ (c) As k increases, the curve approaches a circle centred at O with radius 2. 2(2) 12 (2) T TSS π π∴≈ = ⇒≈ (same as above)

5 5 (a) ( ) ( ) ( ) ( ) ( ) 222 2 2 f, 4 f 2 42 22 2 4 f22 42 f 22 f4 f2 x y xx yy xy x y x y x ax y x a ax y a x ay y ax y y ax a a =+−+ − = −+ − = + − − =− −=− = + = =− (b) ( ) ( ) f 0 2 4 2 0 --- (1) f 0 2 2 0 --- (2) x y x a ax y y ax y =⇒ −+ − = = ⇒− −= From (2): 1 2y ax= Sub 1 2y ax= into (1): ( ) 2 2 2 42 0 12 42 02 120 2 1 122 4 2 x a ax y x a ax ax x a ax ax x a −+ − = −+ − = −+ =    +=  = + 22 14 2 222 aya aa = = ++ Since a∈  , there is unique solution for x and y value. Hence, surface S has exactly one stationary point. ( ) ( ) ( ) ( ) 2 22 22 2 ff f 22 4 2 88 4 4 8 0 for all . xx yy xyD aa aa aa = − = + −− = +− = +> ∈  And 2f 2 2 0 for all xx aa= +> ∈  . The stationary point is minimum point. (c) ( ) ( ) ( )( ) ( )( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) f , f 0,1 f 0,1 0 f 0,1 1 f 0,1 2 0 4 2 0 1 4 2 f 0 , 12 12 014 xy x y xy x y aa a a ≈ + −+ − = −+ − = −− = − −= ( ) ( )( ) ( ) ( )( ) ( ) f , 2 42 0 4 1 2 2 2 4 2 0 4 11 xy a x y h ak h ≈ +−− − + − + ≈ +−− − + +− ( )22 2 42 4h ak h+ ≈ +−− +

6 ( ) ( ) 2 42 2 h ak h ak ≈+ ≈+ (d)(i) At point P, normal vector of the tangent plane is 42 4 1 a−−   − . The tangent line at point P which is parallel to 1 2 1    also lies in the tangent plane, we have 1 42 240 11 42 810 23 3 2 a a a a −−      =     −   −− +−= = =  (d)(ii) Since P lies on the plane 0αx βy γz++= , ( ) ( ) ( )0 1 20 2 --- (1) α βγ βγ ++= =− Since the tangent line also lies in the plane 0αx βy γz++= , we have 1 20 1 2 0 --- (2) α β γ α βγ       ⋅=       + +=  Substitute (1) into (2), ( )22 0 3 α γγ αγ + − += = ( ) ( )32 0 Since 0, the equation of the plane is 3 2 0 γ x γ y γz γ x yz +− + = ≠ − += Alternative Solution: Clearly, O lies on the plane 0αx βy γz++= . Since P(0,1,2) also lies on the plane, 0 1 2 OP  =   is a direction vector of the plane. Hence 10 3 21 2 121      =×= −       n . So cartesian equation of the plane is 32 0x yz− += .

7 6 Suggested Solution (a) 0H : there is no association between the students’ year of study and their Primary study approach 1H : there is association between the students’ year of study and their Primary study approach Level of significance: 5% Assumption: Assume that the 100 students surveyed constitute a random sample. Test statistic: ( ) 2 2 ij ij ij ij OE E χ − =∑∑ Computation: Under 0H , expected frequencies, ij ij rc E N × = (N = grand total) are shown below: Primary study approach Year 1 Year 3 Total Study Alone 20.25 24.75 45 Group Study 13.5 16.5 30 Online Resources 11.25 13.75 25 Total 45 55 100 Calculation of contributions to test statistic: Primary study approach Year 1 Year 3 Study Alone 1.361 1.113 Group Study 3.130 2.561 Online Resources 0.1389 0.1136 2 8.4175χ∴= Degrees of freedom: ( )( )2131 2v=− −= ( ) 2value P 8.4175 0.01486p χ−= ≥= Conclusion: Since p – value = 0.01486 > 0.01 0H∴ is not rejected at the 1% significance level. Hence there is insufficient evidence to conclude that there is association between the students’ year of study and their primary study approach (b) The two cells that contribute the most to the chi-square test statistics are: • Group Study, Year 1 • Group Study, Year 3 These high contributions suggest that more Year 1 students chose Group Study than expected, while fewer Year 3 students did so. A possible explanation is that Year 1 students may prefer collaborative learning environments to adjust to university life, while Year 3 students, being more experienced and focused on final- year projects, may prefer studying alone or using online resources. No, this does not contradict the conclusion in part (a). While these cells contribute significantly to the chi -square statistic, the overall test did not provide enough evidence to conclude an association at the 1% significance level. The high contributions highlight specific deviations but do not change the overall result 7 Suggested Solution

8 (a) Possible conditions are: The average rate of reports being made is constant throughout any 1-hour time interval. Or Each report is independent of any other report. Reason why the condition may not hold: During peak period, there is a higher chance of report being made during the flight arrival period and hence rate of repo