RVHS ACJC EJC NJC 2022 FM Prelim P2 (Solutions)

1 9649/02/2022 Section A: Pure Mathematics [44 marks] Solution 2. (a) det( ) 0A A is invertible but 00 00   is non - invertible. OR 00det 000  = 00 00   is not in the set. Since the set does not contain the zero matrix, it is not a subspace of ( )22M  . (b) Let U denote the set of all 22 matrices B such that T =−BB . 0 0 0 0 0 0 0 0 0 0 0 0 T      = =−           Hence 00 00 U  and U is non-empty. Consider any two elements B , C U . ( ) ( ) ( ) T TT+ = + =− + − =− +B C B C B C B C U + BC (Closure under addition) Consider any element B U and k  . ( ) ( ) ( ) T Tk k k k= = − =−B B B B kUB (Closure under scalar multiplication) Hence U is a subspace of ( )22M  . Solution 3. dsec sec tan .d xx   =  = d d d d sec tan .d d d d y y x y xx ==

2 9649/02/2022 ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 22 23 22 2 2 2 2 3 2 2 2 2 2 2 2 2 4 2 2 2 2 4 2 2 2 d d d d sec tan sec tan secd d d d dd sec tan sec tan secdd dd sec sec 1 sec sec 1 secdd dd sec sec sec 2sec 1dd dd 21dd y y x y xx yy xx yy xx yy xx yyx x x x xx                    = + + = + + = − + − + = − + − = − + − ( ) ( ) 22 32 22 1 d d d 2 1 .d d d y y y x x xx x x  = − + − Substituting into the differential equation in x and y: 2 23 2 22 2 22 2 2 2 1 d 2 d d2 d d2 d sec d 2cos (shown)d y ky x x x y ky x y ky y ky      += += += += Characteristic equation: ( ) 2 0 i 0 mk m k k += =   Complementary function: ( ) ( ) 11 cos sin cos sec sin sec . cy A k B k A k x B k x  −− =+ =+ Since 22cos cos 2 1=+ , particular integral: cos 2 sin 2py C D E = + + 2 sin 2 2 cos 2 4 cos 2 4 sin 2 . p p y C D y C D    =− + =− − Substituting into the new differential equation: ( ) ( ) ( ) 4 cos 2 4 sin 2 cos 2 sin 2 cos 2 1 4 cos 2 4 sin 2 cos 2 1 C D k C D E C k D k kE         − − + + + = + − + − + = + Comparing coefficients:

3 9649/02/2022 1 1 , kE E k = = ( )41 1 ,4 Ck C k −= = − and ( ) ( ) 40 0 4 . Dk Dk −= =  11 cos 24 py kk  = + − . Thus, 11cos sin cos 2 4y A k B k kk  = + + + − . 2 2 2 22 Now, cos 2 2cos 1 2 1sec 22 1. x xx   =− =− −= − = Thus, the general solution is ( ) ( ) ( ) 2 11 2 21cos sec sin sec 4 xy A k x B k x k x k −− −= + + + − where A and B are arbitrary constants. Source: TMJC Promo 2025 Qn 4 (a) ( ) 22 22 22 f, f f x y x y x y x xy y xy =+ = + = + 22 22 f x xy y xy   +=   + At 1, 2,xy== 11f 25 =  Directional derivative in the direction of greatest ascent is in the direction 1 2 = u  1111f f 1 2255 D    = = =        u u  (b) Equation of tangent plane at 00,:x x y y== ( ) ( )( ) ( )( ) ( ) ( ) 0 0 0 0 0 0 0 0 22 00 0 0 0 0 2 2 2 2 0 0 0 0 f , f , + f , xyz x y x y x x x y y y xyz x y x x y y x y x y = + − − = + + − + − ++

4 9649/02/2022 When 0, 0,xy== ( ) ( )22 00 0 0 0 0 2 2 2 2 0 0 0 0 22 22 00 00 22 00 0 xyz x y x y x y x y xyxy xy = + + − + − ++ += + − + = Hence, ( )0,0,0 lies on all tangent planes. Hence, all tangent planes to the surface pass through the origin. (c) Equation of normal to the plane at 00,:x x y y== 0 22 00 0 0 0 22 0022 00 ,. 1 x xyx yy xy xy    +  = +   + + −  r  In parametric form: 0 00 2 2 2 2 0 0 0 0 0 00 2 2 2 2 0 0 0 0 22 00 1 1 xx x x x y x y yy y y x y x y z x y       = + = + ++   = + = + ++  = + − Sub into equation of surface ( ) 22f,z x y x y= = + , we have: ( ) 22 22 0 0 0 0 2 2 2 2 0 0 0 0 2 22 00 22 00 22 00 22 00 22 00 11 1 1 x y x y x y x y xy xy xy xy xy                  + − = + + +       ++        = + +  +  = + +  + = + +

5 9649/02/2022 ( ) ( ) 2 2 2 2 0 0 0 0 2 2 2 2 2 2 2 2 0 0 0 0 0 0 0 0 or or 1 1 (n.a.) 0 x y x y x y x y x y x y        + − = + + + − = + + + − =− + + − = =− = When 0 = , we get the point ( ) 22 0 0 0 0,,x y x y + so the normal only intersects the surface at ( ) 22 0 0 0 0,,x y x y + and does not intersect the surface again. Solution 5. (i) ( ) ( )21 7 π 3π 5π 9π 11π 13π 7 7 7 7 7 7 7 7 i 2 1 π7 i π i i i i i i iπ , 5, 6 128 0 128 128e , 2e , 0,1, 2, 3, 4 2e , 2e , 2e , 2e , 2e , 2e , 2e k k z z zk zk + + = += =− = ==  (ii), (iv) (iii) Considering sum of roots of the equation 7 128 0z += and taking real part,

6 9649/02/2022 2 2 2 π 3π 5π 9π 11π 13πcos cos cos cos π cos cos cos 07 7 7 7 7 7 π 3π 5π 5π 3π πcos cos cos ( 1) cos cos cos 07 7 7 7 7 7 π 3π 5π2 cos cos cos 1 07 7 7 π 3π 5π 1cos cos cos7 7 7 2 π 3π 5π1 2sin 1 2sin 1 2sin14 14 14 + + + + + + = + + + − + + + =  + + − = + + =      − + − + − =           2 2 2 1 2 2 2 2 1 2 π 3π 5π 13 2 sin sin sin14 14 14 2 3π 3π 5π 5sin sin sin14 14 14 2 4       − + + =            −      + + = =           (iv) Area ( ) 2 2 12 π π 1 π2 sin 3 sin or 4 2sin2 7 7 2 7 π4sin or 1.74 units7  =  +    = (v) ( )( ) ( )( ) ( ) ( )( ) ( ) ( )( ) 1 2 3 4 3 2 1 4 1 2 3 2 3 4 1 4 arg arg arg arg arg z z z z z z z z z z z z z z z z  −−  −− = − − − + − − −

7 9649/02/2022 This argument represents the sum of two opposite angles in a cyclic quadrilateral (or equivalently angles in opposite segments) and thus they must add up to  radians (shown). (Note that since 1,z 2,z 3z and 4z are any 4 distinct roots, the quadrilateral shown above is not unique.) Section B: Statistics [50 marks] Solution 6. (i) 0H : The colour preference is independent of gender 1H : The colour preference is dependent on gender Under 0H , the expected frequencies are row total column total grand total ie = White Green Black Blue Red Others Male 92 75.3 65.3 53.3 84 30 Female 46 37.7 32.7 26.7 42 15 Degree of freedom ( ) ( )6 1 2 1 5= −  − = From GC, 2 test 14.02682385 = and -value 0.0154399223p = Since p-value < 0.05, we reject 0H at the 5% significance level and conclude that there is sufficient evidence to claim that preferences for colours of gaming keyboards differ between male and female gamers. (ii) The table of contributions are as follows: White Green Black Blue Red Others M 0.0435 0.00147 2.08 0.00208 2.01 0.533 F 0.0870 0.00295 4.17 0.00417 4.02 1.07 The largest contributions are from the “Black” and “Red” columns. More male prefer black while fewer male prefer red than expected, and more female prefer red while fewer female prefer black than expected. Solution 7. (i) To carry out a t-test on the data will require the condition that the scores (or difference between the scores) to be normally distributed. This condition might not be met by the data and hence it is not appropriate to use the t-test. (ii) Let X = score after module – score before module Assume that the distribution of the population of the differences is symmetrical. 0H : Population median of X is 0 1H : Population median of X is greater than 0

8 9649/02/2022 A B C D E F G H I J K L 1st 71 70 65 42 74 81 56 69 57 77 81 55 2nd 51 72 46 93 95 31 81 84 88 67 70 82 X −20 2 −19 51 21 −50 25 15 31 −10 −11 27 R 6 1 5 12 7 11 8 4 10 2 3 9 From table, the sum of negative ranks, 27Q= . From MF26, at 1% level of significance with 12n= , One-tail critical value of ( )9 or 69T  Since 27 falls beyond the critical region, we do not reject 0H and conclude at 1% level of significance that there is insufficient evidence to claim that the population median for the scores in the writing test has improved (i.e. insufficient evidence to conclude that the module is effective). Solution 8. (i) For a p.d.f., 1 2 0 d11 k