2017 A Level H2 FM 9649 P2 (Qns & Solutions)

1 2017 GCE A Level H2 FM 9649 Paper 2 Solutions [Solution] (i) Recurrence relation: 123r r ru u u −−=− Auxiliary equation: 2 3 1 0mm− + = 35 2m = . Hence, general solution is: 3 5 3 5 22 rr ru A B    +−=+           , r = 0, 1, 2, … (ii) Using the initial conditions, 0 2u = 2 (1)AB + = −−−− 1 3u = 3 5 3 5 3 (2)22AB   +− + = −−−−          ( ) 3 5 3 523 22BB   +− − + =           3 5 5 3 B + − = 1, 1BA = = . Hence, 3 5 3 5 22 rr ru    +−=+          (iii) When 0 2u = , from (1) we have ( )3 5 3 5 222 rr ru A A    +−= + −          3 5 3 5 3 52 2 2 2 r r r A      − + − = + −                  3501 2 − , when ,r→ 35 02 − → . So, when ,r→ 00ruA→  = 352 2 r ru  −=  and 1 35u =− .

2 [Solution] 2 2 21 cos cos 2 cos3 cos 21 2 3 2 2 2 + sin sin 2 sin 3 sin 21 2 3 n n nC iS n n n nin              + = − + − + +                 − + − + +            ( ) ( ) ( ) 221 cos sin cos 2 sin 2 cos 2 sin 212 nnC iS i i n i n         + = − + + + − + +       2 3 22 2 21 e e e e1 2 3 i i i i nn n nC iS         + = − + − + +           Compare with: ( ) 2311 1 2 3 n nn n n nx x x x x n        − = − + − + +               for n + ( ) 2 1 niC iS e  + = − ( ) 2 222 n iii e e e  −=− 2 cos sin cos sin2 2 2 2 n ine i i       = − − +       2 2 sin 2 n inei =−  ( ) ( ) 2 22 22 sin n nin nei =−  ( )( ) 2 2cos sin 4 sin n nn i n = + − Comparing real parts: ( ) 2 24 cos sin n nCn =− Comparing imaginal parts: ( ) 2 24 sin sin n nSn  =−

3 [Solution] 2, 2 , 0x at y at t p= =   (i) Curve surface area 22 0 dd2 dd p xyy dt tt    =+        ( ) ( ) ( ) 22 0 2 2 2 2 p at at a dt=+ 22 0 4 2 1 p a t t dt=+  ( ) 3/22 2 0 1 4 3 / 2 p t a  +=  ( ) 3/2228 113 ap = + −  (ii) When p is small, Curve surface area rl= ( ) 3/2228 113 ap = + −  2283 1132 ap  + −  224 ap= When p is small, the arc of the parabola is almost a line segment joining (0, 0) to (ap2, 2ap), so the approximate surface area 224 ap is the surface area of the cone generated when this line segment is rotated about the x-axis = ( ) ( ) 2 22(2 ) 2ap ap ap+ ( ) 1 1 2 22 2 2 2 2 2 2 22 4 4 1 4 4 pa p p a p a p =  + =  +    x y 0 P(ap2, 2ap) x

4 [Solution] (i) 2 1 1 3 2 1 3 1 2 4 3 1   =   M ~ 2 1 1 0 1 1 0 1 1 0 1 1   −  −  − ~ 2 1 1 0 1 1 0 0 0 0 0 0   −    rank of M = 2  number of independent rows of M = 2  number of independent columns of M = 2 (ii) (a) 20 0 x y z yz + + =  −= ,x z y z =− =  A basis for the null space of M = 1 1 1  −   . (b) Since the column space = range space, and the first two columns are linearly independent,  a basis for the range space of M = 21 32,31 43                     .

5 [Solution] 2 2 2 dd 5 6 2edd xyy yxx − + + = Auxiliary equation: 2 5 6 0mm+ + =  2 or 3m=− − Complimentary Function: 23 ee xx y A B −− =+ where A and B are arbitrary constants. Let particular integral be 2 e x y px − = 22d e 2 ed xxy p pxx −− =− 2 22 2 d 4 e 4 e d xxy p px x −− =− + Substitute into DE: 2 2 2 2 2 2 4 e 4 e 5 e 10 e 6 e 2e x x x x x x p px p px px − − − − − − − + + − + = 22 e 2e 2 xx p p −− = = General Solution is 2 3 2 e e 2 e x x x y A B x − − − = + + 2 3 2 2d 2 e 3 e 2e 4 ed x x x xy A B xx − − − − =− − + − When 0, 0xy== , 0AB+= ---- (1) When d0, 5d yx x== , 2 3 2 5AB− − + = 2 3 3AB+ =− ----- (2) Solving (1) and (2): 3, 3AB= =− Particular solution is ( ) 23 3 2 e 3e xx yx −− = + − 3 2 2d 9e 4e 4 ed x x xy xx − − − = − − When d 0,d y x = 3 2 2 9e 4e 4 e 0 x x x x − − − − − = ( )4e1 9 x x − =+ Since e x y − = is a strictly decreasing function and ( )4 19yx=+ is a strictly increasing function, therefore the two graphs e x y − = and ( )4 19yx=+ intersect only once. Hence d 0d y x = only has one real solution. Thus, ( )0.444,0.808 is the only turning point on the graph of 2 3 2 e e 2 e x x x y A B x − − − = + + . x y 0

6 [Solution] (i) Consider 0.7 0.6 0.4 0.2 0.2 0.1 0.4 0.4    − − = − − AI When  = 1, ( )det A I − ( ) ( )( ) ( )0.3 0.4 1 0.08 0.6 0.2 0.4 1 0.02 0.4 0.08 0.1=− − − − − − − + − −           = 0  A has an eigenvalue equal to 1. And A –  I = 0.3 0.6 0.4 0.2 1 0.2 0.1 0.4 0.6 −  − − Let 0.2 0.1 0.52 1 0.4 0.14 0.2 0.6 0.18            = −  =          −      x , we have 0.3 0.6 0.4 0.52 0 0.2 1 0.2 0.14 0 0.1 0.4 0.6 0.18 0 −         −         −     = Therefore, an eigenvector of A corresponding to eigenvalue 1 is 0.52 0.14 0.18    . (ii) ( ) 1 0.7 0.6 0.4 1 0.3 0 0.2 0.2 0 0 1 0.1 0.4 0.4 1 0.3    −−             − = − =            − − − − +       AI   = 0.3 ( ) 10 00 10          − =       −    AI Therefore, an eigenvector of A corresponding to eigenvalue 0.3 is 1 0 1   − . Similarly,

7 ( ) 2 0.7 0.6 0.4 2 2 0.4 5 0.2 0.2 5 5 1 3 0.1 0.4 0.4 3 0.6 3      − − −             − − = − − = +             − − −       AI   = −0.2 ( ) 20 50 30          − − =           AI Therefore, an eigenvector of A corresponding to eigenvalue −0.2 is 2 5 3  − . 1−A=QDQ 0.7 0.6 0.4 0.2 0 0.2 0.1 0.4 0.4  = A , let 0.2 0 0 0 0.3 0 0 0 1 − = D , and 2 1 0.52 5 0 0.14 3 1 0.18  =− − Q AQ = 0.7 0.6 0.4 0.2 0 0.2 0.1 0.4 0.4    2 1 0.52 5 0 0.14 3 1 0.18  − − 0.4 0.3 0.52 1 0 0.14 0.6 0.3 0.18 − =−− QD = 2 1 0.52 5 0 0.14 3 1 0.18  − − 0.2 0 0 0 0.3 0 0 0 1 −   0.4 0.3 0.52 1 0 0.14 0.6 0.3 0.18 − =−− Thus AQ = QD  1−=A QDQ If A = QDQ–1, A2 = (QDQ–1)(QDQ–1) = QD(Q–1Q)DQ–1 = QD(I)DQ–1 = QD2Q–1 A3 = (QD2Q–1)(QDQ–1) = QD3Q–1  An = QDnQ–1 for n = 0, 1, 2,  When n = −1, A−1 = (QDQ−1)−1 = (Q−1)−1D−1Q−1 = QD−1Q−1 Thus An = QDnQ–1 for n = −1, 0, 1, 2,  ( ) 1 1 0 0 0 0.3 0 0 0 0.2 QQ n n −   =   −

8 [Solution] Let X and Y be the respective time from taking the paracetamol tablets and the new tablet till gaining pain relief. Let x and y be the population mean time of X and Y respectively. Assumptions: X and Y follow two independent normal distribution with a common variance. 1 30n = , 1609 53.63330x== , 2 2 1 160992314 207.516130 1 30 xs = − = −  2 26n = , 1242 47.7692326y== , 2 2 1 124267173 313.744626 1 26 ys = − = −  sp2 = 22( 1) ( 1) 2 x x y y xy n s n s nn − + − +− = 16.02172 H0: 0yx−= H1: 0xy− Level of significance: 5% Under H0, Test statistic: 30 26 2 54 0 ~ 11 30 26 p XY tt S +− −− = + From GC, tcal = 1.366 p-value = 0.0888 > 0.05 Since p-value > level of significance, we do not reject H0. There is insufficient evidence at 5% level of significance to conclude that the new tablet acts more quickly than paracetamol.

9 [Solution] H0: The data are