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 =