2020 A Level H2 FM 9649 P2 (Qns & Solutions) (2 Oct 2024)

Possible Solutions to 2020 FM Paper 2 [Solution] 1 The system of equation can be represented by 1 2 3 4 5 1 1 2 1 5 0 1 3 2 1 1 0 1 0 1 1 0 0 0 1 1 1 2 0 x x x x x −−        −− =            −−     It can be reduced to 1 2 3 4 5 1 0 1 0 3 0 0 1 1 0 1 0 0 0 0 1 1 0 0 0 0 0 0 0 x x x x x         −− =    −             1 3 5 2 3 5 45 30 0 0 x x x x x x xx + + = − − = −= The solution is 1 3 5 2 3 5 3 3 3 5 45 55 3 13 11 10 01 01 x x x x x x x x x x xx xx −− −−               +       = = +                              The two vectors 1 1 1 0 0 −      and 3 1 0 1 1 −      are linearly independent. A basis is 13 11 ,10 01 01  − −                            (ii) dimension = 2

[Solution] (i) Considering z7 = 1 z7 = 1 = 2eik  where k = 0, 1, 2 and 3  2 7e ki z  = 7 6 5 4 1... 1 1 zz z z z z −+ + + + = − , ( ) 2 2 4 4 6 6i i i i i i7 7 7 7 7 7 71 1 e e e e e ez z z z z z z z      − − −      − = − − − − − − −              ( ) 2 2 4 4 6 6i i i i i i222 7 7 7 7 7 71 e e +1 e e +1 e e +1z z z z z z z      − − −        = − − + − + − +                    ( ) 222 2 4 61 2 cos +1 2 cos +1 2 cos +17 7 7z z z z z z z      = − − − −       7 6 5 4 1... 1 1 zz z z z z −+ + + + = − = 222 2 4 6( 2 cos 1)( 2 cos 1)( 2 cos 1)7 7 7z z z z z z   − + − + − + (i) [Method 1] Comparing the coefficient of the z term: 1 = 2 4 62cos 2cos 2cos7 7 7   − − − 2 4 6 1cos cos cos7 7 7 2   + + =−  2 3 1cos cos cos7 7 7 2   − − =− using cos( − ) = −cos .  2 3 1cos cos cos7 7 7 2   − + = [Method 2] The sum of roots of the equation 6 5 4 ... 1 0z z z z+ + + + = is −1 = 2 2 4 4 6 6 7 7 7 7 7 7e e e e e e i i i i i i     − − − + + + + +

 2 4 61 2cos 2cos 2cos7 7 7   − = + +  2 3 1cos cos cos7 7 7 2   − + = (ii) [Method 1]: Let z = i in 7 6 5 4 2 2 2 1 2 4 6... 1 ( 2 cos 1)( 2 cos 1)( 2 cos 1)1 7 7 7 zz z z z z z z z z z z   −+ + + + = = − + − + − +− 7 2 2 2 2 4 61 ( 1)( 2 cos 1)( 2 cos 1)( 2 cos 1)7 7 7i i i i i i i i   − = − − + − + − + −i – 1 = (i – 1) 2 4 6( 2 cos )( 2 cos )( 2 cos )7 7 7iii   −−− −i – 1 = 8i (i – 1) 2 4 6cos cos cos7 7 7     2 4 6 1cos cos cos7 7 7 8    =  2 3 1cos cos cos7 7 7 8    = using cos( − ) = − cos, [Method 2]: Comparing the coefficient of z3 term in 6 5 4 2 2 2 2 4 6... 1 ( 2 cos 1)( 2 cos 1)( 2 cos 1)7 7 7z z z z z z z z z z   + + + + = − + − + − + 1 = 2 4 6 2 4 62( 2cos 2cos 2cos ) 8cos cos cos7 7 7 7 7 7      − − − −  2 4 6 2 4 68cos cos cos 4(cos cos cos ) 17 7 7 7 7 7       =− + + −  2 3 18cos cos cos 4( ) 1 17 7 7 2    =− − − = Thus 2 3 1cos cos cos7 7 7 8    =

[Solution] (i)(a) 1 7 6 6 7 1 2 6 1 12 6 6 − − +      ==      −−       (b) 1 7 5 2 6 14 x y −     =    −     1 1 7 5 16 2 6 14 3 x y − −       ==       −        using a GC Coordinates of the point is (16, 3) (c) 1 7 7 2 6 1 2 6 xx x −−     =    −−     [A 3-mark question. So examiner expects you to say more about this image] 7 7 1 2 6 6 2 x xx −−     =+     −−      , x   The image are position vectors of points that lie on a line that passes through the point (7, −6) and is parallel to the vector 1 2 −  or points on the line y = -2x + 8. (ii) For A’ : 1 7 1 1 2 6 0 2 −−     =    −     B’: 1 7 1 6 2 6 1 4 −     =    −−     and C’ : 1 7 0 7 2 6 1 6 −     =    −−     Area of the parallelogram = OA OC 17 26 00 −       =  −          0 0 8  =  = 8 units2

[Solution] (i) xy = c2  0dyxydx+= At P(cp, c p ), 2 1 c pdy dx cp p −= =− Equation of the normal at P is 2 c py px cp − =−  2()c py p x cp− = − When the normal meets the curve again: 2 2 2 3 2( ) ( )cc p x cp pc cx p x xcpxp− = −  − = − 3 2 4 2 ( ) 0p x x c cp pc+ − − = (x – cp)(xp3 + c) = 0  x = cp (point P) or 3 cx p=− (point Q) Coordinates of Q are ( 3 c p− , −cp3 ) (ii) P(cp, c p ), PQ2 = 2 3 2 3( ) ( ) cccp cppp+ + + = 2 2 2 2 3 2 2 2 32 2( ) ( ) ( ) ( ) 2c c ccp cp c pp p p+ + + + + = 2 2 2 2 3 2 2 2 32 2( ) ( ) ( ) ( ) 2c c ccp cp c pp p p+ + + + + = 2 2 2 3 2 33[( ) ( ) ] ( ) ( )cccp cp pp+ + + = 3OP2 + OQ2

5 The curve C is defined parametrically by 349xt=+ , 24y t t= for 0 ta , where a is a positive constant. When C is rotated about the x-axis, the surface area generated is ( )Xa . When C is rotated about the y-axis, the surface area generated is ( )Ya . (i) Show that ( ) ( )( ) 1.53 9 27X a k a = + − for some integer k to be determined. [7] (ii) Find an expression, in terms of a, for ( )Ya . [5] (iii) Find, correct to 4 decimal places, for all values of a for which ( ) ( )X a Y a= and justify that there are no others. [3] (i) 2d 12d x tt = , 1 2d 36d y tt = 22 2 4 2dd 12 36dd xy tttt    + = +       ( ) 2312 9tt=+ ( ) 22 0 dd2d dd a xyX a y t tt    =+        ( ) 3 2 23 0 2 24 12 9 d a t t t t=+  ( ) 23 0 576 9 d a t t t=+  ( ) 23 0 192 3 9 d a t t t=+  ( ) 3 23 3 2 0 9 192 a t   +=   ( ) 1.53128 9 27a = + − , where k = 128 (ii) ( ) 22 0 dd2d dd a xyY a x t tt    =+        ( ) ( ) 3 2 3 0 2 4 9 12 9 d a t t t t= + + ( )( ) 1 234 0 24 4 9 9 d a t t t t= + + ( ) 3 24 3 2 0 9 24 a tt   +=   ( ) 3 2416 9 aa=+ (iii) When ( ) ( )X a Y a= ,

( ) ( ) 1.5 1.534128 9 27 16 9a a a + − = + ( ) ( ) 1.5 1.53 1.5 38 9 9 216 0a a a+ − + − = ( )( ) 1.51.5 38 9 216 0aa− + − = Solving (refer to graph of ( )( ) 1.51.5 38 9 216y a a= − + − ): ( )0 rej. 0aa= or a = 0.8765 or a = 3.8671. If 3.8671a , then ( ) 1.51.5 38 0 and 9 0aa−  +  ( )( ) 1.51.5 38 9 216 0aa− + −  . for all 0a , a = 0.8765 and a = 3.8671 are the only two solutions of ( ) ( )X a Y a= . Section B: Probability and Statistics [50 marks] 6 Cheng sells insurance policies by phone. He knows from past experience that 17% of the calls he makes will be successful in selling a policy. Cheng denotes by X the number of calls he makes each day up to and including his first successful call. (i) State two conditions under which X can be well modelled using a geometric distribution. [2] You are now given that X follows a geometric distribution. (ii) Find the probability that the fifth call Cheng makes one day is the first to be successful. [1] (iii) Find how many calls Cheng must make in order to be 95% certain that at least one will be successful. [3] (iv) Write down the mean, a, and standard deviation, b, of X. Hence find ( )P X a b+ . [3] (i) Whether a call is successful or not is independent of any other call. The probability of a call being successful is a fixed constant 0.17 for every call. (ii) ( )~ Geo 0.17X