2022 A Level 9649 FM P2 (Solutions)

Paper 2 Remarks 1 2d 32d y ax bxx =+ At the turning points, d 0.d y x = So, 23 2 0 (3 ) 0 20 or . 3 ax bx x ax b bxx a += += = =− When 0,x= .yc= When 2 ,3 bx a=− 32 32 32 3 2 3 2 22 33 84 27 9 84 27 9 4 27 bby a b c aa bba b c aa b ca b ca    = − + − +          = − + +       = − + + =+ For the equation 32 0ax bx c+ + = to have 3 distinct real roots, the y- coordinates of the two turning points must have different signs. Thus ( ) 3 2 33 3 2 3 4 027 4 27 0 27 4 0 (shown) bcc a c b a c a c b c  + + + 2 (a) ( )21 1 2 n n nu u u++ =+ 212 0n n nu u u++− − = The auxiliary equation of the RR is 2 10 (2 1)( 1) 0 2 1 or 12 mm m mm m −= + − = =− − = 1 (1)2 1 2 n n n n n u u cd cd = − + = − + 1 2 1 (1)2 1 (2)4 u u a c d a b c d b = − + = − = + = −

(2) – (1): 4 ()3c b a=− (1) + 2(2): 1 ( 2 )3d a b=+ 4 1 1( ) ( 2 ), 13 2 3 n nu b a a b n  = − − + +  (b)(i) 4 1 1( ) ( 2 ), 13 2 3 n nu b a a b n = − − + +  As , 3 1since 0 as , therefore2 1 ( 2 ) 33 29 n n nu n ab ab → → − → → += += Since a and b are distinct positive integers, hence possible pairs of values of a and b are: 1and 4, 5 and 2, 7 and 1a b a b a b= = = = = = (b)(ii) If ab= , then 4 1 1(0) ( 2 )3 2 3 n nu a a = − + + ,1nu a n= Then the sequence is a constant sequence.

3 (a) 00 ( ln )Let f ( , ) , 1, 1.y x yx y x y x += = =     1 0 0 0 1 0 0 0 1 1 2 1 1 1 2 1 1 1 2 2 1 ln10.5f ( , ) 1 0.5 1.5 1 0.5 1.5(1.5 ln1.5)f ( , ) f ( , ) 1 0.25 12 1.5 1.72637... 0.5f ( , ) 2.90376... 0.5 f ( , ) f ( , ) 3.42793... 3.43 (3sf)2 u y x y y y x y x u u y x y y y x y x u += + = + =  += + + = + +  = = + = = + + = = (b)(i) ,0 πsin xxyx x=   Taking logarithms, we have ln ln ln sin .y x x x=− 1 d 1 cos lnd sin yx xxy x x x = + − At turning point, d 0.d y x = 1 cot 1 ln 1tan 1 ln 1tan (shown)1 ln xx x x x x − =+ = + =  + (b)(ii) 10 g( ), with 0.9nnx x x+ == From GC, 12 0.84095, 0.87994, , converges to 0.864 (3s f).nx x x==  x-coordinate of the turning point is 0.864, y-coordinate is 1.16 (3sf) 4 (a)(i) sin cosy A x B x=+ ( ) ( ) 11 22d 1 1 cos sind 2 2 d2 cos sind y A x x B x xx yx A x B xx −−   =−         =− ( ) 2 2 2 2 dd4 2 sin cosdd dd4 2 0 (shown)dd yyx A x B xxx yyxy xx + =− + + + = ( ) ( ) 1 1 12 2 2 2 2 d d 1 12 sin cosd d 2 2 yyx x A x x B x xxx − − −     + = − −            (a)(ii) Let nP be the statement 21 21 d d d4 (4 2) 0d d d n n n n n n y y yxn x x x ++ ++ + + + = for all integers 0n . LHS of 22 0 22 d d d d4 [4(0) 2] 4 2d d d d y y y yP x y x y x x x x= + + + = + +

RHS of 0 0P = LHS = RHS (by (i)) 0P is true. Assume that kP is true for some integers 0k  , i.e., 21 21 d d d4 (4 2) 0d d d k k k k k k y y yxk x x x ++ ++ + + + = We want to prove that 1kP + is true, i.e., 3 2 1 3 2 1 d d d4 (4 6) 0d d d k k k k k k y y yxk x x x + + + + + ++ + + = Consider kP , 21 21 3 2 2 1 3 2 2 1 3 2 1 3 2 1 d d d4 (4 2) 0d d d d d d d4 4 (4 2) 0d d d d d d d4 (4 6) 0 (shown)d d d k k k k k k k k k k k k k k k k k k k k y y yxk x x x y y y yxk x x x x y y yxk x x x ++ ++ + + + + + + + + + + + + + + + + + = + + + + = + + + = 1 is true is truekkPP + Since (1) 0P is true, and (2) kP is true 1kP + is true, by mathematical induction, nP is true for all integers 0n . (b) 2 2 2 dd4 2 25dd yyx y xxx+ + = − Let 2Q( )x Dx Ex F= + + , then Q ( ) 2 Q ( ) x Dx E xD  =+  = Hence, 22 22 4 2(2 ) ( ) 25 (12 ) (2 ) 25 xD Dx E Dx Ex F x Dx D E x E F x + + + + + = − + + + + = − By comparing coefficients, 2 1, 12 0, 2 25 1, 12 and 1 Q( ) 12 1 D D E E F D E F x x x = + = + =− = =− =−  = − − Let F( ) sin cosx A x B x=+ and Q( ) F( )y x x=+ , then           ( ) 2 2 2 2 ddLHS 4 2 dd 4 Q ( ) F ( ) 2 Q ( ) F ( ) Q( ) F( ) 4 Q ( ) 2Q ( ) Q( ) 4 F ( ) 2F ( ) F( ) 25 0 25 RHS yyxy xx x x x x x x x x x x x x x x x x x = + +    = + + + + +    = + + + + + = − + = − =

Hence, Q( ) sin cosx A x B x++ for any arbitrary constants A and B, is a solution of this differential equation. 5 (a) ( )( ) ( ) ( ) ( ) ( ) ( ) 22 2 2 2 2 2 2 2 2 2 2 2 22 2 22 2 2 2 2 2 2 2 22 2 2 1 2 cos 2 cos sin sin 1 2cos 2cos sin sin 1 2cos 2cos sin sin 1 cos cos 2sin sin2 seccos cos 2 2 tan tan sec 1 1 2 tan tan sec 1 1 tan x xy y r r r r rr r r r r r                       + + = + + = + + = ++ =  + + = + + = + + + = ++ ( ) 2 2 2 2 sec sec (shown) 1 1 tan r    = = ++ (b) ( ) ( ) ( ) 1 4 1 4 π 2 2 0 π1 0 11 1 1 1 secArea d 2 1 1 tan 1 tan 1 tan2 1 π1tan 1 tan tan 1 02 4 2 11 πtan 22 2 4 1 πtan 228    − −− − − = ++ =+  = + − + = −  =−   (c) 2 2 22 sec 1 ...1 (1 tan ) 1 2sin cos cosr     = = =+ + + + 22 22 d 2sin cos 2sin 2cos2 d (1 2sin cos cos ) rr         +−= ++ When d 0d r  = , 22sin cos cos sin 1 sin 2 cos 22 tan 2 2       =− = = Let tanm = 2 2 2 2 1 01 m mmm =  + − =− 15 2m −= (d) Exact area enclosed by E

1 1 1 1 15tan 22 215tan 2 15tan 21 15tan 2 11 11 1 sec4d 2 1 (1 tan ) 2 tan (1 tan ) 1 5 1 52 tan 2 tan22 1 5 1 52 tan tan 22    − − − − −+ −− −+−  −− −− −− = ++ =+     +−=−               +−=−               2 2   = = 6 (a) Let T denote the time spent by Chan waiting at the bus stop in minutes. Thus, ( )~ Exp 0.2T . ( ) 1 51fe 5 t t − = , t > 0 ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 0.2 6 0.2 0.2 6 0.2 0.2 6 0.2 1.2 P 6 and P 6 P P6 P 0.2e d 0.2e d e e 0e 0e e x t x t x t x t t t T t T tT t T t Tt Tt Tt x x  − +  − − + − −+ − −  +  +  =  +=  = −= − −− = −− =   which is independent of t. (b) ( ) ( ) ( ) ( ) 2 2 2 0.8 Var Var 0.8 0.8 Var 0.64 5 16 litres . YT YT T = = = = = 7 (a) The Wilcoxon test is appropriate in this case as it takes into account both the magnitude and sign of the difference in the amount each customer spends on Friday and on Saturday, which are available from the data. Moreover, the distribution of the difference in the amount spent by each customer on Friday and on Saturday is not known and is unlikely to follow a normal distribution. Note ( ) 11E5 5T = =  = Not allowed to quote lack of memory property. Question is basically asking you to show. Need to integrate. Cannot just quote cdf and use cdf to show! NO need to integrate!

(b) Let D = amount spent by customer on Friday − amount spent on Saturday. Let Dm be the population median of D. (NOT differences of median) 0 1 :0 :0 D D Hm Hm =  Customer A B C D E F G H I J Amount spent on Friday 190 152 139 100 90 88 60 55 42 20 Amount spent on Saturday 150 120 132 103 99 83 66 40 32 24 D 40 32 7 −3 −9 5 −6 15 10 −4 Rank 10 9 5 −1 −6 3 −4 8 7 −2 sum of positive ranks = 42P= Q = sum of negative ranks = 13 ( )min , 13T P Q== From MF26, for n = 10, 1-tailed test at 5% significance level, critical region is 10T  . Since T = 13 lies outside the critical region, we do not reject H0 and conclude that there is insufficient evidence at 5% significance level that on average, customers who shop regularly on both Fridays and Saturdays sp