2020 A Level H2 FM 9649 P1 (Qns & Solutions)(1 Oct 2024)

2020 A Level H2 Further Mathematics (9649) Paper 1 (Suggested Solution) [Solution] Using de Movire’s Theorem and Binomial expansion z4 = (cos  + i sin )4 = cos 4 + i sin 4 ----- (1) Also, ( ) 4 cos isin+ ( ) ( ) ( ) ( ) 2 3 44 3 2cos 4cos isin 6cos isin 4cos isin isin       = + + + + 4 3 2 2 3 4cos 4icos sin 6cos sin 4icos sin sin= + − − +        ( ) ( ) 4 2 2 4 3 3cos 6cos sin sin i 4cos sin 4cos sin= − + + −        --- (2) Comparing the real parts of (1) and (2): 4 2 2 4cos 4 cos 6cos sin sin    = − + Comparing the imaginary parts of (1) and (2): 33sin 4 4cos sin 4cos sin    =− Therefore, sin 4tan 4 cos 4  = 33 4 2 2 4 4cos sin 4cos sin cos 6cos sin sin         −= −+ 3 24 4 tan 4 tan 1 6 tan tan   −= −+ (dividing throughout by cos4 )

[Solution] (i) When  = 0, r = 1 + 2 sin 0 = 1 . The point the curve meets the initial line is (1, 0) When 3 2  = , r = 1 + 2sin 3 4  = 1 + 2 . The point is (1 + 2 , 3 2  ) (ii) r = 1 + 2 sin 2  cos 2 dr d   = 2 2 2 2 (1 2sin ) cos22 drr d   + = + + [This is not the simplest yet !] = 221 4sin 4sin cos2 2 2   + + + = 2 + 24sin 3sin22 + Thus L = 2 220 2 (2 4sin 3sin ) d   ++ Using a GC, L = 15.13490477  15.135 (correct to 3 dps)

[Solution] Given Xn = 2nXn – 1 – n(n – 1)Xn – 2 , n  3  122 ( 1) ! ! ! n n nX nX n n X n n n −− −=−  122 ! ( 1)! ( 2)! n n nX X X n n n −−=− −− Let ! n n XU n= . Thus U1 = 1 21 X = and U2 = 2 7 2! 2 X = Recurrence relation becomes: Un = 2Un – 1 – Un – 2 ---- (*) The characteristic equation is m2 – 2m + 1 = 0 (m – 1)2 = 0  m = 1 The general solution for (*) is: Un = A + Bn n = 1, 2 = A + B ----- (1) n = 2, 7 2 = A + 2B  7 = 2A + 4B ------- (2) (2) - 2(1): 3 = 2B  3 2B= and thus A 1 2= Thus Un 1 2= + 3 2 n  1 3 ! (3 1)! 2 2 2 n n X nn X nn = +  = + , n = 1, 2, …

[Solution] (i) f(3) = 33 – 9(3) – 14 = – 14 < 0 f(4) = 43 – 9(4) – 14 = 14 > 0 There is a change of sign in the interval [3, 4] and f is a continuous function. So there is a root (or at least one) in the interval [3, 4] (ii) Linear interpolation is based on drawing a chord joining the points (3, −14) and (4, 14) and the intersection between the chord and the x-axis. So an approximation must be in (3, 4). So we can take  = 3.5 as an approximation of . (iii)(a) f’(x) = 3x2 – 9 f’’(x) = 6x For x  (3, 4) , f’(x) > 0 and f’’(x) > 0. So, f is increasing and concave upwards in (3, 4). Thus  is an under-estimate of  (b) So  <  < 4 2nd stage of linear interpolation: 2 3.5 (4) 4 (3.5) (4) (3.5) ff ff += + = 59.5 16.625 = 3.57894  3.58 (3 sf)

[Possible Solution] (i) z3 = z1z2 = (3 + i)(2 + i) = 5 + 5i Note: The word ‘deduce” forces us to use the diagram to obtain 1111tan tan4 2 3  −−=+ . If not arg z3 = arg z1 + arg z2 will do ! Consider the (z2)* = 2 – i and 1 + 2i In the diagram, OP*2 P1A is a square with diagonal OP1 P1OP*2 = 4  and the angle between OP2* and the positive x-axis is  = 1 1tan 2 −

The angle between OP1 and the positive x-axis is 1 1tan 3 − =  . Thus 1111tan tan4 2 3   −−= + = + (ii) ( ) 1 1 1 1 1 1 tan tantan 1 tan tan tan tan tan tan tan tantan 1 tan tan tan tan tan 1 ABAB AB x A A x y B B y ABAB AB xyxy xy − − − − − − ++= − =  = =  = ++=  −  ++=  − It is known that 1 1 1tan tan tan 1 xyxy xy − − −  ++=  − Similarly, 1 1 1tan tan tan 1 xyxy xy − − −  −−=  + Thus 1 1 1 2 1 1 ( ) 1tan tan tan 11 1 ( ) ( ) 1 k k n k n n k kn k n nk n k n n k − − −  + + + +    +=   + + +     − + + + = 2 2 2 11 22 1 ( ) 2 1tan tan( )( 1) ( ) n nk n k k n nk k n k n nk k n n k n k k −−   + + + + + + + =   + + + − + + + −    2 11 2 ( ) 1 1tan tan (( ) 1) nk n n k n −− ++==  ++ Alternatively use 1 1 1tan tan tan 1 xyxy xy − − −  −−=  + will be less tedious Consider 1 1 1 11 11tan tan tan 111 n n k n n k n n k − − −  − +−=  +  +  + ( ) ( ) ( ) 11 2tan tan 1 1 n k n n n k k n n k n nk n n k −−  +−  + ==  ++ ++ + (shown)

(ii) Using (ii), 1 1 1 1 1 2 1 1 1 1 1tan tan tan tan tan2 2 1 2 2 1 3 7 − − − − −= + = ++ + + 1 1 1 1 1 2 1 1 5 1 1tan tan tan tan tan7 7 5 7 35 1 12 17 − − − − −= + = ++ + + 1 1 1 1 1 2 1 1 15 1 3tan tan tan tan tan17 17 15 17 17(15) 1 32 109 − − − − −= + = ++ + + and 11 1 1 1 1 1 46 173 11 46 173 1 1 219 3tan tan tan tan tan ( )46 173 1 7957 109 − − − − −  + + = = =  −  Thus 1 1 1 1 11 1 1 1 3tan tan tan tan tan4 3 3 12 32 109  − − − − −= + + + + 1 1 1 1 11 1 1 1 12 tan tan tan tan tan4 3 12 32 46 173  − − − − −= + + + + [Solution] Given En = 5(En – 1)3 – 3En – 1 and E0 = 1 (i) E1 = 5 – 3 = 2 E2 = 5(23) – 3(2) = 34 E3 = 5(343) – 3(34) = 196418

(ii) 1 () 5 nn nF =− were 1 5 1 5 and 22 +−== Using a GC, F9 = 9 9 9 91 1 1 5 1 5( ) ( ) ( ) 34 2255   +−− = − =  27 27 27 27 27 1 1 1 5 1 5( ) ( ) ( ) 196418 2255 F   +−= − = − =  (iii) Note that E0 = 1 = F1 E1 = 2 = F3 E2 = 34 = F9 = 23F E3 = 196418 = F27 = 33F Conjecture: En = 3nF for n = 0, 1, 2, 3, … Let P(n) be the statement En = 3nF for n = 0, 1, 2, 3, … where 1 () 5 nn nF =− P(0), P(1), P(2) and P(3) are true. Assume P(k) is true for some positive integer k, that is Ek = 33 3 1 () 5 kk kF =− . 1 5 1 5 1, (1 5) 12 2 4  +−= =  = − =− | To prove P(k+1) is true that is Ek+1 = 11 1 33 3 1 () 5 kk kF  ++ + =− LHS = Ek+1 = 5(Ek)3 – 3Ek = Ek (5(Ek)2 – 3) = 2 33[5( ) 3]kkFF − = 33 3 3 21 ( ) 5( ) 3 55 kk kk   −−−   = 3 3 2(3 ) 2(3 ) 3 31 ( ) 2( ) 3 5 k k k k k k      − + − − = 3 3 2(3 ) 2(3 ) 31 ( ) 2( ) 3 5 k k k k k     − + − − = 3 3 2(3 ) 2(3 )1 ( ) 1 5 k k k k    − + − as 33( ) ( 1) 1 kk  = − =− = 3(3 ) 3 2(3 ) 3 3 2(3 ) 3(3 ) 31 () 5 k k k k k k k k        + − − − + = 113 3 3 3 3 3 3 3 3 31 () 5 k k k k k k k k k k           ++ − + − − + = 113 3 3 3 3 31 () 5 k k k k k k       ++ − − − + + = 11 1 33 3 1 () 5 kk kF ++ +−=

P(k) is true  P(k+ 1) is true. Thus by induction …. 7 Use the substitution cosu y x= to find the general solution of the differential equation 2 2 dd cos 2 sin cos 0dd yy x x ay xxx − + = in each of the cases • 1,a=− • 3a= , giving each answer for y in the form f ( )yx= . [11] cosu y x= ----- (1) dd cos sindd uy x y xxx=− 22 22 d d d d cos sin sin cosd d d d u y y y x x x y xx x x x= − − − 2 2 dd cos 2 sin cosdd yy x x y xxx= − − ----- (2) Substitute eqn (1) and (2) into DE: 2 2 dd cos 2 sin cos 0dd yy x x ay xxx − + = becomes 2 2 dd cos 2 sin cos cos cos 0dd yy x x y x y x ay xxx  − − + + = 2 2 d ( 1) 0d u aux + + = When 1a=− , 2 2 d 0d u x = 1 d d u Ax = 1 1 2du A x A x A= = + 12cosy x A