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 += +