(c) (cos θ + i sin θ) 5 = cos 5 θ + 5 cos 4 θ (i sin θ) + 10 cos 3 θ(i sin θ) cos 2 θ(i sin θ) 3 + 5cos θ (i sin θ) 4 + (i sin θ) 5 (A1)

Transcription

1 . (a) (cos θ + sn θ) = cos θ + cos θ( sn θ) + cos θ(sn θ) + (sn θ) = cos θ cos θ sn θ + ( cos θ sn θ sn θ) (b) from De Movre s theorem (cos θ + sn θ) = cos θ + sn θ cos θ + sn θ = (cos θ cos θ sn θ) + ( cos θ sn θ sn θ) equatng real parts cos θ = cos θ cos θ sn θ = cos θ cos θ ( cos θ) = cos θ cos θ + cos θ = cos θ cos θ Note: Do not award marks f part (a) s not used. (c) (cos θ + sn θ) = cos θ + cos θ ( sn θ) + 0 cos θ( sn θ) + 0 cos θ( sn θ) + cos θ ( sn θ) + ( sn θ) () from De Movre s theorem cos θ = cos θ 0 cos θ sn θ + cos θ sn θ = cos θ 0 cos θ ( cos θ) + cos θ( cos θ) = cos θ 0 cos θ + 0 cos θ + cos θ 0 cos θ + cos θ cos θ = 6 cos θ 0 cos θ + cos θ Note: If compound angles used n (b) and (c), then marks can be allocated n (c) only. (d) cos θ + cos θ + cos θ = (6 cos θ 0 cos θ + cos θ) + ( cos θ cos θ) + cos θ = 0 6 cos θ 6 cos θ + cosθ = 0 cos θ (6 cos θ 6 cos θ + ) = 0 cos θ ( cos θ )( cos θ ) = 0 cos 0; ; ; ; 6 A IB Questonbank Mathematcs Hgher Level rd edton

2 (e) cos θ = 0 7 θ =... ; ; ; ;... 7 θ =... ; ; ; ; Note: These marks can be awarded for verfcatons later n the queston. now consder 6 cos θ 0 cos θ + cos θ = 0 cos θ (6 cos θ 0 cos θ + ) = 0 cos θ = cos θ = 0 00 (6)() ; cos θ = (6)() 0 00 (6)() cos snce max value of cosne angle 0 closest to zero cos 0. () cos 0 8 R []. (a) usng the factor theorem z + s a factor z + = (z + )(z z + ) (b) () METHOD z = z + = (z + )(z z + ) = 0 solvng z z + = 0 z = therefore one cube root of s γ METHOD γ = γ = = IB Questonbank Mathematcs Hgher Level rd edton

3 METHOD γ = e γ = e () METHOD as γ s a root of z z + = 0 then γ γ + = 0 γ = γ Note: Award for the use of z z + = 0 n any way. Award R for a correct reasoned approach. R METHOD γ = γ = () METHOD ( γ) 6 = ( γ ) 6 = (γ) = (γ ) = ( ) = METHOD ( γ) 6 = 6γ + γ 0γ + γ 6γ + γ 6 Note: Award for attempt at bnomal expanson. use of any prevous result e.g. = 6γ + γ + 0 γ + 6γ + = Note: As the queston uses the word hence, other methods that do not use prevous results are awarded no marks. IB Questonbank Mathematcs Hgher Level rd edton

4 IB Questonbank Mathematcs Hgher Level rd edton (c) METHOD A = A A + I = 0 from part (b) γ γ + = 0 ) ( = 0 ) ( = 0 hence A A + I = 0 METHOD A = 0 Note: Award mark for each of the non-zero elements expressed n ths form. verfyng A A + I = 0 (d) () A = A I A = A A = A I A = I Note: Allow other vald methods. () I = A A A = A A A A A = I A Note: Allow other vald methods. [0]

5 . (a) () ω = cos sn = cos sn = cos + sn = () + ω + ω = + cos sn cos sn = + = 0 (b) () e e = e = e e e e e e e e = e θ ( + ω + ω ) = 0 () Note: Award for one pont on the magnary axs and another pont marked wth approxmately correct modulus and argument. Award for thrd pont marked to form an equlateral trangle centred on the orgn. (c) () attempt at the expanson of at least two lnear factors IB Questonbank Mathematcs Hgher Level rd edton

6 (z )z z(ω + ω ) + ω or equvalent () use of earler result F(z) = (z )(z + z + ) = z () equaton to solve s z = 8 z =, ω, ω A Note: Award for correct solutons. [6]. (a) z = e z = e addng or subtractng z = e e () z = e e 9 Notes: Accept equvalent solutons e.g. z = e Award marks as approprate for solvng (a + b) = +. Accept answers n degrees. (b) e = + Note: Accept geometrcal reasonng. [7]. (a) any approprate form, e.g. (cos θ + sn θ) n = cos (nθ) + sn (nθ) (b) z n = cos nθ + sn nθ = cos( nθ) + sn( nθ) n z = cos nθ sn (nθ) therefore z n = sn (nθ) n z (c) z z z z z z z z z z z z () IB Questonbank Mathematcs Hgher Level rd edton 6

7 = z z + 0z 0 z z z (d) z z z 0 z z z z z ( sn θ) = sn θ 0 sn θ + 0 sn θ 6 sn θ = sn θ sn θ + 0 sn θ (e) 6 sn θ = sn θ sn θ + 0 sn θ LHS = 6sn = 6 = RHS = sn sn 0sn = 0 Note: Award for attempted substtuton. = hence ths s true for θ = (f) 0 = sn 6 d (sn sn 0sn )d 6 0 cos cos 0cos = = 0 (g) 0 cos d 8, wth approprate reference to symmetry and graphs.rr Note: Award frst R for partally correct reasonng e.g. sketches of graphs of sn and cos. Award second R for fully correct reasonng nvolvng sn and cos. [] IB Questonbank Mathematcs Hgher Level rd edton 7

8 6. EITHER changng to modulus-argument form r = θ = arctan n n n n cos sn n f sn 0 n = {0, ±, ±6,...} N OR θ = arctan () n n k, k n k, k N [] 7. (a) () x y x y z x y x y x y x y () z + x z x x y for k to be real, y y y = k () x y y 0 y(x + y ) = 0 x y hence, y = 0 or x + y = 0 x + y = () when x + y =, z + z = x x k R (b) () w n = cos( nθ) + sn( nθ) = cos nθ sn nθ w n + w n = (cos nθ + sn nθ) + (cos nθ sn nθ) = cos nθ () (rearrangng) IB Questonbank Mathematcs Hgher Level rd edton 8

9 (w + w ) (w + w ) + = 0 ( cos θ) cos θ + = 0 ( cos θ cos θ + ) = 0 ( cos θ ) cos θ + = 0 6 cos θ cos θ = 0 ( cos θ )( cos θ + ) = 0 cos, cos cos sn cos sn w, Note: Allow FT from ncorrect cos θ and/or sn θ. [] IB Questonbank Mathematcs Hgher Level rd edton 9

10 8. METHOD r, θ ()() cos sn cos sn 8 8 METHOD ( )( ) = (= ) ( )( ) = 8 () 8 METHOD Attempt at Bnomal expanson ( ) = + ( ) + ( ) + ( ) () = 9 + () 8 = 8 [] 9. (a) EITHER w sn cos = cos + sn = Hence w s a root of z = 0 IB Questonbank Mathematcs Hgher Level rd edton 0

11 OR Solvng z = z = cos n sn n, n 0,,,,. n = gves cos sn whch s w (b) (w )( + w + w + w + w ) = w + w + w + w + w w w w w = w Snce w = 0 and w, w + w + w + w + = 0. R (c) + w + w + w + w = cos sn cos sn cos sn cos sn cos sn cos sn cos sn cos sn cos sn cos sn cos sn cos sn Notes: Award for attemptng to replace 6 and 8 by and Award for correct cosne terms and for correct sne terms. cos cos 0 Note: Correct methods nvolvng equatng real parts, use of conjugates or recprocals are also accepted. cos cos Note: Use of cs notaton s acceptable throughout ths queston. [] 0. (a) or () IB Questonbank Mathematcs Hgher Level rd edton

12 arg or arg ( ) = 7 accept () m z n z arg (z ) = m arctan arg (z ) = n arctan () = m 7 n accept n N (b) m n n m m n k, where k s an nteger m n k m m k m k 6 m k The smallest value of k such that m s an nteger s, hence m = n =. N [] IB Questonbank Mathematcs Hgher Level rd edton

13 . (a) z = ( ) Let = r(cos θ + sn θ) r θ = z = = cos sn cos n sn n n n = cos sn = cos sn Note: Award above for ths lne f the canddate has forgotten to add and no other soluton gven. 7 7 = cos sn = cos sn = 9 9 cos sn Note: Award for correct answers. Accept any equvalent form. A IB Questonbank Mathematcs Hgher Level rd edton

14 (b) Note: Award for roots beng shown equdstant from the orgn and one n each quadrant. for correct angular postons. It s not necessary to see wrtten evdence of angle, but must agree wth the dagram. A (c) 8 cos sn z 6 6 z cos sn 6 6 = cos sn () = N ( a = 0, b = ) [] IB Questonbank Mathematcs Hgher Level rd edton

15 . 8 = 8e For n = 0 n 6 ( 8) e = cos sn 6 6 = + For n = ( 8) cos sn 6 6 = + For n = ( 8) cos sn = [8] 6. (a) z = + r e r = θ = arctan = re θ r = θ = arctan e e IB Questonbank Mathematcs Hgher Level rd edton

16 (b) (c) cos nθ + sn nθ = (cos θ + sn θ) n Let n = Left hand sde = cos θ + sn θ = cos θ + sn θ Rght hand sde = (cos θ + sn θ) = cos θ + sn θ Hence true for n = Assume true for n = k cos kθ + sn kθ = (cos θ + sn θ) k cos(k + )θ + sn(k + )θ = (cosθ + sn θ) k (cos θ + sn θ) = (cos kθ + sn kθ)(cos θ + sn θ) = cos kθ cosθ sn kθ sn θ + (cos kθ snθ + sn kθ cosθ) = cos(k + )θ + sn(k + )θ Hence f true for n = k, true for n = k + However f t s true for n = true for n = etc. R hence proved by nducton 8e (d) e e = cos sn = (e) a = 8e β = 8e Snce e and e are the same α = β R IB Questonbank Mathematcs Hgher Level rd edton 6

17 (f) EITHER α = + β = α* = β* = + αβ* = ( + ) ( + ) = = βα* = ( )( ) = + = + αβ* + βα* = OR Snce α* = β and β* = α αβ* = e e e βα* = e e e αβ* + βα* = e e = cos sn cos sn = 8 cos 8 n (g) α n n = e Ths s real when n s a multple of +.e. n = N where N R [] IB Questonbank Mathematcs Hgher Level rd edton 7