exam style q's

Countesthorpe Community College

Q1. The cubic equation

z3 – 4iz2 + qz – (4 – 2i) = 0

where q is a complex number, has roots α, β and γ.

(a) Write down the value of:

(i) α + β + γ;(1)

(ii) αβγ.(1)

(b) Given that α = β + γ, show that:

(i) α = 2i;(1)

(ii) βγ = –(l + 2i);(2)

(iii) q = – (5 + 2i).(3)

(c) Show that β and γ are the roots of the equation

z2 – 2iz – (1 + 2i) = 0(2)

(d) Given that β is real, find β and γ.(3)

(Total 13 marks)


x3 + px2 + qx + r = 0

where p, q and r are real, has roots α, β and γ .

(a) Given that

α + β + γ = 4 and α2 + β2 + γ2 = 20

find the values of p and q.(5)

(b) Given further that one root is 3 + i, find the value of r.(5)

(Total 10 marks)



z3 + pz2 + 6z + q = 0

has roots α, β, and γ.

(a) Write down the value of αβ + βγ + γα.(1)

(b) Given that p and q are real and that α2 + β2 + γ2 = –12:

(i) explain why the cubic equation has two non-real roots and one real root;(2)

(ii) find the value of p.(4)

(c) One root of the cubic equation is – 1 + 3i.

Find:

(i) the other two roots;(3)

(ii) the value of q.(2)

(Total 12 marks)


z3 + qz + (18 – 12i) = 0

where q is a complex number, has roots α, β and γ.


(i) αβγ;(1)

(ii) α + β + γ.(1)

(b) Given that β + γ = 2, find the value of:

(i) α;(1)

(ii) βγ;(2)

(iii) q.(3)

(c) Given that β is of the form ki, where k is real, find β and γ.(4)

Countesthorpe Community College(Total 12 marks)


z3 + iz2 + 3z – (1 + i) = 0

has roots α, β and y.


(i) α + β + γ;(1)

(ii) αβ + βγ + γα;(1)

(iii) αβγ.(1)

(b) Find the value of:

(i) α2 + β2 + γ2;(3)

(ii) α2β2 + β2γ2 + γ2α2;(4)

(iii) α2β2γ2.(2)

(c) Hence write down a cubic equation whose roots are α2, β2 and γ2.(2)

(Total 14 marks)


z3 + pz2 + 25z + q = 0

where p and q are real, has a root α = 2 – 3i.

(a) Write down another non-real root, β, of this equation.(1)

(b) Find:

(i) the value of αβ;(1)

(ii) the third root, γ, of the equation;(3)

(iii) the values of p and q.(3)

(Total 8 marks)


Q7. It is given that α, β and γ satisfy the equations

α + β + γ = 1

α2 + β2 + γ2 = – 5

α3 + β3 + γ3 = – 23

(a) Show that αβ + βγ + γα = 3.(3)

(b) Use the identity

(α + β + γ)(α2 + β2 + γ2 – αβ – βγ – γα) = α3 + β3 + γ3 – 3αβγ

to find the value of αβγ.(2)

(c) Write down a cubic equation, with integer coefficients, whose roots are α, β and γ.(2)

(d) Explain why this cubic equation has two non-real roots.(2)

(e) Given that α is real, find the values of α, β and γ.(4)

(Total 13 marks)

Q8. The roots of the cubic equation

z3 – 2z2 + pz + 10 = 0

are α, β and γ.

It is given that α3 + β3 + γ3 = –4.

(a) Write down the value of α + β + γ.(1)

(b) (i) Explain why α3 – 2α2 + pα + 10 = 0.(1)

(ii) Hence show that

α2 + β2 + γ2 = p + 13(4)

(iii) Deduce that p = –3.(2)

(c) (i) Find the real root α of the cubic equation z3 – 2z2 – 3z + 10 = 0.

Countesthorpe Community College(2)

(ii) Find the values of β and γ.(3)

(Total 13 marks)

Q9. (a) (i) Show that is a root of the equation z7 = 1.(1)

(ii) Write down the five other non-real roots in terms of ω.(2)

(b) Show that

1 + ω + ω2 + ω3 + ω4 + ω5 + ω6 = 0(2)

(c) Show that:

(i) ;(3)

(ii) .(4)

(Total 12 marks)


2z3 + pz2 + qz + 16 = 0

where p and q are real, has roots α, β and γ.

It is given that α = 2 + .

(a) (i) Write down another root, β, of the equation.(1)

(ii) Find the third root, γ.(3)

(iii) Find the values of p and q.(3)

(b) (i) Express α in the form reiθ, where r > 0 and –π < θ ≤ π.(2)

(ii) Show that

(2)

Countesthorpe Community College(iii) Show that

where n is an integer.(3)

(Total 14 marks)


z3 – 2z2 + k = 0 (k ≠ 0)

has roots α, β and γ.

(a) (i) Write down the values of α + β + γ and αβ + βγ + γα.(2)

(ii) Show that α2 + β2 + γ2 = 4.(2)

(iii) Explain why α3 – 2α2 + k = 0.(1)

(iv) Show that α3 + β3 + γ3 = 8 – 3k.(2)

(b) Given that α4 + β4 + γ4 = 0:

(i) show that k = 2;(4)

(ii) find the value of α5 + β5 + γ5.(3)

(Total 14 marks)

Q12. (a) Show that (1 + i)3 = 2i – 2.(2)

(b) The cubic equation

z3 – (5 + i)z2 + (9 + 4i)z + k(1 + i) = 0

where k is a real constant, has roots α, β and γ.

It is given that α = 1 + i.

(i) Find the value of k.(3)

(ii) Show that β + γ = 4.(1)

(iii) Find the values of β and γ.(5)

(Total 11 marks)

Q13. (a) Find the six roots of the equation z6 = 1, giving your answers in the form ei ,


where –π < ≤ π.(3)

(b) It is given that w = eiθ,where θ ≠ nπ.

(i) Show that = 2i sin θ.(2)

(ii) Show that .(2)

(iii) Show that cot θ – i.(3)

(iv) Given that z = cot θ – i, show that z + 2i = zw2.(2)

(c) (i) Explain why the equation

(z + 2i)6 = z6

has five roots.(1)

(ii) Find the five roots of the equation

(z + 2i)6 = z6

giving your answers in the form a + ib.(4)

(Total 17 marks)

Q14. It is given that z = eiθ.

(a) (i) Show that

z + = 2 cos θ(2)

(ii) Find a similar expression for

z2 + (2)

(iii) Hence show that

z2 – z + 2 – + = 4 cos2 θ – 2 cos θ(3)

(b) Hence solve the quartic equation


z4 – z3 + 2z2 – z + 1 = 0

giving the roots in the form a + ib.(5)

(Total 12 marks)

Q15. The complex numbers z1 and z2 are given by

z1 = and z2 = i

(a) Show that z1 = i.(2)

(b) Show that | z1 | = | z2 |.(2)

(c) Express both z1 and z2 in the form reiθ, where r > 0 and –π < θ ≤ π.(3)

(d) Draw an Argand diagram to show the points representing z1, z2 and z1 + z2.(2)

(e) Use your Argand diagram to show that

tan π = 2 + (3)

(Total 12 marks)

Q16. (a) (i) Given that z6 – 4z3 + 8 = 0, show that z3 = 2 ± 2i.(2)

(ii) Hence solve the equation

z6 – 4z3 + 8 = 0

giving your answers in the form reiθ, where r > 0 and – π < θ ≤ π.(6)

(b) Show that, for any real values of k and θ,

(2)

(c) Express z6 – 4z3 + 8 as the product of three quadratic factors with real coefficients.(3)

(Total 13 marks)

Q17. Use De Moivre’s Theorem to find the smallest positive angle θ for which

(cos θ + i sin θ)15 = – i

Countesthorpe Community College(Total 5 marks)

Q18. (a) Find the three roots of z3 = 1, giving the non-real roots in the form eiθ,where –π < θ ≤ π.

(2)

(b) Given that ω is one of the non-real roots of z3 = 1, show that

1 + ω + ω2 = 0(2)

(c) By using the result in part (b), or otherwise, show that:

(i) (2)

(ii) (1)

(iii) where k is an integer.(5)

(Total 12 marks)

Q19. (a) (i) Expand

(1)

(ii) Hence, or otherwise, expand

(3)

(b) (i) Use De Moivre’s theorem to show that if z = cos θ + i sin θ then

(3)


(ii) Write down a corresponding result for .(1)

(c) Hence express cos4 θ sin2 θ in the form

A cos 6θ + B cos 4θ + C cos 2θ + D

where A, B, C and D are rational numbers.(4)

(d) Find .(2)

(Total 14 marks)

Q20. (a) (i) By applying De Moivre’s theorem to (cos θ + i sin θ)3, show that

cos 3θ = cos3 θ – 3 cos θ sin2 θ(3)

(ii) Find a similar expression for sin 3θ.(1)

(iii) Deduce that

(3)

(b) (i) Hence show that tan is a root of the cubic equation

x3 – 3x2 – 3x + 1 = 0(3)

(ii) Find two other values of θ, where 0 < θ < π, for which tan θ is a root of this cubic equation.

(2)

(c) Hence show that

(2)(Total 14 marks)

Q21. (a) Express 4 + 4i in the form reiθ, where r > 0 and –π < θ ≤ π.(3)

(b) Solve the equation

Countesthorpe Community Collegez5 = 4 + 4i

giving your answers in the form reiθ, where r > 0 and –π < θ ≤ π.(5)

(Total 8 marks)

Q22. (a) Prove by induction that, if n is a positive integer,

(cos θ + i sin θ)n = cos nθ + i sin nθ(5)

(b) Hence, given that

z = cos θ + i sin θ

show that

(3)

(c) Given further that , find the value of

(4)(Total 12 marks)

Q23. Given that satisfies the equation

where a is real:

(a) find the value of a;(3)

(b) find the other three roots of this equation, giving your answers in the form reiθ, where r > 0 and –π < θ ≤ π.

(5)(Total 8 marks)

Q24. (a) Show that

(z4 – eiθ) (z4 – e–iθ) = z8 – 2z4 cos θ + 1


(b) Hence solve the equation

z8 – z4 + 1 = 0

giving your answers in the form , where –π < ≤ π.(6)

(c) Indicate the roots on an Argand diagram.(3)

(Total 11 marks)

Q25. (a) (i) Express each of the numbers 1 + i and 1 – i in the form reiθ, where r > 0.(3)

(ii) Hence express

(1 + i)8(1 – i)5

in the form reiθ, where r > 0.(3)


z3 = (1 + i)8 (1 – i)5

giving your answers in the form a eiθ, where a is a positive integer and –π < θ ≤ π.

(4)(Total 10 marks)



(b) Show that

1 + ω + ω2 + ω3 + ω4 + ω5 + ω6 = 0(2)

(c) Show that:

(i) ;(3)


(ii) .(4)

(Total 12 marks)


2z3 + pz2 + qz + 16 = 0

where p and q are real, has roots α, β and γ.

It is given that α = 2 + .

(a) (i) Write down another root, β, of the equation.(1)

(ii) Find the third root, γ.(3)

(iii) Find the values of p and q.(3)

(b) (i) Express α in the form reiθ, where r > 0 and –π < θ ≤ π.(2)

(ii) Show that

(2)

(iii) Show that

where n is an integer.(3)

(Total 14 marks)

Q28. (a) (i) Use de Moivre’s Theorem to show that

cos 5θ = cos5 θ – 10 cos3 θ sin2 θ + 5 cos θ sin4 θ

and find a similar expression for sin 5 (5)

(ii) Deduce that

(3)


(b) Explain why is a root of the equation

t4 – 10t2 + 5 = 0

and write down the three other roots of this equation in trigonometrical form.(3)

(c) Deduce that

(5)(Total 16 marks)

Q29. (a) Express in the form reiθ, where r > 0 and –π < θ ≤ π:

(i) 4(1 + i√3);

(ii) 4(1 – i√3).(3)

(b) The complex number z satisfies the equation

(z3 – 4)2 = –48

Show that .(2)

(c) (i) Solve the equation

(z3 – 4)2 = –48

giving your answers in the form reiθ, where r > 0 and –π < θ ≤ π.(5)

(ii) Illustrate the roots on an Argand diagram.(3)

(d) (i) Explain why the sum of the roots of the equation

(z3 – 4)2 = –48

is zero.(1)

(ii) Deduce that .(3)

(Total 17 marks)

Q30. (a) Use the identity with A = (r + 1)x and B = rx to show that


(4)

(b) Use the method of differences to show that

(5)(Total 9 marks)


Q31. The diagram shows a curve which starts from the point A with coordinates (0, 2). The curve is such that, at every point P on the curve,

where s is the length of the arc AP.

(a) (i) Show that

(3)

(ii) Hence show that

(4)

(iii) Hence find the cartesian equation of the curve.(3)

(b) Show that

y2 = 4 + s2

(2)(Total 12 marks)



(b) Show that

1 + ω + ω2 + ω3 + ω4 + ω5 + ω6 = 0(2)

(c) Show that:

(i) ;


(ii) .(4)

(Total 12 marks)

Q33. (a) Express

5 sinh x + cosh x

in the form Aex + Be–x, where A and B are integers.(2)


5 sinh x + cosh x + 5 = 0

giving your answer in the form ln a, where a is a rational number.(4)

(Total 6 marks)

Q34. (a) Show that

9 sinh x – cosh x = 4ex – 5e–x

(2)

(b) Given that

9 sinh x – cosh x = 8

find the exact value of tanh x.(7)

(Total 9 marks)

Q35. (a) Show that

9 sinh x – cosh x = 4ex – 5e–x

(2)

(b) Given that

9 sinh x – cosh x = 8

find the exact value of tanh x.(7)

(Total 9 marks)


M1. (a) (i)

B11

(ii)

B11

(b) (i)

AG

B11

(ii)

Some method must be shown, eg

M1

AG

A12

(iii) q = αβ + βγ + γα

M1

= α(β + γ) + βγOr α2 + βγ , ie suitable grouping

M1

= 2i.2i − 2i − 1 = −2i − 5

AG

A13

(c) Use of β + γ = 2i and βγ = −2i −1

Countesthorpe Community CollegeM1

z2 −2iz − (1 + 2i) = 0

Elimination of say to arrive at − (1 + 2i) = 0 M1A0 unless also some reference to γ being a root

AG

A12

(d) f (−1) = 1 + 2i − 1 − 2i = 0

For any correct method

M1

β = −1, γ = 1 + 2i

A1 for each answer

A1A13

[13]

M2. (a) p = −4

B1

(α + β + γ)2 = Σα2 + 2Σαβ

M1

16 = 20 + 2Σαβ

A1

Σαβ = −2

A1F

q = −2

A1F5

Countesthorpe Community College(b) 3 − i is a root

B1

Third root is −2

B1F

αβγ = (3 + i) (3 − i)(−2)

M1

= −20

Real αβγ

A1F

r = +20

Real r

A1F5

Alternative to (b)Substitute 3 + i into equation

M1

(3 + i)2 = 8 + 6i

B1

(3 + i)3 = 18 + 26i

B1

r = 20

Provided r is real

A2,1,0[10]

M3. (a)

B11


(b) (i) Sum of squares < 0 ∴ not all real

E1

Coefficients real ∴ conjugate pair

E12

(ii)

A1 for numerical values inserted

M1A1

A1F

p = 0

cao

A1F4

(c) (i) −1 − 3i is a root

B1

Use of appropriate relationship eg

M0 if used unless the root 2 is checked

M1

Third root 2

incorrect p

A1F3

(ii) q = − (−1 − 3i)(−1 + 3i)2

allow even if sign error


= −20

ft incorrect 3rd root

A1F2

[12]

M4. (a) (i) αβγ = – 18 + 12i

accept – (18 – 12i)

B11

(ii) α + β + γ = 0

B11

(b) (i) α = –2

B1F1

(ii)

ft sign errors in (a) or (b)(i) or slips suchas miscopy

M1A1F2

(iii)

M1

Countesthorpe Community College= –2 × 2 + 9 – 6i

ft incorrect βγ or α

A1F

= 5 – 6i

A1F3

(c) β = ki, γ = 2 – ki

B1

ki(2 – ki) = 9 – 6i

M1

2k = – 6 (k2 = 9) k = –3

imaginary parts

m1

β = –3i, γ = 2 + 3i

A14

[12]

M5. (a) (i)

B11

(ii)

B11


(iii)

B11

(b) (i) used

Allow if sign error or 2 missing

M1

= (– i)2 – 2 × 3

A1F

= –7

ft errors in (a)

A1F3

(ii)

Allow if sign error in 2 missing

M1

A1

= 9 – 2(1 + i)(–i)

ft errors in (a)

A1F

= 7 + 2i

ft errors in (a)

A1F4

(iii) α2 β2 γ2 = (1 +i)2 = 2i

M1

ft sign error in αβγ

A1F2


(c) z3 + 7z2 + (7 + 2i)z – 2i = 0

Correct numbers in correct places

B1F

Correct signs

B1F2

[14]

M6. (a) 2 + 3i

B11

(b) (i) αβ = 13

B11

(ii) αβ + βγ + γα = 25

M1A0 for –25 (no ft)

M1

γ(α + β) = 12

A1F

γ = 3

ft error in αβ

A1F3

Alternative


Attempt at (z – 2 + 3i)(z – 2 – 3i)

(M1)

z2 – 4z + 13

(A1)

cubic is (z2 – 4z + 13)(z – 3) γ = 3

(A1)(3)

(iii)

M1 for a correct method for either p or q

M1A1F

ft from previous errorsp and q must be realfor sign errors in p and q allow M1 but A0

A1F3

AlternativeMultiply out or pick out coefficients

(M1)

p = –7, q = –39

(A1,A1)(3)

[8]

M7. (a) Use of

M1


A1

AG

A13

(b) 1(–5 – 3) = –23 – 3αβγFor use of identity

M1

αβγ = –5

A12

(c) z3 – z2 + 3z + 5 = 0

M1

For correct signs and “= 0”

A1F2

(d) α2 + β2 + γ2 < 0 non real roots

B1

Coefficients real conjugate pair

B12

(e) f(–1) = 0 z + 1 is a factor

M1A1

(z + 1)(z2 – 2z + 5) = 0

A1


z = –1, 1 ± 2i

A14

[13]

M8. (a) α + β + γ = 2

B11

(b) (i) α is a root and so satisfies the equation

E11

(ii)

M1A1

Substitution for and

m1

AG

A14

(iii)

do not allow this M mark if used in (b)(ii)

M1

p = –3

AG

Countesthorpe Community CollegeA1

2

(c) (i) f(–2) = 0

M1

α = –2

A12

(ii) (z + 2)(z2 – 4z + 5) = 0

For attempting to find quadratic factor

M1

Use of formula or completing the squarem0 if roots are not complex

m1

= 2 ± i

CAO

A13

[13]

M9. (a) (i)

B11


(ii) Roots are ω2, ω3, ω4, ω5, ω6

OE; M1A0 for incomplete setSC B1 for a set of correct roots in terms of eiθ

M1A12

(b) Sum of roots considered

M1

= 0

or

A12

(c) (i)

M1

Or

A1

AG

A13

(ii)

Allow these marks if seen earlier in the solution

B1,B1

Using part (b)

M1

Result

Countesthorpe Community CollegeAG

A14

[12]

M10. (a) (i)

B11

(ii) αβγ = –8

Allow for +8 but not ±16

M1

αβ = 16

B1

A13

(iii)

SC if failure to divide by 2 throughout,allow M1A1 for either p or q correct ft

M1

p = –7, q = 28

ft incorrect γ

A1F,A1F3


Alternative to (a)(ii) and (a)(iii)(z2 – 4z + 16)(az + b)

(M1)

αβ = 16

(B1)

a = 2, b = +1, γ =

(A1)

Equating coefficients

(M1)

p = –7

(A1F)

q = 28

(A1F)

(b) (i)

B1,B12

(ii)

M1

AG

A12


(iii)

B1

M1

AG

A13

[14]

M11. (a) (i)

B1

B12

(ii)

Used. Watch (M1A0)

M1

= 4

AG

A12


(iii) Clear explanation

eg α satisfies the cubic equation since it is a root.Accept z = α

E11

(iv)

Or

M1

= 8 – 3kAG

A12

(b) (i) α 4 = 2α 3 – kα

B1

Or

M1

= 2 (8 – 3k) – 2k

ft on

A1

k = 2

AG

A14

(ii)

M1

Countesthorpe Community CollegeSubstitution of values

A1

= – 8

A13

[14]

M12. (a) (1 + i)2 = 2i or (1 + i) =

B1

2i(1 + i) = 2i – 2

AG

Alternative method:

(1 + i)3 = 1 + 3i2 + 3i3 + i3 B1

= 2i − 2 B1

B12

(b) (i) Substitute z = 1 + i

M1

Correct expansion

allow for correctly picking out either the realor the imaginary parts

A1

k = −5

A13

(ii) β + γ = 5 + i − α = 4

AG

B1

Countesthorpe Community College1

(iii) αβγ = 5(1 + i)

allow if sign error

M1

βγ = 5

ft incorrect k

A1F

z2 − 4z + 5 = 0

M1

Use of formula or (z – 2)2 = –1

No ft for real roots if error in k

A1F

z = 2 ± i

A1F5

NB allow marks for (b) in whatever order they appear [11]

M13. (a)

M1

OE

M1A1 only if:

(1) range for k is incorrect e.g. 0,1,2,3,4,5

(2) i is missing

A2,1,03


(b) (i)

AG

M1A12

(ii)

M1

AG

A12

(iii)

Or for

M1

A1

= cotθ − i

AG

A13

(iv)

ie any correct method

M1


z + 2i = zw2

AG

A12

(c) (i) No coefficient of z6

E11

(ii) (w2)6 = 1 w2 =

B1

z = cot −i, k = ± 1, ± 2, 3

M1A2, 1, 0

Alternatively:

z + 2i = z B1

z = M1

roots A2, 1, 0

(NB roots are ± − i; ± − i; − i)4

[17]

M14. (a) (i) z + = cos θ + i sin θ + cos (−θ) + i sin (−θ)

Or z + = eiθ + e−iθ

M1


= 2cos θAG

A12

(ii) z2 + = cos 2θ + i sin 2θ + cos(−2θ) + i sin(−2θ)

M1

= 2 cos 2θOE

A12

(iii) z2 − z + 2 − +

= 2cos 2θ − 2cosθ + 2

M1

Use of cos 2θ = 2 cos2θ − 1

m1

= 4cos2 θ − 2cosθAG

A13

(b) z + = 0 z = ±i

M1A1

z + =1 z2 − z + l = 0

M1A1

z +

A1F


Alternative:cosθ = 0 θ = ± π M1

z = ±i A1

cosθ = θ = ± π M1

z = A1A15

Accept solution to (b) if done otherwise

AlternativeIf θ = + π θ = π

M1

z = i z =

A1

Or any correct z values of θ

M1

Any 2 correct answers

A1

One correct answer only

B1[12]

M15. (a)

AG

M1A12


(b) |z2| = = 1 = |z1|

M1A12

(c) r = 1

PI

B1

Deduct 1 mark if extra solutions

B1B13

(d)

Positions of the 3 points, relative to each other, must be approximately correct

B2,1F2

(e) Arg(z1+z2) =

Clearly shown

B1

tan =

Countesthorpe Community CollegeAllow if B0 earned

M1

= 2 +

AG must earn B0 for this

B13

[12]

M16. (a) (i)

M1

= 2 ± 2i

AG

A12

(ii) 2 + 2i = 2 , 2 − 2i = 2

M1 for either result or for one of r = 2 , θ =

M1

A1A1

or

M1 for either

M1


allow A1 for any 3 correct ft errors in

A2,1,0 F6

(b) Multiplication of brackets

M1

Use of eiθ + e–iθ = 2 cos θAG

A12

(c)

PI

M1A1F

Product is

(or z2 + 2z + 2)

A1F3

(z2 − 2 cos z + 2)[13]


M17. (cos θ + i sin θ)15 = cos15 θ + i sin15 θor = e15i θ

M1

cos15 θ = 0

sin15 θ = −1

or −i =

m1A1

m1 for both R&I parts written down

A1F

ft provided the value of 15θ is a correct value

A1F5

SCcos15 θ + i sin15 θ = i

(M1)

sin15 θ = −1

or for cos15 θ = 0

(B1)

(B1)(3)

[5]


M18. (a) 1,

M1 for any method which would lead tothe correct answers

Accept e0 or e0i

Also accept answers written down correctly

M1A12

(b) Any correct method

M1

Shown for one root

AG

A12

(c) (i)

ie use of result in (b)

AG

A12

(ii)

AG

A11


(iii)

M1A1

Use of

m1

A1

AG

A15

[12]

M19. (a) (i)

B11

(ii)

Alternatives for M1A1:

Countesthorpe Community CollegeM1A1

CAO (not necessarily in this form)

A13

(b) (i)

M1A1

= 2 cos nθAGSC: if solution is incomplete and(cos θ + i sin θ)–n is written ascos nθ – i sin nθ, award M1A0A1

A13

(ii) zn – z–n = 2i sin nθ

B11

(c) RHS = 2 cos 6θ + 4 cos 4θ – 2 cos 2θ – 4

ft incorrect values in (a)(ii) provided theyare cosines

M1A1F

LHS = –64 cos4 θ sin2 θ

M1

cos4 θ sin2 θ

A14


(d)

ft incorrect coefficients but not letters A, B, C, D

M1A1F2

[14]

M20. (a) (i) cos 3θ + i sin 3θ = (cos θ + i sin θ)3

M1

= cos3 θ + 3 i cos2 θ sin θ + 3i2 cos θ sin2 θ + i3 sin3 θ

A1

Real parts: cos 3θ = cos3 θ – 3 cos θ sin2 θAG

A13

(ii) Imaginary parts:sin 3θ = 3 cos2 θ sin θ – sin3 θ

A1F1

(iii)

Used

M1

Error in sin 3θ

A1F


AG

A13

(b) (i)

Used (possibly implied)

B1

Must be hence

M1

x3 – 3x2 – 3x + 1 = 0

A13

(ii) Other roots are tan , tan

B1B12

(c)

Must be hence

M1

A12

[14]


M21. (a) Any method for finding r or θ

M1

A1A13

(b)

M1 needs some reference to a +2kπi

M1A1FA1F

Accept r in any form eg

Correct but some answers outside rangeallow A1ft incorrect r, θ in part (a)

A2,1,0F5

[8]

M22. (a) (cos θ + i sin θ)k+1 =(cos kθ + i sin kθ)(cos θ + i sin θ)

M1

Countesthorpe Community CollegeMultiply out

Any form

A1

= cos(k + 1)θ + i sin(k + 1)θClearly shown

A1

True for n = 1 shown

B1

P(k) P(k + 1) and P(1) true

provided previous 4 marks earned

E15

(b)

M1A1

AG

A13

(c)

M1

A1


M0 for merely writing

M1

= 0

A1F4

[12]

M23. (a)

Allow M1 if

M1

OE could be or

A1

ft errors in 24

A1F3

(b) For other roots, r = 2

for realising roots are of form 2 × eiθ.M1 for strictly correct θ

B1


i.e must be

M1A1

Roots are

ft error in or r

A2,1,0F5

[8]

M24. (a) Correct multiplication of brackets

M1

eiθ + e–iθ = 2 cos θClearly shown

A12

(b) 2 cos θ = 1

SC If ‘hence’ not used and, say,z8 – z4 + 1 = 0 is solved by formula, lose

M1

M1A1, but then continue M1m1 etc if is obtained

A1


M1

m1

A1 if 3 roots correct

A2,1,0F6

(c)

B1 for 4 roots indicated correctly on a circle.CAO

B2,1,0

Indication that r = 1

B13

[11]

M25. (a) (i)

B1 both correct

B1


OE

B1B13

(ii) or equivalent single expression

No decimals; must include fractional powers

B1F

Raising and adding powers of e

M1

or equivalent angle

Denominators of angles must be different

A1F3

(b)

M1

CAO

B1

Correct answers outside range: deduct 1 mark only

A2,1F4

[10]


M26. (a) (i)

B11



M1A12


M1

= 0

or

A12

(c) (i)

M1

Or

A1

AG

A13


(ii)


B1,B1

Using part (b)

M1

Result

AG

A14

[12]

M27. (a) (i)

B11

(ii) αβγ = –8

Allow for +8 but not ±16

M1

αβ = 16

B1

A13

(iii)


SC if failure to divide by 2 throughout,allow M1A1 for either p or q correct ft

M1

p = –7, q = 28

ft incorrect γ

A1F,A1F3

Alternative to (a)(ii) and (a)(iii)(z2 – 4z + 16)(az + b)

(M1)

αβ = 16

(B1)

a = 2, b = +1, γ =

(A1)

Equating coefficients

(M1)

p = –7

(A1F)

q = 28

(A1F)

(b) (i)

B1,B12


(ii)

M1

AG

A12

(iii)

B1

M1

AG

A13

[14]

M28. (a) (i) cos 5θ + i sin 5θ = (cosθ + i sinθ )5

Attempt to expand 3 correct terms

M1

Expansion in any form

Correct simplification


Equate real parts:

m1

cos 5θ = cos5 θ – 10 cos3 θ sin2 θ + 5 cosθ sin4 θAG

A1

Equate imaginary parts:

sin 5θ = 5 cos4 θ sinθ – 10 cos2 θ sin3 θ + sin5 θCAO

A15

(ii)

Used

M1

Division by cos5 θ or by cos4 θ

m1

AG

A13

(b)

Or for tan 4 θ – 10 tan 2 θ + 5 = 0

M1

Or for tan 5θ = 0


Other roots

OE

B13

(c) Product of roots = 5

M1

Or

B1

A1

A1

– sign rejected with reason

E15

Alternative

Use of quadratic formula M1

A1

B1

Correct selection of +ve values E1

Multiplied together to get A1[16]


M29. (a) (i)

for either used

M1

If either r or θ is incorrect but the same value in both

(i) and (ii) allow A1 but for θ only if it is given as

A1

(ii)

A13

(b)

taking square root

M1

AG

A12

(c) (i)

for the 2; ft incorrect 8, but no decimals

Countesthorpe Community CollegeB1F

for either, PI

M1

Allow A1 for any 2 roots not +/– each other

Allow A2 for any 3 roots not +/– each other

Allow A3 for all 6 correct roots

Deduct A1 for each incorrect root in the interval;ignore roots outside the intervalft incorrect r

A3,2,1F5

(ii)

Radius 2

clearly indicated; ft incorrect r

allow B1 for 3 correct points condone lines

B1F

Plotting roots

B2,13

(d) (i) Sum of roots = 0 as coefficient of z5 = 0

OE

E11


(ii)

M1

A1

AG

A13

[17]

M30. (a)

M1A1

Multiplying up

A1

Printed result

AG

A14

(b)


At least three lines to be shownAccept if x‘s used

M1A1

.................................

Clear cancellation

m1

Sum

A1

AG

A15

[9]

M31. (a) (i)

Allow M1 for


then A1 for

M1A1

AG

A13

(ii)

For separation of variables; allow withoutintegral sign

M1

Allow if C is missing

A1

C = 0

A1

AG if C not mentioned allow

SC incomplete proof of (a)(ii),differentiating

allow M1A1 only

A14

(iii)

M1


Allow if C is missing

A1

C = 0

Must be shown to be zero and CAO

A13

(b)

Use of cosh2 = 1 + sinh2

M1

= 4 + s2

AG

A12

[12]

M32. (a) (i)

B11



M1A12



M1

= 0

or

A12

(c) (i)

M1

Or

A1

AG

A13

(ii)


B1,B1

Using part (b)

M1

Result

AG

A14

[12]


M33. (a)

M0 if no 2s in denominator

M1

= 3ex – 2e–x

A12

(b) 3ex – 2e–x + 5 = 03e2x + 5ex – 2 = 0

ft if 2s missing in (a)

M1

(3ex – 1)(ex + 2) = 0

A1F

ex ≠ – 2

any indication of rejection

E1

ex = x = In

provided quadratic factorises into real factors

A1F4

[6]

M34. (a)

M0 if cosh x mixed up with sinh x


= 4ex – 5e–x

AG

A12

(b) Attempt to multiply by ex

M1

4e2x – 8ex – 5 = 0

A1

(2ex – 5)(2ex + 1) = 0

ft provided quadratic factorises(or use of formula)

M1

PI but not ignored

E1F

A1F

M1 PI for attempt to use

or equivalent fraction

M1A1F7

[9]


M35. (a)

M0 if cosh x mixed up with sinh x

M1

= 4ex – 5e–x

AG

A12

(b) Attempt to multiply by ex

M1

4e2x – 8ex – 5 = 0

A1

(2ex – 5)(2ex + 1) = 0

ft provided quadratic factorises(or use of formula)

M1

PI but not ignored

E1F

A1F

M1 PI for attempt to use

or equivalent fraction

M1A1F7

[9]

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