HSC 4 Unit Extension 2 Mathematics Dot Point Outline

4u Maths Summary Contents: Graphs Basic Graphs Trigonometric Graphs Addition & Subtraction of Ordinates Reflection in Axis Rules for Graphing Multiplication of Ordinates Division of Ordinates Behaviours at Critical Points Index Rules Implicit Differentiation Inequalities

3 3 7 11 13 15 20 21 23 24 26 30

Complex Numbers Arithmetic of Complex Numbers Geometrical Representation Basic Proofs Vectors De Moivre’s Theorem nth Roots Loci of Complex Numbers Regions

32 32 35 38 41 44 46 47 49

Conics Ellipse Tangents & Normals Equation of a Chord Chord of Contact Basic Proofs Hyperbola Tangents & Normals Equation of a Chord Chord of Contact Basic Proofs Rectangular Hyperbola Equation of a Chord Tangent Normal Chord of Contact Basic Proofs Loci

51 51 56 56 56 57 63 65 68 68 68 69 71 72 73 74 74 76

Conics in Cones Eccentricity Range

78 78

4u Maths Summary

Page 1 of 117

Integration Algebraic Substitutions Trigonometric Integrals Trigonometric Substitutions Integration by Parts Recurrence Formulas Partial Fractions

80 80 80 81 82 82 83

Volumes Volumes Using Limits Volumes by slicing Volumes by shells Volumes with non-circular slices

84 85 86 87 89

Mechanics Projectile Motion Simple Harmonic Motion Mathematical Descriptions of Motion Upwards Motion Downwards Motion Circular Motion Uniform Circular Motion Conical Pendulum Banked Circular Track

90 91 93 95 96 97 99 100 101 104

Polynomials Integer Roots of Polynomials Factoring Polynomials Roots & Coefficients of Polynomials Partial Fractions

107 107 109 111 114

Harder 3 Unit Circle Geometry Induction Inequalities

117 117 117 117

4u Maths Summary

Page 2 of 117

•

SYLLABUS REFERENCE Topic 1 Graphs Basic Curves Graph a linear equation. ax + by + c = 0 y = mx + b

RELATED INFORMATION

ax + by + c = 0 Gradient = – a b

y-intercept: y = – c

b

y = mx + b Gradient = m y-intercept: y = c y

m

10

5

-10

-5

5

10

x

(0, -3) -5

-10

•

Graph a quadratic Function. y = ax2 + bx + c

Axis of symmetry: x = – b

2a

2

Roots: ax + bx + c = 0 If a > 0, then concave up. If a < 0, then concave down. y 10

5

-10

-5

5

10

x

-5

-10

•

Graph a cubic function. 4u Maths Summary

Steps: 1. Find factor of whole equation. Page 3 of 117

y = ax3 + bx2 + cx + d

2. Use polynomial division. 3. Find roots of remaining equation. Steps: 1. Find stationary points and their nature. 2. Find points of infle xion. 3. Find intercepts. 4. Use table of values. Properties: • Has at most three roots. y 10

5

-10

-5

5

10

x

-5

-10

3

2

y = x + 4x – x – 3

•

Graph a quartic function. y = ax4 + bx3 + cx2 + dx + e

4u Maths Summary

Steps: 1. Find factor of whole equation. 2. Use polynomial division. 3. Find roots of remaining equation. Steps: 1. Find stationary points and their nature. 2. Find points of inflexion. 3. Find intercepts. 4. Use table of values. Properties: o Has at most four roots.

Page 4 of 117

y 10

5

-4

-2

2

x

4

-5

-10

4

3

2

y = x – 2x – 2x + 4x – 1

•

Properties: • Asymptote: x ? 0 • Use limit to find horizontal asymptotes.

Graph a rectangular hyperbola. xy = k

y

y= k x

10

5

-10

-5

5

10

-5

-10

• Graph a circle. x2 + y2 + 2gx + 2fy + c =0 (x – h)

2

+ (y – k)

2

2

= r

4u Maths Summary

xy = 4

x2 + y2 + 2gx + 2fy + c = 0 1. Complete the square to get equation into the form below. 2

2

2

(x – h) + (y – k) = r • Centre of circle: (h, k) • Radius of circle = r

Page 5 of 117

x

y 10

5

-10

-5

5

10

x

-5

-10

2

2

(x – 3) + (y + 2) = 9

•

Graph an exponential function. y = ax For both cases: a>1&0 0 • As x à -8, xe–x à -8 • As x à 8, xe–x à 0. This enables a rough shape to be quickly sketched. The exact positions of the stationary points and points of inflexion may be determined by calculus. To graph y = x (x + 1) , the x–2

y = x (x + 1) x–2

graphs of y = x, y = x + 1 and y = x – 2 can be first sketched. The regions in the number plane, in which the graph exists, can be then shaded, discontinuities determined, points of intersection with coordinate axes marked on and the behaviour of the function for x à ± 4u Maths Summary

Page 23 of 117

8 investigated. Exact positions of stationary points and points of inflexion could lastly be determined if required.

y 15

10

5

-2

2

4

6

x

-5

y = x

y = x + 1

y =

1 x – 2 y = x (x + 1) x–2

Properties: • Asymptotes of the inverse function of the denominator are also asymptotes for the overall graph. • Roots occur where the numerator functions’ roots occur. •

To sketch y =

x (x + 1) , a rough sketch of x–2

y = x (x + 1) can first be drawn and then square root x–2

rules (see below) can be used. Drawing graphs of the form y = [f(x)]n • Graph a function

y = [ f(x)]

n

n

by first graphing y = f(x). y = [ f(x)]

4u Maths Summary

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y 10

5

-2

2

4

x

6

-5

y 3

2

1

-1

-0.5

0.5

1

1.5

x

-1

-2

y = (x – 2)

2

– 3 y = ((x – 2)

2

– 3)

2

y = ((x – 2)

2

– 3)

3

Rules:

•

Graph a function y = f(x) by first graphing y = f(x).

4u Maths Summary

•

When f(x) = 0 or 1, y remains at 0 or 1.

•

When f(x) > 1, y becomes larger

•

When 0 > f(x) > 1, y becomes smaller.

•

If n is even, y is always = 0.

• •

If n is odd, y has the same sign as f(x). Its derivative is n [f (x) ]n–1 f '(x), then all stationary points and intercepts on the x-axis of y = f (x) are stationary points of [f(x)]n.

y = ±

f(x)

Page 25 of 117

y 10

5

-10

-5

5

x

10

-5

-10

y = (x – 2)

2

– 3

y =

(x – 2)

2

– 3

y = – (x – 2)

2

– 3

Properties: • For a positive square root, only values above the x-axis are shown. • For a negative square root, only values below the x-axis are shown. • For the graph of y2 = f(x), both values above and below the x-axis are graphed. • Where f(x) < 0, the square root graph does not exist. • Where f(x) > 1, the values of y are decreased. • Where 0 < f(x) < 1, the values of y are increased. • Where f(x) = 0 or 1, the values of y do not change. •

If y =

(f(x) then y' =

f'(x) .This leads to the 2 f(x)

position of stationary points. General approach to curve sketching. • Use implicit differentiation to dy compute dx for

curves given in implicit form.

Implicit Differentiation: •

Differentiate each part separately.

•

dy Any part containing a y, multiply that derivative by dx dy Make dx the subject.

•

Make y the subject in the original equation.

•

• Sub the value for y into the derivative function. Sample Problem: 1 2

1 2

Differentiate x + y = 4 4u Maths Summary

Page 26 of 117

Solution: 1 2

1 2

x + y = 4 – 12

x 2

– 12

+ y × dy = 0 2 dx

– 12

– 12

y × dy = – x 2 dx 2 – 12

dy = – x × 2 – 12 dx 2 y dy = dx

y  –  x  1 2 1 2

Rearrange original to find expression for y½ 1 2

1 2

y = 4 – x Sub into derivative. dy = – dx

•

4 – x     x  1 2

1 2

Curves graphed could include:

4

2

x –1

4

y =

x

y =

x

y

2

x –1 2

y = x e

–x 6

2

y = x ln (x – 1) 2

y

2

x

2

= x – 9x

4

2

+ 2y = 4 sinx y = x y = x cosx

2

-3

-2

-1

1

2

3

x

-2

2

y = x e

4u Maths Summary

–x

Page 27 of 117

y

6

4

2

-2

2

4

6

x

8

-2

2

y = x ln (x – 1) y 10

5

-10

-5

5

10

x

-5

-10

2

y

4u Maths Summary

2

= x – 9x

Page 28 of 117

y 10

5

-5

5

x

10

-5

-10

2

x

2

+ 2y

= 4

y 3

2

1

-3

-2

-1

1

2

3

x

-1

-2

-3

y =

4u Maths Summary

sinx x

Page 29 of 117

y

0.04

0.02

-360

-180

180

360

x

-0.02

-0.04

y = x cosx y 360

180

-360

-180

180

360

x

-180

-360

•

Solve inequalities Inequalities: by sketching an appropriate graph. • For graphs where a function is expressed as being larger or smaller than a fixed value or other graph, graph them as separate functions. Eg: Graph: 2

x – 3 ≥ 2x

4u Maths Summary

Page 30 of 117

y 10

5

-10

-5

5

10

x

-5

-10

•

Find the number of solutions of an equation by graphical considerations.

4u Maths Summary

Rules: • Graph each function separately. • Where the blue function has a higher y-value than the red graph, the inequality is fulfilled. Graph Solutions: For the graph: 2

x – 3 = 2x

The number of solutions can be determined by the number of times the graphs cross. The x-values at which they cross are the solutions to the equation.

Page 31 of 117

Topic 2 Complex Numbers Arithmetic of complex numbers and solving quadratic equations. • Appreciate the necessity of introducing the symbol i, where i2 = -1, in order to solve quadratic equations. •

•

Write down the real part Re(z) and the imaginary part Im(z) of a complex number z = x + iy. Add, subtract and multiply complex numbers written in the form x + iy.

The complex number, i, is used when a quadratic equation has no real roots. For instance the equation x2 + 2x + 3 = 0 has no real roots and therefore i must be introduced in order to solve it. This occurs when the quadratic formula: x = –b±

2

b – 4ac 2a returns a math error.

For the complex number z = x + iy: Real Part = Re(z) = x Imaginary Part = Im(z) = y

Addition of Complex Numbers: The real parts are added together and the imaginary parts are added together. (a + ib) + (c + id) = (a + c) + (b + d)i Subtraction of Complex Numbers: The real parts are subtracted and the imaginary parts are subtracted. (a + ib) – (c + id) = (a – c) + (b – d)i Multiplication of Complex Numbers: Use the same rule as for two pairs of brackets. (Eg: first term by first term, first term by second, etc) (a + ib)(c + id)

2

= ac + iad + ibc + bdi

= ac + iad + ibc – bd

•

= (ac – bd) + (ad + bc)i

Find the complex conjugate z of the number x + iy.

Complex Conjugate: For the equation: z = x + iy

The conjugate is: z = x – iy

•

Divide a complex number a + ib by a 4u Maths Summary

Complex Number Division: Multiply both the numerator and the denominator by the Page 32 of 117

complex number c + id.

conjugate of the denominator. a + ib c + id

= a + ib × c – id c + id c – id

=

ac – iad + ibc + bd 2

2 2

c – id

=

(ac + bd) 2

2

+ i

c + d

•

•

Write down a condition for a + ib to be equal to c + id. Prove that there are always two square roots of a non-zero complex number.

bc – ad c

2

2

+ d

Complex Number Equality: Two complex numbers are equal is their real parts are equal and their imaginary parts are equal. ∴ a + ib = c + id IF a = c & b = d Square Roots of Complex Numbers: Find z given z2 = a + ib. Let z = x + iy

2

∴ (x + iy) = a + ib

2

2 2

x + 2xyi + i y = a + ib

2

2

(x – y ) + (2xy)i = a + ib

Equating Real & Imaginary Parts:

2

2

x – y = a

2xy = b

•

Find the square 4u Maths Summary

Solving these two equations simultaneously gives a quartic equation whereby an expression for x2 can be derived. One of these expressions will be negative and therefore has no solutions. The other remaining expression will be positive and therefore has two solutions, one positive and one negative. Therefore, a non-zero complex number has two square roots. Square Roots of Complex Numbers: Page 33 of 117

roots of a complex Example: number a + ib. Find the square root of 3 + 4i. Solution: Let z

2

= 3 + 4i

Let z = x + iy (x + iy)  2

2

2

x 

– y

2

2

 

x – y

= 3 + 4i + (2xy)i = 3 + 4i

= 3 & 2xy = 4 2

Multiply first equation by x 4

2 2

2

x – x y = 3x

Square the second equation 2 2

xy

= 4

Solve Simultaneously 4

2

x – 3x – 4 = 0  2 x 

 2



– 4   x + 1 = 0

2

x = 4, – 1 ∴x = ±2 ∴ x = 2, y = 1

& x = -2, y = -1

i is a device by which quadratic equations with real coefficients could be always solvable. It could be shown that there exist 2 complex roots for a complex number. This then leads to the discovery that a quadratic equation with complex coefficients will have 2 complex roots. In finding the square roots of a + ib, the statement a + ib = x + iy, where a, b, x, & y are real, leads to the need to solve the equations x2 – y2 = a & 2xy = b. Examining graphs of these curves for various values of a and b will lead to the conclusion that two roots will always exist for a complex number.

4u Maths Summary

Page 34 of 117

y 10

5

-10

-5

5

10

x

-5

-10

2

2

x – y

•

Solve quadratic equations in the form: 2

ax

•

= 5 2xy = 9

Expand

+ bx + c = 0 ,

where a, b & c are complex. Geometrical Representation Appreciate that there exists a one to one correspondence between the complex number a + ib and the ordered pair (a, b).

The complex number a + ib , represents an ordered pair of (a, b) on the argand diagram. y 6

5

4

3

2

1

0.5

•

Plot the point corresponding to a + ib on the Argand diagram. 4u Maths Summary

1

1.5

2

2.5

3

x

1 + 4i For the complex number a + ib. The x-value = a The y-value = b

Page 35 of 117

y 6

5

4

3

2

1

0.5

•

•

Define the modulus ( |z| ) and argument (arg z) of a complex number z. Find the modulus and argument of a complex number.

1

1.5

2

2.5

3

x

1 + 4i The modulus of z is the distance from the origin to the complex number z. It can be referred to |z|.

The argument of z is the angle that a line drawn from the origin to a complex number, z, makes with the x-axis in the positive direction. It can be referred to as arg (z). The modulus of the complex number z = x + iy is: 2

|z| =

2

x +y

The argument of the complex number z = x + iy is: •

Tan θ = y x

Write a + ib in Modulusargument form.

Modulus-argument form is also known as: • Mod-Arg form • Polar form Mod-Arg Form of a complex number: z = r (cos θ + i sin θ) z = r cis θ

Where: r =

2

2

x + y Tan θ = y x

Remember: Mod z = |z| = |x + iy| = Arg z = θ = ph z

4u Maths Summary

2

2

x + y

=

zz

= r

Page 36 of 117

y

x + iy (r, θ ) r

θ

•

Prove basic relations involving modulus and argument.

y

x

x

Arg z is any value of θ for which x = |z| cos θ and y = |z| sin θ . |z1z2| = |z1||z2|  z1  |z1|   =  z2  |z2|

Arg (z1z2) = Arg z1 + Arg z2 ± 2π   Arg  z1  = Arg z1 – Arg z2 ± 2π  z2 

|z| = |z| =

2

2

2

2

x + y

2

zz = |z| = |z| = x + y

2

z + z = 2x (z1z2) = z1 × z2 z – z = 2yi Arg z = – Arg z z

•

Use modulusargument relations to do calculations involving complex numbers.

4u Maths Summary

-1

z = 1 = 2 z |z|

Multiplication in Mod-Arg Form: Let z1 = r1 cis θ1 & z2 = r2 cis θ2 z1z2 = r1r2 cis ( θ1 + θ2) Division in Mod-Arg Form: z 1 = r1 cis ( θ – θ ) 1 2 z2 r2

Page 37 of 117

•

Recognize the geometrical relationships between the point representing z and points representing z , cz (c is real) & iz.

y 10

5

-10

-5

5

10

x

-5

-10

z = 4 + 3i z = 4 – 3 i iz = -3 + 4i cz = 8 + 6i where c = 2

Rules: • •

i

2

i

3

• •

i

4

• • •

Students should be able to prove these relations.

= -1 = –i = 1

Multiplying a complex number by i is equal to a quarter turn about the origin. z is a reflection of z in the x-axis.

Multiplying a complex number by a real value, c, results in both the real and imaginary part being multiplied by that number. Proof #1: |z 1z 2| = |z 1| × |z 2| & arg (z 1z 2) = arg z1 + arg z 2 Let z1z2

z1 z2 = =

= r1cisθ1 = r2cisθ2 (r1cisθ1 )(r2cisθ2) r1r2 (cosθ1 + isinθ1)(cosθ2 + isinθ2) 2

= r1r2 (cosθ1cosθ2 + isinθ2cosθ1 + isinθ1cosθ2 + i sinθ1sinθ2 ) = r1r2 [(cosθ1cosθ2 – sinθ1sinθ2) + i (sinθ1cosθ2 + cosθ1sinθ2)] = r1r2 [cos(θ1 + θ2) + isin(θ1 + θ2)] = r1r2 cis(θ1 + θ2) |z1z2| = r1r2 ∴ |z1z2| = r1r2 = |z1||z2| arg (z1z 2) = θ1 + θ2 ∴ arg ( z1z2) = θ1 + θ 2 = arg z1 + arg z2

4u Maths Summary

Page 38 of 117

Proof #2:  z1  z  |z |   = 1 & arg  1  = arg z 1 – arg z 2  z2  |z 2|  z2  Let z1 z2

z1 = r1 cis θ1 z2 = r2 cis θ2 θ = r1 cis 1 r2 cis θ2 θ θ = r1 cis 1 × r2 cis( – 2) r2 cis θ2 r2 cis( – θ2) r r cis( θ – θ2) = 12 2 1 r2 cis 0°

However, since cis 0° = 1 = r1 cis( θ1 – θ2) r2 z  r |z | ∴   1  = 1 = 1 r2 |z2|  z2  ∴ arg  z1  = θ1 – θ2 = arg z1 – arg z2  z2 

Proof #3: Let

z 1 = a + ib z 1 = a – ib z 2 = c + id z 2 = c – id

z1 + z2

z1 + z2

= = = = =

(a – ib) + (c – id) (a + c) – ib – id (a + c) – i (b + d) a + ib + c + id (a + c) + i (b + d)

= (a + c) – i (b + d) = LHS

Proof #4:

4u Maths Summary

Page 39 of 117

Let

z1 = a + ib z1 = a – ib z2 = c + id z2 = c – id

z1z2

= (a – ib)(c – id)

z1z2

= = = =

ac – adi – bci + bdi ac – bd – adi – bci (ac – bd) – i (ad + bc) (a + ib)(c + id)

= = = LHS = =

ac + adi + bci + b di ac – bd + adi + bci (ac – bd) + i (ad + bc) (ac – bd) – i (ad + bc) RHS

2

2

Proof #5: n

n

|z | = |z| Let z = rcis θ z

n

n

= r cis nθ

n

n

|z | z |z| |z|

= r = rcis θ = r

n

n

= r n

n

∴ |z | = |z|

Proof #6: n

arg(z ) Let z z

= n argz = rcis θ

n

n

= r cis nθ n

arg(z ) z argz n argz

= = = = n

∴ arg(z )

Sample Question Q4 (1987) a. Let OABC be a square on an argand diagram where O is the origin. The points 4u Maths Summary

nθ rcis θ θ nθ

= n argz

Answer: a. Let z = 3 + 2i ∴ iz = -2 + 3i

Page 40 of 117

A and C represent the complex numbers z and iz respectively. Find the complex number represented by B. b. The square is now rotated about O through 45° in an anticlockwise direction to OA’B’C’. Find the complex numbers represented by the points A’, B’ and C’.

?

iz z

O

By graphical solution, B = 1 + 5i 1 + 5i = z + iz ∴ B = z + iz b. To rotate a complex number about the origin by 45° is the equivalent of multiplying a complex number by: |z| × z 2 |z| ∴

•

Appreciate that a complex number can be represented as a vector on the Argand diagram.

A' = z |z| B' = z + iz |z| C' = iz |z|

The complex number z = x + iy can be represented as a vector OZ .

Z

O

•

Appreciate the geometrical 4u Maths Summary

For the addition of two complex numbers. Use vector addition to find the resulting vector. Page 41 of 117

•

•

•

significance of the addition of two complex numbers. Given the points representing z 1 and z 2, find the position of the point representing z, where z = z1 + z2. Appreciate that the vector z = z1 + z2, corresponds to the diagonal of a parallelogram with vectors representing z 1 and z 2 as adjacent sides. Given vectors z 1 and z 2, construct vectors z 1 – z 2 and z 2 – z 1.

For z1 + z2 Where z1 = 5 + 2i & z2 = 1 + 3i

Familiarity with the vector representation of a complex number is extremely useful when work on curves and loci is encountered.

That is: • The gradient of the vector joining z1 to the origin is equal to the gradient of the vector joining (z1 + z2) to z2. • The gradient of the vector joining z2 to the origin is equal to the gradient of the vector joining (z1 + z2) to the point z1.

Subtract the appropriate real parts and separately subtract the appropriate imaginary parts. z2

z1

z2 - z1

z1 – z2 z 1 = 1 + 4i z 2 = 3 + 6i z2 – z1 z 1 – z 2

•

•

4u Maths Summary

We need to be able to interpret the expression | z – (a + ib) | as the magnitude of a vector joining (a, b) to the point representing z. Basically it is the same graph only it is based at (a, b) instead of the origin. Students need to recognise that the expression arg(z – z1) refers to the angle, which a vector joining the point representing z1 to the point representing z, makes with Page 42 of 117

the positive direction of the real axis. Once again, basically the same graph but with its base at z1 instead of the origin. •

Given z 1 and z 2, construct the vector z 1z 2.

z1 z2

z1 z2

z 1 = 1 + 3i

z 2 = 3 + 2i

z1 × z2

•

Prove geometrically that |z 1 + z 2| ≤ |z 1| + |z 2| .

z1 + z2

z2 z1

4u Maths Summary

Page 43 of 117

C

r2 D A r1

O

Using z1 = 2 + 2i = A z2 = 1 + 3i = B z1 + z2 = 3 + 5i = C Let

|z 1| = r1 |z 2| = r2

Let

AD be perpindicular to OC

∴

∆ ADO is a right ∠∆ where OA is the hypotenuse ∆ ACD is also a right ∠∆ where AC is the hypotenuse

∴

∴

•

n

(Pythagoras' theorem)

DC ≤ r2

(Pythagoras' theorem)

|OD| + |DC| ≤ r1 + r2 |z 1 + z 2| ≤ |z 1| + |z 2|

Proof:

Prove, by induction, that (cos θ + i sin θ)

OD ≤ r1

=

cos nθ + i sin nθ

for positive integers n.

4u Maths Summary

Page 44 of 117

(cos θ + isin θ)

n

= cos nθ + isin nθ

Proof by Mathematical Induction: Let n = 1 1

LHS = (cos θ + i sin θ) = cos θ + i sin θ RHS = cos(1) θ + i sin(1) θ = cos θ + i sin θ ∴ True for n = 1 Assume true for n = k (cos θ + i sin θ)

k

= cos kθ + i sin kθ

Let n = k + 1 RHS = cos(k + 1)θ + i sin(k + 1) θ LHS = (cos θ + i sin θ)

k+1 1

k

= (cos θ + i sin θ) (cos θ + i sin θ) = (cos θ + i sin θ)(cos kθ + i sin kθ) = cos θcoskθ – sinθsinkθ + i cosθsinkθ + i sinθcosθ = (cosθcoskθ – sinθsinkθ) + i (cos θsinkθ + sinθcos θ) = cos (kθ + θ) + i sin (kθ + θ) = cos(k + 1)θ + i sin(k + 1) θ = RHS ∴ True for n = k + 1 ∴ True for all positive integer values

•

Be able to reproduce this. n

Prove that n

(cos θ + i sin θ)

=

cos nθ + i sin nθ

for negative integers n.

(cos θ + isin θ) = cos nθ + isin nθ Let n = – m, where m is a positive integer. –m

LHS = (cos θ + isin θ) 1 = m (cos θ + isin θ) 1 = cos mθ + isin mθ 1 × cos mθ – isin mθ = θ θ cos m + isin m cos mθ – isin mθ cos mθ – isin mθ = 2 2 cos mθ + sin mθ θ θ = cos m – isin m 1 = cos( – mθ) + isin( – mθ) = cos nθ + isin nθ = RHS

•

Find any integer 4u Maths Summary

This is known a De Moivre’s theorem: Page 45 of 117

•

n

n

power of a given complex number.

(r cis θ)

Find the complex nth roots of ±1 in modulusargument form.

To find the nth roots of any complex number we use the fact:

= r cis nθ n

= r (cos nθ + isin nθ)

n

n

If R (cos φ + isin φ) = r(cos θ + isin θ) then: 1 n

R = r

θ + 2k π and φ = , where k = 0, 1, 2,...(n-1) n

Therefore the nth roots of the complex number 1 n

r (cos θ + i sin θ) have the modulus r and arguments given by φ = θ + 2k π , where k = 0, 1, 2,…(n – 1). n

Finding nth roots of ±1: n

z = 1 Let n = 5 z

5

= 1 1 n

θ π Using z = r cis  + 2k   n  z1 = cis 0° = 1

where k = 0,1,2,3,4

 π z2 = cis  2  5  π z3 = cis  4  5  

 π z4 = cis  6  5   8π  z5 = cis   5 

Use the same method for zn = -1.

4u Maths Summary

Page 46 of 117

• •

Sketch the nth roots ±1 on an Argand diagram. Illustrate the geometrical relationship connecting the nth roots of ±1.

z2

z3

1

z4

z5 • •

•

The roots represent evenly spaced points on the unit circle. They are also the vertices of an nth sided polygon.

Given equations Re(z) = c, Im(z) = k, sketch lines parallel to the appropriate axis.

y 4

2

-4

-2

2

4

x

-2

-4

Re(z) = 3

•

Given an equation, |z – z 1| = |z – z 2| ,

4u Maths Summary

Im(z) = 2

Rules: • For Re(z) = c, the real part of z becomes the graph. Therefore the locus of the graph is x = c. • For Im(z) = k, the imaginary part of z becomes the graph. Therefore the locus of the graph is y = k . For this type of locus, let z1 and z2 represent points on the argand diagram. If the distance from P to z1 is equal to the distance from P to z2 , then we know from plane geometry P(z)

Page 47 of 117

sketch the corresponding line.

that the locus is the perpendicular bisector of the line joining z1 and z2.

z2 z1 O

z 1 = 2 + 0i z 2 = -1 + i

•

Given equations |z| = R and |z – z 1| = R , sketch the corresponding circles.

4u Maths Summary

Rules: • Let z1 and z2 be normal Cartesian co-ordinates. • Find the midpoint of z1 and z2. • Find the gradient of the line joining z1 to z2 . • Inverse the gradient to find the gradient of the line perpendicular to it. • Use point-gradient formula to find equation of the line representing the point P. These two equations represent the locus of a circle. |z| = R

Is the locus of a circle with a centre at the origin and radius R units.

Page 48 of 117

y 3

2

1

-4

-2

2

4

x

-1

-2

-3

|z| = 2 |z – z 1| = 2 where z 1 = 2 – i

•

Given equations arg z = θ and arg (z – z 1) = θ , sketch the corresponding rays.

Rules: • The value of R becomes the length of the radius. • The point z1 becomes the centre of the circle. The graph is a ray which originates from either the origin or z1, and goes off at angle θ in the positive direction.

z1

θ θ

O arg z =

•

Sketch regions associated with any of the above curves. 4u Maths Summary

π 4

arg (z – z 1) =

π where z1 = -3 + 2i 3

Rules: • The graph is just a ray that starts at z1 • The ray makes an angle θ with the positive direction. The graphs are sketched exactly the same way but the area in which the equation is fulfilled must be determined. Remember: Page 49 of 117

•

• Dotted lines for < or >. • Solid lines for = or =. Give a geometrical Use basic names to describe the locus of a point or the description of any region in which an equation is fulfilled. such curves or Eg: The area inside a circle with radius 3 units and centre at regions. (3, 2). Be able to graph and describe these:

•

Be able to understand the intersection of more than one region or graph.

Examples need only involve replacing z by z = x + iy in relations such as:

4u Maths Summary

Page 50 of 117

•

•

Topic 3 Conics Write down the defining equation of an ellipse with centre at the origin.

Equation of an ellipse with centre at the origin: 2

x

2

a

2

+

y b

= 1

2

This is regarded as the defining equation.

Sketch the ellipse showing points of intersection with the axes of symmetry.

y

(0, b)

(-a, 0)

x

(a, 0)

(0, -b)

2

x

2

2

+

a

•

Find the lengths of the major and minor axes and semi-major and semi-minor axes of an ellipse.

y b

2

= 1

y

x

2b

2a

Rules: • Major axis length = 2a 4u Maths Summary

Page 51 of 117

•

•

Write down the parametric coordinates of a point on an ellipse. Sketch an ellipse using its auxiliary circle.

• Minor axis length = 2b • Semi-major axis length = a • Semi-minor axis length = b Parametric Coordinates: x = a cos θ y = b sin θ y

(acos θ , asin θ )

(acos θ , bsin θ ) x

2

x

2

2

+ y = a

2

x

2

a

2

+

y b

2

= 1

Equation of the Ellipse: x = a cos θ y = b sin θ Equation of the Auxiliary Circle: x = a cos θ y = a sin θ The parametric representation x = a cos θ , y = b sin θ is useful in graphing the ellipse from an auxiliary circle. The shape of an ellipse should be examined as the ratio b a

•

Find the equation of an ellipse from its focus-directrix definition.

4u Maths Summary

varies. The definition of a conic is: PS

= e

PM

Where

P – Is any point on the conic S – The focus of the conic M – A point on the directrix Page 52 of 117

e – The degree of eccentricity For an ellipse, 0 < e < 1 Rearranging the definition of a conic gives: PS = ePM

Therefore, the equation of an ellipse can be calculated by substituting in appropriate values for the focus, directrix and eccentricity.

•

Find the eccentricity from the defining equation of an ellipse.

The focus -directrix definition should be used whenever a focal distance is to be calculated. Using the definition of a conic: PS = ePM 2

(x – ae) + (y – 0) (x – ae)

2

2

2 2 = e x – a   e

2

+ y

2

2 2

2

x – 2aex + a e + y 2

2 2

2

 1 

– e  x + y

2 2

2

2 2

∴b

Given the equation of an ellipse, find the co-ordinates of the foci and equations of the directrices.

2

2 2

= a – ae 2

2

= a  1 – e

 

2

+

a

•

2 2

= e x – 2aex + a 2

x – ex + y

x

 x – a 2 + (y – y) 2   e 

= e

2

y 2

= 1

2

a  1 – e 2

 

2

= a  1 – e

 

For the ellipse with equation: 2 2 x y + 2 = 1 2 a b Foci: (±ae, 0) Directrices: x = ±a e

4u Maths Summary

Page 53 of 117

•

Sketch an ellipse, marking on it the positions of its foci and directrices.

y

D

x = –

D

a e

x =

(0, b)

(-ae, 0)

a e

(ae, 0)

(-a, 0)

O

x

(a, 0)

(0, -b)

D’

D’ 2

x

2

2

+

a

•

y b

2

= 1

The major properties of an ellipse are to be proven for both a general ellipse with centre O and for ellipses with given values of a and b. Use implicit differentiation to find the equations of the tangent and the normal at P(x1, y1) on an ellipse.

2

x

2

+

y

= 1 2 2 a b Implicitly Differentiating Gives: 2x 2y × dy = 0 2 + 2 dx a b 2y

2x × dy = – 2 dx b a 2

2

dy = – 2xb 2 dx 2a y 2

dy = – b x 2 dx ay Let P = (x1, y1) ∴ Gradient = –

2

b x1 2

a y1

4u Maths Summary

Page 54 of 117

∴ Equation of the tangent at P is: 2

y – y1 = –

b x1 2

(x – x1)

a y1 2

2

a y1(y – y1) = – b x1(x – x1) 2

2

2

2

2

a y1y – a y1 = – b x1x + b x1 2

2

2

2

2

2 2

a y1y + b x1x = b x1 + a y1 2 2

Dividing everything by a b gives y y1 b

2

x x1

+

2

=

2

a x

2

+

a

2

Since

2

x

y b

2

2

+

y

2 2 = 1 a b x x1 yy + 21 = 1 2 a b 2

Gradient of the Normal =

a y1 2

b x1 ∴ Equation of the Normal: 2

a y1

y = y1 =

2

(x – x1)

b x1 2

2

b x1 (y – y1) = a y1 (x – x1) 2

2

2

2

b x1y – b x1y1 = a y1x – a x1y1 Dividing everything by x1y1 gives: 2

2

b y – b 2 = a x – a2 y1 y1 2

2

a x – b y = a2 – b 2 x1 y1

4u Maths Summary

Page 55 of 117

•

Find the equations of the tangent and the normal at P ( acos θ , bsin θ ) on an ellipse.

x = acos θ dx = – asin θ dθ y = bsin θ dy = bcos θ dθ ∴ dy = – bcos θ dx asin θ ∴ Equation of the tangent is: θ y – bsin θ = – bcos (x – acos θ) asin θ 2

2

aysin θ – absin θ = – bxcos θ + abcos θ 2

2

aysin θ + bxcos θ = abcos θ + absin θ Dividing Everything by ab ysin θ + xcos θ = cos2θ + sin2θ b a ∴ Tangent is: xcos θ + ysin θ = 1 a b θ Gradient of the Normal = asin bcos θ θ y – bsin θ = asin (x – acos θ) b cos θ 2

2

bycos θ – b sinθcosθ = axsin θ – a sinθcos θ Dividing everything by sinθcos θ by – b 2 = ax – a2 sin θ cos θ ax – by = a2 – b 2 cos θ sin θ

• •

Find the equation of a chord of an ellipse. Find the equation of a chord of contact.

4u Maths Summary

The equation for the chord of an ellipse is: x cos  θ + φ  + y sin  θ + φ  = cos  θ – φ        a  2  b  2   2  Let PT and PQ be tangents of an ellipse.

Page 56 of 117

Let

P = (x1, y1) Q = (x2, y2) T = (x0, y0) Equation of tangent PT: x0 x1 y0 y1 2 + 2 = 1 a b Equation of tangent PQ: x0 x2 y y + 0 22 = 1 2 a b Hence both P & Q satisfy: x x0 yy + 20 = 1 2 a b

•

The chord of contact is useful as a tool in the proof of a number of properties of an ellipse. Prove that the sum of the focal lengths is constant.

P = (acos θ, bsin θ)

Let

S = (ae, 0) S' = ( – ae, 0) PS = PS

2

(acos θ – ae)

2

2

2

2

b b

2 2

PS PS PS

2

2

2 2

2 2

2

+ b sin θ

 

= a – ae

2 2

PS

2

= a 1 – e 2

ae

PS

2 2

= a cos θ – 2a ecos θ + a e

Using

PS

+ (bsin θ – 0)

= a – b

2

= a cos θ – 2a ecos θ + a – b

2

= a cos θ – 2a ecos θ + a – b  1 – sin θ 

2

= a cos θ – 2a ecos θ + a – b cos θ

2

=

2

2

2

2

2

2

2

2

2

2

 2 a 

2

2

2

2

+ b sin θ

2

2

2 

2

2

2

2

2

2

– b cos θ – 2a ecos θ + a

2

= a e cos θ – 2a ecosθ + a

2 2

2

= a (1 – ecosθ)

2

2

2

2

2

2

PS = a (1 – ecosθ) PS' = a (1 + ecosθ) PS + PS' = 2a (A constant)

The focus -directrix definition leads to a simple proof that the sum of the focal lengths is constant (2a).

4u Maths Summary

Page 57 of 117

•

Prove the reflection property, namely that the tangent to an ellipse at a point P on it is equally inclined to the focal chords through P.

4u Maths Summary

Page 58 of 117

4u Maths Summary

Page 59 of 117

 tan φ =  b  θ aesin  

4u Maths Summary

Page 60 of 117

tan θ =  – b   aesin θ  ∴

tan θ = tan φ θ = φ ∴ ∠ RPS' = ∠ TPS

•

4u Maths Summary

The reflection property of the ellipse may be approached by using the result that the bisector of an angle of a triangle divides the opposite side into two intervals, whose lengths are in the same ratio as the lengths of the other two sides.

Page 61 of 117

•

Prove that the chord of contact from a point on a directrix is a focal chord.

∴ QR is a focal chord

•

Prove that part of the tangent between the point of contact and the directrix subtends a right angle at the 4u Maths Summary

Page 62 of 117

•

corresponding focus. Prove simple properties for both the general 2

ellipse

x

2

2

+

a

•

•

y

2

=1

b

and for ellipses with given values of a and b. Write down the defining equation of a hyperbola with centre at the origin. Sketch the hyperbola 2

x

2

Students are not expected to do proofs, under examination conditions, which are more difficult than those involved in the Contents and Skills objectives. Locus problems on the ellipse are not included.

General Equation of a Hyperbola: 2

x a

2

–

y

2

b

2

= 1

2

–

y

2

=1

, showing points of intersection with axes of symmetry and positions of asymptotes. a

b

(-a, 0)

(a, 0)

Rules: • Cuts the x-axis at (-a, 0) & (a, 0)

•

•

•

The asymptotes are: y = ± bx

•

The shape of the hyperbola should be examined as

a

b varies. a

Find the length of the major and minor axes and semi-major and semi-minor axes of a hyperbola. Write down the 4u Maths Summary

Rules: • Length of major axis = 2a • Length of minor axis = 2b • Length of semi-major axis = a • Length of semi-minor axis = b Parametric Coordinates: Page 63 of 117

•

parametric coordinates of a point on the hyperbola. Find the equation of a hyperbola from its focusdirectrix definition.

x = a sec θ y = b tan θ

The definition of a conic is: PS

= e

PM

Where

P – Is any point on the conic S – The focus of the conic M – A point on the directrix e – The degree of eccentricity

For a hyperbola , e > 1 Rearranging the definition of a conic gives: PS = ePM

Therefore, the equation of a hyperbola can be calculated by substituting in appropriate values for the focus, directrix and eccentricity.

•

Find the eccentricity from the defining equation of a hyperbola.

The focus -directrix definition should be used whenever a focal distance is to be calculated. Using the definition of a conic: PS = ePM 2

(x – ae) + (y – 0) (x – ae)

2

2

2 2 = e x – a   e

2

+ y

2

2 2

2

x – 2aex + a e + y 2

2 2

2

 1 

– e  x + y

2 2

2

2 2

2 2

2

= e x – 2aex + a 2

x – ex + y

x

 x – a 2 + (y – y) 2   e 

= e

2 2

= a – ae 2

2

= a  1 – e

 

2

+

a

y 2

2

a  1 – e

From before b

 

2

= 1 2

2

= a 1 – e

 

2

Since e > 1, 1 – e < 0 ∴b

•

Given the equation of the hyperbola, find the coordinates of its foci and equations of its directrices. 4u Maths Summary

2

2 2



= a  e – 1 

For the hyperbola with equation: 2 2 x y – 2 = 1 2 a b Foci: (±ae, 0) Directrices: Page 64 of 117

•

x = ±a e

Sketch a hyperbola, marking on it the positions of its foci and directrices.

y= –a

e

(-ae, 0)

•

y= a

e

(ae, 0)

Use implicit differentiation to find the equations of the tangent and normal at P(x1, y1) on a hyperbola.

Gradient of Tangent

4u Maths Summary

Page 65 of 117

Equation of tangent to hyperbola: x x1 yy – 21 = 1 2 a b

Equation of normal to hyperbola: 2

2

a x + b y = a2 + b2 x1 y1

4u Maths Summary

Page 66 of 117

•

Find the equations of the tangent and normal at P (asec θ , btan θ ) on the hyperbola.

Gradient of Tangent

Equation of tangent to hyperbola xsec θ – ytan θ = 1 a b

Equation of normal to hyperbola by + ax = a2 + b2 tan θ sec θ

4u Maths Summary

Page 67 of 117

•

•

Find the equation of a chord of a hyperbola.

The equation of a chord from P (asec θ, btan θ) to Q(asec φ , btan φ ) is: x cos  θ – φ  – y sin  θ + φ  = cos  θ + φ        a  2  b  2   2 

Find the equation of a chord of contact.

Let the point on directrix = T(x0 , y0) Let P(x1, y1) & Q(x2 , y2) Therefore the chord formulas for both are: x0 x1 2

a

–

y0 y1 b

2

= 1

&

x0 x2 2

a

–

y0 y2 b

2

= 1

Both equations satisfy the relation: x x0 y y0 – = 1 a

•

Prove that the difference of the focal lengths is a constant.

4u Maths Summary

2

b

2

The chord of contact is useful in proving some properties. Proving PS – PS’ = k

Page 68 of 117

• •

•

Prove the reflection property for a hyperbola. Prove that the chord of contact from a point on the directrix is a focal chord. Prove simple properties for the general hyperbola and also hyperbolae with given values of a and b.

The same geometry theorem, as used in the case of the ellipse, is useful in proving the reflection property of the hyperbola. This proof is exactly the same for the hyperbola as it is with the ellipse.

The major properties of the hyperbola are to be proven for both the general hyperbola with centre O and for hyperbolae with given values of a and b. Students are not e xpected to do proofs, under examination conditions, which are more difficult than those involved in the skills objectives. Locus problems, on a hyperbola with the general equation, are not in the course.

The Rectangular Hyperbola • Prove that the hyperbola with equation

xy =

1 2

2

a is the

hyperbola 2

2

2

x – y = a referred to different axes.

4u Maths Summary

Page 69 of 117

•

•

Write down the eccentricity, coordinates of foci and vertices, equations of directrices and equations of asymptotes. Sketch the hyperbola

xy =

1 2

A definition needs to be given for a rectangular hyperbola. It quickly follows, from seeing the connection between x2 – y2 = a2 and xy = ½a2 , that the eccentricity is 2 . Eccentricity = 2 Foci = ± (a, a) Directrices = x + y = ±a a a  Vertices = ±  ,  or ± (c, c) 2   2 Asymptotes = x = 0, y = 0

D

2

a , for varying values of a, marking on vertices, foci, directrices and asymptotes.

D’

S

S’

4u Maths Summary

Page 70 of 117

•

Write down the parametric coordinates for the rectangular hyperbola 2

xy = c , for

•

varying values of c. Find the equation of the chord joining P  cp, c   p  to Q  cq, c  .  q 

• Asymptotes are the x and y axes. When the value of c changes, substitute the new value into the following formulas: x = ct y = c t ∴ P  ct, c  t  

Gradient of PQ

4u Maths Summary

Page 71 of 117

•

Find the equation of the tangent at P.

4u Maths Summary

Page 72 of 117

•

Find the equation of the normal at P.

•

Find the equation of the chord joining P(x1, y 1) to Q(x2, y 2).

4u Maths Summary

Page 73 of 117

Therefore, the equation of the chord from P to Q is:

•

Find the equation of the chord of contact from T(x0, y0).

•

Find the point of Use the equations for tangents and normals. intersection of tangents and of Solve these simultaneously to find the points of intersection. normals. Area of the Triangle: Prove simple geometrical properties of the rectangular hyperbola T including: o The area of the triangle P bounded by a tangent and the asymptotes is a constant. o The length of R the intercept, cut off a tangent by the asymptotes, equals twice

•

4u Maths Summary

Page 74 of 117

the distance of the point of contact from the intersection of the asymptotes.

Length of Intercept: T

P

O

4u Maths Summary

R

Page 75 of 117

•

Find loci of points including: o Loci determined by intersection points of tangents. o Loci determined by intersection points of normals o Loci determined by midpoints of intervals.

4u Maths Summary

Intersection of tangents:

This example is for a problem whereby the points P & Q must join to form a chord that passes through a given point, in this case (0, 4). Also, for this example, c = 3.

Page 76 of 117

Intersection of Normals: Midpoints of intervals:

4u Maths Summary

Page 77 of 117

The locus is therefore a hyperbola that shares the same asymptotes as the original equation. It is not intended that locus problems should include sophisticated techniques for elimination of parameters. Students are expected to be able to proceed from a pair of parametric equations to obtain a locus expressible by a linear equation (perhaps with constraints on x or y). In cases where the resulting locus is not expressible in terms of a linear equation, it will be given in algebraic o r geometric form and students will verify that this form is satisfied (perhaps with additional constraints). •

•

•

Appreciate that the various conic sections (circle, ellipse, parabola, hyperbola and pairs of intersecting lines) are indeed the curves obtained when a plane intersects a (double) cone Relate the various ranges of values of the eccentricity e to the appropriate conic and to understand how the shape of a conic varies as its eccentricity varies. Appreciate that the equations of all conic sections 4u Maths Summary

For e = 1: • The locus is a parabola • PS = PM For e < 1: • The locus in an ellipse. • PS < PM For e > 1: • The locus is a hyperbola • PS > PM All conic sections deal with just equations in x and y.

Page 78 of 117

involve only quadratic expressions in x and y.

4u Maths Summary

Page 79 of 117

• •

•

Topic 4 Integration Use a table of standard integrals. Change an integrand into an appropriate form by use of algebra. Evaluate integrals using algebraic substitutions.

The table of standard integrals is supplied in an exam and can be referred to for integration. Look for constant values that can be removed, trig substitutions, etc.

Let u equal some portion of the integrand so that the integrand can be rearranged to give an integrand in terms of u and du. Eg: ⌠  u du ⌡ •

•

Evaluate simple trigonometric integrals.

Only simple substitutions are needed, eg u = 1 + x2, 2 4 ⌠ ⌠ x v2 = 1 – x in  x(1 + x ) dx ,  dx . ⌡ ⌡ 1–x The effect on limits of integration is required, and definite integrals are to be treated. Trigonometric Identities: 2

2

sin θ + cos θ = 1 2

2

1 + cot θ = cosec θ 2

2

tan θ + 1 = sec θ

Sums & Differences: sin( θ + φ ) = sinθcosφ + sinφ cosθ sin( θ – φ ) = sinθcosφ – sinφ cosθ cos(θ + φ) = cos θcos φ – sinθsinφ cos(θ – φ) = cos θcos φ + sinθsinφ θ φ tan( θ + φ) = tan + tan θ 1 – tan tan φ θ φ tan( θ – φ) = tan – tan 1 + tan θtan φ

Double Angles:

sin 2θ = 2sinθcosθ 2

2

cos 2θ = cos θ – sin θ 2

= 2cos θ – 1 2

= 1 – 2sin θ 2tan θ tan 2θ = 2 1 – tan θ

Transformations: 4u Maths Summary

Page 80 of 117

asin θ + bcosθ asin θ – bcosθ acos θ + bsinθ acos θ – bsinθ

= = = =

rsin( θ + α ) rsin( θ – α ) rcos( θ – α ) rcos( θ + α )

2

where r =

a + b tan α = b a

2

Integration of Trigonometric Functions: Basic Trigonometric Integration: ⌠  sinax dx = – 1 cosax + c ⌡ a ⌠  cosax dx = 1 sinax + c ⌡ a 2 ⌠  sec ax dx = 1 tanax + c ⌡ a Integration of Squared Trig Functions: 2 ⌠  sin ax = 1 x – 1 sin 2ax + c 2 ⌡ 4a ⌠ 2  cos ax = 12 x + 1 sin 2ax + c ⌡ 4a Inverse Trigonometric Functions: -1  1 ⌠   dx = sin  x  + c ⌡ 2 2 a  a –x

⌠  – ⌡

1 2

2

a –x

-1 dx = cos  x  + c a 

-1 ⌠ a dx = tan  x  + c  2 2 ⌡ a + x a 

•

Evaluate Integrals using trigonometric substitutions.

The Substitution Method:

2

1 + t

t

θ 2

1

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Let t = tan  θ  2 

θ Ratios: 2

tan  θ  = t 2  θ  sin   = 2 

t 2

1 + t

cos  θ  = 2 

1 2

1 + t

Basic Ratios: sinθ =

2t 2

1+t

2

1–t cosθ = 2 1+t 2t tan θ = 2 1–t

Derivative: dθ = 2 2 dt 1+t

•

•

Evaluate integrals using integration by parts.

This works for an integrand whereby one part is also to be differentiated and the other part capable of being integrated.

Derive and use recurrence relations.

Recurrence Formula: n ⌠  cos x dx ⌡ ⌠ n In =  cos x dx ⌡ ⌠ n –1 =  cos x . cosx dx ⌡

⌠ ⌠  u dv dx = uv –  v du dx ⌡ dx ⌡ dx or ⌠ ⌠  uv' dx = uv –  vu' dx ⌡ ⌡

⌡ n– 1

Let u = cos x du = – sinx (n – 1) cos n – 2x dx dv = cosx dx v = sinx

n– 1

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n– 2 ⌠ x +  (n – 1)sinxcos x . sinx dx ⌡ n– 1 ⌠ 2 n– 2 = sinxcos x + (n – 1)  sin x cos x dx ⌡ n– 1 2 n –2 ⌠ = sinxcos x + (n – 1)  (1 – cos x)cos x dx ⌡ n– 1 n –2 n ⌠ = sinxcos x + (n – 1)  cos x – cos x dx ⌡

In = sinxcos

= sinxcos In = sinxcos

n– 1

n– 1

n– 1

n –2 n ⌠ ⌠ x + (n – 1)  cos x dx – (n – 1)  cos x dx ⌡ ⌡

x + (n – 1)In – 2 – (n – 1)In

In + (n – 1)I n = sinxcos

n –1

x + (n – 1)In – 2

n –1

nIn = sinxcos x + (n – 1)In – 2 n – 1 In = 1 sinxcos x + n – 1 In – 2 n n

Then use this new-found equation to solve an integral such as: 6 ⌠  cos x dx ⌡

•

•

Integrate rational functions by completing the square in a quadratic denominator. Integrate rational functions whose denominators have simple linear or quadratic factors.

Completing a square will often result in an inverse tan result.

Partial Fractions: Used in situations whereby the denominator is broken into parts so it can be integrated. Example: ⌠ 5x + 1  ⌡ (x – 1)(x + 2) 5x + 1 = a + b (x – 1)(x + 2) x– 1 x+2 ∴ 5x + 1 = a(x + 2) + b(x – 1) Then solve for x = -2 & 1 to find values of a & b Then integrate: Let

⌠ a ⌠  +  b ⌡ x–1 ⌡ x+ 2

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Integrals involving the Log Function: ⌠ 1  dx = lnx + c ⌡ x ⌠ h'(x)  dx = ln h(x) + c ⌡ h(x) •

If one of the factors in the denominator is a quadratic, then instead of b being part of the equivalence, use a basic linear function. Example: ⌠ 5x + 1 = a + 2bx + c  2 ⌡ (x + 3x + 2)(x – 2) x– 2 x + 3x + 2

•

4u Maths Summary

Note that sometimes polynomial division is needed to solve an integrand.

Page 84 of 117

•

Topic 5 Volumes Appreciate that, by dividing a solid into a number of slices or shells, whose volumes can be simply estimated, the volume of the solid is the value of the definite integral obtained as the limit of the corresponding approximating sums.

A solid’s volume can be calculated by dividing it into slices or shells. Let us say ∆V = A(x). ∆x By summing these together in a series and taking the limit as x à 0.

V =

lim

∆x → 0

A(x). ∆x

a

b

⌠ V =  A(x) dx ⌡a This can then be adjusted so a definite integral is reached. •

•

•

4u Maths Summary

Σ b

The purpose of this topic is to provide practical examples of the use of a definite integral to represent a quantity (in this case, a volume) whose value can be regarded as the limit of an appropriate approximating sum. Emphasis is to be placed on understanding the various approximation methods given, deriving the relevant approximate expression for the corresponding element of volume and proceeding from this to expressing the volume as a definite integral. The evaluations of infinite series by a definite integral, or of integrals by summation of series, are not included in this topic. Volumes of revolution could lead, from questions involving rotation about a coordinate axis, to rotation about a line parallel to a coordinate axis, eg find the volume of the solid formed when the region bounded by y = 2 x , the xaxis and x = 4 is rotated about the line x = 4. Always draw a sketch

Page 85 of 117

•

•

Find the volume of a solid of revolution by summing the volumes of slices with circular crosssections. Find the volume of a solid of revolution by summing the volumes of slices with annular crosssections.

2

x = y From this revolution we take a typical slice, which in this case is a spherical slice. ∆x radius = y

A(x)

So, therefore the area of the circle can be expressed as: 2 A(x) = π y Meaning the volume of this slice is: 2 ∆V = π y ∆x By summing together and taking the limit, the volume of the whole solid is:

Σ 5

V =

lim

∆x → 0

A(x) ∆x

0

5

⌠ V =  A(x) dx ⌡0 5

2 ⌠ V =  ( π y ) dx ⌡0 5

2 ⌠ V = π  y dx ⌡0

Since y2 = x

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5

⌠ V = π  x dx ⌡0

 2 5

V = π x   2 0 3 V = 25π units 2

This method can also be applied to rotations around the yaxis.

•

Find the volume of a solid of revolution by summing the volumes of cylindrical shells.

Notes: • Examples involving annular shells should include questions as difficult as the following. The region R, 2 4 bounded by: 0 ≤ x ≤ 2 , 0 ≤ y ≤ 4x – x , is rotated about the y-axis. The solid so formed is sliced by planes perpendicular to the y-axis. Express the areas of the cross-sections so formed as a function of y, the distance of the plane from the origin. Use this result to calculate the volume of the solid. This is used when a graph is being rotated about the y-axis between x = a & x = b. Take for instance, this example, where we rotate the area enclosed between the line y = -x + 2 a nd the x & y-axes.

We firstly take our typical slice. This is a cylinder with: Radius = x Height = y Width = ?x This is then unfolded to give a rectangle. ?x

4u Maths Summary

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y

2π x

Therefore: ∆V = Volume of a cylindrical shell . 2

2

∆V = [ π (x + ∆x) – π x ] y 2

2

2

∆V = [ π (x + 2x∆x + ∆x ) – π x ] y 2

∆V = π y (2x∆x + ∆x ) 2

However, ∆x is negligible, so: ∆V = 2 π x y. ∆x

Taking the limit and integrating, gives:

Σ 2

V =

lim

∆x → 0

∆V

0

2

⌠ V = 2π  xy dx ⌡0 Since y = – x + 2 2

⌠ V = 2π  x ( – x + 2) dx ⌡0 2

2 ⌠ V = 2π  ( – x + 2x ) dx ⌡0

Finding definite integral gives: 2  x3 2 V = 2π  – + x   3 0 3 π V = 8 units 3

Notes: • A formula for summing by cylindrical shells should not be learnt. Each problem should rather be developed from first principles.

4u Maths Summary

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•

Find the volume of a solid which has parallel cross-sections of similar shape. 4

5 4

3 We are given the dimensions of the base and told the height is 4m. That is, 4m right through the middle of the pyramid from top to bottom. For this our typical slice is:

a b

∆x

c

Since the sizes of a, b and c will vary proportional to how far along the pyramid we go, we break them down into three single triangles. This way, each of the values for a, b and c can be calculated in terms of x. Since each of these vary proportionally with x:

a

5 a = x 5 4 5x a = 4

4u Maths Summary

b

c

4 b = x 4 4

b = x

3 c = x 3 4 3x c = 4

Page 89 of 117

We now have values for our typical slice.

5x 4

x

∆x

3x 4

We then use the area of a triangle. A(x) = 1 bh 2 A(x) =

1 2

× 3x × x 4

Meaning the volume of the typical slice is: 2

∆V = 3x ∆x 8

By summing and taking the limit:

Σ 4

V =

lim

∆x → 0

0

2

3x ∆x 8

4

2 ⌠ V =  3x dx ⌡0 8

 3x3  4 V =    24  0 3

V = 8 units

Notes: • The process of writing the limiting sum as an integral should be extended to cases where cross-sections are other than circular. These cases should only involve problems in which the geometrical shape is able to be visualised, eg prove that the volume of a pyramid of height h on a square base of side a is

4u Maths Summary

1 3

2

a h

Page 90 of 117

Topic 6 Mechanics Projectile Motion • Derive the equations of motion of a projectile

Horizontally:

∴

x = vt cosθ . x = v cosθ .. x = 0 Vertically:

2

y = vt sinθ – gt 2 . y = v sinθ – gt .. y = –g Cartesian Equation of Motion: ∴

4u Maths Summary

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2

y =

– gx 2

2

(1 + tan θ) + x tan θ

2V Maximum Height: . Max height occurs when y = 0

2

2

θ h = V sin 2g Range: Since projectile motion is a parabola, the time taken for the entire journey will be double that taken to reach the maximum height.

4u Maths Summary

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2

θ R = V sin 2 g Maximum Range: Max range occurs when the projectile is fired at 45°. 2

•

Use equations for horizontal and vertical components of velocity and displacement to answer harder problems on projectiles. Simple Harmonic Motion • Write down equations for displacement, velocity and acceleration given that a motion is simple harmonic.

4u Maths Summary

R = V g Time of Flight: Since projectile motion is a parabola, the time taken for the entire journey will be double that taken to reach the maximum height. T = 2Vsinθ g Use the above equations and substitute in values that are known in order to find those that aren’t.

Simple Harmonic Motion:

Page 93 of 117

x . x . x .. x

= a cos(nt + ∈) = d (a cos(nt + ∈) dx = – an sin(nt + ∈ ) = d ( – an sin(nt + ∈ ) dx

.. 2 x = – an cos(nt + ∈) .. 2 x = – n [acos(nt + ∈)] .. 2 x = –n x

•

•

Use relevant formulae and graphs to solve harder problems on simple harmonic motion. Use Newton’s laws to obtain equations of motion of a particle in situations other than projectile motion and simple harmonic motion.

Substitute in known values to the formulas above to find unknown values.

Newton’s Laws: • Every object in a state of uniform motion tends to remain in that state of motion unless an external force is applied to it. • The relationship between an object's mass m, its acceleration a, and the applied force F is F = ma. Acceleration and force are vectors (as indicated by their symbols being displayed in slant bold font); in this law the direction of the force vector is the same as the direction of the acceleration vector. • For every action there is an equal and opposite reaction. This translates to a few necessary formulas: F = ma where

F - Force in Newtons m - Mass in kg a - Acceleration in ms

p = mv where

–2

p - Momentum m - Mass v - Velocity in ms

–1

1 kg wt = 9.8 N • 4u Maths Summary

The classical statement of Newton’s first and second laws Page 94 of 117

•

Describe mathematically the motion of particles in situations other than projectile motion and simple harmonic motion. Mathematical Descriptions of Motion .. • Given x = f(x) and initial conditions 2

derive v = g(x) and describe the resultant motion. • Recognise that a motion is simple harmonic given an equation for either acceleration, velocity or displacement, and describe the resultant motion. Resisted Motion along a horizontal line • Derive, from Newton’s laws of motion, the equation of motion of a particle moving in a single direction under a resistance proportional to a power of the speed.

4u Maths Summary

of motion should be given as an illustration of the application of calculus to the physical world. Resolution of forces, accelerations and velocities in horizontal and vertical directions is to be used to obtain the appropriate equations of motion in two dimensions. Use the above formulas and represent situations using them. •

If

Students should be able to represent mathematically, motions described in physical terms. They should be able to explain, in physical terms, features given by mathematical descriptions of motion in one or two dimensions.

.. x = f(x), use

d  1 v2  dx  2 

2 f(x) = d  1 v  dx  2  2 ⌠ v =  f(x) dx ⌡ Look for equations that look like the ones above. 1 2

Also, if the motion needs to be proven, integrate or differentiate to show that: .. 2 x = –n x

Remembering F = ma, n If resistance is – kv ∴

n

a = – kv

n

F = – mkv

Page 95 of 117

•

Derive an expression for velocity as a function of time

.. If x = f(t), use dv dt dv = f(t) dt

⌠ v =  f(t) dt ⌡ .. If x = f(v), use dv if initial conditions are (t,v) dt dv = f(v) dt dt = 1 dv f(v)

•

Derive an expression for velocity as a function of displacement.

⌠ t =  1 dv ⌡ f(v) Then rearrange .. 2 If x = f(x), use d  12 v   dx  d  1 v2  = f(x)   dx  2  1 2

•

Derive an expression for displacement as a function of time.

Motion Vertically Upwards • Derive, from Newton’s laws of motion, the equation of motion of a particle, moving vertically upwards in a medium, with a resistance R proportional to the first or second power of its speed.

2 .. d x If x = f(t), use 2 dt 2

d x = f(t) 2 dt Then intergrate twice

Remember F = ma, .. n x = – (g + kv ) n

F = – m (g + kv )

• •

• • •

4u Maths Summary

2 ⌠ v =  f(x) dx ⌡

Typical cases to consider include those in which the resistance is proportional to the speed and to the square of the speed. Analysis of the motion of a particle should include consideration of the behaviour of the particle as t becomes large. Graphs can offer assistance in understanding the behaviour of the particle. The origin should be placed at the point of projection. The maximum height reached by the particle can be obtained from the expression relating speed and displacement. The time taken to reach this maximum height can be Page 96 of 117

• • •

Derive an expression for velocity as a function of time

obtained from the expression relating speed and displacement. The time taken to reach this maximum height can be obtained from the expression relating speed and time. Problems should include cases where the magnitude of

the resistance is given. (eg: R = .. If x = f(t), use dv dt dv = f(t) dt

1 10

2

v )

⌠ v =  f(t) dt ⌡ .. If x = f(v), use dv if initial conditions are (t,v) dt dv = f(v) dt dt = 1 dv f(v)

•

Derive an expression for displacement as a function of time.

•

Solve problems by using the expressions derived for acceleration, velocity and displacement. Motion of a Particle Falling Downwards • Derive, from Newton’s laws of motion, the equation of motion of a particle falling in a medium with a resistance R proportional to the first or 4u Maths Summary

⌠ t =  1 dv ⌡ f(v) Then rearrange 2 .. d x If x = f(t), use 2 dt 2

d x = f(t) 2 dt Then intergrate twice Use the above equations and substitute in values.

Remember, F = ma n

a = g – kv F = ma

n

F = m (g – kv ) • •

Cases, other than where the resistance is proportional to the first or second power of the speed, are not required to be investigated. Students should place the origin at the point from which Page 97 of 117

second power of its speed.

•

Determine the terminal velocity of a falling particle, from its equation of motion.

the particle initially falls. If the motion of a particle both upwards and then downwards is considered then the position of the origin should be changed as soon as the particle reaches its maximum height. Care must then be taken in determining the correct initial conditions for the downward motion. • The time taken for the particle to reach the ground should be found . • Problems should include a study of the complete motion of a particle, projected vertically upwards, which then returns to its starting point. For specific resistance functions, comparisons should be made between the times required for its upward and downward journeys and between the speed of projection and the speed of its return. Terminal velocity occurs when acceleration has ceased. n g – kv = 0 n

g = kv n v = g k v =

•

Derive expressions for velocity as a function of time and for velocity as a function of displacement.

n

g k

Time: .. If x = f(t), use dv dt dv = f(t) dt

⌠ v =  f(t) dt ⌡ .. If x = f(v), use dv if initial conditions are (t,v) dt dv = f(v) dt dt = 1 dv f(v) ⌠ t =  1 dv ⌡ f(v) Then rearrange Displacement: .. 2 If x = f(x), use d  12 v   dx  d  1 v2  = f(x)   dx  2  1 2

4u Maths Summary

2 ⌠ v =  f(x) dx ⌡

Page 98 of 117

•

Derive an expression for displacement as a function of time.

•

Solve problems by using the expressions derived for acceleration, velocity and displacement. Circular Motion • Define angular velocity of a point moving about a fixed point.

•

•

•

Deduce, from this definition of angular velocity, expressions for angular acceleration of a point around a fixed point. Prove that the instantaneous velocity of a particle moving in a circle of radius R, with angular velocity ω , is Rω .

2 .. d x If x = f(t), use 2 dt 2

d x = f(t) 2 dt Then intergrate twice Use above equations and substitute in known values.

Angular Velocity:

. ω = dθ = θ dt Through differentiation ω = dθ dt 2 . .. θ ω = d2 = θ dt

Velocity: v = Change in arc AB Change in time v = d Rθ dt θ v = Rd dt v = Rω

Prove that the tangential and normal components of the force acting on a particle 4u Maths Summary

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moving in a circle of radius R, with angular velocity ω . , need to be Mr ω and 2 – mRω respectively. Uniform Circular Motion • Write down the formula appropriate for a particle moving around a circle with uniform angular velocity.

4u Maths Summary

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•

Apply these formulae to the solution of simple problems Conical Pendulum • Use Newton’s law to analyse the forces acting on the bob of a conical 4u Maths Summary

Use above equations, substitute in known values.

Conical Pendulum:

Page 101 of 117

pendulum.

Vertically: T cosθ – mg = 0 Radially: 2

•

2 T sinθ = mv = mr ω r

Derive results

2

2

Tension = 4π mn l

4u Maths Summary

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h =

4u Maths Summary

g 2

ω

Page 103 of 117

2

•

Discuss the behaviour of the pendulum as its features vary. • Apply derived formulae to the solution of simple problems. Banked Circular Track • Use Newton’s laws to analyse the forces acting on a body, represented by a particle, moving at constant speed around a banked circular track. •

tan θ = v rg • The vertical depth of the bob below the pivot point is independent of the length of the string and the mass of the bob. • As the speed of the particle increases, it rises upwards. Use above formulas and substitute in known values.

Vertically: N cosθ – F sinθ – mg = 0 Radially: 2

N sinθ + F sinθ = mv r

Derive results

4u Maths Summary

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2

h = vd Rg 2

•

tan θ = v Rg

Calculate the optimum speed around a banked track given the construction specifications.

v =

4u Maths Summary

Rg tan θ

Page 105 of 117

•

Calculate the forces acting on a body, travelling around a banked track, at a speed other than the optimum speed.

2

F = mv cosθ – mg sinθ r

4u Maths Summary

Page 106 of 117

Topic 7 Polynomials Integer roots of polynomials with integer coefficients. • Prove that, if a polynomial has integer coefficients and if a is an integer root, then a is a divisor of the constant term. • Test a given polynomial with integer coefficients for possible integer roots.

The general form of a polynomial is: n

n– 1

a nx + a n – 1x

n –2

+ a n – 2x

+ ..... + a1x + a0

a 0 is the constant term

If a is an integer root, then a is a factor of a0 . This is known as the remainder theorem. If P(a) = 0 then: • (x – a) is a factor of P(x) • a is a root of P(x) Example: 2

P(x) = x – 6x + 5 = 0 Testing x = 5 2

– 6×5+5 = 0

P(5) = 5

Since P(5) = 0 , (x – 5) is a factor of P(x)

•

All possible integer roots of polynomials lie among the positive and negative integer divisors of its constant term.

However, not all polynomials contain integer coefficients. If P  b  = 0 then: • •

a 

(ax – b) is a factor of P(x) b is a factor of P(x) a

Where b is a factor of the constant term and a is a factor of the leading term. Example: P(z) = 2z

3

– 3z

2

+ 2z – 3 = 0

z could potentially be any of the following: ±3 , ±1 , ± 32 , ±12 In this case, P 3  = 0.  2

∴ (2z – 3) is a factor of P(z)

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Multiple Roots • Define a multiple root

A polynomial of degree n has n roots, but they are not necessarily all different. We say that ‘a’ is a root of multiplicity ‘r’ when the factor (z – a) occurs ‘r’ times. Therefore: r

P(x) = (z – a) .Q(x)

•

•

Write down the order (multiplicity) of a root. Prove that if r

P(x) = (x – a) .S(x)

, where r > 1 and S(a) ≠ 0 , then P'(x) has a root a of multiplicity (r – 1).

The order of a root is the number of times it appears as a factor.

r

where r > 0 and S(a) ≠ 0

P(x) = (x – a) .S(x) r

u = (x – a)

du = r (x – a)r – 1 dx v = S(x) dv = S'(x) dx r

r– 1

P'(x) = (x – a) .S'(x) + r (x – a) r– 1

= (x – a)

r– 1

•

•

•

= (x – a)

Solve simple problems involving multiple roots of a polynomial. State the fundamental theorem of algebra. Deduce that a polynomial of degree n > 0, with real or complex coefficients, has exactly n complex roots, allowing for multiplicities.

4u Maths Summary

.S(x)

[(x – a)S'(x) + r S(x)] Q(x)

These can include examples whereby the derivative of a function must first be established.

The Fundamental Theorem of Algebra asserts that every polynomial P(x) of degree ‘n’ over the complex plane has at least one complex root. n

n –1

Let P(z) = a nz + a n – 1z

+ .... + a 1z + a 0 = 0

Let z1 be a root. So P(z1) = 0 ∴ P(z) = (z – z1)Qn – 1(z) where Qn – 1(z) is a polynomial of degree (n – 1) By continuing this pattern we get: P(z) = a n(z – z1)(z – z2)....(z – zn) where an ≠ 0

Using this result, the factor theorem should now be used to prove (by induction on the degree) that a polynomial of degree n > 0 with real or complex coefficients has exactly n complex roots. Page 108 of 117

•

•

Factoring Polynomials Recognise that a real polynomial of degree n can be written as a product of real linear and real quadratic factors. Factor a real polynomial into a product of real linear and real quadratic factors.

A real polynomial can be written as a product of real linear and quadratic factors. Example: 3

2

P(z) = 2z – 3z + 2z – 3 Using the remainder theorem: P  3  = 0 2 

∴ 2z – 3 is a factor of P(z) 2

z + 0z + 1 3

2

3

2

2z – 3 |2z – 3z + 2z – 3 2z – 3z

2

0z + 2z 2

0z + 0z 2z – 3 2z – 3 0 2

P(z) = (2z – 3)(z + 1)

•

•

Recognise that a complex polynomial of degree n can be written as a product of n complex linear factors. Factor a polynomial into a product of complex linear factors.

A complex polynomial of degree n can be written as a product of n complex linear factors. Using the example from above: 2 P(z) = (2z – 3)(z + 1) 2

2

= (2z – 3)(z – i ) = (2z – 3)(z – i)(z + i)

•

As you can see above, the complex roots of real polynomials occur in conjugate pairs.

That means that if (z – i) is a factor of P(x), (z + i) is also a factor. •

Students should be able to factor cubic and quartic polynomials over both the real and complex planes.

There are instances where a polynomial cannot be factorised using the remainder theorem. There are a number of alternatives that need to be familiarised: • Difference of cubes: 3

z –1 =0 2

(z – 1)(z + z + 1) = 0

4u Maths Summary

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•

Difference of two squares: 4

z =1 4

z –1=0 2

2

(z – 1)(z + 1) = 0 z = ±1, ±i Completing the square:

•

4

z = -1 4

z +1=0 4

2

2

z + 2z + 1 – 2z = 0 2

2

2

(z + 1) – 2z = 0 2

2

2

2

(z + 1 – 2 z)(z + 1 + 2 z) = 0

•

•

(z – 2 z + 1)(z + 2 z + 1) = 0 • Any combination of two or more of the above methods. If given the roots, say a, b & c.

Write down a polynomial given P(x) = (x – a)(x – b)(x – c) a set of properties sufficient to define it. Solve Solve as shown above. polynomial There is, however, another method that can be employed. equations over the real and complex planes. De Moivre’s Theorem: If z = r(cosθ + isinθ) then z z

n

= r (cosθ + isinθ)

n

n

n

= r (cosnθ + isinnθ)

n

n

n

n

∴ r (cosθ + isinθ) = r (cosnθ + isinn θ)

Example: Use De Moivre's theorem to express cos 3θ in terms of cosθ and sin 2θ in terms of sinθ. cos 3θ + i sin 3θ 3

= (cosθ + i sinθ) 3

2

2

3

= cos θ + 3 i cos θsinθ – 3cosθsin θ – i sin θ Equating Real & Imaginary parts: cos 3θ

3

2

= cos θ – 3cosθsin θ 3

2

= cos θ – 3cosθ(1 – cos θ) 3

= 4cos θ – 3cosθ

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2

sin 3θ

3

= 3cos θsinθ – sin θ 2

3

= 3(1 – sin θ)sinθ – sin θ 3

= 3sinθ – 4sin θ

Roots & Coefficients of Polynomials • Write down the relationships between the roots and coefficients of polynomial equations of degrees 2, 3 & 4.

Quadratic Equations: 2

If α & β are roots of ax + bx + c = 0 then: α +β = –b a αβ = c a

Cubic Equations: 3

2

If α , β & γ are roots of ax + bx + cx + d = 0 then: α +β+γ = –b a αβ + βγ + αγ = c a αβγ = – d a

Quartic Equations: 4

3

2

If α , β , γ & δ are roots of ax + bx + cx + dx + e = 0 then: α +β+γ+δ = –b a αβ + αγ + αδ + βγ + βδ + γδ = c a αβγ + αβδ + βγδ + αγδ = – d a αβγδ = e a

•

Use these relationships to form a polynomial equation given its roots.

Quadratic: 2

P(x) = x – (sum of roots )x + (Product of roots )

Cubic: 3

2

P(x) = x – (Sum of roots )x + (Sum of roots 2 at a time )x – (Product of roots )

Quartic: 4

3

2

P(x) = x – (Sum of roots )x + (Sum of roots 2 at a time )x – (Sum of roots 3 at a time )x + (Product of roots )

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•

Form an equation, whose roots are a multiple of the roots of a given equation.

Roots are: mα, mβ & mδ m(α + β + δ ) = – mb a 2

2 m ( αβ + αδ + βδ ) = m c a 3

3 m αβδ = – m d a 2

3

3 2 ∴ x + mb x + m c x + m d = 0 a a a 3

•

Form an equation, whose roots are the reciprocals of the roots of a given equation.

2

2

3

ax + mbx + m cx + m d = 0 Roots are: 1 , 1 & 1 α β γ 1 +1+1 α β γ = βγ + αγ + αβ αβγ =c× –a a d ∴ 1 +1+1 = – c α β γ d 1 + 1 + 1 αβ αγ βγ

α +β+γ αβγ

=

= –b× –a a d ∴ 1 + 1 + 1 = b αβ αγ βγ d 1 = –a αβγ d 3 2 ∴x + cx + bx + a = 0 d d d 3

2

dx + cx + bx + a = 0

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•

Form an equation, whose roots differ by a constant from the roots of a given equation.

Roots are: α + k , β + k & γ + k α + β + γ + 3k = – b + 3k = – b + 3ak a a (α + k )( β + k ) + ( α + k )( γ + k ) + ( β + k )( γ + k ) 2

2

= αβ + αk + βk + k + αγ + αk + γk + k + βγ + βk + γk + k = αβ + αγ + βγ + 2k ( α + β + γ) + 3k

2

2

2 = c + 2k  – b  + 3k a  a 2 = c – 2bk + 3k = c – 2bk + 3ak a a a

2

∴ Sum of Roots Two at a time = c – 2bk + 3ak a (α + k )( β + k )( γ + k ) 2

2

2

2

= αβγ + αβk + αγk + αk + βγk + βk + γk + k 2

= αβγ + k ( αβ + αγ + βγ ) + k ( α + β + γ) + k

3 3

2   3 = – d + kc +  – k b  + k a a  a  2

= – d + kc – k b + ak a

3

  = –  d – kc + k b – ak  a   2

3

2

2

3 2 ∴ x + b – 3ak x + c – 2bk + 3ak x + d – kc + k b – ak a a a 3

2

2

2

3

= 0 3

ax + ( b – 3ak)x + ( c – 2bk + 3ak )x + (d – kc + k b – ak ) = 0

•

Form an equation, whose roots are the squares of the roots of a given equation.

2

2

Roots are: α & β 2

α +β

2 2

= (α + β ) – 2 αβ 2 =  – b  – 2c a  a 2

=

b

– 2c a a 2 2

=

b – 2ac 2

a

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2 2

α β

2 =  c  a  2

=

c a

2 2

2

2

2

2 b c ∴ x + 2ac – x + 2 = 0 2 a a 2 2

a x + (2ac – b )x + c = 0

Partial Fractions • Write a fraction in terms of quotient and remainder. 2

f(x) = 2x + 3x + 8 x+ 2 2x – 1 2

x + 2 2x + 3x + 8 2

2x + 4x –x+8 –x–2 10 R(x) f(x) = Q(x) + B(x) f(x) = (2x – 1) +

•

10 x+ 2

Write a fraction in terms of distinct linear factors.

f(x) =

x+ 3 2

x – 6x + 5 x+3 ≡ a + b 2 x– 5 x– 1 x – 6x + 5 x + 3 = a(x – 1) + b(x – 5) Let x = 1 4 = -4b b = -1 Let x = 5 8 = 4a a = 2

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x+3

∴

=

2

x – 6x + 5

•

Write a fraction in terms of distinct linear factors and a simple quadratic factor.

2 – 1 x– 5 x– 1

2

6x – 53x + 42 2

≡

(2x – 3)(2x – 5x – 3) 2

bx + c a + 2 2x – 3 2x – 5x – 3

2

6x – 53x + 42 = a(2x – 5x – 3) + (bx + c)(2x – 3) Let x = 1.5 -24 = -6a a = 4 Let x = 0 42 = -12 – 3c c = -18 Let x = 1 -5 = (-6 × 4) + (b – 18)(-1) -19 = b – 18 b = -1 x + 18 4 f(x) = – 2 2x – 3 2x – 5x – 3

•

Write a fraction in terms of the product of two different simple quadratic factors.

3

2

f(x) = 2x 2+ 7x –24x + 3 (x – 1)(x + 1) 3

2

3

2

2x + 7x – 4x + 3 ≡ ax + b + cx + d 2 2 2 2 (x – 1)(x + 1) x –1 x +1 2

2

2x + 7x – 4x + 3 = (ax + b)(x + 1) + (cx + d)(x – 1) Let x = 1 8 = 2(a + b) a +b = 4 (1) Let x = 0 3 = b–d (2)

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Let x = 2 39 = 5(2a + b) + 3(2c + d) Let x = -1 12 = 2(b – a) b–a = 6 (4) (1) into (4) 4 –a –a =6 -2a = 2 a = -1 b = 5 d = 2 Sub a, b & d into (3) 39 = 15 + 6c + 6 6c = 18 c = 3 3x + 2 x–5 f(x) = 2 – 2 x +1 x –1

•

(3)

Apply these Refer to integration summaries. partial fraction decompositions to the integration of corresponding functions.

4u Maths Summary

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•

•

•

•

•

Topic 8 Harder 3 Unit Circle Geometry Solve more difficult problems in geometry. Induction Carry out proofs by mathematical induction in which S(1), S(2)…S(k) are assumed to be true in order to prove S(k+1) is true. Use mathematical induction to prove results in topics which include geometry, inequalities, sequences and series, calculus and algebra. Inequalities Prove simple inequalities by use of the definition of a > b for real a and b. Prove further results involving inequalities by logical use of previously obtained inequalities.

4u Maths Summary

See attached sheet.

Mathematical induction occurs in four steps. STEP 1) Prove the statement is true for the lowest possible integer value, usually n = 1. STEP 2) Assume the result is true for n = k. STEP 3) Use algebraic manipulation to prove the result is true for n = k + 1. STEP 4) Have a concluding statement. Since the formulas was proven true for n=1, it was assumed true for n=k. It was then proven true for n=k+1, meaning it is true for all n = 1.

If a > b then: 2

(a – b) > 0 2

a – 2ab + b 2

a +b

2

2

> 0

> 2ab

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