c1c2 revision notes

AS Mathematics

Revision Notes

y = sin θ

-1

-0.75

-0.5

-0.25

0

0.25

0.5

0.75

1

0 30 60 90 120 150 180 210 240 270 300 330 360

θ

yRotating vector

-1

-0.5

0

0.5

1

-1 -0.5 0 0.5 1x

y

Bob Francis 2004

2 Revision Notes Core Mathematics 1

Basic Algebra Topic Examples References Manipulating Algebraic Expressions You should be familiar with: • Collecting like terms • Removing (expanding) brackets • Factorising with one bracket • Multiplication • Algebraic Fractions

9x – 5y – 2x + 4y + x = 8x – y 3x(4 – 5y) = 12x – 15xy 35ab + 14a – 28b = 7(ab + 2a – 4b) (3xy)2 × 5x = 9x2y2 × 5x = 45x3y2

3 4 45 9 15x x

y y× = ; 2 3 8 9

3 4 12x y x y−

− =

MEI Core 1

Pages 2 to 5

Linear Equations and Rearranging Formulae You should be able to: • Solve linear equations using any of the

above techniques • Rearrange a formula (change the subject

of an equation) using algebraic techniques

3 2 15 8 20 5 84 5x x x x x+ = ⇒ + = ⇒ = −

1.6x⇒ = −

(9 9 532 32 325 5 9

f c c f c f= + ⇒ = − ⇒ = − )

MEI Core 1

Pages 7 to 9

Pages 11 to 12

Quadratic Equations You should be able to: • Factorise three different types of

quadratic expression

• Solve quadratic equations using three different methods:

1) Factorising

2) Completing the square

3) Quadratic formula:

ax2 + bx + c = 0 ⇒ 2 4

2b b acx

a− ± −

=

• Recognise the type of solution to a

quadratic equation, dependent on the sign of the discriminant b2 – 4ac

b2 – 4ac > 0 2 distinct roots ⇒b2 – 4ac = 0 2 equal roots ⇒b2 – 4ac < 0 0 real roots ⇒

2x2 – 10x = 2x(x – 5) 3x2 – 48 = 3(x2 – 16) = 3(x – 4)(x + 4) 2x2 – x – 15 = (2x + 5)(x – 3) 1) x2 – x – 12 = 0 ⇒ (x + 3)(x – 4) = 0 x = –3 or x = 4 ⇒

2) x2 – 2x – 10 = 0 ⇒ (x – 1)2 – 1 – 10 = 0 (x – 1)⇒ 2 = 11 x = 1 ⇒ 11±

3) 2x2 – x – 9 = 0⇒21 ( 1) 4 2 ( 9)2 2

x± − − × × −

=×

⇒1 73

4x ±

= ⇒ x = –1.89 or 2.39

For the equation x2 + 6x + c = 0:

Discriminant = 62 – 4×1×c = 36 – 4c

36 – 4c > 0 c < 9 2 distinct roots ⇒ ⇒

36 – 4c = 0 c = 9 2 equal roots ⇒ ⇒

36 – 4c < 0 c > 9 0 real roots ⇒ ⇒

MEI Core 1

Pages 13 to 24

Interactive Spreadsheet

Parabolas

Simultaneous Equations You should be able to: • Solve a pair of linear simultaneous

equations by elimination or substitution • Solve a pair of (one linear and one non-

linear) simultaneous equations by substitution

Solve 6x + y = 40 (1) and 4x + 3y = 36 (2): Eliminate y by multiplying equation (1) by 3

then subtract equation (2), etc. Make y the subject of (1) and substitute in

equation (2), solve for x, etc.

Solve 6x + y = 40 (1) and x2 + y2 = 64 (2): Make y the subject of (1), subst. in equation

(2), solve a quadratic equation. in x etc.

MEI Core 1

Pages 28 to 31

Pages 55 to 56

Pages 68 to 72

Core Mathematics 1 Revision Notes 3

Co-ordinate Geometry Topic Examples References Gradients of Straight Lines For a line AB, where A(x1, y1), B(x2, y2)

Gradient of AB = m is

2 1

2 1

y yx x

−−

Parallel lines have equal gradients. If a line AB has gradient m1, then a perpendicular line PQ has gradient m2, s.t.

21

1mm−

= or m1 m2 = −1

Gradient of line from A (-2, 3) to B (4, -1) is

m = ( 1) 34 ( 2)− −− −

= 46

− = − 23

Any line with gradient which simplifies to − 2

3 is parallel to AB.

If line AB has gradient m1 = − 23 , then

perpendicular line PQ has gradient m2 s.t.

21

1mm−

= = 32

MEI Core 1 Pages 34 to 37

Interactive Spreadsheet Co-ordinate Geometry

Distances and Mid-points For a line AB, where A (x1, y1), B (x2, y2) Distance from A to B is

2 22 1 2 1( ) ( )x x y y− + −

Co-ordinates of mid-point M are

1 2 1 2,2 2

x x y y+ +⎛ ⎞⎜ ⎟⎝ ⎠

Find the distance from A (-2, 3) to B (4, -1) and the co-ordinates of mid-point M

Length of AB = 2( 1 3) (4 ( 2))− − + − − 2

= 16 36+ = 52 = 7.21 (3 sf)

Mid-point M = ( 2) 4 3 ( 1),

2 2− + + −⎛ ⎞

⎜ ⎟⎝ ⎠

= (1, 1)

MEI Core 1

Pages 37 to 39

Equations of Straight Lines Equation of a straight line:

• passing through (0, c) with gradient m is y = mx + c

• passing through (x1, y1) with gradient m is y – y1 = m(x – x1)

or y = mx + c where c = y1 − mx1

A straight line with equation ax + by = c passes through (c/a, 0) and (0, c/b)

A straight line passing through (0, 7) with gradient –2 has equation y = 7 – 2x

A straight line passing through (5, −8) with gradient 3 has equation

y + 8 = 3(x – 5) ⇒ y = 3x – 23 or y = 3x + c where c = −8 − 3×5

⇒ y = 3x – 23

A straight line with equation 3x + 4y = 30 passes through (10, 0) and (0, 7.5)

MEI Core 1

Pages 42 to 51

Circles and Other Curves A circle with centre (0, 0) and radius r has equation

x2 + y2 = r2

A circle with centre (h, k) and radius r has equation

(x – h)2 + (y – k)2 = r2

Points of intersection with circles and other lines or curves may be found by solving a pair of simultaneous equations. Other curves of note include

y = x−n, for n = 2, 3, 4, −1, -2, −3.

A circle with centre (0, 0) and radius 5 has equation

x2 + y2 = 25 A circle with centre (−3, 2) and radius 8 has equation (x + 3)2 + (y – 2)2 = 49

x⇒ 2 + 6x + y2 – 4y – 36 = 0 To find points of intersection of

x2 + 6x + y2 – 4y – 36 = 0 and y = 4 – x • substitute y = 4 – x in circle equation, • solve resulting quadratic in x, • find corresponding y values

MEI Core 1

Pages 60 to 66


-10

-5

0

5

10

-4 -3 -2 -1 0 1 2 3x

y

Polynomials Topic Examples References

Operations with Polynomials Polynomial functions may be added, subtracted and multiplied. A polynomial function may be divided by a linear function to give a quotient with or without a remainder.

(5x3 – 2x + 4) + (2x2 + 7x – 5) = 5x3 + 2x2 + 5x – 1

(5x3 – 2x + 4) – (2x2 + 7x – 5) = 5x3 – 2x2 – 9x + 9

(5x3 – 2x + 4)(2x2 + 7x – 5) = 5x3(2x2 + 7x – 5) – 2x(2x2 + 7x – 5) + 4(2x2 + 7x –

5) = 10x5 + 35x4 – 29x3 – 6x2 + 38x – 20

Given that (x – 1) is a factor of (5x3 – 2x – 3) find (5x3 – 2x – 3) ÷ (x – 1): Let 5x3 – 2x – 3 ≡ (ax2 + bx + c)(x – 1) Compare coefficients to show a = 5, b = 5, c = 3 or use long division


Polynomial Functions A polynomial function of x has terms in positive integer powers of x, and may have a constant term. The order of a polynomial is the highest power of x appearing in the polynomial.

Name Order Example

Quadratic 2 f(x) = 2x2 – 5x + 7 Cubic 3 f(x) = x3 + 2x2 – 5x + 4 Quartic 4 f(x) = 3x4 – 5x3 + 2x2 + x


Factor and Remainder Theorems Let f(x) be any polynomial function in x Factor Theorem

• f(a) = 0 (x – a) is a factor of f(x) ⇔

• f(a/b) = 0 (bx – a) is a factor of f(x) ⇔

Remainder Theorem • The remainder when f(x) is divided by (x – a) is f(a)

Let f(x) = x3 − 5x2 – 2x + 24 f(3) = 33 − 5×32 – 2×3 + 24 = 0

⇔ (x – 3) is a factor of f(x)

Let f(x) = 3x3 + x2 + x − 2 f( 2

3 ) = 3× ( 23 )3 + ( 2

3 )2 + 23 − 2 = 0

⇔ (3x – 2) is a factor of f(x) f(2) = 3×23 + 22 + 2 − 2 = 28 ⇒ when f(x) is divided by (x – 2) remainder is 28


Solving equations / Graph sketching Let y = f(x) = ax3 + bx2 + cx + d

• Graph of y = f(x) intersects y-axis at (0, d)

• Factorise f(x) to solve f(x) = 0

• Use roots of f(x) = 0 to find co-ordinates of points of intersection with x-axis

• Sketch graph

Sketch the graph of y = f(x) = x3 + x2 – 6x Graph passes through (0, 0) Factorising: f(x) = x(x2 + x – 6) = x(x + 3)(x – 2) hence f(x) = 0 x = 0, x = −3 or x = 2 ⇒

MEI Core 1 Pages 85 to 88 Pages 91 to 92

Binomial Expansions For any natural number n (a + b)n = an + nC1an-1b + nC2an-2b2 + … …+ nCn-2a2bn-2 + nCn-1abn-1 + bn

where nCr = !

!( )!

n

r n r−

(2 + x)4 = 1× 24 + 4× 23x + 6× 22x2 + 4× 2x3 + 1× x4 = 16 + 32x + 24x2 + 8x3 + x4

Term in x5 in expansion of (3 – x)8 = 8C5 × 33 × (−x)5 = 56× 27× (−x)5 = −1512x5

coefficient of x⇒ 5 is −1512

MEI Core 1 Pages 108 to

115


-2

0

2

4

6

8

10

-5 -4 -3 -2 -1 0 1x

y

-40

-30-20

-100

1020

30

-5 -4 -3 -2 -1 0 1 2 3x

y

-60-50-40-30-20-10

0102030

-4 -3 -2 -1 0 1 2 3 4 5x

y

Curves and Translations Topic Examples References

Quadratic Functions Any quadratic function may be given by f(x) = ax2 + bx + c

a(x + p)≡ 2 + q [completing the square]

where p = 2ba

and q = c – ap2

has vertex (–p, q) and line of symmetry x = –p and y-intercept is at (0, c)

y = 2x2 + 8x + 7 ≡ 2(x + 2)2 – 1

p = 2ba

= 82 2×

= 2

q = 7 – 2× 22 = –1

⇒ vertex is (−2, −1); ⇒ line of symmetry is x = −2

and y-intercept is at (0, 7)


Cubic Functions Any cubic function may be given by f(x) = ax3 + bx2 + cx + d

sign of a [coefficient of highest power of x] determines behaviour for large x

May have 0 or 2 turning points May have 1 stationary point of inflection

y = 8 + 14x – 5x2 – 2x3 ≡ (2 – x)(1 + 2x)(4 + x)


Quartic Functions Any quartic function may be given by f(x) = ax3 + bx3 + cx2 + dx + e

sign of a [coefficient of highest power of x] determines behaviour for large x

May have 1 or 3 turning points May have 1 stationary point of inflection

y = (x + 3)(x – 1)2(x – 4) MEI Core 1 Pages 82 to 91

Translations A translation of a function may be given by

f(x – t) + s

This is given by the vector which

represents a two-way shift:

ts

⎛ ⎞⎜ ⎟⎝ ⎠

t units horizontally, s units vertically

y = f(x) = x2 – 5 ⇒ vertex is at (0, −5);

Vector gives new vertex (4, 2) and 47

⎛ ⎞⎜ ⎟⎝ ⎠

f(x – 4) + 3 = [(x – 4)2 – 5] + 7 = x2 – 8x + 18


105

-10

-5

0

5

10

15

20

-6 -4 -2 0 2 4 6 8 10x

y


-30

-20

-10

0

10

20

-8 -6 -4 -2 0 2 4 6 8 10 12x

y

Uncertainty Topic Examples References

Errors Absolute error

Difference between true and measured values

Relative error = absolute error / true value Percentage error = relative error × 100

The radius of a circle is measured as 5 cm, but its true length = 5.2 cm. Consider its area: • Absolute error in area = π5.22 – π52 = 6.4088 • Relative error in area = 6.4088 / 84.9487

= 0.075 (2 s.f.) • Percentage error in volume = 7.5% (2 s.f.)

MEI Pure 1 Pages 119 to

122

Inequalities Linear inequalities may be handled just like linear equations, except that when both sides are multiplied or divided by a negative number, the inequality must be reversed.

Quadratic inequalities may be handled by solving the corresponding quadratic equation, then use a sketch graph or table to identify the solution set(s).

Set f(x) = (x – α)(x – β) = 0

• 4x + 5 ≤ 23 4x ⇒ ≤ 18 x ≤ 4.5 ⇒

• 15 – 2x > 43 −2x > 28 ⇒ x < 14 ⇒

• Solve x2 – 4x < 21 ⇒ x2 – 4x – 21 = 0 x2 – 4x – 21 = 0 ⇒ (x + 3)(x – 7) = 0

x = −3 or x = 7 ⇒

hence x2 – 4x < 21 ⇒ −3 < x < 7


125

e

Ei

TopiWora1 =(ab)n

nab

⎛ ⎞⎜ ⎟⎝ ⎠

You multi Ratio• If

m• If

m

Law

a-m ≡

am ×

amn ≡

x = α or x = β [ α < β] ⇒

x < α a < x < β x > β ther f(x) < 0 f(x) > 0 f(x) < 0 or f(x) > 0 f(x) < 0 f(x) > 0

Indices c Examples References king with powers and roots a a0 = 1 a≡ nbn n ab ≡ n a n b

n

n

ab

≡ n

nn

a ab b

≡

should be able to add, subtract and ply expressions of form a b± c

nalising the denominator: the denominator of a fraction is b , ultiply top and bottom by b

the denominator of a fraction is a+ b , ultiply top and bottom by a− b

51 = 5 70 = 1 (5×4)2 ≡ 52×42 7x ≡ 7 x

3 3

5 125x x⎛ ⎞ ≡⎜ ⎟

⎝ ⎠ 3

6 2

64 4x x

≡

(3 2 5) (2 7 5) 5 5 5 5(1 5)+ + − = − = − (3 2 5) (2 7 5) 1 9 5+ − − = + (3 5)(2 5) 6 3 5 2 5 5 1 5+ − = − + − = −

1 5 2 (1 5 2) 3 3 5 6155 3 5 3 3

− − × −≡ ≡

×

2

1 5 2 (1 5 2)(3 2)3 2 (3 2)(3 2)

3 15 2 2 5 2 13 16 23 2 7

− − −≡

+ + −

− − + × −≡ ≡

−


129

s of indices

1ma

1na ≡ n a

an a≡ m+n am a÷ n a≡ m−n

( )nma or ( ) mna

mna ≡ n ma or ( )m

n a

24− = 214

= 116

13125 = 3 125 = 5

2x2 ×5x3 ≡ 10x5 2

5

96

xx

≡ 332

x− ≡ 3

1.5x

x6 ≡ ( )32x or ( )23x 23x ≡ 3 2x or ( )2

3 x

MEI Core 1

Pages 130 to 135


Sequences and Series Topic Examples References Sequences and series A sequence is an ordered set of numbers

u1, u2, u3, … , uk, … , un where uk is the general term. A series is the sum of the terms of a sequence:

u1 + u2 + u3 + … + un = 1

n

kk

u=

∑Different ways to define a sequence: Inductive definition: uk+1 = f(uk) with first term u1 Deductive definition: uk = f(k) for k = 1, 2, 3, …

Hybrid: uk+1 = f(uk, k) for k = 1, 2, 3, ... A periodic sequence is one such that for some fixed integer p, uk+p = uk for all k An oscillating sequence alternates either side of a ‘middle’ value.

The sequence 1, 2, 5, 14, 41, … may be defined using the inductive definition:

uk+1 = 3uk + 1 with first term u1 = 1

The sequence 1, 4, 9, 16, 25, … may be defined using the deductive definition:

uk = k2 for k = 1, 2, 3, … and the hybrid definition:

uk+1 = uk + 2k + 1 with first term u1 = 1 The sum of the first 10 terms is given by:

2

1

n

kk

=∑ = 1 + 4 + 9 + … + 100

The sequence defined by the formula

uk = 15 ( 1)k +× − for k = 1, 2, 3, …

gives terms 5, −5, 5, −5, 5, … and is periodic with p = 2, i.e. uk+2 = uk for all k and oscillating


166

Arithmetic Progressions A sequence in which there is a constant difference (d) between successive terms. Inductive definition: uk+1 = uk + d with first term a Deductive definition: uk = a + (k – 1)d The last term, un = l is given by l = a + (n – 1)d The sum of the n terms, Sn is given by Sn = 1

2 n(a + l)

= 12 n[2a + (n – 1)d]

An arithmetic progression has terms

3, 7, 11, … and Sn = 210 Find the number of terms and the last term. First term a = 3, constant difference d = 4, hence Sn = 1

2 n[2a + (n – 1)d] = 210

⇒ 12 n[6 + 4(n – 1)] = 210

2n⇒ 2 + n − 210 = 0 (n – 10)(2n + 21) = 0 ⇒ n = 10 [ignore n = −10.5] ⇒Hence last term l = 3 + 9×4 = 39


173

Interactive Spreadsheet Sequences and

Series

Geometric Progressions A sequence in which there is a constant ratio (r) between successive terms. Inductive definition: uk+1 = r uk with first term a Deductive definition: uk = ark-1

The last term, un = l is given by l = arn-1

The sum of the n terms, Sn is given by

Sn = (1 )1

na rr

−−

or ( 1

1

na rr

−−

)

Provided –1 < r < 1, Sn converges to a limit known as the sum to infinity:

= S∞ 1a

r−

A geometric progression has 2nd term 60 and 5th term 12.96. Find the sum of the first 10 terms to and the sum to infinity. 2nd term = 60 ar = 60 ⇒ 5th term = 12.96 ar⇒ 4 = 12.96 Solving simultaneously:

4312.96 0.216 0.6

60ar r rar

= ⇒ = ⇒ =

− substituting for r in ar = 60 gives a = 100

− hence = 10S10100(1 0.6 )

1 0.6−

− = 248.488…

S∞ = 1001 0.6−

= 250


183

Interactive Spreadsheet Sequences and

Series


-30

-20

-10

0

10

20

30

-2 -1 0 1 2 3 4 5x

y

Differentiation Topic Examples References Gradient functions

y = f(x) = k xn ⇒ddyx

= f ’(x) = k n xn-1

for all rational numbers n

y = x3 − 5x2 + 7x − 8 ⇒ddyx

= 3x2 – 10x + 7

f(x) = 4x(5 – x) = 20x – 4x2 ⇒ f’(x) = 20 – 8x

MEI Core 2 Pages 191 to 202 Pages 339 to 340

Tangents and Normals Tangent and normal to y = f(x) at (x1, y1)

Let m1 = f ’(x1) = ddyx

at x = x1 and m2 = −1

1m

Equation of tangent is y – y1 = m1(x – x1)

Equation of normal is y – y1 = m2(x – x1)

Find equation of tangent and normal to the curve

y = f(x) = x3 − 5x2 + 7x − 8 at (3, −5):

f ’(x) = 3x2 – 10x + 7 ⇒ m1 = f ’(3) = 4 and m2 = 1

4− = −0.25 Equation of tangent is

y + 5 = 4(x – 3) y = 4x − 19 ⇒Equation of normal is

y + 5 = −0.25 (x – 3) ⇒ y = −0.25x − 4.25 or x + 4y + 19 = 0


207

Stationary Points

y = f(x) ⇒ ddyx

= f ’(x) = 0 for stationary

Nature of stationary points:

• maximum point

• minimum point

• point of inflection

Test gradient either side to determine the nature of the stationary point.

Find the coordinates of the stationary points (determining their nature) for the curve

y = f(x) = x4 – 4x3 ⇒ f ‘(x) = 4x3 – 12x2

Stationary points occur where:

f ‘(x) = 0 ⇒ 3 2 24 12 4 ( 3)x x x x 0− = − = x = 0 or x = 3 ⇒Testing gradients either side of stationary points:

• Point of inflection at (0, 0) • Minimum point at (3, −27)


220

Second derivatives

Second derivative of y = f(x) is 2

2

dd

yx

= f”(x)

To find the nature of stationary points: 2

2

dd

yx

= f”(x) > 0 Minimum point ⇒

2

2

dd

yx

= f”(x) < 0 Maximum point ⇒

but if 2

2

dd

yx

= f”(x) = 0, check gradients

either side for maximum point, minimum point or point of inflection

The volume, V m3, of a box is given by 9x2 − 2x3

For turning points: ddVx

= 18x – 6x2 = 0

6x(3 – x) = 0 ⇒ x = 0 or x = 3 ⇒

The second derivative 2

2

dd

Vx

= 18 – 12x

When x = 0, 2

2

dd

Vx

= 18 > 0 Minimum point ⇒

When x = 3, 2

2

dd

Vx

= –18 < 0 ⇒ Maximum point

Hence box has maximum volume when x = 3

MEI Pure 1 Pages 221 to

226


Integration Topic Examples References

Reversing Differentiation ddyx

= f’(x) = k xn y = ⇒ 1k

n+ xn+1 + c

for all rational numbers n (n –1) ≠

Find equation of curve with gradient given by 10 + 4x – x2, which passes through (0, 7): ddyx

= 10 + 4x – x2 ⇒ y = 2(10 4 ) dx x x+ −∫

y = 10x + 2x⇒ 2 – 13 x2 + c

y = 10x + 2x⇒ 2 – 13 x2 + 7 [x = 0 ⇒ y = 7]


236

Indefinite integrals

y = f(x) = k xn ⇒ f ( )dx x∫ = 1k

n+ xn+1 + c

for all rational numbers n (n –1) ≠

f(x) = x3 + 3x2 – 5x + 4 ⇒ f ( )dx x∫ = 0.25x4 + x3 – 2.5x2 + 4x + c

MEI Core 2 Page 247 Page 347

Definite integrals

y = f(x) = k xn ⇒ f ( )db

ax x∫ = 1

1bn

ak

n x ++⎡⎣ ⎤⎦

= ( ) ( )1 11 1

n nk kn nb a+ +

+ +−

4

1( 3)(2 1)dx x x− +∫ =

4 2

1(2 5 3)dx x x− −∫

= 43 2

152

3 2 3x x x⎡ ⎤− −⎣ ⎦

= 1283( 40 12)− − − 52

3 2( 3)− − = −4.5

MEI Core 2 Page 246 Page 348

Areas and graphs Area enclosed by a curve and the x-axis

The integral f ( )db

ax x∫ represents the area

enclosed by the graph of y = f(x), the lines x = a, x = b, and the x-axis.

Area enclosed between two curves

The integral [f ( ) g( )]db

ax x x−∫ represents

the area enclosed by the graph of y = f(x), the lines x = a, x = b, and the graph of y = g(x).

Area enclosed by a curve and the y-axis

The integral represents the area

enclosed by the graph of x = g(y), the lines y = p, y = q, and the y-axis.

g( )dq

py y∫

Area enclosed by y = 9 – x2 and the x-axis: 3 2

3(9 ) dx x

−−∫ =

33

3139x x

−−⎡ ⎤⎣ ⎦

= (27 – 9) – (−27 + 9) = 36

Area enclosed by y = 9 – x2 and y = x + 3:

( ){ }2 2

3(9 ) 3 dx x x

−− − +∫ =

2 2

3(6 ) dx x x

−− −∫

= 22 3

31 12 36x x x

−− −⎡ ⎤⎣ ⎦ = 22

3 − 272(− ) = 20 5

6

Area enclosed by y = 5 x , the y-axis & y = 10:

x = 2

25y ⇒ Area =

210

0d

25y

y∫ = 3 10

0175 y⎡ ⎤⎣ ⎦

= 100075 – 0 = 13 1

3


259 Pages 348 to

349

Numerical Integration The Trapezium Rule estimates the value of a definite integral (area under a graph), using n trapezoidal strips, each of width h:

[ ]0 1 2 112 2( ... )n nA h y y y y y−= + + + +

Area between graph of y = 2x, the y-axis, the x-axis and x = 3, using 3 strips, is approximately

0 1 212 1 2 2(2 2 ) 2× × + + + 3⎡ ⎤⎣ ⎦

= [ ]12 1 1 2(2 4) 8× × + + + = 10.5 units2


264

e

e


-1

-0.5

00 30 60 90 120 150 180 210 240 270 300 330 360

θ

0.5

1

-1

-0.5

0

5

1

0 30 60 90 120 150 180 210 240 270 300 330 360θ

0.

-4

-2

00 30 60 90 120 150 180 210 240 270 300 330 360

θ

2

4

TrigonometryTopic Examples References Trigonometrical functions of special angles Angle 0° 30° 45° 60° 90°

Cosine 1 √3/2 1/√2 1/2 0 Sine 0 1/2 1/√2 √3/2 1 Tangent 0 1/√3 1 √3 −


274

Trigonometrical functions of any angles (Circular Functions) x = cosθ y = sinθ z = tanθ

Solve the equations (0o ≤ θ ≤ 360o): 3 cosθ = 2 cosθ = ⇒ 2

3 θ = 48.2⇒ o (from calculator) or θ = 360o – 48.2o = 311.8

5 sinθ + 2 = 0 ⇒ sinθ = − 2

5 = −0.4 [θ = −23.6o (from calculator)] θ = 180⇒ o + 23.6o = 203.6o

or θ = 360o – 23.6o = 311.8o

2 tanθ + 5 = 0 ⇒ sinθ = − 5

2 = −2.5 [θ = −68.2o (from calculator)] θ = −68.2⇒ o + 180o = 111.8o

or θ = −68.2o + 360o = 291.8o


283

Interactive Spreadsheet

Circular Functions

Sine Rule and Cosine Rule If ABC is any triangle with sides a, b, c, then

sin sin sina b c

A B= =

C

= + −

2 2 2 2 cosa b c bc A ;2 2 2

cos2

b c aAbc

+ −=

Find a where c = 10, A = 50°, C = 75° [7.93]

Find ∠A where a = 7, b = 10, B = 35°[39.4°] Find a where b = 8, c = 11, A = 50° [8.48]

Find ∠A where a = 7, b = 8, c = 10 [44.0°]


293

Trigonometrical Identities sintancos

θθθ

= sin2θ + cos2θ = 1

Solve: 4 sin θ = 5 cos θ (0o ≤ θ 360≤ o)

Solve: 2 sin2θ = 3 cos θ (0o ≤ θ 360≤ o) Use sin2θ = 1− cos2θ, solve quadratic in θ

MEI Core 2 Page 276

Area of any triangle For triangle ABC, area is given by:

Area = 12 sinbc A = 1

2 sinac B = 12 sinab C

For ∆ ABC: A = 58, b = 10 cm, c = 7 cm: Area = 1

2 sinbc A = 12 ×70×sin 58o

= 29.7 cm2 (3 s.f.)


297

Circular measure Angle measurement: π radians = 180o

For a sector of a circle with radius r and angle at centre θ radians: Arc length = rθ Area of sector = 1

2 r2θ

A sector of a circle has radius 5 cm and angle at the centre 1.5 radians.

Arc length = 5×1.5 = 7.5 cm

[Perimeter = 5 + 5 + 5×1.5 = 17.5 cm ]

Area of sector = 12 ×25×1.5 = 18.75 cm2


304

21

1 11

2 23


Logarithms and Exponentials Topic Examples References

Logarithms y = loga x x = a⇔ y a > 0, x > 0

where a is the base of the logarithm• log 1a a ≡• log 1 0a ≡• for base 10: log10 x is often written log x

4

34 log 81 81 3= ⇔ =

Solve: 2log 32 2 32xx= ⇔ = ⇒ x = 5

Solve: ⇒ a = 7 3log 343 3 343a a= ⇔ =


320

Laws of logarithms (for any base)

• log xy ≡ log x + log y

• log xy

⎛ ⎞⎜ ⎟⎝ ⎠

log x − log y ≡

• log xn n log x ≡

N.B. 1log log yy

⎛ ⎞≡ −⎜ ⎟

⎝ ⎠; 1log lnn x x

n≡

2 log a + 3 log b – 4 log c = 2 3

4log a bc

⎛ ⎞⎜ ⎟⎝ ⎠

3 2

3

2log log 3log log3

x x y zy z

⎛ ⎞≡ − −⎜ ⎟⎜ ⎟

⎝ ⎠

10 101 1log log

2a

a⎛ ⎞ ≡ −⎜ ⎟⎝ ⎠


324

Exponential equations

To solve , “take logs” of both sides: xa b=xa b= log a = x log b x = ⇒ ⇒

loglog

ab

Solve: 5x = 14 : take logs10 of both sides

5x = 14 log(5⇒ x) = log 14

x log 5 = log 14 ⇒

x = ⇒ log14log 5

= 1.64 (to 3 s.f.)

MEI Core 2 Page 321

Modelling Curves Non linear functions can be transformed to linear form by “taking logs” of both sides: (1) Power functions:

• y = kxn ⇒ log y = n log x + log k • plot log y against log x • gradient is n • intercept is log k

(2) Exponential functions:

• y = kax ⇒ log y = x log a + log k • plot log y against x • gradient is log a • intercept is log k

(1) The variables x and y are believed to be connected by the relationship y = kxn: Taking logs10: log y = n log x + log k

Plot log y against log x gives • gradient n = 1.5 • intercept log k = 0.6 k = 10⇒ 0.6 4 ≈

hence relationship is given by

y = 4x1.5 or y = 4 3x

(2) The variables x and y are believed to be connected by the relationship y = kax: Taking logs10: log y = x log a + log k

Plot log y against x gives • gradient log a = 0.3 ⇒ a = 100.3 2 ≈• intercept log k = 0.49 k = 10⇒ 0.49 3 ≈

hence relationship is given by y = 3× 2x

x 2 3 4 5 6 y 11.3 20.8 32.0 44.7 58.8 log x 0.30 0.48 0.60 0.70 0.78 log y 1.05 1.32 1.51 1.65 1.77

x 1.5 2.7 3.4 8.1 10 y 9 19 32 820 3100 log y 0.95 1.28 1.51 2.91 3.49


331


-3

-2

-1

0

1

2

3

-3 -2 -1 0 1 2 3x

y

-3

-2

-1

0

1

2

3

0 90 180 270 360x

y

-5

-4

-3

-2

-1

0

1

2

3

4

5

-5 -4 -3 -2 -1 0 1 2 3 4 5x

y

Curves, Stretches and Reflections Topic Examples References

Logarithmic and Exponential Functions A logarithmic function may be given by

f(x) = loga x for x > 0 where a is the base of the logarithm

The graph y = loga x passes through (1, 0) and has a positive gradient throughout.

An exponential function may be given by

f(x) = ax

The graph y = ax passes through (0, 1) and has a positive gradient throughout.

MEI Core 2

Pages 323 to 324

Stretches A one way stretch of the graph y = f(x) with scale factor a parallel to the y-axis is given by

y = a f(x)

A one way stretch of the graph y = f(x) with scale factor 1/a parallel to the x-axis is given by

y = f(ax)

Starting with the curve y = sin(x) Stretch factor 3 parallel to the y-axis gives y = 3 sin(x) Stretch factor 1

2 parallel to the x-axis gives y = sin(2x)


315

Reflections A reflection of the graph y = f(x) in the x-axis is given by

y = – f(x)

A reflection of the graph y = f(x) in the y-axis is given by

y = f(–x)

Starting with the curve y = (x + 2)2

Reflection in x-axis gives y = – (x + 2)2

Reflection in y-axis gives y = (–x + 2)2


315

y = ax

y = loga x

y = f(–x)

y = –f(x)

c1c2 revision notes

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