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AS Pure 1 Notes

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TABLE OF CONTENTS
2
CHAPTER 1
2
CHAPTER 2
3
CHAPTER 3
3
CHAPTER 4
4
CHAPTER 5
4
CHAPTER 6
5
CHAPTER 7
6
CHAPTER 8
6
CHAPTER 9
Quadratics
Functions
Coordinate Geometry
Circular Measure
Trigonometry
Vectors
Series
Differentiation
Integration
© Copyright 2015 by Z Notes
First edition © 2015, by Emir Demirhan, Saif Asmi & Zubair Junjunia.
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CIE A LEVEL- MATHEMATICS [9709]
2.4 Finding Inverse
1. QUADRATICS
ο‚·
1.1 Completing the square
2.5 Relationship of Function & its Inverse
2
2
𝑛
𝑛
π‘₯ 2 + 𝑛π‘₯ ⟺ (π‘₯ + ) − ( )
2
2
2
π‘Ž(π‘₯ + 𝑛) + π‘˜
Where the vertex is (−𝑛, π‘˜)
ο‚·
i.
𝑦-intercept
π‘₯-intercept
Vertex (turning point)
ii.
iii.
1.3 Discriminant
𝑏 2 − 4π‘Žπ‘
If 𝑏 2 − 4π‘Žπ‘ = 0, real and equal roots
If 𝑏 2 − 4π‘Žπ‘ < 0, no real roots
If 𝑏 2 − 4π‘Žπ‘ > 0, real and distinct roots
Solution:
First pull out constant, 4, from π‘₯ related terms:
4(π‘₯ 2 − 6π‘₯) + 11
Use following formula to simplify the bracket only:
𝑛 2
𝑛 2
(π‘₯ − ) − ( )
2
2
4[(π‘₯ − 3)2 − 32 ] + 11
4(π‘₯ − 3)2 − 25
(π‘₯ − 𝑑)(π‘₯ − 𝛽) < 0 ⟹ 𝑑 < π‘₯ < 𝛽
(π‘₯ − 𝑑)(π‘₯ − 𝛽) > 0 ⟹ π‘₯ < 𝑑 or π‘₯ > 𝛽
Part (ii)
1.5 Solving Equations in Quadratic Form
To solve an equation in some form of quadratic
Substitute 𝑦
E.g. 2π‘₯ 4 + 3π‘₯ 2 + 7, 𝑦 = π‘₯ 2 , ∴ 2𝑦 2 + 3𝑦 + 7
2. FUNCTIONS
Domain = π‘₯ values
&
Range = 𝑦 values
ο‚· One-one functions: one π‘₯-value gives one 𝑦-value
2.1 Find Range
ο‚·
ο‚·
ο‚·
ο‚·
Complete the square or differentiate
Find min/max point
If min then, 𝑦 ≥ min 𝑦
If max then, 𝑦 ≤ max 𝑦
2.2 Composition of 2 Functions
ο‚·
E.g. 𝑓𝑔(π‘₯) ⟹ 𝑓(𝑔(π‘₯))
2.3 Prove One-One Functions
ο‚·
ο‚·
One π‘₯ value substitutes to give one 𝑦 value
No indices
Question 10:
𝑓(π‘₯) = 4π‘₯ 2 − 24π‘₯ + 11, for π‘₯ ∈ ℝ
𝑔(π‘₯) = 4π‘₯ 2 − 24π‘₯ + 11, for π‘₯ ≤ 1
Express 𝑓(π‘₯) in the form π‘Ž(π‘₯ − 𝑏)2 + 𝑐,
hence state coordinates of the vertex of the
graph 𝑦 = 𝑓(π‘₯)
State the range of 𝑔
Find an expression for 𝑔−1 (π‘₯) and state its
domain
Part (i)
1.4 Quadratic Inequalities
ο‚·
ο‚·
ο‚·
The graph of the inverse of a function is the
reflection of a graph of the function in 𝑦 = π‘₯
{W12-P11}
1.2 Sketching the Graph
ο‚·
ο‚·
ο‚·
Make π‘₯ the subject of formula
Observe given domain, π‘₯ ≤ 1.
Substitute highest value of π‘₯
𝑔(π‘₯) = 4(1 − 3)2 − 25 = −9
Substitute next 3 whole numbers in domain:
π‘₯ = 0, −1, −2
𝑔(π‘₯) = 11, 23, 75
Thus they are increasing
∴ 𝑔(π‘₯) ≥ −9
Part (iii)
Let 𝑦 = 𝑔(π‘₯), make π‘₯ the subject
𝑦 = 4(π‘₯ − 3)2 − 25
𝑦 + 25
= (π‘₯ − 3)2
4
𝑦 + 25
π‘₯ =3+√
4
Can be simplified more
1
π‘₯ = 3 ± √𝑦 + 25
2
Positive variant is not possible because π‘₯ ≤ 1 and
using positive variant would give values above 3
1
∴ π‘₯ = 3 − √𝑦 + 25
2
1
∴ 𝑔−1 (π‘₯) = 3 − √π‘₯ + 25
2
Domain of 𝑔−1 (π‘₯) = Range of 𝑔(π‘₯) ∴ π‘₯ ≥ −9
Page 2 of 6
CIE A LEVEL- MATHEMATICS [9709]
3. COORDINATE GEOMETRY
Vector change from (−1,3) to (3,9) is the vector
change from (3,9) to 𝑅
Finding the vector change:
πΆβ„Žπ‘Žπ‘›π‘”π‘’ 𝑖𝑛 π‘₯ = 3 − −1 = 4
πΆβ„Žπ‘Žπ‘›π‘”π‘’ 𝑖𝑛 𝑦 = 9 − 3 = 6
Thus 𝑅
𝑅 ′ 𝑠 π‘₯ = 3 + 4 = 7 and 𝑅 ′ 𝑠 𝑦 = 9 + 6 = 15
𝑅 = (7,15)
3.1 Length of a Line Segment
Length = √(π‘₯2 − π‘₯1 )2 + (𝑦2 − 𝑦1 )2
3.2 Gradient of a Line Segment
π‘š=
𝑦2 − 𝑦1
π‘₯2 − π‘₯1
4. CIRCULAR MEASURE
3.3 Midpoint of a Line Segment
π‘₯1 + π‘₯2 𝑦1 + 𝑦2
(
,
)
2
2
4.1 Radians
πœ‹ = 180° and 2πœ‹ = 360°
3.4 Equation of a Straight Line
ο‚·
ο‚·
𝑦 = π‘šπ‘₯ + 𝑐
𝑦 − 𝑦1 = π‘š(π‘₯ − π‘₯1 )
Radians to degrees:
4.2 Arc length
3.5 Special Gradients
ο‚·
ο‚·
ο‚·
ο‚·
Parallel lines: π‘š1 = π‘š2
Perpendicular lines: π‘š1 π‘š2 = −1
The gradient at any point on a curve is the
gradient of the tangent to the curve at that point
The gradient of a the tangent at the vertex of a
curve is equal to zero – stationary point
{Wxx-Pxx}
πœ‹
180
180
× πœ‹
Degrees to radians: ×
𝑠 = π‘Ÿπœƒ
4.3 Area of a Sector
1
𝐴 = π‘Ÿ2πœƒ
2
{S11-P11}
Question 9:
Question 10:
Point 𝑅 is a reflection of point (−1,3) in the line
3𝑦 + 2π‘₯ = 33.
Find by calculation the coordinates of 𝑅
Solution:
Find equation of line perpendicular to 3𝑦 + 2π‘₯ = 33
intersecting point (−1,3)
2
3𝑦 + 2π‘₯ = 33 ⇔ 𝑦 = 11 − 3 π‘₯
2
π‘š=−
3
3
π‘š. π‘š1 = −1 and so π‘š1 = 2
Perpendicular general equation:
3
𝑦 = π‘₯+𝑐
2
Substitute known values
3
9
3 = 2 (−1) + 𝑐 and so 𝑐 = 2
Final perpendicular equation:
2𝑦 = 3π‘₯ + 9
Find point of intersection by equating two equations
2
3π‘₯ + 9
11 − π‘₯ =
3
2
13
13 =
π‘₯
3
π‘₯ = 3,
𝑦=9
Triangle 𝑂𝐴𝐡 is isosceles, 𝑂𝐴 = 𝑂𝐡 and 𝐴𝑆𝐡 is a
tangent to 𝑃𝑆𝑇
i.
Find the total area of shaded region in terms
of π‘Ÿ and πœƒ
ii.
1
When πœƒ = 3 and π‘Ÿ = 6, find total perimeter of
shaded region in terms of √3 and πœ‹
Part (i)
Solution:
Use trigonometric ratios to form the following:
𝐴𝑆 = π‘Ÿ tan πœƒ
Find the area of triangle 𝑂𝐴𝑆:
π‘Ÿ tan πœƒ × π‘Ÿ 1 2
𝑂𝐴𝑆 =
= π‘Ÿ tan πœƒ
2
2
Use formula of sector to find area of 𝑂𝑃𝑆:
1
𝑂𝑃𝑆 = π‘Ÿ 2 πœƒ
2
Area of 𝐴𝑆𝑃 is 𝑂𝐴𝑆 − 𝑂𝑃𝑆:
1
1
1
∴ 𝐴𝑆𝑃 = π‘Ÿ 2 tan πœƒ − π‘Ÿ 2 πœƒ = π‘Ÿ 2 (tan πœƒ − πœƒ)
2
2
2
Page 3 of 6
CIE A LEVEL- MATHEMATICS [9709]
Multiply final by 2 because 𝐡𝑆𝑇 is the same and
shaded is 𝐴𝑆𝑃 and 𝐡𝑆𝑇
1
π΄π‘Ÿπ‘’π‘Ž = 2 × π‘Ÿ 2 (tan πœƒ − πœƒ) = π‘Ÿ 2 (tan πœƒ − πœƒ)
2
5.3 Tangent Curve
Part (ii)
Use trigonometric ratios to get the following:
πœ‹
6
cos ( ) =
3
𝐴𝑂
∴ 𝐴𝑂 = 12
Finding 𝐴𝑃:
𝐴𝑃 = 𝐴𝑂 − π‘Ÿ = 12 − 6 = 6
Finding 𝐴𝑆:
πœ‹
𝐴𝑆 = 6 tan ( ) = 6√3
3
Finding arc 𝑃𝑆:
5.4 When sin, cos and tan are positive
π΄π‘Ÿπ‘ 𝑃𝑆 = π‘Ÿπœƒ
πœ‹
𝑃𝑆 = 6 × = 2πœ‹
3
Perimeter of 1 side of the shaded region:
𝑃𝑒1 = 6 + 6√3 + 2πœ‹
Perimeter of entire shaded region is just double:
2 × π‘ƒπ‘’1 = 12 + 12√3 + 4πœ‹
5. TRIGONOMETRY
5.5 Identities
sin πœƒ
tan πœƒ ≡ cos πœƒ
5.1 Sine Curve
sin2 πœƒ + cos 2 πœƒ ≡ 1
6. VECTORS
ο‚·
Forms of vectors
π‘₯
(𝑦 )
𝑧
5.2 Cosine Curve
Page 4 of 6
π‘₯𝐒 + 𝑦𝐒 + 𝑧𝐀
βƒ—βƒ—βƒ—βƒ—βƒ—
𝐴𝐡
𝒂
ο‚·
Position vector: position relative to origin βƒ—βƒ—βƒ—βƒ—βƒ—
𝑂𝑃
ο‚·
Magnitude = √π‘₯ 2 + 𝑦 2
ο‚·
βƒ—βƒ—βƒ—βƒ—βƒ—
Unit vectors: vectors of magnitude 1 = |𝐴𝐡| 𝐴𝐡
ο‚·
ο‚·
βƒ—βƒ—βƒ—βƒ—βƒ— = 𝑂𝐡
βƒ—βƒ—βƒ—βƒ—βƒ— − 𝑂𝐴
βƒ—βƒ—βƒ—βƒ—βƒ—
𝐴𝐡
Dot product: (π‘Žπ’ + 𝑏𝐣). (𝑐𝐒 + 𝑑𝐣) = (π‘Žπ‘π’ + 𝑏𝑑𝐣)
ο‚·
cos 𝐴 = |π‘Ž||𝑏|
1
π‘Ž.𝑏
CIE A LEVEL- MATHEMATICS [9709]
{S03-P01}
Question 8:
Points 𝐴, 𝐡, 𝐢, 𝐷 have position vectors 3π’Š + 2π’Œ, 2π’Š −
2𝒋 + 5π’Œ, 2𝒋 + 7π’Œ, −2π’Š + 10𝒋 + 7π’Œ respectively
i.
Use a scalar product to show that 𝐡𝐴 and
𝐡𝐢 are perpendicular
ii.
Show that 𝐡𝐢 and 𝐴𝐷 are parallel and find
the ratio of length of 𝐡𝐢 to length of 𝐴𝐷
7. SERIES
7.1 Binomial Theorem
(π‘₯ + 𝑦)𝑛 = 𝑛𝐢0 π‘₯ 𝑛 + 𝑛𝐢1 π‘₯ 𝑛−1 𝑦 + 𝑛𝐢2 π‘₯ 𝑛−2 𝑦 2 + β‹―
+ 𝑛𝐢𝑛 𝑦 𝑛
𝑛
πΆπ‘Ÿ =
Solution:
Part (i)
First find the vectors representing 𝐡𝐴 and 𝐡𝐢:
𝐡𝐴 = 𝑂𝐴 − 𝑂𝐡
𝐡𝐴 = 3π’Š + 2π’Œ − (2π’Š − 2𝒋 + 5π’Œ)
−1
𝐡𝐴 = −1π’Š − 2𝒋 + 3π’Œ = (−2)
3
𝐢𝐡 = 𝑂𝐡 − 𝑂𝐢
𝐢𝐡 = 2π’Š − 2𝒋 + 5π’Œ − (2𝒋 + 7π’Œ)
2
𝐢𝐡 = 2π’Š − 4𝒋 − 2π’Œ = (−4)
−2
Now use the dot product rule:
𝐡𝐴. 𝐢𝐡 = 0
−1
2
(−2) . (−4)
3
−2
= (−1 × 2) + (−2 × −4) + (3 × −2) = 0
Thus proving they are perpendicular since cos 90 = 0
Part (ii)
Find the vectors representing 𝐡𝐢 and 𝐴𝐷:
𝐡𝐢 = −𝐢𝐡
2
𝐡𝐢 = − (−4)
−2
−2
−1
𝐡𝐢 = ( 4 ) = 2 ( 2 )
2
1
𝐴𝐷 = 𝑂𝐷 − 𝑂𝐴
𝐴𝐷 = −2π’Š + 10𝒋 + 7π’Œ − (3π’Š + 2π’Œ)
−5
−1
𝐴𝐷 = −5π’Š + 10𝒋 + 5π’Œ = ( 10 ) = 5 ( 2 )
5
1
Direction vector shows that they are parallel
Calculate lengths of each:
|𝐡𝐢| = 2 (√(−1)2 + 22 + 12 ) = 2√6
|𝐴𝐷| = 5 (√(−1)2 + 22 + 12 ) = 5√6
∴ |𝐴𝐷|: |𝐡𝐢| = 5: 2
𝑛(𝑛 − 1)(𝑛 − 2) … (𝑛 − (π‘Ÿ − 1))
π‘Ÿ!
7.2 Arithmetic Progression
π‘’π‘˜ = π‘Ž + (π‘˜ − 1)𝑑
1
𝑆𝑛 = 𝑛[2π‘Ž + (𝑛 − 1)𝑑]
2
7.3 Geometric Progression
π‘’π‘˜ = π‘Žπ‘Ÿ π‘˜−1
𝑆𝑛 =
{W05-P01}
π‘Ž(1−π‘Ÿ 𝑛 )
(1−π‘Ÿ)
𝑆∞ =
π‘Ž
1−π‘Ÿ
Question 6:
A small trading company made a profit of $250 000 in
the year 2000. The company considered two different
plans, plan 𝐴 and plan 𝐡, for increasing its profits.
Under plan 𝐴, the annual profit would increase each
year by 5% of its value in the preceding year. Under
plan 𝐡, the annual profit would increase each year by
a constant amount $𝐷
i.
Find for plan 𝐴, the profit for the year 2008
ii.
Find for plan 𝐴, the total profit for the 10
years 2000 to 2009 inclusive
iii.
Find for plan 𝐡 the value of 𝐷 for which the
total profit for the 10 years 2000 to 2009
inclusive would be the same for plan 𝐴
Solution:
Part (i)
Increases is exponential ∴ it is a geometric sequence:
2008 is the 9th term:
∴ 𝑒9 = 250000 × 1.059−1 = 369000 (3s.f.)
Part (ii)
Use sum of geometric sequence formula:
250000(1 − 1.0510 )
𝑆10 =
= 3140000
1 − 1.05
Part (iii)
Plan B arithmetic; equate 3140000 with sum formula
1
3140000 = (10)(2(250000) + (10 − 1)𝐷)
2
𝐷 = 14300
Page 5 of 6
CIE A LEVEL- MATHEMATICS [9709]
8. DIFFERENTIATION
Part (ii)
Rate of increase in time can be written as:
𝑑π‘₯
𝑑𝑑
We know the following:
𝑑𝑦 4
𝑑𝑦
=
π‘Žπ‘›π‘‘
= 0.02
𝑑π‘₯ 3
𝑑𝑑
Thus we can formulate an equation:
𝑑𝑦 𝑑𝑦 𝑑π‘₯
=
÷
𝑑π‘₯ 𝑑𝑑 𝑑𝑑
Rearranging the formula we get:
𝑑π‘₯ 𝑑𝑦 𝑑𝑦
=
÷
𝑑𝑑 𝑑𝑑 𝑑π‘₯
Substitute values into the formula
𝑑π‘₯
4
= 0.02 ÷
𝑑𝑑
3
𝑑π‘₯
3
= 0.02 × = 0.015
𝑑𝑑
4
𝑑𝑦
When 𝑦 = π‘₯ 𝑛 , 𝑑π‘₯ = 𝑛π‘₯ 𝑛−1
𝑑𝑦
ο‚·
1st Derivative = 𝑑π‘₯ = 𝑓 ′ (π‘₯)
ο‚·
2nd Derivative = 𝑑π‘₯ 2 = 𝑓 ′′ (π‘₯)
ο‚·
Increasing function: 𝑑π‘₯ > 0
ο‚·
Decreasing function: 𝑑π‘₯ < 0
ο‚·
Stationary point: 𝑑π‘₯ = 0
𝑑2 𝑦
𝑑𝑦
𝑑𝑦
𝑑𝑦
8.1 Chain Rule
𝑑𝑦 𝑑𝑦 𝑑𝑒
=
×
𝑑π‘₯ 𝑑𝑒 𝑑π‘₯
8.2 Nature of Stationary Point
ο‚·
ο‚·
ο‚·
ο‚·
Find second derivative
Substitute π‘₯-value of stationary point
If value +ve → min. point
If value –ve → max. point
9. INTEGRATION
∫ π‘Žπ‘₯ 𝑛 𝑑π‘₯ =
8.3 Connected Rates of Change
∫(π‘Žπ‘₯ + 𝑏)𝑛 =
𝑑𝑦 𝑑𝑦 𝑑π‘₯
=
×
𝑑𝑑 𝑑π‘₯ 𝑑𝑑
{W05-P01}
Question 6:
The equation of a curve is given by the formula:
6
𝑦=
5 − 2π‘₯
i.
Calculate the gradient of the curve at the
point where π‘₯ = 1
ii.
A point with coordinates (π‘₯, 𝑦) moves along a
curve in such a way that the rate of increase of
𝑦 has a constant value of 0.02 units per
second. Find the rate of increase of π‘₯ when
π‘₯=1
Solution:
Part (i)
π‘Žπ‘₯ 𝑛+1
+𝑐
𝑛+1
(π‘Žπ‘₯ + 𝑏)𝑛+1
+𝑐
π‘Ž(𝑛 + 1)
Definite integrals: substitute coordinates and find ‘c’
9.1 To Find Area
ο‚·
ο‚·
ο‚·
Integrate curve
Substitute boundaries of π‘₯
Subtract one from another (ignore c)
𝑑
∫ 𝑦 𝑑π‘₯
𝑐
9.2 To Find Volume
Differentiate given equation
6(5 − 2π‘₯)−1
𝑑𝑦
= 6(5 − 2π‘₯)−2 × −2 × −1
𝑑π‘₯
= 12(5 − 2π‘₯)−2
Now we substitute the given π‘₯ value:
𝑑𝑦
−2
= 12(5 − 2(1))
𝑑π‘₯
𝑑𝑦 4
=
𝑑π‘₯ 3
4
Thus the gradient is equal to at this point
ο‚·
ο‚·
ο‚·
Square the function
Integrate and substitute
Multiply by πœ‹
𝑑
∫ πœ‹π‘¦ 2 𝑑π‘₯
𝑐
3
Page 6 of 6
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