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# AS Pure Maths Topic 1

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```CIE AS LEVEL
PURE MATHEMATICS 1
TOPICS
1
4
7
Quadratics
𝜃𝜃
Circular
Measure
𝑑𝑑𝑑𝑑
= 𝑓𝑓𝑓(𝑥𝑥)
𝑑𝑑𝑑𝑑
Differentiation
𝑦𝑦
2
5
𝑓𝑓(𝑥𝑥)
Functions
Trigonometry
𝑏𝑏
� 𝑓𝑓 𝑥𝑥 𝑑𝑑𝑑𝑑
𝑎𝑎
8 Integration
3 Co-ordinate
Geometry
6
𝑎𝑎
𝑆𝑆∞ =
1 − 𝑟𝑟
Series
𝑥𝑥
AS
Quadratics
• Solving quadratic equations by factorisation
O
B
J
E
C
T
I
V
E
S
• Completing the square
• The quadratic formula
• Solving simultaneous equations (one linear and one quadratic)
• Solving more complex quadratic equations
• Maximum and minimum values of a quadratic function
• Solving quadratic inequalities
• The number of roots of a quadratic equation
• Intersection of a line and a quadratic curve
AS
1
Ways of Solving Quadratic Equations
Factorisation
2
Completing the
square
3
Quadratic
Equation
AS
Factorisation
Solve the equation:
𝑥𝑥 2 + 3𝑥𝑥 − 10 = 0
−2 &times; 5 = −10 (constant term)
−2 + 5 = 3 (coefficient of x)
𝑥𝑥 2 + 3𝑥𝑥 − 10 = 0
𝑥𝑥 − 2 𝑥𝑥 + 5 = 0
𝑥𝑥 = 2 𝑜𝑜𝑜𝑜 − 5
AS
Factorisation
Solve:
6𝑥𝑥 2 + 5 = 17𝑥𝑥
6𝑥𝑥 2 − 17𝑥𝑥 + 5 = 0
2𝑥𝑥 − 5 3𝑥𝑥 − 1 = 0
5
1
𝑥𝑥 = 𝑜𝑜𝑜𝑜
2
3
Write in the form 𝑎𝑎𝑥𝑥 2 + 𝑏𝑏𝑏𝑏 + 𝑐𝑐 = 0
Factorise &amp; Solve
AS
Factorisation
Solve:
9𝑥𝑥 2
9𝑥𝑥 2 − 39𝑥𝑥 − 30 = 0
− 39𝑥𝑥 − 30 = 0
3𝑥𝑥 2 − 13𝑥𝑥 − 10 = 0
3𝑥𝑥 + 2 𝑥𝑥 − 5 = 0
2
𝑥𝑥 = − 𝑜𝑜𝑜𝑜 5
3
Divide both sides by common
factor of 3
Factorise &amp; Solve
AS
Factorisation
Solve:
21
2𝑥𝑥
−
2
𝑥𝑥+3
=1
21 𝑥𝑥 + 3 − 4𝑥𝑥 = 2𝑥𝑥 𝑥𝑥 + 3
2𝑥𝑥 2 − 11𝑥𝑥 − 63 = 0
2𝑥𝑥 + 7 𝑥𝑥 − 9 = 0
7
𝑥𝑥 = − 𝑜𝑜𝑜𝑜 9
2
Multiply both sides by
2𝑥𝑥(𝑥𝑥 + 3)
Expand brackets &amp; rearrange
Solve
AS
Factorisation
3𝑥𝑥 2 +26𝑥𝑥+35
Solve:
𝑥𝑥 2 +8
3𝑥𝑥 2 + 26𝑥𝑥 + 35
=0
2
𝑥𝑥 + 8
3𝑥𝑥 2 + 26𝑥𝑥 + 35 = 0
3𝑥𝑥 + 5 𝑥𝑥 + 7 = 0
5
𝑥𝑥 = − 𝑜𝑜𝑜𝑜 − 7
3
=0
Multiply both sides by
(𝑥𝑥 2 + 8)
Factorise
Solve
AS
Factorisation Questions
LEVEL 1
LEVEL 2
AS
Factorisation Questions
LEVEL 3
LEVEL 4
AS
Factorisation Questions
LEVEL 5
AS
Completing the square
• The method of completing the square aims to rewrite a quadratic
expression using only one occurrence of the variable, making it an
easier expression to work with.
𝑥𝑥 2 + 2𝑑𝑑𝑑𝑑 = 𝑥𝑥 + 𝑑𝑑
𝑥𝑥 2 − 2𝑑𝑑𝑑𝑑 = 𝑥𝑥 − 𝑑𝑑
2
2
− 𝑑𝑑2
− 𝑑𝑑2
AS
Completing the square
Complete the square: 𝑥𝑥 2 + 8𝑥𝑥 − 7
8
=4
2
𝑥𝑥 + 4
2
Divide the coefficient
of x by 2
− 42 − 7
𝑥𝑥 2 + 8𝑥𝑥 − 7 = 𝑥𝑥 + 4
2
−23
AS
Completing the square
Express 2𝑥𝑥 2 − 12𝑥𝑥 + 3 in the form 𝑝𝑝 𝑥𝑥 − 𝑞𝑞
be found
2𝑥𝑥 2 − 12𝑥𝑥 + 3 = 𝑝𝑝 𝑥𝑥 − 𝑞𝑞
2
2
+ 𝑟𝑟 where p, q and r are constants to
+ 𝑟𝑟
2𝑥𝑥 2 − 12𝑥𝑥 + 3 = 𝑝𝑝𝑥𝑥 2 − 2𝑝𝑝𝑞𝑞𝑥𝑥 + 𝑝𝑝𝑞𝑞 2 + 𝑟𝑟
Comparing coefficients:
Substituting p = 2 in equation (2) gives q = 3
𝑥𝑥 2 ∶ 2 = 𝑝𝑝
Substituting p = 2 and q = 3 in equation (3)
therefore gives r = −15
𝑥𝑥 ∶
(1)
−12 = −2𝑝𝑝𝑞𝑞 (2)
constant: 3 =
𝑝𝑝𝑞𝑞 2
+ 𝑟𝑟 (3)
2𝑥𝑥 2 − 12𝑥𝑥 + 3 = 2 𝑥𝑥 − 3
2
− 15
AS
Completing the square
LEVEL 1
LEVEL 2
AS
Completing the square
LEVEL 3
AS
Quadratic Formula
We can solve quadratic equations using the quadratic formula.
If 𝑎𝑎𝑎𝑎 2 + 𝑏𝑏𝑏𝑏 + 𝑐𝑐 = 0, where a, b and c are constants and 𝑎𝑎 ≠ 0, then;
−𝑏𝑏 &plusmn; 𝑏𝑏 2 − 4𝑎𝑎𝑎𝑎
𝑥𝑥 =
2𝑎𝑎
AS
Quadratic Formula
Solve the equation: 6𝑥𝑥 2 − 3𝑥𝑥 − 2 = 0
Using a = 6, b = −3 and c = −2 in the quadratic formula gives:
𝑥𝑥 =
− −3 &plusmn;
−3 2 − 4 &times; 6 &times; −2
2&times;6
3 + 57
3 − 57
𝑜𝑜𝑜𝑜
𝑥𝑥 =
12
12
𝑥𝑥 = 0.879 𝑜𝑜𝑜𝑜 − 0.379
AS
Quadratic Formula Questions
AS
Solving simultaneous equations (one
linear and one quadratic)
The diagram shows the graphs of y = x2 − 4
and y = 2x − 1.
The coordinates of the points of
intersection of the two graphs are (−1, −3)
and (3, 5).
It follows that x = −1, y = −3 and x = 3, y = 5
are the solutions of the simultaneous
equations y = x2 − 4 and y = 2x − 1.
AS
Solving simultaneous equations (one
linear and one quadratic)
The solutions can also be found algebraically:
𝑦𝑦 = 𝑥𝑥 2 − 4 … (1)
𝑦𝑦 = 2𝑥𝑥 − 1 … (2)
Substitute for y from equation (2) into equation (1):
2𝑥𝑥 − 1 = 𝑥𝑥 2 − 4
𝑥𝑥 2 − 2𝑥𝑥 − 3 = 0
𝑥𝑥 + 1 𝑥𝑥 − 3 = 0
𝑥𝑥 = −1 𝑜𝑜𝑜𝑜 3
Substituting x = −1 into equation (2) gives y = −2 − 1 = −3.
Substituting x = 3 into equation (2) gives y = 6 − 1 = 5.
AS
Solving simultaneous equations (one
linear and one quadratic)
AS
Solving simultaneous equations (one
linear and one quadratic)
AS
Solving more complex quadratic
equations
Solve the equation: 4𝑥𝑥 4 − 37𝑥𝑥 2 + 9 = 0
Method 1: Substitution Method
4𝑥𝑥 4 − 37𝑥𝑥 2 + 9 = 0
Let 𝑦𝑦 = 𝑥𝑥 2
4𝑦𝑦 2 − 37𝑦𝑦 + 9 = 0
4𝑦𝑦 − 1 𝑦𝑦 − 9 = 0
1
𝑦𝑦 = 𝑜𝑜𝑜𝑜 9
4
AS
Solving more complex quadratic
equations
Solve the equation: 4𝑥𝑥 4 − 37𝑥𝑥 2 + 9 = 0
Method 2: Factorise Directly
4𝑥𝑥 4 − 37𝑥𝑥 2 + 9 = 0
4𝑥𝑥 2 − 1 𝑥𝑥 2 − 9 = 0
𝑥𝑥 2
1
= 𝑜𝑜𝑜𝑜 9
4
1
𝑥𝑥 = &plusmn; 𝑜𝑜𝑜𝑜 &plusmn; 3
2
AS
Solving more complex quadratic
equations
Solve the equation: 𝑥𝑥 − 4 𝑥𝑥 − 12 = 0
Let 𝑦𝑦 = 𝑥𝑥
𝑦𝑦 2 − 4𝑦𝑦 − 12 = 0
𝑦𝑦 − 6 𝑦𝑦 + 2 = 0
𝑦𝑦 = 6 𝑜𝑜𝑜𝑜 − 2
𝑥𝑥 = 6
𝑥𝑥 = 36
𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟 𝑥𝑥 = −2
AS
Solving more complex quadratic
equations
Solve the equation: 3 9𝑥𝑥 − 28 3𝑥𝑥 + 9 = 0
3 9𝑥𝑥 − 28 3𝑥𝑥 + 9 = 0
3 32𝑥𝑥 − 28 3𝑥𝑥 + 9 = 0
3𝑦𝑦 2 − 28𝑦𝑦 + 9 = 0
3𝑦𝑦 − 1 𝑦𝑦 − 9 = 0
𝑦𝑦 =
1
3
𝑜𝑜𝑜𝑜 9
⇒
𝑥𝑥
3 =
1
3
Let 𝑦𝑦 = 3𝑥𝑥
𝑜𝑜𝑜𝑜 9 ⇒ 𝑥𝑥 = −1 𝑜𝑜𝑜𝑜 𝑥𝑥 = 2
AS
Solving more complex quadratic
equations
AS
Maximum and minimum values of a
quadratic function
• The shape of the graph of the function 𝑓𝑓 𝑥𝑥 = 𝑎𝑎𝑥𝑥 2 + 𝑏𝑏𝑏𝑏 + 𝑐𝑐 is called a
parabola.
• The orientation of the parabola depends on the value of a, the coefficient of 𝑥𝑥 2 .
Stationary
Point
If a &gt; 0, the curve has a minimum
point that occurs at the lowest point
of the curve.
If a &lt; 0, the curve has a maximum
point that occurs at the highest point
of the curve.
AS
Maximum and minimum values of a
quadratic function
If 𝑓𝑓 𝑥𝑥 = 𝑎𝑎𝑎𝑎 2 + 𝑏𝑏𝑏𝑏 + 𝑐𝑐 is written in the form 𝑓𝑓 𝑥𝑥 = 𝑎𝑎 𝑥𝑥 − ℎ
then:
If a &gt; 0, there is a minimum point at (h, k)
If a &lt; 0, there is a maximum point at (h, k).
2
+ 𝑘𝑘 ,
AS
Quadratic Inequalities
• If 𝑥𝑥 − 𝑎𝑎 𝑥𝑥 − 𝑏𝑏 &lt; 0 , then a &lt; 𝑥𝑥 &lt; 𝑏𝑏 if b &gt; a
• If 𝑥𝑥 − 𝑎𝑎 𝑥𝑥 − 𝑏𝑏 &gt; 0 , then 𝑥𝑥 &lt; 𝑎𝑎 𝑜𝑜𝑜𝑜 𝑥𝑥 &gt; 𝑏𝑏 if b &gt; a
• If we multiply or divide both sides of an inequality by a
negative number, then the inequality sign must be
reversed.
AS
Quadratic Inequalities
Solve 𝑥𝑥 2 − 5𝑥𝑥 − 14 &gt; 0
𝑥𝑥 − 7 𝑥𝑥 + 2 &gt; 0
Critical values: 7 and -2
Solution: 𝑥𝑥 &lt; − 2 𝑜𝑜𝑜𝑜 𝑥𝑥 &gt; 7
𝑥𝑥 &lt; −2
−2 &lt; 𝑥𝑥 &lt; 7
𝑥𝑥 &gt; 7
_
+
-2
+
7
AS
Quadratic Inequalities
Solve 2𝑥𝑥 2 + 3𝑥𝑥 ≤ 27
2𝑥𝑥 2
+ 3𝑥𝑥 − 27 ≤ 0
(2𝑥𝑥 + 9)(𝑥𝑥 − 3) ≤ 0
Critical values: - 4.5 and 3
Solution: −4.5 ≤ 𝑥𝑥 ≤ 3
𝑥𝑥 ≤ −4.5 −4.5 ≤ 𝑥𝑥 ≤ 3
𝑥𝑥 ≥ 3
_
+
- 4.5
+
3
AS
Quadratic Inequalities
AS
Quadratic Inequalities
AS
Quadratic Inequalities
AS
The number of roots of a quadratic
equation
If f(x) is a function, then we call the solutions to the equation f(x) = 0
the roots of f(x).
Consider solving the following three quadratic equations:
The part of the quadratic formula underneath the square root sign is
called the discriminant.
AS
The number of roots of a quadratic
equation
The sign (positive, zero or negative) of the discriminant tells us how many
roots there are for a particular quadratic equation
AS
The number of roots of a quadratic
equation
AS
The number of roots of a quadratic
equation
Find the values of k for which the equation 4𝑥𝑥 2 + 𝑘𝑘𝑘𝑘 + 1 = 0 has
two equal roots.
For two equal roots:
𝑘𝑘 2 − 4 4 1 = 0
𝑘𝑘 2 − 16 = 0
𝑘𝑘 = &plusmn;4 (4 𝑜𝑜𝑜𝑜 − 4)
𝑏𝑏 2 −4𝑎𝑎𝑎𝑎 = 0
AS
The number of roots of a quadratic
equation
Find the values of k for which the equation 𝑥𝑥 2 − 5𝑥𝑥 + 9 = 𝑘𝑘(5 − 𝑥𝑥) has
two equal roots.
For two equal roots:
𝑥𝑥 2 − 5𝑥𝑥 + 9 = 𝑘𝑘(5 − 𝑥𝑥)
𝑏𝑏 2 −4𝑎𝑎𝑎𝑎 = 0
𝑥𝑥 2 + 𝑘𝑘 − 5 𝑥𝑥 + 9 − 5𝑘𝑘 = 0
Rearrange in the form 𝑎𝑎𝑥𝑥 2 + 𝑏𝑏𝑏𝑏 + 𝑐𝑐 = 0
(𝑘𝑘 − 5)2 −4 1 9 − 5𝑘𝑘 = 0
𝑘𝑘 2 − 10𝑘𝑘 + 25 − 36 + 20𝑘𝑘 = 0
𝑘𝑘 2 + 10𝑘𝑘 − 11 = 0 ⇒ 𝑘𝑘 + 11 𝑘𝑘 − 1 = 0 ⇒ 𝑘𝑘 = −11 𝑜𝑜𝑜𝑜 1
AS
The number of roots of a quadratic
equation
AS
The number of roots of a quadratic
equation
AS
Intersection of a line and a quadratic
curve
When considering the intersection of a straight line and a parabola,
there are three possible situations.
AS
Intersection of a line and a quadratic
curve
• The discriminant of the resulting equation then enables us to say
how many points of intersection there are.
• The three possible situations are shown in the following table.
AS
Intersection of a line and a quadratic
curve
Find the value of k for which 𝑦𝑦 = 𝑥𝑥 + 𝑘𝑘 is a tangent to the
curve 𝑦𝑦 = 𝑥𝑥 2 + 5𝑥𝑥 + 2
𝑥𝑥 2 + 5𝑥𝑥 + 2 = 𝑥𝑥 + 𝑘𝑘
𝑥𝑥 2 + 4𝑥𝑥 + 2 − 𝑘𝑘 = 0
Rearrange in the form 𝑎𝑎𝑥𝑥 2 + 𝑏𝑏𝑏𝑏 + 𝑐𝑐 = 0
Since the line is a tangent to the curve, the discriminant of the
quadratic must be zero, so:
42 − 4 1 2 − 𝑘𝑘 = 0
16 − 8 + 4𝑘𝑘 = 0
4𝑘𝑘 = −8 ⇒ 𝑘𝑘 = −2
AS
Intersection of a line and a quadratic
curve
AS
Intersection of a line and a quadratic
curve
END OF QUADRATICS SECTION
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