CIE Pure-maths 1 Notes

CIE A LEVEL- MATHEMATICS [9709]

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SECTION 1: PURE MATHEMATICS 1

1. QUADRATICS

1.1 Completing the square

2 + ( +

2)2

(

2)2

( + )2 +

Where the vertex is (, )

1.2 Sketching the Graph -intercept

-intercept

Vertex (turning point)

1.3 Discriminant 2 4

If 2 4 = 0, real and equal roots

If 2 4 < 0, no real roots

If 2 4 > 0, real and distinct roots

1.4 Quadratic Inequalities ( )( ) < 0 < <

( )( ) > 0 < or >

1.5 Solving Equations in Quadratic Form

To solve an equation in some form of quadratic

Substitute

E.g. 24 + 32 + 7, = 2, 22 + 3 + 7

2. FUNCTIONS Domain = values & Range = values

One-one functions: one -value gives one -value

2.1 Find Range Complete the square or differentiate

Find min/max point

If min then, min

If max then, max

2.2 Composition of 2 Functions E.g. () (())

2.3 Prove One-One Functions One value substitutes to give one value

No indices

2.4 Finding Inverse

Make the subject of formula

2.5 Relationship of Function & its Inverse The graph of the inverse of a function is the

reflection of a graph of the function in =

{W12-P11} Question 10:

() = 42 24 + 11, for () = 42 24 + 11, for 1

i. Express () in the form ( )2 + , hence state coordinates of the vertex of the graph = ()

ii. State the range of iii. Find an expression for 1() and state its

domain Solution:

Part (i)

First pull out constant, 4, from related terms: 4(2 6) + 11

Use following formula to simplify the bracket only:

(

2)2

(

2)2

4[( 3)2 32] + 11 4( 3)2 25

Part (ii)

Observe given domain, 1. Substitute highest value of

() = 4(1 3)2 25 = 9 Substitute next 3 whole numbers in domain:

= 0,1,2 () = 11, 23, 75 Thus they are increasing

() 9 Part (iii)

Let = (), make the subject = 4( 3)2 25 + 25

4= ( 3)2

= 3 + + 25

4

Can be simplified more

= 3 1

2 + 25

Positive variant is not possible because 1 and using positive variant would give values above 3

= 3 1

2 + 25

1() = 3 1

2 + 25

Domain of 1() = Range of () 9


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3. COORDINATE GEOMETRY

3.1 Length of a Line Segment

Length = (2 1)2 + (2 1)2

3.2 Gradient of a Line Segment

=2 12 1

3.3 Midpoint of a Line Segment

(1 + 2

2 ,1 + 2

2)

3.4 Equation of a Straight Line

= +

1 = ( 1)

3.5 Special Gradients Parallel lines: 1 = 2

Perpendicular lines: 12 = 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} Question 10:

Point is a reflection of point (1,3) in the line 3 + 2 = 33.

Find by calculation the coordinates of Solution:

Find equation of line perpendicular to 3 + 2 = 33 intersecting point (1,3)

3 + 2 = 33 = 11 2

3

= 2

3

.1 = 1 and so 1 =3

2

Perpendicular general equation:

=3

2 +

Substitute known values

3 =3

2(1) + and so =

9

2

Final perpendicular equation: 2 = 3 + 9

Find point of intersection by equating two equations

11 2

3 =

3 + 9

2

13 =13

3

= 3, = 9

Vector change from (1,3) to (3,9) is the vector change from (3,9) to Finding the vector change:

= 3 1 = 4 = 9 3 = 6

Thus = 3 + 4 = 7 and = 9 + 6 = 15

= (7,15)

4. CIRCULAR MEASURE

4.1 Radians = 180 and 2 = 360

Degrees to radians:

180

Radians to degrees: 180

4.2 Arc length =

4.3 Area of a Sector

=1

22

{S11-P11} Question 9:

Triangle is isosceles, = and is a

tangent to

i. Find the total area of shaded region in terms

of and

ii. When =1

3 and = 6, find total perimeter of

shaded region in terms of 3 and Solution:

Part (i)

Use trigonometric ratios to form the following:

= tan Find the area of triangle :

= tan

2=

1

22 tan

Use formula of sector to find area of :

=1

22

Area of is :

= 1

22 tan

1

22 =

1

22(tan )


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Multiply final by 2 because is the same and

shaded is and

= 2 1

22(tan ) = 2(tan )

Part (ii)

Use trigonometric ratios to get the following:

cos (

3) =

6

= 12 Finding :

= = 12 6 = 6

Finding :

= 6 tan (

3) = 63

Finding arc :

=

= 6

3= 2

Perimeter of 1 side of the shaded region:

1 = 6 + 63 + 2 Perimeter of entire shaded region is just double:

2 1 = 12 + 123 + 4

5. TRIGONOMETRY

5.1 Sine Curve

5.2 Cosine Curve

5.3 Tangent Curve

5.4 When sin, cos and tan are positive

5.5 Identities

tan sin

cos sin2 + cos2 1

6. VECTORS Forms of vectors

() + +

Position vector: position relative to origin

Magnitude = 2 + 2

Unit vectors: vectors of magnitude 1 =1

||

=

Dot product: ( + ). ( + ) = ( + )

cos =.

||||


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{S03-P01} Question 8:

Points ,,, have position vectors 3 + 2, 2 2 + 5, 2 + 7,2 + 10 + 7 respectively

i. Use a scalar product to show that and are perpendicular

ii. Show that and are parallel and find the ratio of length of to length of

Solution:

Part (i)

First find the vectors representing and : =

= 3 + 2 (2 2 + 5)

= 1 2 + 3 = (123

)

= = 2 2 + 5 (2 + 7)

= 2 4 2 = (2

42

)

Now use the dot product rule: . = 0

(123

) . (2

42

)

= (1 2) + (2 4) + (3 2) = 0 Thus proving they are perpendicular since cos 90 = 0 Part (ii)

Find the vectors representing and : =

= (2

42

)

= (242

) = 2(121

)

= = 2 + 10 + 7 (3 + 2)

= 5 + 10 + 5 = (5105

) = 5(121

)

Direction vector shows that they are parallel Calculate lengths of each:

|| = 2 ((1)2 + 22 + 12) = 26

|| = 5 ((1)2 + 22 + 12) = 56

||: || = 5: 2

7. SERIES

7.1 Binomial Theorem ( + ) = 0

+ 1 1 + 2

22 +

+

=

( 1)( 2) ( ( 1))

!

7.2 Arithmetic Progression = + ( 1)

=1

2[2 + ( 1)]

7.3 Geometric Progression =

1

=(1)

(1) =

1


A small trading company made a profit of $250 000 in the year 2000. The company considered two different plans, plan and plan , for increasing its profits. Under plan , the annual profit would increase each year by 5% of its value in the preceding year. Under plan , the annual profit would increase each year by a constant amount $

i. Find for plan , the profit for the year 2008 ii. Find for plan , the total profit for the 10

years 2000 to 2009 inclusive iii. Find for plan the value of for which the

total profit for the 10 years 2000 to 2009 inclusive would be the same for plan

Solution: Part (i)

Increases is exponential it is a geometric sequence: 2008 is the 9th term:

9 = 250000 1.0591 = 369000 (3s.f.)

Part (ii)

Use sum of geometric sequence formula:

10 =250000(1 1.0510)

1 1.05= 3140000

Part (iii)

Plan B arithmetic; equate 3140000 with sum formula

3140000 =1

2(10)(2(250000) + (10 1))

= 14300


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8. DIFFERENTIATION

When = ,

= 1

1st Derivative =

= ()

2nd Derivative = 2

2= ()

Increasing function:

> 0

Decreasing function:

< 0

Stationary point:

= 0

8.1 Chain Rule

=

8.2 Nature of Stationary Point Find second derivative

Substitute -value of stationary point

If value +ve min. point

If value ve max. point

8.3 Connected Rates of Change

=


The equation of a curve is given by the formula:

=6

5 2

i. Calculate the gradient of the curve at the point where = 1

ii. A point with coordinates (, ) moves along a curve in such a way that the rate of increase of has a constant value of 0.02 units per second. Find the rate of increase of when = 1

Solution: Part (i)

Differentiate given equation

6(5 2)1

= 6(5 2)2 2 1

= 12(5 2)2 Now we substitute the given value:

= 12(5 2(1))

2

=

4

3

Thus the gradient is equal to 4

3 at this point

Part (ii)

Rate of increase in time can be written as:

We know the following:

=

4

3

= 0.02

Thus we can formulate an equation:

=

Rearranging the formula we get:

=

Substitute values into the formula

= 0.02

4

3

= 0.02

3

4= 0.015

9. INTEGRATION

=+1

+ 1+

( + ) =( + )+1

( + 1)+

Definite integrals: substitute coordinates and find c

9.1 To Find Area Integrate curve

Substitute boundaries of

Subtract one from another (ignore c)

9.2 To Find Volume Square the function

Integrate and substitute

Multiply by

2

CIE Pure-maths 1 Notes

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