# Extension 1 Mathematics

## Notes about these Notes

These notes are from the 2008 NSW HSC course so some parts may not be relevant to the new course.Title: 12 November 2007 Ext 1 Maths

Date: 12 November 2007 7:03 PM

Category: School

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__Counting Techniques__

If one event happens p ways then another event can happen q times then the total number of ways they can happen together is pq.

eg. A 3 digit 3 letter combintaion. (Like old number plates). How many can be made?

10 x 10 x 10 x 26 x 26 x 26 = 17576000

eg. 4 number combination lock. No digit can be repeted. How many combinations?

10 x 9 x 8 x 7

The number of ways n objects can be arranged is (n!)

eg. Arrange 5 people in a straight line

5!

Now in a circle

4! (1st person sits anywhere then everyone else (4 people) is arranged around them)

eg. n beads on a necklace. How many combinations?

__(n-1)!__

2

(necklace can be flipped over)

__Permutations__

n objects r at a time

nPr = (n!)/(n-r)!

eg. Given 4 letters how many 2 letter arrangements?

4P2 = 12

Title: 14 November 2007 Ext 1 Maths

Date: 14 November 2007 7:14 PM

Category: School

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8 women and 6 men. Trying to form a committee of 2 men 3 women.

How many combinations?

8C3 x 6C2 = 840

If 2 women refuse to serve together?

(_ _) _ _ _

1 ways to warrange the 2 women x 6 for the remaining 6 women and 6C2 for the men. Take this away from the original and get the answer

750 ways.

Title: 5 February 2008 Ext 1 Maths

Date: 5 February 2008 3:01 PM

Category: School

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__Inverse Functions__

The derivative of sin-1x =

____1 ___. x’ (x’ is the derivative of x)

√1-x2

cos-1x is the same but -1 instead of 1 on top

tan-1x is

__1__ . x’

1+x2

Title: 11 February 2008 Ext 1 Maths

Date: 11 February 2008 5:51 PM

Category: School

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__Integration__

derivative of:

sin-1: 1/(root(a2-x2))

tan-1: a/(a2 + x2)

Title: 12 February 2008 Ext 1 Maths

Date: 12 February 2008 5:54 PM

Category: School

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State the domain and range of y = -2sin-1(1-4x)

y/2 sin-1(1-4x)

π/2 ≤ y/2 ≤ π/2 solve

-1 ≤ (1-4x) ≤ 1 solve

Evaluate

sin-1[sin-1(3/2)-cos-1(4/5)]

Ok, try drawing the triangle for this.

You will find the thing in the brackets is an angle take away itself. Therefore the answer is zero.

Title: 25 February 2008 Ext 1 Maths

Date: 25 February 2008 5:00 PM

Category: School

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Prove that tan-1 (1/4) + tan-1 (3/5) = π/4

Treat each tan-1 as if it is an angle (because it is). If these two angles add to π/4 and tan(π/4) = 1 then do this: tan(tan-1 (1/4) + tan-1 (3/5))

It should equal 1 and thus proves the top bit = π/4

__Mutually inverse functions__

f-1(f(x)) = f(f-1(x))

f(x) = x3 is mutually inverse but f(x)= x2 isn’t. You can find out why by putting it in the above equation.

__Methods of Integration__

If you are integrating something complicated then substitute something in.

2∫√(2x+1) dx

u = 2x+1

so 2∫√(u) dx

but

__du__= 2

dx

so 2∫√(u) du/2

= ∫√(u) du

= solve the integration and substitute u = 2x +1 back in.

Title: 3 March 2008 Ext 1 Maths

Date: 3 March 2008 7:23 PM

Category: School

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So when doing that integration by substitution thing we have only been doing indefinite integrals.

To do a definite integral:

The numbers at the top and bottom of the S after the substitution has taken place, are now wrong.

To fix these numbers, substitute the bottom number into the equation for u. e.g. u = x2 - 5 and the bottom number is 0 then u = -5

This number you get for u is the new bottom number. Do the same for the top number but this time using the top number.

PLUS

cos2ø + sin2ø = 1

cos2ø - sin2ø = cos2ø

2cos2ø = 1 + cos2ø

Title: 4 March 2008 Ext 1 Maths

Date: 4 March 2008 6:38 PM

Category: School

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Using what I learnt today in the 4th March 2U file:

Consider a circle, radius r and centre at origin. A triangle is constructed such that the hypotenuse is r and the bottom side lies on x. The hypotenuse joins origin and a point P(x,y)

y = rsinø

x = rcosø

r is the hypotenuse and ø is the turn of the line. If you had a line r=2 and turn π/3 then using the above equations P = (1, √3)

Parametric definition of a parabola, vertex at origin with a focal length ‘a’

x = 2at

y = at2

Derivative of x = t

Title: 10 March 2008 Ext 1 Maths

Date: 10 March 2008 7:29 PM

Category: School

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You will be asked to find tangents to the parabola. Also, normals to this tangent.

A line that joins to parts of the parabola: a chord.

One that goes through the focus: a focal chord.

A focal chord that is parallel to the directrix: the latus rectum.

The tangents at the end of a focal chord are perpendicular.

Title: 29 April 2008 Ext 1 Maths

Date: 29 April 2008 12:30 PM

Category: School

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To recap on parabolas

Equation of tangent at P(2ap,ap2):

y = px - ap2

Equation of the chord of contact when tangents are drawn from the point x1,y1:

xx1 = 2a(y + y1)

Title: 5 May 2008 Ext 1 Maths

Date: 5 May 2008 10:03 PM

Category: School

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Finding the locus of a point to do with parabolas.

Find the co-ordinates of the point in terms of other variables then eliminate the variables down to x and y.

Title: 12 May 2008 Ext 1 Maths

Date: 12 May 2008 7:37 PM

Category: School

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MATHEMATICAL INDUCTION

To prove stuff by induction:

1. Prove that the equation holds for the lowest number in the sum (the number on the bottom of the sigma) (put the number under the sigma into the equation and show that you get the correct result)

2. Now substitute in the value of k and assume that sum for this number is true. (e.g you should find something like 1+2+3+4+...+2k = Something)

3. Prove that the sum also holds for k + 1 (now you will get something like 1+2+3+...+2(k+1) = somethingElse. The first bit can be changed into the thing in the above step so now you get Something+2(k+1) = somethingElse. Simplify)

Title: 13 May 2008 Ext 1 Maths

Date: 13 May 2008 6:33 PM

Category: School

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At the end of an induction thingy you must put at the end (substituting in pronumerals and values that you use):

Therefore true for n = k+1 Therefore If true for n=1 then also true for n=2,3 etc. all integer values n≥1

DIVISIBILITY TESTS

For example; prove that 34n - 1 is divisible by 80 for n≥1 and n is an integer

Use exactly the same steps: Prove true for n=1, assume true for n=k, prove true for n=k+1.

But for the assume n=k bit, do this:

34k - 1 = 80M (where M is an integer)

In n=k+1, substitute the 80M number in and factor out the 80 to show that the statement is divisible by that number.

Title: 26 May 2008 Ext 1 Maths

Date: 26 May 2008 7:01 PM

Category: School

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APPLICATIONS OF CALCULUS TO THE REAL WORLD

Every derivative describes the rate of change.

If you had a cube that had edges increasing at 0.01 mm per second then what is the rate of change of the volume?

dV/dt = dV/dx . dx/dt

where

V = x3

x = (0.01)t

use the derivatives to find dV/dt

Title: 2 June 2008 Ext 1 Maths

Date: 2 June 2008 7:06 PM

Category: School

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Further Growth and Decay:

Sometimes a population will grow towards an asymptote.

dN/dt = k(N-P)

N = P + Aekt

By differentiating the N= equation you get the thing above it. You can integrate the thing above it to get the N=

N is the population (or thing that is changing)

P is the asymptote or the number the population is approaching

A is a constant

k is a constant

t is the time (in any unit)

Title: 3 June 2008 Ext 1 Maths

Date: 3 June 2008 3:04 PM

Category: School

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Motion (I think this is a 2 unit concept)

The rate of change of displacement(x) with respect to time is a measure of velocity, or:

x = f(t)

dx/dt = f ’(t) = v = x’

The rate of change of velocity is called acceleration

dv/dt = v’(t) = a

Title: 11 June 2008 Ext 1 Maths

Date: 11 June 2008 8:19 PM

Category: School

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In projectile motion:

Equations of motion:

Horizontal movement in 2 dimensions:

acceleration = 0 (x-double-dot)

velocity = c (x-dot)(integral of zero is a constant)

velocity = vcosø

x-displacement = vtcosø

y-displacement = vtsinø - 1/2dt2

Title: 16 June 2008 Ext 1 Maths

Date: 16 June 2008 11:34 AM

Category: School

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Cartesian Equation of trajectory

Use x = vtcosø and y = vtsinø - 1/2gt2 as parametric definition of the line.

You get something like:

y = (tanø)x - (gsec2ø/2v2)x2

Greatest height occurs when y-dot=0

Time of flight = twice the time taken to reach biggest height.

Title: 23 June 2008 Ext 1 Maths

Date: 23 June 2008 11:38 AM

Category: School

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v = dx/dt = x-dot therefore ∫v dt = x

a = dv/dt = d2x/dt2 = x-doubledot therefore ∫a dt = v

a = v(dv/dx)

a = (d/dx) (1/2v2)

v2 = u2 + 2ax (you might recognise this from physics)

Title: 24 June 2008 Ext 1 Maths

Date: 24 June 2008 11:43 AM

Category: School

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__Simple Harmonic Motion__

At origin: v is maximum

a is minimum (o)

At extreme left and right: v is minimum (0)

a is maximum

__Acceleration is always in the opposite direction to the displacement__

x-doubledot = α-x

x-doubledot = -n2x where n2 is a constant

x = asin(nt + α) where is a α constant

Title: 22 July 2008 Ext 1 Maths

Date: 22 July 2008 3:15 PM

Category: School

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S.H.M.

Displacement of a particle is given by x = 2sin(3t-π/4)

This is simple harmonic because a = -n2x where n is a constant

Differentiate x twice and you will get x-doubledot = -9x

Amplitude is 2 because sin gives a value of 1 to -1 only.

The time of greatest displacement occurs when v=0 or x is maximum. So x-dot = 0 Solve.

Title: 23 July 2008 Ext 1 Maths

Date: 23 July 2008 3:19 PM

Category: School

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Stuff we missed.

If deg P(x) ≥ deg A(x)

P(x)/A(x) = Q(x) + R(x)/A(x)

arcsin(-x) = -arcsin(x) (Show that... (draw a graph))

arccos(-x) = π - arccos(x)

arctan(-x) = -arctan(x)

arcsin(x) + arccos(x) = π/2

lim (h -> 0) (sinh)/h = 1

Sketch y = x/(x2 +1)

Title: 28 July 2008 Ext 1 Maths

Date: 28 July 2008 7:04 PM

Category: School

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IT’S THE HALVING THE INTERVAL METHOD! (yay)

Find the root on a Polynomial P(x) which is between x=1 and x=2. Find it to one decimal place.

Halve 1 and 2 is 1.5. Test: does it work as a root to 1 decimal point?

Halve that interval again on the side that the root would be on. Continue etc etc.

Title: 29 July 2008 Ext 1 Maths

Date: 29 July 2008 11:33 AM

Category: School

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The formula for the angle between two lines is

tan-1

__|m__1

__-m__2|

|1+m1m2|

That is:

the tan inverse of the absolute value of ( (gradient one take away gradient two) divided by (one plus gradient one times gradient two) )