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Maths Core 3

Lots more graphs, trig and calculator questions in Core 3... But something nice to start as usual:

Algebraic Fractions

Simplifying Fractions

Back in the day when Maths was all numbers, you'll remember simplifying fractions.

E.g.

If you think about it, it can be written like this:

There's a 4 on the top and the bottom, and these can be cancelled (i.e. taking out the common factor) to leave 3/4.

In algebra, it's just the same. For example:

You need to remember the rules of factorising here. Notice the difference of two squares here, and taking an x out of the equation. The question will usually require you to factorise both top and bottom and then cancel any brackets that appear on top and bottom.

Multiplication and Division

Again, easy stuff. Remember to simplify the two fractions first, and then times the tops and bottoms.

E.g.

For division, you need to multiply the first fraction by the reciprocal of the second. (In other words, flip it on its head).

Adding and Subtracting

Things get a little more interesting when adding and subtracting in algebra. If you think back to numbers, you needed to find a common denometer.

E.g.

In algebra, it works just the same, though to find a common denometer, you multiply to two denometers. Then, to balance things, the numerators need multiplying by the corresponding denometer...

If you have a repeated root, you only use it once:

(x-3) appears in both these fractions, so when multiplying, there's no point in including it twice:

Remember to check to see if it will factorise:

Improper Fractions

Improper fractions occur when the numerator is larger than the denominator: 7/4, 16/11, etc.

In algebra, they occur when the highest power of x is equal or bigger on the numerator than the denometer:

What can we do with them? It's quite simple: divide.

You'll probably remember algebraic division from Core 2. This is one method, there is another.

We can divide by thinking about the remainder. Look at:

We could write this:

If we then multiply each side by (x + 1), we end up with:

If we put x = -1, we get:

1 + 1 - 7 = C
C = -5

Next, we can equate coefficients. Looking at x2, there's a 1 before it. On the other side of the equation, Ax times by x makes x2. If we ignore all the x and just looking at co-efficients, we get:

1 = A

Looking at x on the LHS, there's -1 before the x. On the RHS, Ax times 1 and B times x make x.

-1 = B + A
B = -2

Using these, we see that:

Questions get harder, but essentially, that's all there is to it!

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Functions

When we look at a graph, we see that thee y values are often connected to the x values by a formula... y=x2 for example.

A mapping links the values of one set of numbers (the domain) with another set of numbers (the range).

Example

b = a + 2

This type of mapping is called one to one mapping, because one domain number is turning into one range number.

y = x

This type of mapping is called one to many.

y = x2

This type of mapping is called many to one.

A final type is many to many. For example, a circle: x2 + y2 = 9. You can imagine how difficult it would be to draw out like the others though... very messy!

A function is a mapping where every element of the domain is mapped into just one range value.

In the four examples above, the functions are the one to one mapping and the many to one.

An easy way to tell if a mapping is function, is to draw a graph. If a vertical line crosses the graph just once, it's a function. For example:

This is a function.

ONE TO ONE mapping.

A function.

MANY TO ONE mapping.

This is not a function.

ONE TO MANY mapping.

Simple enough, hm? Well, they try to make it harder with strange notation. Learn the following:

                          

- Natural numbers: 0, 1, 2, 3...
- Integers numbers: 0, 1,2,3...
- Rational numbers: 2, , 1.23...
- Real numbers: All above plus π, 2
- Complex numbers: All real and all imaginary

Example

If f(x) = 2x + 1 with all real values of x, find f(3).

Pretty obviously, you substitute x for 3. Therefore: (2 x 3) + 1 = 7.

Find the range.

The range is all the possible y-values. Draw out this graph and you'll see a infinite straight line, and so therefore, the range is all real values of y.

If f(a) = 0, find a.

Put the equation equal to 0: 2a + 1 = 0.

A simple bit of algebra will rearrange to make a the subject of the formula:
a = 1/2.

Composite functions

These have more than one function.

  • fg(x) means do the "g" function first and then the "f" function.
  • gf(x) means do the "f" function first and then the "g" function.
  • ff(x) means do the "f" function and then the "f" function again.

Example:

f : x --> 2x
g : x --> x2 - 3

Find fg(2)

Stick x=2 into x2 - 3 = 1.
Stick x=1 into 2x = 2.

Find gf(2)

Stick x=2 into 2x = 4.
Stick x=4 into x2 - 3 = 13.

As you can see, changing around the functions produces very different answers!

Find fg(x)

Stick x=x into x2 - 3 = x2 - 3
Stick x=(x2 - 3) into 2x = 2(x2 - 3)

Find gf(x)

Stick x=x into 2x = 2x
Stick x=2x into x2 - 3 = (2x)2 - 3 = 4x2 - 3

A question might also ask you the opposite of this, giving you an end equation and asking you to find the order of the functions involved.

For example:

p(x) = x - 5
q(x) = x2 + 1
r(x) =

Find in terms of p, q and r the functions:

Think of x. What do you need to do to change it into the above? Well, first step is take away 5. That matches p(x). Then, we need the reciprocal, so r(x).
Put it together and we have: rp(x).

This doesn't look much like any of the functions, so we need to complete the square:

(x - 5)2 + 26 - 25

(x - 5)2 + 1

Now we can see that x needs -5, and so p(x) and then everything is squared and + 1, so q(x).
Put them together to get: qp(x).

Inverse functions

These only exist if the original function is one-to-one. However, a domain can be limited to make a function one-to-one.

f(x) is a function. f'(x) is its inverse.

In order to find f'(x), we make x the subject of the formula.

For example:

f(x) = 3x - 1

Say f(x) is y, to make it look easier:

y = 3x - 1

Now rearrange to make x the subject of the formula:

y + 1 = 3x

x =

Next, we replace x with f'(x) and y with x. Therefore:

f'(x) =

If we draw out a function and it's inverse function, you might notice something interesting...

f(x) = x2 when x > 0 (limiting domain to make it one-on-one).

f'(x) = +x

As you can see, the inverse function is symmetrical to the function in the line y=x.

The is true for all functions and inverse functions: there is reflection in the y=x line.

Another thing to realise is that the range of an inverse function is the same as the domain of a function. And, vice versa, the domain of an inverse function is the same as the range of the original function.

And that's all you need to know for functions!

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Exponential & Logarithmic Functions

There exists a number between 2 and 3 called "e" such that y=ex is a graph where y= at all points.

"e" is an irrational number = 2.7182818...

ex is called the exponential function and graphed out is:

As with any graph, ex can be translated, stretched, reflected... etc.

Question: The price of a computer varies according to P = 100 + 900e-t/2 where P is a value in £ and t is time in years.

What is its value when new?

When t=0, e0 = 1, and therefore the price is £1000.

Find its value 2 years after purchase.

Simply sub in t=2, 100 + 900e-1 = £431.09.

Sketch the graph

We know that when x=0, y=1000 and P ---> £100 when t ---> as e-t/2 ----> 0. The graph is reflected in x-axis due to -t/2.

The inverse of ex is logex, more commonly known as lnx.

As with all inverse graphs, lnx is a reflection of y=ex in y=x and the domain is the range, and vice versa.

You need to be able to switch between lnx and ex in order to solve equations. A few examples to show how it works:

Solve:

e(x-3) = 2
x - 3 = ln2
x = ln2 + 3 (exact answer)
x = 3.69 (3 s.f.)

Solve:

3lnx - 1 = 4
3lnx = 5
lnx = 5/3
x = e5/3 (exact)
x = 5.29 (3 s.f.)

That's really all these is to these two functions! It may be short, but they appear from now on... so make sure you understand them!

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Numerical Methods

If we solve f(x) for something x2 - 5x + 6 = 0, we get x=3 and x=2. On a graph it would look like:

Notice how there's a change of sign on either side of each root? Using this idea, we can calculate where roots lie.

Show that f(x) = ex - x - 3 has a root between 2 and 3, and another root near -3.

Taking 2 and 3, we can say:

f(2) = e2 - 2 - 3 = -1.28. This is less than 0. <0.
f(3) = e3 - 3 -3 = +2.39. This is greater than 0. >0.

Because the graph is continuous and there is a change of sign, we know that there is a root.

If there's a root near -3, we know the max range is -3.5 to -2.5. We do exactly the same using these figures:

f(-3.5) = e-3.5 + 3.5 - 3 = +0.53 > 0.
f(-2.5) = e-2.5 + 2.5 - 3 = -0.43 < 0.

There is a change of sign and a continuous graph, so again, there must be a root near -3.

Iterative Methods

The question might ask you to rearrange the equation into another form.

Example: Rearrange x3 + 4x - 3 = 0 into the form x = .

x3 + 4x - 3 = 0
4x = 3 - x3
x =

Next, it might ask you to use the iterative formula xn+1 = to find roots to a certain level of accuracy (we'll do 2 d.p.), starting with x0 = 1.

So x0 = 1 is subbed into the formula to get x1, x1 is subbed in to get x2, etc...

x0 = 1
x1 = (3 - 1) 4 = 0.5
x2 = (3 - 0.5) 4 = 0.71875
x3 = (3 - 0.718755) 4 = 0.65717
x4 = 0.67904
x5 = 0.6717
x6 = 0.6742

When doing this, you must use at least 3 d.p. and when the numbers start to converge, so that rounded to 2 d.p. we get the same value, this is our answer. Here it is: 0.67.

To prove this, we use the earlier proof with the original equation:

f(0.665) = 0.6653 + 4x0.665 - 3 = +0.00754 > 0.
f(0.675) = 0.6753 + 4x0.675 - 3 = -0.046 < 0.

There's a change of sign and it's a continuous graph (draw it on your calculator to check that) and so, we know there is a root at 0.67, correct to 2 d.p.

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Transforming Graphs

T...transforming graphs? Didn't we do this in Core 1... even at GCSE? Well, yes, but there are some new rules now: modulus or absolute value.

|5| = 5
|-5| = 5

A modulus makes the numerical values all positive. In graphs:

|f(x)| = f(x) for f(x) ≥ 0
|f(x)| = -f(x) for f(x) ≤ 0

To draw a |f(x)| we draw the normal graph first, and then anything below the x-axis is reflected in the x-axis.

For example: y=|x|

First, draw y=x:
Next, reflect anything below the x-axis to get y=|x|:

For a random y=|f(x)| that an exam gives, it's exactly the same:

 
Reflect anything below the x-axis to get y=|f(x)|:

If, however, we have y=f|(x)|, the same rule can't apply. Because the modulus makes anything positive, both f|(2)| and f|(-2)| have the same value, for example. For these, we sketch x≥0 and then reflect in the y-axis.

Example:

y = 2|x| - |x|2

y = 2x - x2 looks like this:
Therefore, y = 2|x| - |x|2 would look like:

Example: Solve |x-1| = 2

Solving equations that involve modulus is like solving equations for any other graph. First, we draw out the two graphs, y = 2 and y = |x-1| and see where they intersect:

There is an intersection on the normal y=x-1 line, so we can say:

x-1 = 2
x=3

There is also an intersection on the reflection, so:

-(x-1) = 2
-x + 1 = 2
x = -1

The other rules

To remind you:

  1. -f(x) is a reflection in the x-axis.
  2. f(-x) is a reflection in the y-axis.
  3. af(x) is a stretch scale factor "a" in the y-direction.
  4. f() is a stretch scale factor "a" in the x-direction.
  5. f(x + a) is a translation of -a in the x-direction.
  6. f(x) + a is a translation of a in the y-direction.

Using these, we can manipulate coordinates (a lot easier than drawing graphs, in my opinion!)

Example:

y = f(x) has coordinates A (-1,4), B (0,3) and C (-2,0). Find the corresponding coordinates for y = 3f(x-2)

For this, we need to times y by 3 and add 2 to x. So:

A = (1,12)
B = (2,9)
C = (0,0)

Find y = -f(x) + 4.

Change the sign for y and and 4 to y:

A = (-1,0)
B = (0,1)
C = (-2,4)

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Trigonometry

Remember basic trig from a triangle, back at GCSE? ...silly old Harry... SOH CAH TOA...

sinΘ =

cosΘ =

tanΘ =

We know what sin(e), cos(ine) and tan(gent) are, but there are some new trig identities...

sec x =
cosec x =
cot x =

Graphs:

For example:

sec 30 = 1cos 30 = 1 =

cot 220 = 1tan 220 = 1.192

We know that tan x = and we know now that cot x = . Subbing the first into the second, we can say:

cot x =

Armed with these new identities, we can simplify a whole lot of trig or prove that one thing is the same as another.

Example: Prove that tan x + cot x sec xcosec x

We only work with one side of the equation and state which side we're going to use. E.g. LHS:

tan x + cot x

Now we use our trig identities and see what we can get:

+

Adding these together we get:

Now we think, "Oh, I recognise the top line!" because sin2x + cos2x 1. Replacing the top line, we get:

Almost there now; you'll see that if we covert this back to cosecs and secs, we get the RHS:

sec xcosec x as required

Find x in radians to 2d.p. for 0 ≤ x ≤ 2π when cosec x = 2.1.

We know cosec x is , so:

= 2.1
sin x =
x = 0.50c

Then using the grid:

x = 0.50c or 2.65c

Sec, cosec and cot also open up a load more identities by thinking of the identity:

sin2x + cos2x = 1

If we divide everything by cos2x, we don't change anything, but we get the identity:

tan2x + 1 = sec2x

Similarly, if we divide by sin2x, we get:

1 + cot2x = cosec2x

With these, we can change and simplify a lot more trig.

Example: Simplify (sec2x - 1)cot x.

Notice the sec2x... this means we need the identity: tan2x + 1 = sec2x.
If we rearrange it, we'll get: tan2x = sec2x - 1, like in the question. Substitute in to get

tan2x.cot x

cot x = , so one of the tans on the top will be cancelled out, leaving:

tan x

Using tan x = , find cos x and sin x.

The simplest thing to do here, is think about the triangle that tan comes from: TOA. We draw the triangle and label 3 and 4 as opposite and adjacent:

From this, using Pythagoras, we can see that H=5.

Next, we draw the grid to see where the positive tan angle is:

We want cos x, which is CAH, but remember, in the tan quadrant, cos is negative, so:

cos x = -4/5

The same with sin: SOH and negative again:

sin x = -3/5

Solve tan 2x - 2secx + 1 = 0 in the range -π ≤ x ≤ π.

Right, so we have a tan2x, so we'll need to change it using the identity: tan2x + 1 = sec2x:

sec2x - 2secx = 0

Taking out a sec, and converting it to s, we get:

( - 2) = 0

We have two equations that we solve separately, remembering the grid for cos:

= 0
cos x = 0

No solutions in range.

- 2 = 0
= 2
cos x =

- or

Inverse Trig functions

As with all inverse functions, the inverse trig are a reflection in y=x. Because inverse can only occur in one-to-one functions, we need to restrict the domain... it must also be in radians.

For sin x, the inverse is sin-1x or arcsin x, and the domain is -1 ≤ x ≤ 1:

For arccos x, the cos x graph is restricted to 0 ≤ x ≤ π:

For arctan x, the domain of tan x is limited to - x ≤ :

Example:

Find: arcsin(sin)

= arcsin
=

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Further Trig

Even more trig? No wonder they say Core 3 is a lot of Trig...

The next bunch of identities are called the addition formulae. Nicely, they're given in the formula booklet, so we don't have to remember them... just as long as we know how to use them.

sin(A + B) = sinAcosB + cosAsinB
sin(A - B) = sinAcosB - cosAsinB

cos(A + B) = cosAcosB - sinAsinB
cos(A - B) = cosAcosB + sinAsinB

tan(A + B) =
tan(A - B) =

Example: Prove tanA + tanB sin(A+B)secAsecB

As usual with proofs, we state which side we'll be working with: LHS.

tanA + tanB

+

sin(A + B) x x

sec(A + B)secAsecB as required.

Double Angles

These are special cases:

sin2A = 2sinAcosA

cos2A = cos2A - sin2A
cos2A = 1 + 2cos2A
cos2A = 1 - 2sin2A

We can prove them by thinking about the addition formulae:

sin2A sin(A + A)
sin(A + A) = sinAcosA + cosAsinA
sinAcosA + cosAsinA = 2sinAcosA

cos2A cos(A + A)
cos(A + A) = cosAcosA - sinAsinA
cosAcosA - sinAsinA = cos2A - sin2A

From cos2A - sin2A, if we think about sin2x + cos2x = 1, we can substitute in to get 1 - 2sin2A or 1 + 2cos2A.

Example: Solve the equation sin2x = sinx in the range 0 ≤ x ≤ 2π.

Replace sin2x with 2sinxcox:

2sinxcosx = sin x

Subtract and then factorise out sin x:

2sinxcosx - sin x = 0
sinx(2cosx - 1) = 0

sin x = 0
x = 0c, π, 2π
2cosx - 1 = 0
cos x =
x = ,

Sum and Product Formulae

From the addition rules, we can say that:

sin(A + B) + sin(A - B) = 2sinAcosB
sin(A + B) - sin(A - B) = 2cosAsinB

Now if we let A + B = P
and let A - B = Q
we get:

sinP + sin Q = 2sincos

sinP - sin Q = 2sincos

Similarly, we get:

cosP + cosQ = 2coscos
cosP - cosQ = -2sinsin

Notice the minus sign on the last cos formulae -- be careful! These are also kindly provided in the formula booklet.

Example: Factorise and hence evaluate: sin75o - sin15o.

Looking at the formula, it clearly needs: sinP - sin Q = 2sincos.

sin75 - sin15 = 2sincos
2sin60cos45
2 x x
=

A final type of trig question (yes, I really said final!) might ask you to arrange some trig in the form Rsin(x α) or Rcos(x α).

Example. Find R and α when 3cosx + 4sinx Rcos(x - α)

First off, we expand the cos(x - α) using the addition rules:

3cosx + 4sinx Rcosxcosα + Rsinxsinα

Now we equate up coefficients for both cosx and sinx, we get:

3 = Rcosα
4 = Rsinα

If we divide the bottom by the top, we get:

tanα = 4/3
α = 51.3o (3 s.f.)

If we square 3 = Rcosα and 4 = Rsinα and then add them, we get:

32 + 42 = R2cos2α + R2sin2α

Next, factorise out R2:

25 = R2(cos2α + sin2α)

We think: "Ah ha! cos2α + sin2α = 1, so we replace that to get:

25 = R2
R = 5

Now we have R and α, we can write them in:

3cosx + 4sinx 5cos(x - 53.1o)

That's all there is to it! You need to learn this method, however, because it's not in the formula booklet...

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Differentiation

Differentiation was introduced back in Core 1... Basically all that's changed is that we can differentiate harder algebra.

To differentiate, you need to use this form:

y = [f(x)]n

= nf'(x)[f(x)]n-1

It may look kind of scary, but it's much cleaner that using the chain rule, another way. What the algebra means is: the power (n) times the differential (f'(x)) times the original to the power of 1 less ([f(x)]n-1).

So, let's take (2x3 - 1)6 as an example:

First we take the power: 6.

Times this by the differential of the brackets: 6x2.

Finally times by the original, to the power of 1 less: (2x3 - 1)5.

All those together:

6 x 6x2 x (2x3 - 1)5

36x2(2x3 - 1)5

Easy, hm?

The exam will give you all the usual graph questions:

Find the gradient of the curve at the point... this means, find the and sub in the coordinates.

Find the perpendicular line at the point... This means find the resiprical the number and times by -1. Finally use the formula: y - y1 = m(x - x1) where x1 and y1 are the coorinates of the point.

Product and Quotient Rules

You need to learn these. They're not given in the formula booklet:

Product Rule:

= v + u

Quotient Rule:

=

For example, find the of y=x4(2x + 1)3.

First, we notice two things times together. We can't practically expand the brackets, however, so we need the product rule. We call the first thing "u" and the second "v" and differentiate them separately, and then stick them into the product rule.

y=x4(2x + 1)3

= v + u

= (2x + 1)3.4x3 + x4.6(2x+1)2

= 4x3(2x + 1)3 + 6x4(2x+1)2

Factorising this, we get:

= 2x3(2x + 1)2(7x + 2)

u = x4 v = (2x + 1)3
= 4x3 = 3x2x(2x+1)2
=6(2x+1)2

The hardest part is probably the factorising at at the end.

Example: Differentiate .

This one requires the quotient because the bottom line (2x2 - 1)-1 cannot be differentiated by any other method. It begins in the same way as the product rule, finding and :

=

=

=

u = 3x v = 2x2 - 1
= 3 = 4x

If we needed to differentiate again, say to find a max or minimum point, we would use the product rule. This is because it can be written as: (-6x2 - 3)(2x2 - 1)-2. The -2 means it can be differentiated normally.

Log and Exponential Functions

For ex, we have another differentiation rule that you'll need to know:

y = ef(x)

= f'(x)ef(x)

It's identical to the original, but times by the differential of x...

Example: y = e2x2

= 4x(e2x2)

Ln x:

y = ln f(x)

=

For any form of lnx, the differential of the x function is divided by the original x function.

Example: y = 3ln3x

=

These, like anything else, can also be used in the product or quotient rules.

Trig Functions

These only apply when in radians. Most of them are found in the formula booklet, but the two most important need to be learned:

y = sin x
= cos x

y = sin f(x)
= f'(x)cosf(x)

y = cos x
= - sin x

y = cos f(x)
= - f'(x)sin(fx)

For example:

y = sin6x

The differential of 6x is 6, so: 6xsin6x.

= 6sin6x

Another example:

y = cos4x

This is the same as (cos x)4 and so, we need to find the power: 4.

Times it by the differential of cos x = - sin x.

And finally times by the original to 1 less power: (cos x)3.

All together, this is: 4 x - sinx x (cos x)3:

= - 4sinx(cosx)3

From the formula booklet, we find the other trig and their equivalent differentiations:

f(x)
f'(x)
tan kx
k sec2 kx
sec x
sec x tan x
cot x
-cosec2x
cosec x
-cosec x cot x

All these can be proved by converting back into sin and cos, and then differentiating. For example, Prove tan x differentiates to sec2x.

y = tan x

y =

From this, we see we need the quotient rule:

y =

=

=
(remember cos2x + sin2x = 1)

=

= sec2x as required.

u = sin x v = cos x
= cos x = - sin x

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And with that, Core 3 is finished! Take a deep breath before plunging into Core 4...