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A Level Mathematics C3 and C4 Revision Notes

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TSR Wiki > Study Help > Subjects and Revision > Revision Notes > Mathematics > A Level Mathematics C3 and C4 Revision Notes

Some exam boards put certain topics in the Core 3 module, while others will have the same topics in the Core 4 module, and vice versa. We have therefore put the topics for Core 3 and 4 together.

Contents

Algebra, Series and Functions

Functions

Binomial Expansions

Coordinate Geometry

Trigonometry

Trigonometric identities to learn:

\mathrm{cosec} \theta = \frac{1}{\sin \theta}

\sec \theta = \frac{1}{\cos \theta}

\cot \theta = \frac{1}{\tan \theta} = \frac{\cos \theta}{\sin \theta}

\sin^2 \theta + \cos^2 \theta = 1

1 + \cot^2 \theta = \mathrm{cosec}^2 \theta

\tan^2 \theta + 1 = \sec^2 \theta

\sin 2 \theta = 2 \sin \theta \cos \theta

\cos 2 \theta = \cos^2 \theta - \sin^2 \theta = 2\cos^2 \theta - 1 = 1 - 2\sin^2 \theta

\tan 2 \theta = \frac{2 \tan \theta}{1 - \tan^2 \theta}

C3 Differentiaton

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 dy/dx= 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 (2x^3 - 1)^6 as an example:

First we take the power: 6.

Times this by the differential of the brackets: 6x^2.

Finally times by the original, to the power of 1 less: (2x^3 - 1)^5.

All those together:

6 \times 6x^2 x (2x^3 - 1)^5

36x^2(2x^3 - 1)^5

Easy, hm? if your struggling pm kevlar for more help!

The exam will give you all the usual graph questions:

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

Find the perpendicular line at the point... This means find the reciprical the dy/dx 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: To find \frac{dy}{dx}=uv we use the product rule which is uv'+vu'

Quotient Rule:

To find \frac{dy}{dx}=\frac{u}{v} we use the quotient rule which is \frac{dy}{dx}=\frac{vu'-uv'}{v^2}

NOTE: Quotient rule will be written in the AQA formulae book, but in the form g(x) and f(x) i've written both in there simplest forms for revision just remember prodUct U is cap as it starts with u then ends in u and remember the + sign, and quotient rule starts with v ends with v with a minus sign all over v^2

Examples

Product Rule : For example, find the dy/dx where y=x^4(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.

u = x^4 v = (2x + 1)^3 u'=4x^3 v'=2 \times 3(2x+1)^2 = 6(2x+1)^2 now we plug this into the formula

\frac{dy}{dx}=uv'+vu'

\frac{dy}{dx}=[x^4 \times 6(2x+1)^2] + [(2x+1)^3 \times 4x^3]

which simplifies to

4x^3(2x + 1)^3 + 6x^4(2x+1)^2


This factorises to 2x^3(2x + 1)^22(7x + 2) [This is the hardest part but don't worry if you can't do it in the exam as it's usually worth ONLY 1 mark]

Quotient rule

Integration

The following indefinite integrals follow naturally from the differentiation work

\\ \displaystyle \int e^{x} \, \mathrm{d}x = e^{x} + c \\ \\ \displaystyle \int \frac{f'(x)}{f(x)} \, \mathrm{d}x = \ln |f(x)| + c \\ \displaystyle \int \cos x \, \mathrm{d}x = \sin x + c \\

Numerical Methods

When using iterration to find x1, x2 etc...

Type the given value for x0 into your calculator and press '='

Then enter the iterrative formula into your calculator and for xn, simply put 'Ans'. Then to find the value for x1, just press '='. Then, to find x2, simply press '=' again without any further work. Likewise for x3, x4 etc...

Vectors

|(x\mathbf{i},y\mathbf{j},z\mathbf{k})| = \sqrt{x^2 + y^2 + z^2}

\mathbf{a}\cdot\mathbf{b} = |\mathbf{a}| |\mathbf{b}| \cos \theta

If two vectors are perpendicular then \mathbf{a}\cdot\mathbf{b} = 0

If two vectors are parallel then \mathbf{a}\cdot\mathbf{b} = |\mathbf{a}| |\mathbf{b}|

Two vectors are skew (cross on only 2 dimensions but not 3) when the three equations formed by the two lines, in terms of ks,js,is and kt,jt,it do not have common solutions of s and t for all three equations.

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