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1. 1. Formulae in other document but otherwise should be all covered...
2. 2. From additional mathematics for OCR book
3. 3. <ul><li>Linear expressions </li></ul><ul><li>Quadratic expressions </li></ul><ul><li>Completing the square </li></ul><ul><li>Simultaneous equations </li></ul>
4. 4. <ul><li>This is basically GCSE if not SAT revision </li></ul><ul><li>When simplifying remember to: </li></ul><ul><ul><li>Collect like terms </li></ul></ul><ul><ul><li>Remove brackets </li></ul></ul><ul><ul><li>Factorise </li></ul></ul><ul><ul><li>Find a common denominator when involving fractions </li></ul></ul><ul><li>When solving an equation remember to: </li></ul><ul><ul><li>Simplify </li></ul></ul><ul><ul><li>Do the same on both sides so that it remains the same equation </li></ul></ul><ul><li>They sometimes ask to rearrange an expression in which case be careful to do so correctly </li></ul>Ex 1A, B & C
5. 5. <ul><li>This is when the highest power is 2 </li></ul><ul><li>It often involves expanding and factorising </li></ul><ul><li>Before starting to solve a quadratic equation, make sure that all terms of the quadratic are on the left hand side of the equation </li></ul><ul><li>There are 3 ways to solve a quadratic equation: </li></ul><ul><ul><li>Factorise </li></ul></ul><ul><ul><li>Completing the square </li></ul></ul><ul><ul><li>Using the Quadratic Formula </li></ul></ul><ul><li>Remember that the formula is: </li></ul><ul><ul><ul><li>You use this when you are in a calculator test and cannot factorise </li></ul></ul></ul>Ex 1D
6. 6. <ul><li>It is used when you cannot factorise a quadratic </li></ul><ul><li>It is also useful when sketching a graph as it identifies the line of symmetry and the vertex </li></ul><ul><li>Method: </li></ul><ul><ul><li>X²-8x=-3* </li></ul></ul><ul><ul><ul><li>Take the coefficient of x-8 </li></ul></ul></ul><ul><ul><ul><li>Half it -4 </li></ul></ul></ul><ul><ul><ul><li>Square the answer +16 </li></ul></ul></ul><ul><ul><ul><li>Add this to both sides </li></ul></ul></ul><ul><ul><ul><li>Factorise where possible </li></ul></ul></ul><ul><ul><ul><li>Take the square root of both sides </li></ul></ul></ul><ul><ul><ul><li>Add the constant to both sides </li></ul></ul></ul><ul><ul><ul><li>Find the answer </li></ul></ul></ul><ul><li>Method: </li></ul><ul><ul><li>Y = x²-8x +3* </li></ul></ul><ul><ul><ul><li>This can be written as y=(x-4)²-13 </li></ul></ul></ul><ul><ul><ul><ul><li>See previous example and its answer </li></ul></ul></ul></ul><ul><ul><ul><li>Therefore the line of symmetry is x=4 and the vertex is (4,-13) </li></ul></ul></ul>Ex 1E
7. 7. <ul><li>This is when there is more than one variable </li></ul><ul><li>...by substitution </li></ul><ul><ul><li>Suitable for when y is the subject </li></ul></ul><ul><ul><li>Take the expression for y from the equation and substitute it in the other equation – then solve as before </li></ul></ul><ul><li>...by elimination </li></ul><ul><ul><li>Suitable for when y is not the subject or either equation </li></ul></ul><ul><ul><li>Multiply the equations as so that when they are subtracted/added from the other they eliminate variables </li></ul></ul><ul><ul><li>Substitute this into the first equation and thus solve </li></ul></ul>Ex 1F
8. 8. <ul><li>Linear inequalities </li></ul><ul><li>Quadratic in equalities </li></ul><ul><li>Algebraic fractions </li></ul><ul><li>Expressions containing a square root </li></ul>
9. 9. <ul><li>Like simplifying linear expressions, you do the same to both sides </li></ul><ul><li>However, remember to have the inequality sign the right way and whether it is equal to or not </li></ul><ul><li>You may also be asked to show the answer on a number line </li></ul><ul><li>In this case, remember that open circles at the end of the line show that the number is not included </li></ul><ul><li>Closed circles mean the figure is included in your answer </li></ul><ul><li>Both can be used in a single answer </li></ul>Ex 2A
10. 10. <ul><li>There are two methods: </li></ul><ul><ul><li>Sketching a graph to show the answer </li></ul></ul><ul><ul><li>Or drawing up a table showing the values of x </li></ul></ul><ul><li>But remember that if it has terms on both sides these must be collected to one side </li></ul><ul><li>These quadratic inequalities will be able to be factorised </li></ul><ul><li>Remember to be careful in reading and working the question especially when using a graph </li></ul>Ex 2B
11. 11. <ul><li>Algebraic fractions follow the same rules as the fractions in arithmetic </li></ul><ul><li>The common denominator should be the lowest common multiple of the original denominators </li></ul><ul><li>Other than being asked to simplify an algebraic fraction you may be asked to solve an equation involving fractions </li></ul><ul><li>This is done in the same way as before but also having to simplify fractions </li></ul><ul><li>Remember that when you multiply a fraction you only multiply its numerator </li></ul>Ex 2C & D
12. 12. <ul><li>It is often easier to use surds when working with square roots to get a more accurate answer than just working out the numerical value </li></ul><ul><li>You should try to make the number that is under the square root sign as small as possible or as easy to work with as possible </li></ul><ul><li>Rationalising the denominator is an important technique to be aware of </li></ul>Ex 2E
13. 14. <ul><li>Triangles </li></ul><ul><li>Sine rule </li></ul><ul><li>Cosine rule </li></ul>
14. 15. <ul><li>Just remember the hyp, app and adj. And that θ is used for the angle </li></ul><ul><li>The Trigonometric ratios are: </li></ul><ul><li>Remember: </li></ul><ul><ul><li>sin θ (etc.) will give you the ratio </li></ul></ul><ul><ul><li>sin known side (etc.) will give you the angle </li></ul></ul><ul><ul><li>side known X sin θ (etc.) will give you the side’s length </li></ul></ul>-1
15. 16. <ul><li>Sin is Opposite divided by Hypotenuse. </li></ul><ul><li>Opposite is a helpful way of remembering it. </li></ul>
16. 17. <ul><li>Tan is Opposite divided by Adjacent </li></ul><ul><li>An easy way to remember is it doesn’t have the hyp and opp is always on top </li></ul><ul><li>Opposite is a helpful way of remembering it. </li></ul>
17. 18. <ul><li>Cos is Adjacent divided by Hypotenuse </li></ul><ul><li>Opposite is a helpful way of remembering it. </li></ul>
18. 19. <ul><li>It is an extension of Pythagoras’ theorem which allows it to be applied to any triangle </li></ul>
19. 20. <ul><li>It is based on that fact that in any triangle the length of any edge is proportional to the sine of the angle opposite to that edge </li></ul>= =
20. 21. <ul><li>This formula (which is cyclic) is for finding the area of a triangle when the lengths of 2 edges are known and also the size of the angle between them </li></ul>= = =
21. 22. <ul><li>3d work </li></ul>
22. 23. <ul><li>A ll </li></ul><ul><li>S ilver </li></ul><ul><li>T ea </li></ul><ul><li>C ups </li></ul><ul><li>Anti-clockwise is always positive </li></ul><ul><li>Clockwise is negative </li></ul><ul><li>Always go from the x axis </li></ul><ul><li>Cosine and Sine are between -1 and 1 whereas Tangent is over 1 </li></ul>All Silver Tea Cups Sin Cos Tan Sin Cos Tan Sin Cos Tan Sin Cos Tan + + + - + - + - - - - + 1 1 -1 -1 It is like have a circle of one unit
23. 24. <ul><li>It is really important to draw good diagrams </li></ul><ul><li>There are two types: </li></ul><ul><ul><li>Representations of 3D objects </li></ul></ul><ul><ul><li>True shape diagrams of 2D sections in a 3D object </li></ul></ul>
24. 25. <ul><li>The two main identities that need to be learnt: </li></ul>
25. 26. <ul><li>Introduction – Curves, Tangents, and Normals </li></ul><ul><li>Gradient of a curve </li></ul><ul><li>Differentiation </li></ul><ul><li>Tangents and normals </li></ul><ul><li>Stationary points and Higher Derivatives </li></ul>
26. 27. <ul><li>Cord : joins two points on the curve </li></ul><ul><li>Tangent : touches the curve at a point of contact </li></ul><ul><li>Normal : perpendicular to the tangent at the point of contact </li></ul><ul><li>The tangent to a curve can be considered as the limit position of a chord </li></ul>Curved line
27. 28. <ul><li>As B gets closer to A we can say that B tends to A (written as B  A) </li></ul><ul><li>The gradient of the cord AB  the gradient of the tangent at A </li></ul><ul><li>E.g. y = x² </li></ul><ul><ul><li>A = (x A ,y A ) B = (x B , y B ) </li></ul></ul><ul><li>From the table, we can assume that the gradient of the tangent to the graph y = x² at A(2,4) is 4 </li></ul>Yb - Ya Xa Ya = (Xa)² Xb Yb = (Xb)² Mab = Xb - Xa 2 4 3.5 12.25 5.5 2 4 3 9 5 2 4 2.5 6.25 4.5 2 4 2.25 5.0625 4.25 2 4 2.1 4.41 4.1 2 4 2.05 4.2025 4.05 2 4 2.001 4.004001 4.001
28. 29. y = x ³ y = x ² y = x ⁴ X Gradient 1 2 2 4 3 6 4 8 5 10 x 2x X Gradient 1 4 2 32 3 108 4 256 5 500 x 4x³ X Gradient 1 3 2 12 3 27 4 48 5 75 x 3x²
29. 30. <ul><li>The value of the gradient of the chord AB as B tends to A is called the differential coefficient of y with respect to x or the derivative of y with respect to x . The limit is denoted by the symbol (read as ‘dy by dx’) </li></ul><ul><li>The process of obtaining the differential coefficient or derivative of a function is called differentiation . </li></ul><ul><li>Note that ‘d’ has no independent meaning and must never be regarded as a factor. The complete symbol means ‘the derivative with respect to x of [previous expression]’ </li></ul><ul><li>We may also write when y is a function of x as f’(x) or y’ </li></ul>dy dx dy dx d dx
30. 32. d dx d dx d dx d dx 1 2 √ x
31. 33. <ul><li>Let y=c </li></ul><ul><li>Graphically this is a horizontal straight line and its gradient is zero </li></ul><ul><li>Therefore differentiating a constant will give you zero </li></ul><ul><ul><li>i.e. (c) = 0 </li></ul></ul>y x y = c 0 d dx
32. 34. <ul><li>(axⁿ) = a (xⁿ) = anxⁿ⁻¹ </li></ul><ul><ul><li>Where ‘a’ is a constant </li></ul></ul><ul><ul><ul><li>i.e. (axⁿ) = anxⁿ⁻¹ </li></ul></ul></ul><ul><li>For example: </li></ul><ul><ul><li>(3x⁶) = 18x⁵ </li></ul></ul>d dx d dx d dx d dx
33. 35. <ul><li>We differentiate each term and then add or/and subtract the terms as necessary </li></ul><ul><li>For example: </li></ul><ul><ul><li>(x⁷ + 5x² - 3x + 4) </li></ul></ul><ul><ul><ul><li>= 7x⁶ + 10x - 3 </li></ul></ul></ul>d dx
34. 36. <ul><li>The gradient of the chord AB as it tends to the point A, is the value of the derivative at that point A. </li></ul><ul><li>We can use this to find the equation of the tangent and/or of the normal to a curve at a given point </li></ul>
35. 37. <ul><li>Q. Find the equation of the tangent and of the normal to the curve: </li></ul><ul><li>y = x² + 3x – 10 at the point (1, -6) </li></ul><ul><li>First differentiate the equation to give: </li></ul><ul><ul><li>= 2x + 3 at x = 1 </li></ul></ul><ul><li>Thus: </li></ul><ul><ul><li>m = 2 X 1 + 3 = 5 </li></ul></ul><ul><li>Using: y – y₁ = m(x – x₁)  substitute the known values </li></ul><ul><ul><li>y + 6 = 5(x – 1) </li></ul></ul><ul><ul><li>y = 5x – 5 – 6 </li></ul></ul><ul><ul><li>y = 5x – 11  equation of tangent </li></ul></ul><ul><li>Then to find the equation of the normal: </li></ul><ul><ul><li>m = 5 so m¹ = -⅕ </li></ul></ul><ul><ul><li>y + 6 = -⅕ (x-1)  use previous method but using -⅕ instead of 5 </li></ul></ul><ul><ul><li>5y + 30 = -x + 1 </li></ul></ul><ul><ul><li>x + 5y + 29 = 0  equation of the normal </li></ul></ul>d dx
36. 38. <ul><li>This is basically doing the second derivative </li></ul><ul><li>This is just differentiating what you already have differentiated </li></ul><ul><li>It can be used to find stationary points in increasing and decreasing functions </li></ul>
37. 39. <ul><li>Increasing is from A to B and from C – represented with the + </li></ul><ul><ul><li>This means that is positive </li></ul></ul><ul><li>Decreasing is from B to C – represented with the - </li></ul><ul><ul><li>This means that is negative </li></ul></ul><ul><li>Stationary point are A, B and C – represented by the zero </li></ul><ul><ul><li>This means that = 0 </li></ul></ul>y x A 0 B C + + + + + + + + + + - - - - - - - - 0 0 d dx d dx d dx
38. 40. <ul><li>This I where the gradient is zero </li></ul><ul><ul><li>They can be maximum points, minimum points, or points of inflection </li></ul></ul><ul><li>To find stationary points: </li></ul><ul><ul><li>Differentiate and find the value(s) of when this = 0 </li></ul></ul><ul><ul><li>Substitute these values into the original equation to find y </li></ul></ul><ul><ul><li>To find the nature of the stationary points work out the second derivative and then substitute the value(s) of x found before to decide if they are a min/max points or points of inflection </li></ul></ul>
39. 41. dy dx dy dx dy dx d²y dx² d²y dx² d²y dx² d³y dx³ = 0 = 0 = 0 = 0 < 0 > 0 ≠ 0 At point P Remember to physically do and say each step in a question including saying if a certain point is a max., min. or point of inflection. Maximum point Minimum point and (doesn’t change sign on either said of P) Point of inflection
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