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Maths Revision - GCSE And Additional Notes

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Maths Revision - GCSE And Additional Notes

1. 1. Straight from My Revision Book By the way, this is probably repeating the other one a lot (Special thanks to Moji who is a star  )
2. 2. <ul><li>Transformations of Graphs </li></ul><ul><li>Congruency </li></ul><ul><li>Completing the square </li></ul><ul><li>Calculus </li></ul><ul><ul><li>Differentiation </li></ul></ul><ul><ul><li>Integrals </li></ul></ul><ul><ul><li>Finding the constant </li></ul></ul><ul><ul><li>Stationary points </li></ul></ul><ul><ul><li>Motion with Variable Acceleration </li></ul></ul><ul><li>Binomial </li></ul><ul><ul><li>Distribution </li></ul></ul><ul><ul><li>Expansion </li></ul></ul><ul><li>The Discriminate </li></ul><ul><ul><li>What is the Discriminate </li></ul></ul><ul><ul><li>Using the Discriminate </li></ul></ul><ul><ul><li>Finding Minimum or Maximum Value </li></ul></ul>
3. 4. <ul><li>When we have to stretch a line along the x-axis, we have to stretch to ¹⁄ n where ‘n’ is the number given </li></ul><ul><ul><li>For example: </li></ul></ul><ul><ul><ul><li>y=sin2x becomes y=sin½x </li></ul></ul></ul><ul><li>When stretching/compressing a line along the y-axis, the stretch is to the number given </li></ul><ul><ul><li>For example </li></ul></ul><ul><ul><ul><li>y=3sinx becomes y=3sinx </li></ul></ul></ul><ul><li>For example: </li></ul><ul><ul><li>Curve y=x² translated to give y=f(x) </li></ul></ul><ul><li>Move right first f(x-2) </li></ul><ul><ul><li>It is -2 because x is squared in the question and moved to the right </li></ul></ul><ul><li>Move down -4 </li></ul><ul><li>Therefore: y=(x-2)²-4 </li></ul><ul><li>Again it is squared because the question has x² </li></ul>y x 0 x A (2,-4) y=x²
4. 5. <ul><li>Similarity : when the angles are the same but the lengths are different </li></ul><ul><li>Triangles are congruent if: </li></ul><ul><ul><li>3 sides are the same (SSS) </li></ul></ul><ul><ul><li>2sides + included angle are the same (SAS) </li></ul></ul><ul><ul><li>2angles + 1side are the same (ASA or AAS) </li></ul></ul><ul><ul><li>In a right angled triangle, the hypotenuse +1side are the same (RHS) </li></ul></ul><ul><li>You will always be given the graph and an equation </li></ul><ul><li>For example (p408 ex17g) </li></ul><ul><ul><li>2a) x²+4x-5  first draw graph in book </li></ul></ul><ul><ul><li>To solve the equation make a grid </li></ul></ul><ul><ul><ul><ul><ul><li>=0 therefore y=0 when x=-5 and 1 </li></ul></ul></ul></ul></ul><ul><li>Note : if this doesn’t work y=0 </li></ul>x -6 -5 -4 -3 -2 -1 0 1 2 x² 36 25 16 9 4 1 0 1 4 +4x -24 -20 -16 -12 -8 -4 0 4 8 -5 -5 -5 -5 -5 -5 -5 -5 -5 -5 y 7 0 -5 -8 -9 -8 -5 0 7
5. 6. <ul><li>When you are told to complete the square: </li></ul><ul><ul><li>Take the co-efficient of x at be, for example in: </li></ul></ul><ul><ul><li> a b c </li></ul></ul><ul><ul><ul><li>x² + 6x + 1 </li></ul></ul></ul><ul><ul><li>Half this and then square it. Both add and subtract it from the equation (so the equation doesn’t change) like so: </li></ul></ul><ul><ul><ul><li>x² + 6x + (6/2)² – (6/2)² + 1 </li></ul></ul></ul><ul><ul><li>This can be factorised into: </li></ul></ul><ul><ul><ul><li>(x+3)² + 8 </li></ul></ul></ul><ul><ul><li>To check you can expand to get: </li></ul></ul><ul><ul><ul><li>x² +6x + 9 + (-9 +1) </li></ul></ul></ul><ul><ul><li>Remember: that if it is 4x² + ... it has to be factorised into 4(x + ...)² rather than (4x + ...)² </li></ul></ul>
6. 8. <ul><li>Differentiation </li></ul><ul><li>Integrals </li></ul><ul><li>Finding the constant </li></ul><ul><li>Stationary points </li></ul><ul><li>Kinematics </li></ul>
7. 9. <ul><li>To differentiate: </li></ul><ul><ul><li>E.g. y=3x⁵+4x⁴-3x²+2 </li></ul></ul><ul><li>For long equations, differentiate in parts and then add/subtract as needed </li></ul><ul><ul><li>1 st – (3x⁵)  multiply the power by the multiple of x  =15x </li></ul></ul><ul><ul><li>Then minus 1 from the power  =15x⁴ </li></ul></ul><ul><ul><li>Repeat for all parts, remembering the signs: </li></ul></ul><ul><ul><ul><li> = 15x⁴+16x³ - 6x [+2] ...(continued) </li></ul></ul></ul>d dx Multiple of x Power
8. 10. <ul><li>For a whole number i.e. The ‘2’, it will =0 </li></ul><ul><li>Therefore the answer is: </li></ul><ul><ul><li> 15x⁴+16x³ - 6x </li></ul></ul><ul><li>Remember: </li></ul><ul><ul><li>For things like 5x – they become 5 </li></ul></ul><ul><ul><li>x²¯¹ = x not x¹ </li></ul></ul><ul><ul><li>x⁽¹⁾¯¹ = x⁰ = 1 </li></ul></ul><ul><ul><li>x¯¹ = ¹ ⁄ x </li></ul></ul><ul><li>...(continued) </li></ul>
9. 11. <ul><li>The second derivative tells us if a point is a maximum point, minimum point or a possible point of inflection </li></ul><ul><li>This is just differentiating what you already have differentiated </li></ul><ul><li>The third derivative tells us if a point is a point of inflection if it does not equal zero. If it does equal zero it is not possible at this level. </li></ul><ul><ul><li>It shouldn’t equal zero but if it does just remember to put not possible at this level as there may be a mistake in the question </li></ul></ul>
10. 12. <ul><li>Integration is the inverse process of differentiation </li></ul><ul><li>The symbol for this is ‘ ∫ ’ </li></ul><ul><li>To integrate: </li></ul><ul><ul><li>E.g. y=3x⁵+4x⁴-3x²+2 </li></ul></ul><ul><li>Do it in parts: </li></ul><ul><ul><li>∫ ( 3x⁵+4x⁴-3x²+2) </li></ul></ul><ul><ul><li>Take the multiple, like 3, and divide it by the power + 1 </li></ul></ul><ul><ul><li>In this case you would get 3/6. You then take the x, x⁵ here, and add 1 to the power again, x⁶ </li></ul></ul><ul><ul><li>You place this over the multiple in the fraction to give you: [(3x⁶)/6] </li></ul></ul><ul><ul><li>Continue like this, adding and subtracting according to the equation given </li></ul></ul><ul><ul><li>For a constant, like 2, simply at x, thus 2 = 2x </li></ul></ul><ul><ul><li>Then simplify if possible </li></ul></ul><ul><ul><ul><li>The example above becomes 2x⁶ </li></ul></ul></ul><ul><ul><li>Then add ‘c’ where c is the constant </li></ul></ul>
11. 13. <ul><li>There are two formulae that need to be learnt: </li></ul><ul><li>Note: Finding the original function given its derivative is the same as finding the integral </li></ul>1. 2. Where ‘a’ is a constant
12. 14. <ul><li>For definite integrals, numbers are given </li></ul><ul><ul><li>E.g. </li></ul></ul><ul><li>You must integrate as normal </li></ul><ul><li>Then you take the number for the upper limit and substitute it in for x, x=3 </li></ul><ul><li>Do the same for the lower limit, x=1, and subtract this from the previous value. Giving: </li></ul><ul><li>Note: As ‘c’ vanishes when you work the previous expression, it is common to replace ‘c’ altogether and write: </li></ul><ul><li>It can also be found using the formula: </li></ul><ul><li>This can be used to find areas of curves </li></ul>Upper limit Lower limit
13. 15. <ul><li>To find the constant, you will be given co-ordinates which then can be substituted into an expression which will be used to find the constant </li></ul><ul><ul><li>For example: </li></ul></ul>Integrate Substitute
14. 16. <ul><li>DON’T FORGET TO PUT IT INTO THE PREVIOUS EXPRESSION! </li></ul><ul><ul><li>So the answer is: </li></ul></ul><ul><li>You could lose marks for leaving it just like that! </li></ul>
15. 17. <ul><li>Explain why is a possible expression for the gradient of the curve and give an alternative expression for </li></ul><ul><ul><li>For the first part: </li></ul></ul><ul><ul><ul><ul><li>because </li></ul></ul></ul></ul><ul><ul><ul><ul><li>So x=0 or x=2  making them stationary points thus a possible expression </li></ul></ul></ul></ul><ul><ul><li>For the second part: </li></ul></ul><ul><ul><ul><li>Where ‘k’ is a constant which is an alternative expression </li></ul></ul></ul><ul><li>The curve passes through the point (3,2). Taking as , find the equation of the curve </li></ul><ul><ul><li>Find the integral: </li></ul></ul><ul><ul><li>Find the constant: </li></ul></ul><ul><ul><li>Substitute this into the equation from the integration: </li></ul></ul><ul><ul><ul><li>Or </li></ul></ul></ul>
16. 18. <ul><li>The 2 nd derivative tells us if a point is a maximum, a minimum, or a possible point of inflection </li></ul><ul><li>The 3 rd derivative tells us if a point is a point of inflection if it doesn’t equal zero </li></ul><ul><li>If it does equal zero it is not possible at this level </li></ul><ul><li>However this should only occur if there was a mistake in workings or the question </li></ul>
17. 19. <ul><li>For </li></ul><ul><li>Find the first derivative </li></ul><ul><ul><li>So, or </li></ul></ul><ul><li>Then find the 2 nd derivative </li></ul><ul><li>Substitute one possible value for ‘x’ into this equation </li></ul><ul><ul><li>So at x=2 a minimum </li></ul></ul><ul><ul><li>So at x=-1 a maximum </li></ul></ul><ul><li>You will often be asked to find the possible points (co-ordinates) where the stationary points are. To do this, simply substitute the possible values of ‘x’ into the equation given </li></ul><ul><li>Therefore: </li></ul><ul><ul><li>At x=2 is a minimum </li></ul></ul><ul><ul><li>At x=-1 is a maximum </li></ul></ul><ul><li>It is important to say that the co-ordinate is a max/min/point of inflection </li></ul>
18. 20. <ul><li>Key: </li></ul><ul><ul><li>v = velocity </li></ul></ul><ul><ul><li>u = initial velocity </li></ul></ul><ul><ul><li>a = acceleration </li></ul></ul><ul><ul><li>t = time </li></ul></ul><ul><ul><li>s = displacement </li></ul></ul><ul><li>Equations: </li></ul><ul><ul><li>v = u + at </li></ul></ul><ul><ul><li>s = [(u + v)/2] X t </li></ul></ul><ul><ul><li>v² = u² + 2as </li></ul></ul><ul><li>To find an equation for v you can do the derivative of s (ds/dt) </li></ul><ul><li>To find an equation for a you can do the derivative of v (dv/dt) </li></ul><ul><li>Example: </li></ul><ul><ul><li>Find an equation for v with the following equation </li></ul></ul><ul><ul><li>s = 5t² - t + 3 </li></ul></ul><ul><ul><li>ds/dt = 10t – 1 </li></ul></ul><ul><ul><li>Thus v = 10t – 1 </li></ul></ul><ul><ul><li>Find an equation for a with answer above </li></ul></ul><ul><ul><li>dv/dt = 10 so a = 10 </li></ul></ul><ul><li>Similar to doing the 2 nd derivative </li></ul>
19. 21. <ul><li>v = ds/dt  s = ∫ v dt </li></ul><ul><li>a = dv/dt  v = ∫ a dt </li></ul><ul><li>Area under the line which also given the velocity in a time-distance graph </li></ul><ul><li>Example: </li></ul><ul><ul><li>a = 2 – 6t (find v) </li></ul></ul><ul><ul><li>v = ∫ a dt </li></ul></ul><ul><ul><li>∫ (2-6t) = 2t – [(6t²)/2] + c </li></ul></ul><ul><ul><li>v = 2t – 3t² + c </li></ul></ul><ul><ul><ul><li>At t = 0; v = 1 </li></ul></ul></ul><ul><ul><li>1 = 0 – 0 + c </li></ul></ul><ul><ul><li>c = 1 </li></ul></ul><ul><ul><li>v = 2 – 3t² + 1 </li></ul></ul><ul><ul><li>s = ∫ v dt </li></ul></ul><ul><ul><li>∫ (2 – 3t² + 1 ) </li></ul></ul><ul><ul><ul><li>At t = 0; and s = 0 </li></ul></ul></ul><ul><ul><li>= [(2t²)/2] – [(3t²)/3] + t </li></ul></ul><ul><ul><li>s = t² - t³ + t + c </li></ul></ul><ul><ul><li>0 = 0 – 0 + 0 + c </li></ul></ul><ul><ul><li>c = 0 </li></ul></ul>
20. 22. <ul><li>Distribution </li></ul><ul><li>Expansion </li></ul><ul><ul><li>Pascal’s Triangle </li></ul></ul>
21. 23. <ul><li>Two mutually exclusive events (meaning that if something happens, something else cannot happen) </li></ul><ul><li>If a trial is conducted ‘n’ times, </li></ul><ul><ul><li>‘ P’ is the probability of ‘success’ in every trial </li></ul></ul><ul><ul><li>‘ q’ is the probability of ‘failure’ </li></ul></ul><ul><ul><ul><li>Note: </li></ul></ul></ul><ul><li>The probability of exactly ‘r’ successes is: </li></ul><ul><ul><li>Where ‘x’ is number of success and r=0,1,2... </li></ul></ul>
22. 24. <ul><li>Remember: </li></ul><ul><ul><li>Indices of a term = the sum of indices of expression </li></ul></ul><ul><ul><ul><li>For example: </li></ul></ul></ul><ul><li>However, Pascal’s Triangle is long to use so the button on the calculator is used instead </li></ul><ul><li>So, if we want to work out the co-efficient of in the expression , we need </li></ul><ul><li>Another way is: </li></ul><ul><li>In general this is: </li></ul><ul><li>Remember: </li></ul><ul><li>Using the co-efficients of each expansion we can get Pascal’s Triangle </li></ul><ul><li>It’s easy to work out as you just add the previous numbers </li></ul><ul><ul><li>For example: 1 2 1 1 3 3 1 </li></ul></ul><ul><li>The triangle is symmetrical and always has ‘1’ at the beginning and end of each line </li></ul>Power of binomial Power of ‘x’ and
23. 26. <ul><li>What is the Discriminate </li></ul><ul><li>Using the Discriminate </li></ul><ul><li>Finding Minimum or Maximum Value </li></ul>
24. 27. <ul><li>Key: ∆ = discriminate </li></ul><ul><li>When using the equation the ‘ ’ decides the answer you get </li></ul><ul><li>If , you get a non-real number as it is a minus </li></ul><ul><li>If , you get two real distinct answers </li></ul><ul><li>If , you get one real equal answer </li></ul><ul><li>It is important to know the correct terminology here </li></ul>
25. 28. <ul><li>For example: </li></ul><ul><ul><li>a b c </li></ul></ul><ul><ul><li>Use to find the discriminate so it will be: </li></ul></ul><ul><ul><li>Then, when , you can factorise to get </li></ul></ul><ul><ul><ul><li>So, </li></ul></ul></ul><ul><ul><ul><li>Or </li></ul></ul></ul><ul><li>When or the answers are above the x-axis therefore the answer is: </li></ul>
26. 29. <ul><li>If given an equation where the constant is negative to find the minimum, the only possible answer is for the equation to equal zero </li></ul><ul><li>For example: </li></ul><ul><ul><li>As cannot be any less than 0, when this is multiplied by 4 it can be no less than zero </li></ul></ul><ul><ul><li>Therefore -25 has o be the minimum possible answer </li></ul></ul>