AQA Further Maths Level 2 Need-To-Know Booklet PDF

AQA Level 2 Certificate in Further Mathematics The Need-To-Know Book for Further Maths Level 2 Everything you need to know for Further Maths Level 2 Examination Board: AQA Brief This document is intended as an aid for revision. Although it includes s...

AQA Level 2 Certificate in Further Mathematics The Need-To-Know Book for Further Maths Level 2 Everything you need to know for Further Maths Level 2 Examination Board: AQA Brief This document is intended as an aid for revision. Although it includes some examples and explanation, it is primarily not for learning content, but for becoming familiar with the requirements of the course as regards formulae and results. It cannot replace the use of a text book, and nothing produces competence and familiarity with mathematical techniques like practice. This document was produced as an addition to classroom teaching and textbook questions, to provide a summary of key points and, in particular, any formulae or results you are expected to know and use for this qualification. Note: The Further Maths Level 2 course is intended for those who have achieved, or are expecting to achieve, an A or A* grade in GCSE Mathematics. As such, a thorough knowledge of the GCSE course is required as a prerequisite to this course. Contents Page Topic 2 Section 1 – Number 5 Section 2 – Algebra 14 Section 3 – Coordinate Geometry 17 Section 4 – Calculus 20 Section 5 – Matrix Transformations 23 Section 6 – Geometry Page 1 of 27 Section 1 – Number Fractions When adding or subtracting fractions, write over a common denominator, then add or subtract the numerators. Answers should be simplified where possible. Note: If adding fractions in mixed number form, it is often easiest to add the whole number and fraction parts separately. Eg: 4 3 4 3 28 15 43 8 𝟖 4 + 5 = (4 + 5) + ( + ) = 9 + ( + ) = 9 + ( ) = 9 + (1 ) = 𝟏𝟎 5 7 5 7 35 35 35 35 𝟑𝟓 When multiplying fractions, there is no need to find a common denominator. Simply multiply the numerators and multiply the denominators. Note: If multiplying fractions in mixed number form, first write as improper fractions. Note: For large numbers it can help to identify common factors in the numerators and denominators and pre-cancel before multiplying together. Eg: 8 9 1 3 𝟑 × = × = 15 16 5 2 𝟏𝟎 Here the 8 and 16 were both divided by 8, and the 9 and 15 were both divided by 3. When dividing fractions, multiply by the reciprocal of the fraction you want to divide by. Note: If dividing fractions in mixed number form, first write as improper fractions. When finding a percentage of an amount, convert to a decimal and multiply. Eg: 26% 𝑜𝑓 £400 = 0.26 × 400 = £𝟏𝟎𝟒 To increase or decrease by a percentage, multiply by the appropriate decimal. Eg: Increase 50kg by 15%: 1.15 × 50 = 𝟓𝟕. 𝟓𝒌𝒈. Decrease 80cm by 8%: 0.92 × 80 = 𝟕𝟑. 𝟔𝒄𝒎. Page 2 of 27 To reverse a percentage change, divide by the relevant decimal multiplier. Eg: A coat is marked down 30% in a sale. If the sale price is £43.40, what was the original price? £43.4 ÷ 0.7 = £𝟔𝟐 To apply compound interest for 𝑛 years, raise the decimal multiplier to the power 𝑛. Eg: James borrows £2500 at an annual interest rate of 17%. How much will he owe after 8 years? £2500 × 1.178 = £𝟖𝟕𝟕𝟖. 𝟔𝟑 To share an amount in a ratio, find the total number of parts, then scale up to the required total. Eg: Julie and Kate work 8 hours and 14 hours respectively each week at a restaurant, and any tips received are shared in this ratio. One week they receive £55 in tips. How much does Kate receive? 8 ∶ 14 ⟺ 4∶7 Julie Kate Total 4 : 7 11 ×5 ×5 ×5 20 : 35 55 𝑲𝒂𝒕𝒆 𝒓𝒆𝒄𝒆𝒊𝒗𝒆𝒔 £𝟑𝟓 Ratios can be combined by making the parts equal. This can be done by forming equivalent ratios. Eg: There is twice as much sugar as flour in a recipe, and 2 parts butter to 3 parts sugar. What is the ratio of sugar to flour to butter? 𝑆∶𝐹 = 2∶1 𝐵∶𝑆 = 2∶3 𝑆∶𝐹 = 6∶ 3 𝐵∶𝑆 = 4∶6 𝐹∶𝑆 = 3∶6 𝑆∶𝐵 = 6∶4 𝐹∶𝑆∶𝐵 = 3∶6∶4 𝑆∶𝐹∶𝐵 =𝟔∶𝟑∶𝟒 Page 3 of 27 Surds A surd is an irrational number involving a root. An irrational number is one that cannot be written as a fraction with whole numbers forming the numerator and denominator. 25 25 √25 5 Eg: √5 is a surd. √16 = 4 so it is not a surd. Neither is √ 4 as √ 4 = = 2. √4 Multiplication or division within a root can be brought outside the root. In general: 𝑎 √𝑎 √𝑎𝑏 = √𝑎 × √𝑏 𝑎𝑛𝑑 √ = 𝑏 √𝑏 Note: Remember that addition does not work in the same way. For instance, √9 + 16 ≠ √9 + √16. Eg: A right-angled triangle has sides of 12𝑐𝑚 and 8𝑐𝑚. Find the possible lengths of the third side. 𝐵𝑦 𝑃𝑦𝑡ℎ𝑎𝑔𝑜𝑟𝑎𝑠 ′ 𝑡ℎ𝑒𝑜𝑟𝑒𝑚: 𝑂𝑝𝑡𝑖𝑜𝑛 1 (𝑚𝑖𝑠𝑠𝑖𝑛𝑔 𝑠𝑖𝑑𝑒 ℎ𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒): √122 + 82 = √208 = √16√13 = 𝟒√𝟏𝟑 𝑂𝑝𝑡𝑖𝑜𝑛 2 (𝑚𝑖𝑠𝑠𝑖𝑛𝑔 𝑠𝑖𝑑𝑒 𝑛𝑜𝑡 ℎ𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒): √122 − 82 = √80 = √16√5 = 𝟒√𝟓 If a fraction has a surd in the denominator, it is often useful to rewrite the fraction in such a way as to have only rational numbers in the denominator. This is called rationalising the denominator. This is done, in simple cases, by simply multiplying top and bottom by the surd in the denominator: 𝑎 𝑎 √𝑏 𝑎√𝑏 = × = √𝑏 √𝑏 √𝑏 𝑏 5 √80 Eg: Simplify the following expression as far as possible: − √18 + √8 √5 5 √160 5√8 4√10√5 10√2 4√2√5√5 5√2 𝟗√𝟐 − √18 + = − 3√2 + = − 3√2 + = − 3√2 + 4√2 = √8 √5 8 5 8 5 4 𝟒 To rationalise the denominator of a more complex fraction, use the difference of two squares: 𝑎 𝑎(𝑏 − √𝑐) 𝑎𝑏 − 𝑎√𝑐 = = 𝑏 + √𝑐 (𝑏 + √𝑐)(𝑏 − √𝑐) 𝑏2 − 𝑐 Eg: 6 6(3 + 2√5) 18 + 12√5 18 + 12√5 𝟔(𝟑 + 𝟐√𝟓) = = = =− 3 − 2√5 (3 − 2√5)(3 + 2√5) 9 − 20 −11 𝟏𝟏 Page 4 of 27 Section 2 – Algebra Expressions To multiply out brackets, multiply every term within the bracket by the multiplier outside: 𝑎(𝑏 + 𝑐) = 𝑎𝑏 + 𝑎𝑐 Note: You will also need to be familiar with rules of indices when dealing with these questions. Eg: Expand the expression 3𝑥 4 𝑦(2𝑥𝑦 − 5𝑥 3 ) 3𝑥 4 𝑦(2𝑥𝑦 − 5𝑥 3 ) = 3𝑥 4 𝑦 × 2𝑥𝑦 − 3𝑥 4 𝑦 × 5𝑥 3 = 𝟔𝒙𝟓 𝒚𝟐 − 𝟏𝟓𝒙𝟕 𝒚 When simplifying an algebraic expression where terms are added, you can collect like terms. Eg: Simplify 3𝑥(2 − 5𝑥) − 7(𝑥 − 5) 3𝑥(2 − 5𝑥) − 7(𝑥 − 5) = 6𝑥 − 15𝑥 2 − (7𝑥 − 35) = 6𝑥 − 15𝑥 2 − 7𝑥 + 35 = 𝟑𝟓 − 𝒙 − 𝟏𝟓𝒙𝟐 To multiply brackets together, every term in each bracket must be multiplied: (𝑎 + 𝑏)(𝑐 + 𝑑) = 𝑎𝑐 + 𝑎𝑑 + 𝑏𝑐 + 𝑏𝑑 Note: To multiply out more than two brackets, it is usually easiest to take one pair at a time. Eg: Multiply out (3𝑥 − 1)(𝑥 + 4)(2 − 𝑥) (3𝑥 − 1)(𝑥 + 4)(2 − 𝑥) = (3𝑥 − 1)(2𝑥 − 𝑥 2 + 8 − 4𝑥) = (3𝑥 − 1)(8 − 2𝑥 − 𝑥 2 ) = 24𝑥 − 6𝑥 2 − 3𝑥 3 − 8 + 2𝑥 + 𝑥 2 = −𝟖 + 𝟐𝟔𝒙 − 𝟓𝒙𝟐 − 𝟑𝒙𝟑 To factorise an expression fully, it is necessary to find terms (numbers, letters or a combination) which are factors of every term in the bracket. Eg: Fully factorise 28𝑥 3 𝑦 2 − 12𝑥 5 𝑦 7 28𝑥 3 𝑦 7 − 12𝑥 5 𝑦 2 = 𝟒𝒙𝟑 𝒚𝟐 (𝟕𝒚𝟓 − 𝟑𝒙𝟐 ) Page 5 of 27 Quadratics A quadratic function can be written in the form 𝑦 = 𝑎𝑥 2 + 𝑏𝑥 + 𝑐, and the shape of the graph produced is known as a parabola. It is symmetrical, and resembles either ∪ (for 𝑎 > 0) or ∩ (for 𝑎 < 0). If it crosses the 𝑥-axis, the equation 𝑎𝑥 2 + 𝑏𝑥 + 𝑐 = 0 has two distinct solutions. If it only touches at one point this will be at its maximum or minimum and the equation will have one (repeated) solution. If the graph doesn’t touch the 𝑥-axis at all, the equation will have no real solutions. To factorise a quadratic, it is often necessary to find more than a single term, forming two brackets. Note: If the quadratic has only an 𝑥 2 and an 𝑥 term, 𝑥 will be a factor, so factorising becomes simpler. To factorise expressions with a single 𝒙𝟐 term, write out two brackets with 𝑥 at the start of each. To determine the numbers to go alongside these, find two numbers that multiply to make the constant term but add to make the 𝑥 coefficient. Eg: Solve 𝑥 2 − 5𝑥 + 4 = 0 𝑥 2 − 5𝑥 + 4 = (𝑥 … )(𝑥 … ) and since −1 and −4 multiply to make 4 but add to make −5: 𝑥 2 − 5𝑥 + 4 = 0 ⟹ (𝑥 − 1)(𝑥 − 4) = 0 ⟹ 𝒙 = 𝟏 𝒐𝒓 𝒙 = 𝟒 To factorise expressions with a number in front of 𝒙𝟐 , before splitting into brackets, find two numbers that multiply to make the product of the 𝑥 2 coefficient and the constant, but add to make the 𝑥 coefficient. Then split the 𝑥 term into these two parts, factorise the first two terms of the expression separately, then the last two terms, and finally factorise the whole thing. Eg: Solve: 2𝑥 2 + 𝑥 − 3 = 0 Two numbers which multiply to make 2 × −3 = −6 and add to make 1: 3 and −2, so: 2𝑥 2 + 𝑥 − 3 = 2𝑥 2 + 3𝑥 − 2𝑥 − 3 = 𝑥(2𝑥 + 3) − 1(2𝑥 + 3) = (2𝑥 + 3)(𝑥 − 1) 𝟑 ⟹ (2𝑥 + 3)(𝑥 − 1) = 0 ⟹ 𝒙=− 𝒐𝒓 𝒙 = 𝟏 𝟐 Page 6 of 27 Another way to solve a quadratic is completing the square. This will work for any quadratic, but can take longer than factorising for simpler equations. Note also that completing the square is a technique that can be used to determine a maximum or minimum point on a quadratic graph without solving the equation. The aim is to write the quadratic expression in the form p(x + q)2 + r, as shown below. Eg: Solve 𝑥 2 + 6𝑥 + 10 = 0 by completing the square. Step 1: Halve the 𝑥 coefficient to (𝑥 + 3)2 find the number to go with 𝑥 in We know that if we multiply this out we would get 𝑥 2 + 6𝑥 + 9 the squared bracket: Step 2: Subtract the square of this (𝑥 + 3)2 − 9 number from the squared By subtracting the 9 we make an expression equal to 𝑥 2 + 6𝑥 bracket: Step 3: Add the constant from the (𝑥 + 3)2 − 9 + 10 = (𝒙 + 𝟑)𝟐 + 𝟏 original expression, and simplify: This is equal to 𝑥 2 + 6𝑥 + 10, but written as a completed square Step 4: We can then solve the 𝑥 2 + 6𝑥 − 5 = (𝑥 + 3)2 − 9 + 1 = (𝑥 + 3)2 − 8 equation by rearranging (notice ⟹ (𝑥 + 3)2 − 8 = 0 that now 𝑥 occurs just once): ⟹ (𝑥 + 3)2 = 8 ⟹ 𝑥 + 3 = ±√8 = ±2√2 ⟹ 𝒙 = −𝟑 ± 𝟐√𝟐 The quadratic formula is the result of completing the square with the general form of a quadratic, 𝑎𝑥 2 + 𝑏𝑥 + 𝑐 = 0. By dividing by 𝑎, completing the square and making 𝑥 the subject, we get: −𝑏 ± √𝑏 2 − 4𝑎𝑐 𝑥= 2𝑎 Note: Some quadratic equations have no solutions, some have exactly one, and some have two. By considering the value of 𝑏 2 − 4𝑎𝑐 we can discriminate between these three cases. If 𝑏 2 − 4𝑎𝑐 is positive, there are two distinct solutions, if 0, only one (a ‘repeated root’), and if negative, no real solutions. Formulae To change the subject of a formula, first isolate the term required, then reverse the operations applied to it. Note: If the term appears twice, it is often necessary to collect them and factorise the expression. Eg: Make 𝑥 the subject of the formula: 2𝑥 + 3 5𝑦 − 𝑥 = 2 ⟹ 7𝑞 2 (5𝑦 − 𝑥) = 2𝑥 + 3 ⟹ 35𝑞 2 𝑦 − 7𝑞 2 𝑥 = 2𝑥 + 3 7𝑞 2 2 2 2) 35𝑞 2 𝑦 − 3 ⟹ 35𝑞 𝑦 − 3 = 2𝑥 + 7𝑞 𝑥 ⟹ 35𝑞 𝑦 − 3 = 𝑥(2 + 7𝑞 ⟹ =𝑥 2 + 7𝑞 2 Page 7 of 27 Rational expressions To simplify, add or subtract rational expressions, simply apply the rules of fractions. You can cancel terms if they are factors of both the numerator and denominator (that is, divide top and bottom by the same thing), and to add or subtract fractions, write with a common denominator then combine the numerators by adding or subtracting. To divide by a fraction, multiply by its reciprocal. 𝑥 2 −100 Eg: Simplify 𝑥 2 +8𝑥−20 𝑥 2 − 100 (𝑥 + 10)(𝑥 − 10) 𝒙 − 𝟏𝟎 = = 𝑥 2 + 8𝑥 − 20 (𝑥 + 10)(𝑥 − 2) 𝒙−𝟐 𝑥+4 𝑥−8 Eg: Simplify − 𝑥 2 −4𝑥 2𝑥 𝑥+4 𝑥−8 𝑥+4 𝑥−8 (𝑥 + 4)(𝑥 − 4) 2(𝑥 − 8) − 2 = − = − 2𝑥 𝑥 − 4𝑥 2𝑥 𝑥(𝑥 − 4) 2𝑥(𝑥 − 4) 2𝑥(𝑥 − 4) (𝑥 + 4)(𝑥 − 4) − 2(𝑥 − 8) 𝑥 2 − 16 − 2𝑥 + 16 = = 2𝑥(𝑥 − 3) 2𝑥(𝑥 − 3) 𝑥 2 − 2𝑥 𝑥(𝑥 − 2) 𝒙−𝟐 = = = 2𝑥(𝑥 − 3) 2𝑥(𝑥 − 3) 𝟐(𝒙 − 𝟑) Functions A function is a mapping which can be either one-to-one or many-to-one. Eg: 𝑓(𝑥) = 3𝑥 − 4 is a one-to-one function, 𝑓(𝑥) = 𝑥 2 + 5 is a many-to-one function. The domain of a function is the set of input values it can take. The range of a function is the set of output values it can generate. 5 Eg: 𝑓(𝑥) = 𝑥 2 has the domain 𝑥 ≠ 0 and the range 𝑓(𝑥) > 0. A function is fully defined by both a rule and a domain. 2𝑥 Eg: 𝑓(𝑥) = (𝑥−3) 𝑓𝑜𝑟 𝑥 ≠ 3. Factor theorem The factor theorem: (𝑥 − 𝑎) 𝑖𝑠 𝑎 𝑓𝑎𝑐𝑡𝑜𝑟 𝑜𝑓 𝑝𝑜𝑙𝑦𝑛𝑜𝑚𝑖𝑎𝑙 𝑃(𝑥) ⟺ 𝑃(𝑎) = 0 (𝑡ℎ𝑎𝑡 𝑖𝑠, 𝑎 𝑖𝑠 𝑎 𝑟𝑜𝑜𝑡) Note: This means that if we know a factor of a polynomial, we can find the corresponding root (a root is defined as the solution to 𝑓(𝑥) = 0; that is, the 𝑥 coordinates of points where the graph 𝑦 = 𝑓(𝑥) crosses the 𝑦-axis). The factor theorem works both ways, so if we know a root already, we can determine the corresponding factor. Page 8 of 27 To factorise a cubic, first find a root by trial and error, then use the factor theorem to generate a linear factor. Finally, use inspection to find the remaining (quadratic) factor, and factorise this, where possible. Eg: Factorise fully 2𝑥 3 + 3𝑥 2 − 16𝑥 + 15 𝑃(𝑥) = 2𝑥 3 − 𝑥 2 − 16𝑥 + 15 𝑃(0) = 15 ⟹ 0 𝑖𝑠 𝑛𝑜𝑡 𝑎 𝑟𝑜𝑜𝑡 𝑃(1) = 0 ⟹ 1 𝑖𝑠 𝑎 𝑟𝑜𝑜𝑡 𝐵𝑦 𝑡ℎ𝑒 𝑓𝑎𝑐𝑡𝑜𝑟 𝑡ℎ𝑒𝑜𝑟𝑒𝑚: 𝑃(1) = 0 ⟹ (𝑥 − 1) 𝑖𝑠 𝑎 𝑓𝑎𝑐𝑡𝑜𝑟 2𝑥 3 − 𝑥 2 − 16𝑥 + 15 = (𝑥 − 1)(… ) By inspection, we need 2𝑥 2 in the second bracket to give the 2𝑥 3 term in the cubic. We also need −15 in the second bracket to give the 15 term in the cubic. 𝑃(𝑥) = (𝑥 − 1)(2𝑥 2 + ⋯ − 15) To get the 𝑥 2 term we need to examine the 𝑥 terms from each, as well as the combination of the 𝑥 2 and constant terms. Since we already have −1 × 2𝑥 2 giving −2𝑥 2 , we need 𝑥 2 to give the result – 𝑥 2. This must be achieved by 𝑥 multiplied by the second bracket’s 𝑥 term which therefore must be simply 𝑥. 𝑃(𝑥) = (𝑥 − 1)(2𝑥 2 + 𝑥 − 15) Finally, factorise the quadratic if possible: 𝑃(𝑥) = (𝑥 − 1)(2𝑥 − 5)(𝑥 + 3) Sketching curves When sketching a quadratic, include: The correct shape: A positive quadratic (one with a positive 𝑥 2 coefficient) will have a single turning point – a minimum – and a negative quadratic will have a maximum point. The graph will be symmetrical about its vertex (the max/min point), and should be a smooth curve with no sharp turns which gets steeper as it increases but never vertical. The 𝒚-intercept: The point at which the curve crosses the 𝑦-axis. There will always be one, and only one, for a quadratic, and it is easy to identify because the curve crosses the 𝑦-axis when 𝑥 = 0. Substituting 𝑥 = 0 into 𝑦 = 𝑎𝑥 2 + 𝑏𝑥 + 𝑐 gives 𝑦 = 𝑐, so the intercept is (0, 𝑐). Any 𝒙-axis crossing points: These are points where the curve 𝑦 = 𝑎𝑥 2 + 𝑏𝑥 + 𝑐 crosses the line 𝑦 = 0, so they are solutions to 𝑎𝑥 2 + 𝑏𝑥 + 𝑐 = 0 and can be found – if any exist – by either factorising, completing the square or applying the quadratic formula. There may be 0, 1 or 2 roots. [Sometimes required] The position of the vertex: By completing the square, it is possible to determine the coordinates of the maximum or minimum point. This allows a more precise sketch. Page 9 of 27 When sketching a cubic, include: The correct shape: A positive cubic decreases without limit as 𝑥 decreases, and increases without limit as 𝑥 increases (so it tends to go from bottom left to top right). A negative does the opposite. Either can have a point of inflection in the middle (a ‘wiggle’), like 𝑦 = 𝑥 3 , or a maximum point and a minimum point. The 𝒚-intercept: Like with the quadratic, since this is simply the value of the cubic expression when 𝑥 = 0, it is just the constant term. Any 𝒙-axis crossing points: These are not always straightforward to find, and a cubic may have one, two (in this case one is also a maximum or a minimum) or three. Use factor theorem to determine the roots. [Sometimes required] The position of any stationary points: If there are stationary points (some cubics have no points where the gradient is zero, some have one, some two), these can be found 𝑑𝑦 using differentiation and solving 𝑑𝑥 = 0. Their nature (maximim, minimum or point of inflection) 𝑑𝑦 can be determined by considering the height ( 𝑦 ) or gradient ( 𝑑𝑥 ) on either side. Simultaneous equations One method for solving simultaneous linear equations is the elimination method. Coefficients of one of the variables are made equal and then the equations are effectively added or subtracted from one another to eliminate this variable. Eg: (1) 5𝑥 − 2𝑦 = 6 (2) 2𝑥 + 4𝑦 = 9 (1)×2 10𝑥 − 4𝑦 = 12 21 7 (2)+(1’) 12𝑥 = 21 ⟹ 𝑥 = 12 = 4 = 1.75 7 7 11 11 Sub into (2) 2 (4) + 4𝑦 = 9 ⟹ 4𝑦 = 9 − 2 = ⟹ 𝑦= = 1.375 2 8 More generally, a wider range of simultaneous equations can be solved using substitution. Eg: (1) 𝑦 − 1 = 2𝑥 (2) 𝑦 2 − 3𝑥 2 = 6 Rearrange (1) 𝑦 = 2𝑥 − 1 Sub into (2) (2𝑥 − 1)2 − 3𝑥 2 = 6 Rearrange (2’) 𝑥 2 − 4𝑥 − 5 = 0 Solve (2’) (𝑥 + 1)(𝑥 − 5) = 0 ⟹ 𝑥 = −1 𝑜𝑟 𝑥 = 5 Sub into (1’) 𝑦 = 2(−1) − 1 = −3 𝑜𝑟 𝑦 = 2(5) − 1 = 9 Write out solutions 𝑥 = −1 𝑦 = −3 𝑜𝑟 𝑥 = 5 𝑦 = 9 Note: Take care to pair up the correct values of 𝑥 and 𝑦 – they represent the coordinates of the crossing points for the two equations. Page 10 of 27 Inequalities When solving a linear inequality, if you multiply or divide by a negative, you reverse the sign. Note: The reason for this is clear when you consider 3 < 5. By subtracting 8 from each side we get −5 < −3, which is of course perfectly correct. But notice that this is equivalent to −3 > −5 which not only has different signs to the original statement but also has the inequality sign reversed. To solve a quadratic inequality, it is necessary first to find the critical values (that is, the solutions to the related equation), then use a sketch to interpret these values as solution regions. Eg: Solve 𝑥 2 − 5𝑥 + 6 ≥ 0 Finding critical values: 𝑥 2 − 5𝑥 + 6 = 0 Sketching the curve: ⟹ (𝑥 − 2)(𝑥 − 3) = 0 ⟹ 𝑥 = 2 𝑎𝑛𝑑 𝑥 = 3 Interpreting the sketch: 𝑥 ≤ 2 𝑜𝑟 𝑥 ≥ 3 Indices The multiplication rule: 𝑎𝑚 × 𝑎𝑛 = 𝑎𝑚+𝑛 The division rule: 𝑎𝑚 ÷ 𝑎𝑛 = 𝑎𝑚−𝑛 The power rule: (𝑎𝑚 )𝑛 = 𝑎𝑚𝑛 The negative index rule: 1 𝑎 −𝑛 = 𝑎𝑛 The zero index result: 𝑎0 = 1 The root rule: 1 𝑚 𝑛 𝑚 𝑛 𝑛 𝑎 𝑛 = √𝑎 𝑎𝑛𝑑, 𝑚𝑜𝑟𝑒 𝑔𝑒𝑛𝑒𝑟𝑎𝑙𝑙𝑦: 𝑎 𝑛 = √𝑎𝑚 = ( √𝑎 ) Page 11 of 27 Eg: 4 √𝑥 2 𝑦3 Simplify: 𝑥 𝑎−3 4 √𝑥 2 𝑦 3 2 3 1 −3 3 − 𝟓 𝟑 = 𝑥 4 𝑦 4 𝑥 −3 = 𝑥 2 𝑦 4 = 𝒙 𝟐 𝒚𝟒 𝑥3 Eg 2: 33𝑥 1 Solve: = 92𝑥+1 27 33𝑥 1 33𝑥 = 2𝑥+1 ⟹ 3 = 9−2𝑥−1 27 9 3 ⟹ 33𝑥−3 = (32 )−2𝑥−1 ⟹ 33𝑥−3 = 3−4𝑥−2 1 ⟹ 3𝑥 − 3 = −4𝑥 − 2 ⟹ 7𝑥 = 1 ⟹ 𝑥= 7 Algebraic proof To prove a result, it is necessary to formulate your assumptions and define your variables. This is usually in the form of an algebraic statement such as an equation. Each line of your proof should then be the next line of a reasoned argument, with each statement being a necessary consequence of the previous one, finishing with a conclusion. Eg: Prove that the difference between two consecutive cube numbers is not a multiple of 3. The difference between two consecutive cube numbers can be written as (𝑛 + 1)3 − 𝑛3 (𝑛 + 1)3 − 𝑛3 = 𝑛3 + 3𝑛2 + 3𝑛 + 1 − 𝑛3 = 3𝑛2 + 3𝑛 + 1 = 3𝑛(𝑛 + 1) + 1 Since 3𝑛(𝑛 + 1) has 3 as a factor, 3𝑛(𝑛 + 1) is necessarily a mutliple of 3. Therefore 3𝑛(𝑛 + 1) + 1 is not a multiple of 3. Sequences To find the 𝑛𝑡ℎ term of a linear sequence, identify the common difference (the difference between any consecutive terms), and compare the sequence to the related multiplication table. Add or subtract the required value to transform this sequence into yours. Eg: Prove that 340 is not in the linear sequence 6, 10, 14, 18, …. 𝑇(𝑛) = 4𝑛 ⟹ 4, 8, 12, … ⟹ 𝑇ℎ𝑖𝑠 𝑠𝑒𝑞𝑢𝑒𝑛𝑐𝑒 𝑚𝑢𝑠𝑡 𝑏𝑒 𝑇(𝑛) = 4𝑛 + 2 338 4𝑛 + 2 = 340 ⟹ 𝑛= = 84.5. 𝑁𝑜𝑡 𝑎 𝑤ℎ𝑜𝑙𝑒 𝑛𝑢𝑚𝑏𝑒𝑟 ⟹ 340 𝑛𝑜𝑡 𝑖𝑛 𝑠𝑒𝑞𝑢𝑒𝑛𝑐𝑒 4 Page 12 of 27 To find the 𝑛𝑡ℎ term of a quadratic sequence, find the sequence of differences, then the sequence of second differences. Halve the second difference to find the 𝑛2 coefficient. Find the difference between this simple quadratic sequence and your original sequence. Find the 𝑛𝑡ℎ term of this (linear) sequence and add it to the 𝑛2 part. Eg: Find the 𝑛𝑡ℎ term rule of the quadratic sequence 12, 32, 62, 102, …. Sequence of differences: 20, 30, 40, … Sequence of second differences: 10, 10, 10, … Quadratic part: 𝑇(𝑛) = 5𝑛2 ⟹ 5, 20, 45, 80, … Difference: 7, 12, 17, 22, … ⟹ 𝑇(𝑛) = 5𝑛 + 2 Complete sequence: 𝑻(𝒏) = 𝟓𝒏𝟐 + 𝟓𝒏 + 𝟐 Some sequences tend towards a limit (that is, sucessive terms get increasingly close to, but never surpass, a particular value). Depending on the 𝑛𝑡ℎ term rule for these sequences, the limit can be 1 found through algebraic manipulation, using the fact that 𝑛 tends to 0 as 𝑛 ‘tends to ∞’ (aka ‘increases without limit’). Eg: 8𝑛−4 Find the limit of the sequence defined by 𝑇(𝑛) = 4𝑛+20 Step 1: Divide through (top and bottom of the fraction) by 𝑛: 4 8−𝑛 𝑇(𝑛) = 20 4+ 𝑛 1 Step 2: Use the fact that 𝑛 → 0 as 𝑛 → ∞: 4 8−𝑛 8−0 → = 2 𝑎𝑠 𝑛 → ∞ 𝑡ℎ𝑒𝑟𝑒𝑓𝑜𝑟𝑒 𝒕𝒉𝒆 𝒍𝒊𝒎𝒊𝒕 𝒐𝒇 𝒕𝒉𝒆 𝒔𝒆𝒒𝒖𝒆𝒏𝒄𝒆 𝒊𝒔 𝟐 20 4+0 4+ 𝑛 Page 13 of 27 Section 3 – Coordinate Geometry Straight lines The distance between two points can be calculated by constructing a right-angled triangle between the coordinates and applying Pythagoras’ Theorem: 𝐷𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑏𝑒𝑡𝑤𝑒𝑒𝑛 (𝑥1 , 𝑦1 ) 𝑎𝑛𝑑 (𝑥2 , 𝑦2 ): √(𝑥2 − 𝑥1 )2 + (𝑦2 − 𝑦1 )2 The midpoint of the line between the points (𝑥1 , 𝑦1 ) and (𝑥2 , 𝑦2 ) is given by: 𝑥1 + 𝑥2 𝑦1 + 𝑦2 ( , ) 2 2 Note: This is simply the average of the 𝑥 coordinates and the average of the 𝑦 coordinates. If you are required to find a point which is a given proportion of the way between two points, it can be helpful to think in terms of adding a certain fraction of the journey to the starting point. 1 1 1 Eg: The point 3 of the way between (𝑥1 , 𝑦1 ) and (𝑥2 , 𝑦2 ) is: (𝑥1 + 3 (𝑥2 − 𝑥1 ), 𝑦1 + 3 (𝑦2 − 𝑦1 )) 2 1 2 1 Note: This can be simplified, in this case, to (3 𝑥1 + 3 𝑥2 , 3 𝑦1 + 3 𝑦2 ), or, alternatively, the midpoint 1 1 formula can be rewritten to say: (𝑥1 + (𝑥2 − 𝑥1 ), 𝑦1 + (𝑦2 − 𝑦1 )). 2 2 The gradient of the line between the points (𝑥1 , 𝑦1 ) and (𝑥2 , 𝑦2 ) is given by: 𝑦2 − 𝑦1 𝑚= 𝑥2 − 𝑥1 Note: This is often described as ‘𝑦 step over 𝑥 step’, or ‘rise over run’. Lines with gradients 𝑚1 and 𝑚2 are parallel if 𝑚1 = 𝑚2. They are perpendicular if 𝑚1 𝑚2 = −1. Note: The concepts behind the above results are more important (and more easily memorable) than the formulae used to describe them mathematically. The result below is the only one where memorising the formula gives a definite advantage to quickly solving problems, especially since simplification into a specific form is not always required. Given the gradient, 𝑚, and a single point, (𝑥1 , 𝑦1 ), the equation of a line can be generated using: 𝑦 − 𝑦1 = 𝑚(𝑥 − 𝑥1 ) Page 14 of 27 Given two points, the equation of a line can be generated using: 𝑦 − 𝑦1 𝑥 − 𝑥1 = 𝑦2 − 𝑦1 𝑥2 − 𝑥1 Note: The above result is simply a combination of the definition of gradient and the gradient & point formula above. As such it is not necessary to memorise this form if you are already confident with finding the gradient between two points and can recall 𝑦 − 𝑦1 = 𝑚(𝑥 − 𝑥1 ). The crossing point of two lines directly corresponds with the values of 𝒙 and 𝒚 which satisfy both equations. This can be found either by reading off the graph, or – more precisely – by solving the equations simultaneously. Circles The equation of a circle with centre (0,0) and radius 𝑟 is given by: 𝑥2 + 𝑦2 = 𝑟2 Note: This can be understood by considering any point on the circle and constructing a right-angled triangle, with the line from the origin to the point as the hypotenuse. By Pythagoras’ Theorem, the square of the radius must be equal to the sum of the squares of the 𝑥 and 𝑦 coordinates. The general equation of a circle with centre (𝑎, 𝑏) and radius 𝑟 is given by: (𝑥 − 𝑎)2 + (𝑦 − 𝑏)2 = 𝑟 2 𝑎 Note: This is simply a translation from the original circle – centre (0,0) – by vector [ ]. 𝑏 Eg: The circle (𝑥 − 3)2 + 𝑦 2 = 25: Note: The most common mistakes to watch out for when interpreting a circle equation are getting the signs wrong for the centre coordinates or forgetting to square root the number on the right to get the radius. Page 15 of 27 To sketch a circle: Step 1: Find the radius and centre from the equation. Step 2: Mark the centre. Step 3: Use the radius to mark on the four points directly above, below and to either side. Step 4: Draw the circle through these four points, indicating where the circle crosses the axes, if applicable. Note: you may also have to rely on other circle facts such as how to calculate the circumference or area, or the circle theorems (see section 6 for details). To find the equation of a circle given the end points of the diameter, calculate the midpoint and the distance between the points. This will give you the centre and the diameter. Halve the diameter to get the radius, then put into the form (𝑥 − 𝑎)2 + (𝑦 − 𝑏)2 = 𝑟 2. Like any other curve, the crossing points of a line and a circle can be found by solving simultaneously (using the substitution method). A tangent to the circle at a particular point is a line which touches the circle only at that point. It is always perpendicular to the radius. A normal to the circle at a particular point is a line passing through a circle which is perpendicular to the tangent at that point. Note: The normal line at any point on a circle will pass through the centre, since it is perpendicular to the tangent which is perpendicular to the radius. To find the equation of a tangent or normal at a particular point: Step 1: Find the centre of the circle. Step 2: Calculate the gradient of the line segment from the centre to your point. Step 3i: For a normal, use this gradient and your point in the formula 𝑦 − 𝑦1 = 𝑚(𝑥 − 𝑥1 ). 1 Step 3ii: For a tangent, first find the tangent gradient by using 𝑚1 = − 𝑚 , then use the formula. 2 Note: You may need to complete the square to convert a circle equation into the preferred form first. Eg: Find the equation of the tangent to the circle 𝑥 2 + 2𝑥 + 𝑦 2 − 6𝑦 = 25 at the point (2,7). 𝑥 2 + 2𝑥 + 𝑦 2 − 6𝑦 = 25 ⟹ (𝑥 + 1)2 − 1 + (𝑦 − 3)2 − 9 = 25 ⟹ (𝑥 + 1)2 + (𝑦 − 3)2 = 35 7−3 4 1 3 𝐶𝑒𝑛𝑡𝑟𝑒: (−1,3) ⟹ 𝐺𝑟𝑎𝑑 𝑜𝑓 𝑟𝑎𝑑𝑖𝑢𝑠 𝑙𝑖𝑛𝑒: = ⟹ 𝐺𝑟𝑎𝑑 𝑜𝑓 𝑡𝑎𝑛𝑔𝑒𝑛𝑡: − =− 2 − (−1) 3 4 4 3 3 3 17 𝐸𝑞𝑢𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑡𝑎𝑛𝑔𝑒𝑛𝑡: 𝑦 − 7 = − (𝑥 − 2) ⟹ 𝑦=− 𝑥+ 4 4 2 Page 16 of 27 Section 4 – Calculus Differentiation is a method for finding the gradient of a curve at any given point. It can be thought of as the rate of change of 𝑦 with respect to 𝑥. For a small change in 𝑥 and a corresponding small change in 𝑦, the gradient of a chord can be written as: 𝛿𝑦 𝛿𝑥 Note: Here, ‘chord’ refers to a straight line segment drawn between two nearby points on a curve. Gradient We define the gradient of a curve at a particular point as the gradient of the tangent to the curve at that point. Note: As the end-points get closer together, the gradient of the chord approaches the gradient of the tangent at each point (ie, the gradient of the curve). This is the basis for differentiation – the limit to which the gradients of the ever-decreasing chords tends is the gradient of the curve at that point. 𝛿𝑦 𝑑𝑦 The limit of 𝛿𝑥 as 𝛿𝑦, 𝛿𝑥 → 0 is written as 𝑑𝑥 and is known as the differential of 𝑦 with respect to 𝑥. Note: The proof of this idea involves the concept of shrinking a small quantity until it is essentially of zero size. This is effectively finding the gradient of a chord connecting a point to itself. While this idea is worth being aware of, it is not necessary to memorise a proof of it. 𝑑𝑦 𝑦 = 𝑥𝑛 ⟹ = 𝑛𝑥 𝑛−1 𝑑𝑥 Note: This can be thought of as “bring the power down in front, then reduce the power by one”. Note: If there is a number multiplied by the 𝑥 𝑛 term, this is not affected by differentiating. Eg: 𝑑𝑦 𝑦 = 5𝑥 4 ⟹ = 20𝑥 3 𝑑𝑥 𝑑𝑦 𝑦 = 𝑓(𝑥) ± 𝑔(𝑥) ⟹ = 𝑓 ′ (𝑥) ± 𝑔′ (𝑥) 𝑑𝑥 Eg: The differential of 2𝑥 4 − 8𝑥 2 + 4 is 8𝑥 3 − 16𝑥 (note that any constants differentiate to 0). 𝑑(𝑓(𝑥)) Note: 𝑓 ′ (𝑥) is sometimes used to denote the derivative of 𝑓(𝑥). It is equivalent to. 𝑑𝑥 Page 17 of 27  𝑑𝑦 To find the gradient of a curve at a particular point, calculate 𝑑𝑥 and substitute in the 𝑥 coordinate. Eg: Find the gradient of the curve 𝑦 = (2𝑥 + 3)(𝑥 2 − 5) at the point (2, −7). Multiply out: 𝑦 = (2𝑥 + 3)(𝑥 2 − 5) = 2𝑥 3 + 3𝑥 2 − 10𝑥 − 15 Differentiate: 𝑑𝑦 = 6𝑥 2 + 6𝑥 − 10 𝑑𝑥 Substitute in 𝑥 = 2: 𝑑𝑦 𝐴𝑡 𝑥 = 2 = 6(22 ) + 6(2) − 10 = 26 ⟹ 𝐺𝑟𝑎𝑑𝑖𝑒𝑛𝑡 = 26 𝑑𝑥 To find a point on a curve with a given gradient, differentiate then set your expression equal to the gradient and solve for 𝑥. Finally, substitute into the original equation for a corresponding value for 𝑦. Eg: Find any points on the curve 𝑦 = 𝑥 3 − 3𝑥 where the gradient is 45. Differentiate: 𝑑𝑦 = 3𝑥 2 − 3 𝑑𝑥 𝑑𝑦 Rewrite using 𝑑𝑥 = 45: 45 = 3𝑥 2 − 3 ⟹ 𝑥 2 − 1 = 15 ⟹ 𝑥 2 = 16 ⟹ 𝑥 = ±4 Substitute back into the original equation: 𝐹𝑜𝑟 𝑥 = 4: 𝑦 = 43 − 3(4) = 52 ⟹ (4,52) 𝐹𝑜𝑟 𝑥 = −4: 𝑦 = (−4)3 − 3(−4) = −52 ⟹ (−4, −52) Given the gradient of a curve at a particular point, we can find the equation of the tangent using 𝑦 − 𝑦1 = 𝑚(𝑥 − 𝑥1 ) where (𝑥1 , 𝑦1 ) is the point on the curve and 𝑚 is the gradient. Eg: Find the equation of the tangent to the curve 𝑦 = 5𝑥 3 − 6 at the point (1, −1). 𝑑𝑦 = 15𝑥 2 𝐴𝑡 𝑥 = 1: 𝑚 = 15(12 ) = 15 ⟹ 𝑢𝑠𝑖𝑛𝑔 𝑦 − 𝑦1 = 𝑚(𝑥 − 𝑥1 ) 𝑑𝑥 𝑦 − (−1) = 15(𝑥 − 1) ⟹ 𝑦 = 15𝑥 − 16 Page 18 of 27 Stationary points 𝑑𝑦 Recall that 𝑑𝑥 means the rate of change of 𝑦 with respect to 𝑥. 𝑑𝑦 >0 ⟹ 𝑦 𝑖𝑛𝑐𝑟𝑒𝑎𝑠𝑒𝑠 𝑎𝑠 𝑥 𝑖𝑛𝑐𝑟𝑒𝑎𝑠𝑒𝑠 𝑑𝑥 𝑑𝑦