AQA IGCSE Further Maths Revision · PDF file 1 AQA IGCSE Further Maths Revision Notes Formulas...
Transcript of AQA IGCSE Further Maths Revision · PDF file 1 AQA IGCSE Further Maths Revision Notes Formulas...
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AQA IGCSE Further Maths Revision Notes
Formulas given in formula sheet:
Volume of sphere: 4
3ππ3 Surface area of sphere: 4ππ2
Volume of cone: 1
3ππ2β Curve surface area: πππ
Area of triangle: 1
2ππ sin πΆ
Sine Rule: π
sin π΄=
π
sin π΅=
π
sin πΆ Cosine Rule: π2 = π2 + π2 β 2ππ cos π΄
Quadratic equation: ππ₯2 + ππ₯ + π = 0 β π₯ =βπΒ±βπ2β4ππ
2π
Trigonometric Identities: tan π β‘sin π
cos π sin2 π + cos2 π β‘ 1
1. Number
Specification Notes What can go ugly
1.1 Knowledge and use of numbers and the number system including fractions, decimals, percentages, ratio, proportion and order of operations are expected.
A few possibly helpful things:
If π: π = π: π (i.e. the ratios are the same), then π
π=
π
π
βFind the value of π after it has been increased by π%β If say π was 4, weβd want to Γ 1.04 to get a a 4% increase.
Can use 1 +π
100 as the multiplier, thus answer is:
π (1 +π
100)
βShow that π% of π is the same as π% of πβ π
100Γ π =
ππ
100
π
100Γ π =
ππ
100
1.2 Manipulation of surds, including rationalising the denominator.
GCSE recap:
Laws of surds: βπ Γ βπ = βππ and βπ
βπ= β
π
π
But note that π Γ βπ = πβπ not βππ
Note also that βπ Γ βπ = π
To simplify surds, find the largest square factor and put this first:
β12 = β4β3 = 2β3
β75 = β25β3 = 5β3
5β2 Γ 3β2 Note everything is being multiplied here. Multiply surd-ey things and non surd-ey things separately.
= 15 Γ 2 = 30
β8 + β18 = 2β2 + 3β2 = 5β2 To βrationalise the denominatorβ means to make it a non-
surd. Recall we just multiply top and bottom by that surd:
6
β3β
6
β3Γ
β3
β3=
6β3
3= 2β3
The new thing at IGCSE FM level is where we have more complicated denominators. Just multiply by the βconjugateβ: this just involves negating the sign between the two terms:
3
β6 β 2β
3
β6 β 2Γ
β6 + 2
β6 + 2=
3(β6 + 2)
2
A trick to multiplying out the denominator is that we have
the difference of two squares, thus (β6 β 2)(β6 + 2) =
6 β 4 = 2 (remembering that β6 squared is 6, not 36!)
2β3β1
3β3+4 β
2β3β1
3β3+4Γ
3β3β4
3β3β4=
(2β3β1)(3β3β4)
27β16=
18β3β3β8β3+4
11=
22β11β3
11= 2 β β3
Iβve seen students inexplicably reorder the terms in the denominator before they multiply by the conjugate, e.g.
(2 + β3)(β3 β 2)
Just leave the terms in their original order! Iβve also seen students forget to negate the sign, just
doing (2 + β3)(2 +
β3) in the
denominator. The problem here is that it wonβt rationalise the denominator, as weβll still have surds! Silly error:
6β6
2β 3β3
(rather than 3β6)
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2. Algebra
Specification Notes What can go ugly
2.2 Definition of a function
A function is just something which takes an input and uses some rule to produce an output, i.e. π(ππππ’π‘) = ππ’π‘ππ’π‘ You need to recognise that when we replace the input π₯ with some other expression, we need to replace every instance of it in the output, e.g. if π(π₯) = 3π₯ β 5, then π(π₯2) = 3π₯2 β 5, whereas π(π₯)2 = (3π₯ β 5)2. See my Domain/Range slides. e.g. βIf π(π₯) = 2π₯ + 1, solve π(π₯2) = 51
2π₯2 + 1 = 51 β π₯ = Β±5
2.3 Domain and range of a function.
The domain of a function is the set of possible inputs. The range of a function is the set of possible outputs. Use π₯ to refer to input and π(π₯) to refer to output. Use βfor allβ if any value possible. Note that < vs β€ is important.
π(π₯) = 2π₯ Domain: for all π₯ Range: for all π(π₯)
π(π₯) = π₯2 Domain: for all π₯ Range: π(π₯) β₯ 0
π(π₯) = βπ₯ Domain: π₯ β₯ 0 Range: π(π₯) β₯ 0
π(π₯) = 2π₯ Domain: for all π₯ Range: π(π₯) > 0
π(π₯) =1
π₯ Domain: for all π₯ except 0.
Range: for all π(π₯) except 0.
π(π₯) =1
π₯β2
Domain: for all π₯ except 2 (since weβd be dividing by 0) Range: for all π(π₯) except 0 (sketch to see it)
π(π₯) = π₯2 β 4π₯ + 7 Completing square we get (π₯ β 2)2 + 3 The min point is (2,3). Thus range is π(π₯) β₯ 3
You can work out all of these (and any variants) by a quick sketch and observing how π₯ and π¦ values vary.
Be careful if domain is βrestrictedβ in some way. Range if π(π₯) = π₯2 + 4π₯ + 3, π₯ β₯ 1 When π₯ = 1, π(1) = 8, and since π(π₯) is increasing after this value of π₯, π(π₯) β₯ 8.
To find range of trigonometric functions, just use a suitable sketch, e.g. βπ(π₯) = sin(π₯)β β Range: β1 β€ π(π₯) β€ 1 However be careful if domain is restricted: βπ(π₯) = sin(π₯) , 180 β€ π₯ < 360β. Range: β1 β€ π(π₯) β€ 0 (using a sketch)
For βpiecewise functionβ, fully sketch it first to find range. βThe function π(π₯) is defined for all π₯:
π(π₯) = {4 π₯ < β2
π₯2 β2 β€ π₯ β€ 212 β 4π₯ π₯ > 2
Determine the range of π(π₯).β From the sketch it is clear
π(π₯) β€ 4 You may be asked to construct a function given information
about its domain and range. e.g. βπ¦ = π(π₯) is a straight line. Domain is 1 β€ π₯ β€ 5 and range is 3 β€ π(π₯) β€ 11. Work out one possible expression for π(π₯).β Weβd have this domain and range if line passed through points (1,3) and (5,11). This gives us π(π₯) = 2π₯ + 1
Not understanding what π(π₯2) actually means. Not sketching the graph! (And hence not being able to visualise what the range should be). This is particularly important for βpiecewiseβ functions. Not being discerning between < and β€ in the range. e.g. For quadratics you should have β€ but for exponential graphs you should have <. Writing the range of a function in terms of π₯ instead of the correct π(π₯).
2.4 Expanding brackets and collecting like terms.
Deal with brackets with more than two things in them. e.g. (π₯ + π¦ + 1)(π₯ + π¦) = π₯2 + π¦2 + 2π₯π¦ + π₯ + π¦ Just do βeach thing in first bracket times each in secondβ
Deal with three (or more) brackets. Just multiply out two brackets first, e.g.
(π₯ + 2)3 = (π₯ + 2)(π₯ + 2)(π₯ + 2) = (π₯ + 2)(π₯2 + 4π₯ + 4) = π₯3 + 4π₯2 + 4π₯ + 2π₯2 + 8π₯ + 8 = π₯3 + 6π₯2 + 12π₯ + 8
Classic error of forgetting that two negatives multiply to give a positive. E.g. in (π¦ β 4)3
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2.5 Factorising GCSE recap: You should know how to factorise difference of two squares, quadratics of form π₯2 + ππ₯ + π, form ππ₯2 + ππ₯ + π and where you have a common factor. 1. Sometimes multiple factorisation steps are required. π₯4 β 25π₯2 = π₯2(π₯2 β 25) = π₯2(π₯ + 5)(π₯ β 5) 2. Sometimes use βintelligent guessingβ of brackets, e.g. 15π₯2 β 34π₯π¦ β 16π¦2 β (5π₯ + 2π¦)(3π₯ β 8π¦) 3. If the expression is already partly factorised, identify common factors rather than expanding out and starting factorising from scratch, e.g. βFactorise (2π₯ + 3)2 β (2π₯ β 5)2β While we could expand, we might recognise we have the difference of two squares:
= ((2π₯ + 3) + (2π₯ β 5))((2π₯ + 3) β (2π₯ β 5))
= (4π₯ β 2)(8) = 16(2π₯ β 1)
Or βFactorise (π₯ + 1)(π₯ + 3) + π¦(π₯ + 1)β = (π₯ + 1)(π₯ + 3 + π¦)
2.6 Manipulation of rational expressions: Use of + βΓΓ· for algebraic fractions with denominators being numeric, linear or quadratic.
βSimplify π₯2+3π₯β10
π₯2β9Γ·
π₯+5
π₯2+3π₯β
Factorise everything first: (π₯ + 5)(π₯ β 2)
(π₯ + 3)(π₯ β 3)Γ·
π₯ + 5
π₯(π₯ + 3)
Flip the second fraction and change Γ· to Γ:
=(π₯ + 5)(π₯ β 2)
(π₯ + 3)(π₯ β 3)Γ
π₯(π₯ + 3)
π₯ + 5
=π₯(π₯ + 5)(π₯ β 2)(π₯ + 3)
(π₯ + 3)(π₯ β 3)(π₯ + 5)
=π₯(π₯ β 2)
π₯ β 3
βSimplify π₯3+2π₯2+π₯
π₯2+π₯β
=π₯(π₯2 + 2π₯ + 1)
π₯(π₯ + 1)=
π₯(π₯ + 1)(π₯ + 1)
π₯(π₯ + 1)= π₯ + 1
2.7 Use and manipulation of formulae and expressions.
Same skills as GCSE. Isolate subject on one side of equation, and factorise it out if necessary.
e.g. βRearrange 1
π=
1
π’+
1
π£ to make π£ the subject.β
Multiplying everything by ππ’π£: π’π£ = ππ£ + ππ’ π’π£ β ππ£ = ππ’ π£(π’ β π) = ππ’
π£ =ππ’
π’ β π
2.8 Use of the factor theorem for integer values of the variable including cubics.
In: 7 Γ· 3 = 2 πππ 1, the 7 is the βdividendβ, the 3 is the βdivisorβ, the 2 is the βquotientβ and the 1 is the βremainderβ. Remainder Theorem: For a polynomial π(π₯), the remainder when π(π₯) is divided by (π₯ β π) is π(π). Factor Theorem: If π(π) = 0, then by above, the remainder is 0. Thus (π₯ β π) is a factor of π(π₯). e.g. When π₯2 β 3π₯ + 2 is divided by π₯ β 2, remainder is π(2) = 22 β3(2) + 2 = 0 (thus π₯ β 2 is a factor as remainder is 0). Note that we negated the -2 and subbed in 2 into the original equation.
βShow that (π₯ β 2) is a factor of π₯3 + π₯2 β 4π₯ β 4β π(2) = 23 + 22 β 4(2) β 4 = 0
βIf (π₯ β 5) is a factor of π₯3 β 6π₯2 + ππ₯ β 20, determine the value of πβ
π(5) = 53 β 6(52) + 5π β 20 = 0 5π β 45 = 0 β π = 9
Harder questions might involve giving you two factors, and two unknowns β just use factor theorem to get two equations, then solve simultaneously.
βFully factorise π₯3 β 3π₯2 β 4π₯ + 12β Step 1: Try a few values of π₯ until you stumble upon a factor. π(1) = 13 β 3(12) β 4(1) + 12 = 6 (so not a factor) π(2) = 23 β 3(22) β 4(2) + 12 = 0 (so (π₯ β 2) a factor)
If you find π(2) = 0, then the factor is (π₯ β2) not (π₯ + 2).
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Step 2: Since a cubic, must factorise like: (π₯ β 2)(π₯+? )(π₯+? )
Since all the constants must multiply to give +12 (as in the original cubic), the two missing numbers must multiply to give -6. This rounds down what we need to try with the remainder theorem: π(β2) = (β2)3 β 3(β2)2 + β― = 0 (therefore a factor) Since (π₯ + 2) is a factor, last factor must be (π₯ β 3)
βSolve π₯3 + π₯2 β 10π₯ + 8 = 0β Using above technique, factorisating gives:
(π₯ β 1)(π₯ β 2)(π₯ + 4) = 0 π₯ = 1, 2 ππ β 4
2.9 Completing the square.
Recap of GCSE: Halve number in front of π₯ for number inside bracket. Square this and βthrow it awayβ. e.g. π₯2 + 4π₯ β 3 β (π₯ + 2)2 β 4 β 3 = (π₯ + 2)2 β 7
If coefficient of π₯2 (i.e. the constant on the front of the π₯2 term) is not 1, always factorise this number out first, even if other numbers donβt have this as a factor:
3π₯2 + 12π₯ β 5
= 3 (π₯2 + 4π₯ β5
3)
= 3 ((π₯ + 2)2 β 4 β5
3)
= 3(π₯ + 2)2 β 12 β 5 = 3(π₯ + 2)2 β 17
In the penultimate line we expanded out the outer 3(β¦ ) bracket. You can check your answer by expanding it back out and seeing if you get the original expression.
Rearrange terms into form ππ₯2 + ππ₯ + π first, e.g. βExpress 2 β 4π₯ β 2π₯2 in the form π β π(π₯ + π)2β
= β2π₯2 β 4π₯ + 2 = β2(π₯2 + 2π₯ β 1) = β2((π₯ + 1)2 β 1 β 1) = β2(π₯ + 1)2 + 4 = 4 β 2(π₯ + 1)2
We always subtract when we βthrow awayβ, even if number we halved was negative, e.g. π₯2 β 6π₯ + 1 =(π₯ β 3)2 β 9 + 1 Often students add by mistake. When coefficient of π₯2 is not 1, be careful to maintain your outer bracket until the last step.
2.10 Sketching of functions. Sketch graphs of linear and quadratic functions.
Quadratics:
e.g. βSketch π¦ = π₯2 β π₯ β 2β a. Factorise and set π¦ to 0 to find x-intercepts (known as βrootsβ).
(π₯ + 1)(π₯ β 2) = 0 β (β1,0), (2,0)
b. Set π₯ = 0 to find π¦-intercept. (0, β2) c. π₯2 term is positive therefore βsmiley faceβ shape.
Sketch π¦ = βπ₯2 + 5π₯ β 4 Roots: βπ₯2 + 5π₯ β 4 = 0
π₯2 β 5π₯ + 4 = 0 (π₯ β 1)(π₯ β 4) = 0 β (1,0), (4,0)
But note frowney face shape as π₯2 term is negative.
You may have to go backwards: find the equation given the sketch. In left example, brackets of
π¦ = (4π₯ + 1)(3π₯ β 2) would work as if 4π₯ + 1 = 0 then
π₯ = β1
4 and similarly with (3π₯ β 2)
GCSE recap: We can complete the square to find the
minimum/maximum point (or use ππ¦
ππ₯= 0!)
π¦ = π₯2 β 6π₯ + 10 = (π₯ β 3)2 + 1
Min point is (3,1) (If π¦ = (π₯ + π)2 + π then minimum point is (βπ, π))
Typical mistake is to do x-intercepts of say 2 and 3 if
π¦ = (π₯ + 2)(π₯ + 3) Similarly if π¦ = (π₯ + 2)2 + 3, an incorrect minimum point would be (2,3) We might similarly make sign errors if doing the reverse: finding the equation from the graph. If one
of the x-intercepts is 3
4,
then one of brackets is (4π₯ β 3), not (4π₯ + 3) or (3π₯ + 4) or (3π₯ β4). Check by setting bracket to 0 and solving.
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Cubics
When cubic is in factorised form then: If (π₯ + π) appears once, curve crosses π₯-axis at βπ If (π₯ + π)2 appears, curve touches at π₯ = βπ As before, can get π¦-intercept by setting π₯ = 0. If π₯3 term is positive, uphill zig-zag shape, otherwise downhill. e.g.
You may have to factorise yourself, e.g. π¦ = π₯3 β 12π₯2 β π₯2(π₯ β 12) which crosses at π₯ = 12 and touches at π₯ = 0. And again you may need to suggest an equation. Suitable equation for graph on right:
π¦ = (π₯ + 2)2(π₯ β 3) (Note that reciprocal graphs are in
C1 only β you can find this in my IGCSEFM Sketching Graphs slides)
Piecewise Functions
Just a function defined in βpiecesβ. So in the above example, the function to draw between 0 β€ π₯ < 1 is π¦ = π₯2 Note: You do not need to know about graph transforms for IGCSEFM (but do for C1).
2.11 Solution of linear and quadratic equations
GCSE recap. Solve π₯2 + 2π₯ β 3 = 0β¦
β¦by factorisation. (π₯ + 3)(π₯ β 1) = 0 β π₯ = β3, 1
β¦by completing the square (π₯ + 1)2 β 4 = 0 (π₯ + 1)2 = 4 π₯ + 1 = Β±2 π₯ = +2 β 1 = 1, π₯ = β2 β 1 = β3
β¦by formula π = 1, π = 2, π = β3
π₯ =β2 Β± β4 β (4 Γ 1 Γ β3)
2= β―
β¦by graph Sketch π¦ = π₯2 + 2π₯ β 3. Comparing to original equation, weβve substituted π¦ for 0. So interested in values of π₯ for which π¦ = 0. We could similarly βlook upβ values of π₯ for other values of π¦, e.g. π₯2 + 2π₯ β 3 = 5.
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2.12 Algebraic and graphical solution of simultaneous equations in two unknowns where the equations could both be linear or one linear and one second order.
To βgraphicallyβ solve simultaneous equations, sketch both lines, and look at the points where they intersect.
Equations might not be in the usual ππ₯ + ππ¦ = π form; if not, rearrange them! e.g.
2π¦ + 3π₯ + 4 2π₯ = β3π¦ β 7
becomes: 3π₯ β 2π¦ = β4 2π₯ + 3π¦ = β7
then solve in usual GCSE manner. Could also solve by substitution. Similarly:
π₯ β 1
π¦ β 2= 3
π₯ + 6
π¦ β 1= 4
becomes: π₯ β 1 = 3π¦ β 6 π₯ + 6 = 4π¦ β 4
For βone linear, one quadraticβ, solve as per GCSE method: rearrange linear equation to make π₯ or π¦ the subject, then sub into quadratic equation and solve, e.g.
π₯ + π¦ = 4 π¦2 = 4π₯ + 5
Then: π¦ = 4 β π₯ (4 β π₯)2 = 4π₯ + 5 16 β 8π₯ + π₯2 = 4π₯ + 5 π₯2 β 12π₯ + 11 = 0 (π₯ β 11)(π₯ β 1) = 0 π₯ = 11 β π¦ = β7 π₯ = 1 β π¦ = 3
Suppose that π₯ = π¦ +4 and π₯2 + π¦2 = 20 Common error is to accidentally drop the +π¦2 after subbing in the π₯, i.e.
π¦2 + 8π₯ + 16 = 20 (extra +π¦2 has gone!) Other common error is squaring brackets. e.g. (4 β π₯)2 = 16 β8π₯ + π₯2 Incorrect variants:
16 β π₯2 16 + π₯2 16 β 8π₯ β π₯2 Donβt forget to find the values of the other variable at the end, and make sure itβs clear which π₯ matches up to which π¦.
2.13 Solution of linear and quadratic inequalities
When solving linear inequalities, just remember that dividing or multiplying by a negative number reverses the direction of the inequality. You can avoid this by putting π₯ on the side which is positive.
To solve quadratic inequalities. 2π₯2 + 5π₯ β€ 3
2π₯2 + 5π₯ β 3 β€ 0 Get 0 on one side. (2π₯ β 1)(π₯ + 3) β€ 0 Factorise
This gives us βcritical valuesβ of π₯ =1
2, π₯ = β3
Then you MUST SKETCH. Since on the left we sketched π¦ = (2π₯ β 1)(π₯ + 3) weβre interested where π¦ β€ 0 This is in indicated region on left, i.e. where
β3 β€ π₯ β€1
2
Had we wanted
(2π₯ β 1)(π₯ + 3) β₯ 0, this would have given us the two βtailsβ of the
graph, and weβd write βπ₯ < β3 ππ π₯ β₯1
2β
When solving quadratic inequalities, students usually get the βcritical valuesβ right but stumble at the last hurdle because they donβt sketch their quadratic, and therefore guess which way the inequality is supposed to go. Use the word βorβ when you want the two tails (and not βandβ or comma)
2.14 Index laws, including fractional and negative indices.
GCSE recap: π₯βπ =1
π₯π π₯0 = 1
82
3 = 22 = 4 25β1
2 =1
2512
=1
5
(27
8)
β23
= (8
27)
23
= (2
3)
2
=4
9
Example: βSolve π₯3
4 = 27β Just raise both sides to the reciprocal of the power to βcancel it outβ.
(π₯34)
43
= 2743
π₯ = 81 Convert any mixed numbers to improper fractions first.
βSolve π₯β2
3 = 27
9β
(π₯β23)
β32
= (25
9)
β32
π₯ = (9
25)
32
= (3
5)
3
=27
125
You might get confused with straight line equations and raise both sides to the negative reciprocal rather than just the reciprocal?
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GCSE recap: To raise an algebraic term to a power, simply do each part of the term to that power, e.g.
(3π₯2π¦3)2 β 9π₯4π¦6
(9π₯4π¦)12 β 3π₯2π¦
12
2.15 Algebraic Proof
βProve that the difference between the squares of two consecutive odd numbers is a multiple of 8.β Let two consecutive odd numbers be 2π + 1 and 2π + 3
(2π + 3)2 β (2π + 1)2 = 4π2 + 12π + 9 β 4π2 β 4π β 1 = 8π + 8 = 8(π + 1)
which is divisible by 8. (The factoring out of 8 makes the divisibility explicit) βProve that π₯2 β 4π₯ + 7 > 0 for all π₯β (Just complete the square!)
(π₯ β 2)2 β 4 + 7 = (π₯ β 2)2 + 3
(π₯ β 2)2 β₯ 0 thus (π₯ β 2)2 + 3 > 0 for all π₯ βIn this identity, β and π are integer constants.
4(βπ₯ β 1) β 3(π₯ + β) β‘ 5(π₯ + π) Work out the values of β and πβ The β‘ means the left-hand-side and right-hand-side are equal for all values of π₯ (known as an identity). Compare the coefficients of π₯ and separately compare constants:
4βπ₯ β 4 β 3π₯ β 3β = 5π₯ + 5π
Comparing π₯ terms: 4β β 3 = 5 β β = 2
Comparing constant terms: β4 β 3β = 5π β π = β2
2.16 Sequences: nth terms of linear and quadratic sequences. Limiting value of a sequence as π β β
Linear sequences recap: 4, 11, 18, 25β¦ β nth term 7π β 3 For quadratic sequences, i.e. where second difference is constant:
Limiting values:
βShow that the limiting value of 3π+1
6πβ5 is
1
2 as π β ββ
π β β means βas π tends towards infinityβ.
Write βAs π becomes large, 3π+1
6πβ5β
3π
6π=
1
2β
The idea is that as π becomes large, the +1 and -5 become
inconsequential, e.g. if π = 1000, then 3001
5995β
3000
6000=
1
2
Make sure you check your formula against the first few terms of the sequence by using π = 1, 2, 3.
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3. Co-ordinate Geometry (2 dimensions only)
Specification Notes What can go ugly
3.1 Know and use the definition of gradient
Gradient is the change in π¦ for each unit increase in π₯.
π =Ξπ¦
Ξπ₯ (change in π¦ over change in π₯)
e.g. If a line goes through (2,7) and (6,5)
π = β2
4= β
1
2
Doing Ξπ₯
Ξπ¦ accidentally, or
getting one of the two signs wrong.
3.2 Know the relationship between the gradients of parallel and perpendicular lines.
Parallel lines have the same gradient. For perpendicular lines:
One gradient is the negative reciprocal of the other.
e.g. 2 β β1
2 β 4 β
1
4
1
5β β5
2
3β β
3
2
Remember that the reciprocal of a fraction flips it.
To show two lines are perpendicular, show the product of the gradients is -1:
β1
4Γ 4 = β1
Example: βShow that π΄(0,0), π΅(4,6), πΆ(10,2) form a right-angled triangle.β Gradients are:
ππ΄π΅ =6
4=
3
2 ππ΄πΆ =
2
10=
1
5, ππ΅πΆ =
β4
6= β
2
3
Since 3
2Γ β
2
3= β1, lines π΄π΅ and π΅πΆ are perpendicular so
triangle is right-angled.
Doing just the reciprocal rather than the βnegative reciprocalβ.
3.3 Use Pythagorasβ Theorem to calculate the distance between two points.
π = βΞπ₯2 + Ξπ¦2 e.g. If points are (3,2) and (6, β2), then
π = β32 + 42 = 5 Note that it doesnβt matter if the βchangeβ is positive or negative as weβre squaring these values anyway.
3.4 Use ratio to find the coordinates of a point on a line given the coordinates of two other points.
βTwo points π΄(1,5) and π΅(7,14) form a straight line. If a point πΆ(5, π) lies on the line, find π.β Method 1 (implied by specification on left): On the π₯ axis, 5 is 4 6ths of the way between 1 and 7. So β4 6thsβ of the way between 5 and 14 is
π = 5 +4
6Γ 9 = 11
Method 2 (easier!): Find equation of straight line first. Using π¦ β π¦1 = π(π₯ β π₯1):
π =9
6=
3
2
π¦ β 5 =3
2(π₯ β 1)
Thus if π₯ = 5 and π = 5:
π β 5 =3
2(5 β 1)
π = 11
3.5 The equation of a straight line in the forms π¦ = ππ₯ + π and π¦ β π¦1 =π(π₯ β π₯1)
βA line goes through the point (4,5) and is perpendicular to the line with equation π¦ = 2π₯ + 6. Find the equation of the line. Put your answer in the form π¦ = ππ₯ + πβ For all these types of questions, we need (a) the gradient and (b) a point, in order to use π¦ β π¦1 = π(π₯ β π₯1):
π = β1
2
π¦ β 5 = β1
2(π₯ β 4)
π¦ β 5 = β1
2π₯ + 2
π¦ = β1
2π₯ + 7
βDetermine the coordinate of the point where this line crosses the π₯ axisβ
0 = β1
2π₯ + 7 β π₯ = 14 β (14,0)
Donβt confuse π₯ and π₯1 in the straight line equation. π₯1 and π¦1 are constants, representing the point (π₯1, π¦1) the line goes through. π₯ and π¦ meanwhile are variables and must stay as variables. Be careful with negative values of π₯ or π¦, e.g. if π = 3 and (β2,4) is the point, then:
π¦ β 4 = 3(π₯ + 2)
3.6 Draw a straight line from given information.
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3.7 Understand the equation of a circle with any centre and radius.
Circle with centre (π, π) and radius π is: (π₯ β π)2 + (π¦ β π)2 = π2
Examples:
βA circle has equation (π₯ + 3)2 + π¦2 = 25. What is its centre and radius?β Centre: (β3,0) π = 5
βDoes the circle with equation π₯2 + (π¦ β 1)2 = 16 pass through the point (2,5)?β In general a point is on a line if it satisfies its equation.
22 + (5 β 1)2 = 16 20 = 16
So no, it is not on the circle.
βA circle has centre (3,4) and radius 5. Determine the coordinates of the points where the circle intercepts the π₯ and π¦ axis.β Firstly, equation of circle: (π₯ β 3)2 + (π¦ β 4)2 = 25 On π₯-axis: π¦ = 0:
(π₯ β 3)2 + (0 β 4)2 = 25 (π₯ β 3)2 = 9 π₯ β 3 = Β±3 β (0,0), (6,0)
On π¦-axis, π₯ = 0: (0 β 3)2 + (π¦ β 4)2 = 25 (π¦ β 4)2 = 16 π¦ β 4 = Β±4 β (0,0), (0,8)
βπ΄(4,7) and π΅(10,15) are points on a circle and π΄π΅ is the diameter of the circle. Determine the equation of the circle.β We need to find radius and centre. Centre is just midpoint of diameter: (7,11) Radius using (4,7) and (7,11):
βΞπ₯2 + Ξπ¦2 = β32 + 42 = 5 Equation: (π₯ β 7)2 + (π¦ β 11)2 = 25 See slides for harder questions of this type.
βπ₯2 β 2π₯ + π¦2 β 6π¦ = 0 is the equation of a circle. Determine its centre and radius.β Need to complete the square to get in usual form.
(π₯ β 1)2 β 1 + (π¦ β 3)2 β 9 = 0 (π₯ β 1)2 + (π¦ β 3)2 = 10
Centre: (1,3) π = β10 Using Circle Theorems
Angle in semicircle is 90Β°: which means that the two chords will be perpendicular to each other (i.e. gradients will multiply to give -1).
The perpendicular from the centre of the chord passes through the centre of the circle. Example: βTwo points on the circumference of a circle are (2,0) and (0,4). If the centre of the circle is (6, π), determine π.β
Gradient of chord: β4
2= β2 Midpoint of chord: (1,2)
Gradient of radius =1
2
Equation of radius: π¦ β 2 =1
2(π₯ β 1)
If π₯ = 6: π¦ β 2 =1
2(6 β 1)
π¦ = 4.5
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The tangent to a circle is perpendicular to the radius. Example: βThe equation of this circle is π₯2 + π¦2 = 20. π(4,2) is a point on the circle. Work out the equation of the tangent to the circle at π, in the form π¦ = ππ₯ + πβ As always, to find an equation we need (i) a point and (ii) the gradient.
Point: (4,2) Gradient of radius is 2
4=
1
2
β΄ Gradient of tangent = β2 π¦ β 2 = β2(π₯ β 4) π¦ = β2π₯ + 10
4. Calculus
Specification Notes What can go ugly
4.1 Know that the gradient
function ππ¦
ππ₯ gives the
gradient of the curve and measures the rate of change of π¦ with respect to π₯.
Whereas with say π¦ = 3π₯ + 2 the gradient is constant (π =
3), with curves, the gradient depends on the point. ππ¦
ππ₯ is the
gradient function: it takes an π₯ value and gives you the gradient at that point.
e.g. If ππ¦
ππ₯= 2π₯, then at (5,12), the gradient is 2 Γ 5 = 10.
Technically this the gradient of the tangent at this point.
Another way of interpreting ππ¦
ππ₯ is βthe rate of change of π¦
with respect to π₯.β
4.2 Know that the gradient of a function is the gradient of the tangent at that point.
4.3 Differentiation of ππ₯π where π is a positive integer or 0, and the sum of such functions.
Multiply by power and then reduce power by 1.
π¦ = π₯3 β ππ¦
ππ₯= 3π₯2
π¦ = 5π₯2 β ππ¦
ππ₯= 10π₯
π¦ = 7π₯ β ππ¦
ππ₯= 7
π¦ = β3 β ππ¦
ππ₯= 0
Put expression in form ππ₯π first, and split up any fractions. Then differentiate.
π¦ = (2π₯ + 1)2 = 4π₯2 + 4π₯ + 1
ππ¦
ππ₯= 8π₯ + 4
π¦ = βπ₯ = π₯12
ππ¦
ππ₯=
1
2π₯β
12
π¦ =1 + π₯
βπ₯= π₯β
12 + π₯
12
ππ¦
ππ₯= β
1
2π₯β
32 +
1
2π₯β
12
Donβt forget that constants disappear when differentiated. Common mistake is to reduce power by 1 then multiply by this new power.
Donβt forget that 1
βπ₯=
π₯β1
2 with a negative power.
4.4 The equation of a tangent and normal at any point on a curve.
Use ππ¦
ππ₯ to find gradient at specific point (ensuring you use
the negative reciprocal if we want the normal). You may need to use the original equation to also find π¦. Then use π¦ β π¦1 = π(π₯ β π₯1) Example: βWork out the equation of the tangent to the curve π¦ = π₯3 + 5π₯2 + 1 at the point where π₯ = β1.β
ππ¦
ππ₯= 3π₯2 + 10π₯
π = 3(β1)2 + 10(β1) = β7 π¦ = (β1)3 + 5(β1)2 + 1 = 5
Therefore: π¦ β 5 = β7(π₯ + 1)
βWork out the equation of the normal to the curve π¦ = π₯3 +5π₯2 + 1 at the point where π₯ = β1.β Exactly the same, except we use negative reciprocal for the gradient:
π¦ β 5 =1
7(π₯ + 1)
Donβt mix up the tangent to the curve and the normal to a curve (the latter which is perpendicular to the tangent).
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4.5 Use of differentiation to find stationary points on a curve: maxima, minima and points of inflection.
At min/max points, the curve is flat, and the gradient therefore 0. Use gradient value just before and after turning point to work out what type it is.
Example: βA curve has equation π¦ = 4π₯3 + 6π₯2 + 3π₯ + 5. Work out the coordinates of any stationary points on this curve and determine their nature.β
ππ¦
ππ₯= 12π₯2 + 12π₯ + 3 = 0
4π₯2 + 4π₯ + 1 = 0 (2π₯ + 1)2 = 0
π₯ = β1
2
Find the π¦ value of the stationary point:
π¦ = 4 (β1
2)
3
+ 6 (β1
2)
2
+ 3 (β1
2) + 5 =
9
2
So stationary point is (β1
2,
9
2).
Look at gradient just before and after:
When π₯ = β0.51,ππ¦
ππ₯= 0.0012
When π₯ = β0.49,ππ¦
ππ₯= 0.0012
Both positive, so a point of inflection.
Common error is to forget to find the π¦ value of the stationary point when asked for the full coordinate
4.6 Sketch a curve with known stationary points.
Self-explanatory. Just plot the points and draw a nice curve to connect them.
5. Matrix Transformations
Specification Notes What can go ugly
5.1 Multiplication of matrices
Do each row of the first matrix βmultipliedβ by each column of the second. And by βmultiplyβ, multiply each pair of number of numbers pairwise, and add these up. See my slides for suitable animation! e.g.
(1 23 4
) (56
) = (1 Γ 5 + 2 Γ 63 Γ 5 + 4 Γ 6
) = (1739
)
(1 23 4
) (1 02 10
) = (5 20
11 40)
Important: When we multiply by a matrix, it goes on the front. So π¨ multiplied by π© is π©π¨, not π¨π©.
When multiplying matrices, doing each column in the first matrix multiplied by each row in the second, rather than the correct way.
5.2 The identity matrix, π° (2 Γ 2 only).
π° = (1 00 1
)
Just as β1β is the identity in multiplication of numbers, as π Γ1 = π and 1 Γ π = π (i.e. multiplying by 1 has no effect), π° is the same for matrices, i.e. π¨π° = π°π¨ = π¨.
5.3 Transformations of the unit square in the π₯ β π¦ plane.
Matrices allow us to represent transformations such as enlargements, rotations and reflections. Example: βFind the matrix that represents the 90Β° clockwise rotation of a 2D point about the origin.β
Easiest way to is to consider some arbitrary point, say (13
),
and use a sketch to see where it would be after the
transformation, in this case (3
β1). Thus more generically
weβre looking for a matrix such that:
( ) (π₯π¦) = (
π¦βπ₯
)
It is easy to see this will be (0 1
β1 0)
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Using the same technique we can find:
Rotation 90Β° anticlockwise about the origin: (0 β11 0
)
Reflection 180Β° about the origin: (β1 00 β1
)
Reflection in the line π¦ = π₯: (0 11 0
)
Reflection in the line π₯ = 0: (β1 00 1
)
Reflection in the line π¦ = 0: (1 00 β1
)
Enlargement scale factor 2 centre origin: (2 00 2
)
Note that a rotation is anticlockwise if not specified.
The βunitβ square consists of the points (00
), (10
), (11
), (01
).
To find the effect of a transformation on a unit square, just transform each point in turn. e.g. βOn the grid, draw the image of the unit square after it is
transformed using the matrix (3 00 3
).β
Transforming the second point for example we get:
(3 00 3
) (10
) = (30
)
5.4 Combination of transformations.
The matrix π΅π΄ represents the combined transformation of π΄ followed by π΅. Example:
βA point π is transformed using the matrix (β1 00 1
), i.e. a
reflection in the line π₯ = 0, followed by (0 11 0
), i.e. a
reflection in the line π¦ = π₯. (a) Give a single matrix which represents the
combined transformation. (b) Describe geometrically the single transformation
this matrix represents.β
(a) (0 11 0
) (β1 00 1
) = (0 1
β1 0)
(b) Rotation 90Β° clockwise about the origin.
It is easy to accidentally multiply the matrices the wrong way round. It does matter which way you multiply them!
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6. Geometry
Specification Notes What can go ugly
6.1 Perimeter and area of common shapes including
area of triangle 1
2ππ sin πΆ
and volumes of solids. Circle Theorems.
6.2 Geometric proof: Understand and construct geometric proofs using formal arguments.
Examples: βTriangle π΄π΅πΆ is isosceles with π΄πΆ = π΅πΆ. Triangle πΆπ·πΈ is isosceles with πΆπ· = πΆπΈ. π΄πΆπ· and π·πΈπΉ are straight lines. (a) Prove that angle π·πΆπΈ = 2π₯ and (b) Prove that π·πΉ is perpendicular to π΄π΅β Make clear at each point what the angle is youβre calculating, with an appropriate reason. It may help to work out the angles on the diagram first, before writing out the steps.
(a) β πΆπ΅π΄ = π₯ (base angles of isosceles triangle are equal) β π΄πΆπ΅ = 180 β 2π₯ (angles in Ξπ΄π΅πΆ add to 180) β π·πΆπΈ = 2π₯ (angles on straight line add to 180)
(b) β π·πΈπΆ =180β2π₯
2= 90 β π₯ (base angles of
isosceles triangle are equal) β π·πΉπ΄ = 180 β (90 β π₯) β π₯ = 90Β°
β΄ π·πΉ is perpendicular to π΄π΅. (In general with proofs itβs good to end by restating the thing youβre trying to prove) βπ΄, π΅, πΆ and π· are points on the circumference of a circle such that π΅π· is parallel to the tangent to the circle at π΄. Prove that π΄πΆ bisects angle π΅πΆπ·.β β π΅πΆπ΄ = β π΅π΄πΈ (by Alternate Segment Theorem) β π΅π΄πΈ = β π·π΅π΄ (alternate angles are equal) β π·π΅π΄ = β π΄πΆπ· (angles in the same segment are equal) So β π΅πΆπ΄ = β π΄πΆπ·. π΄πΆ bisects β π΅πΆπ·.
Not given reasons for each angle. Angles (and their reasons) not being given in a logical sequence. Misremembering Circle Theorems! (learn the wording of these verbatim)
6.3 Sine and cosine rules in scalene triangles.
Sine rule: π
sin π΄=
π
sin π΅=
π
sin πΆ (recall from GCSE that if have a
missing angle, put sinβs at top). Cosine rule: π2 = π2 + π2 β 2ππ cos π΄ (use when missing side is opposite known angle, or all three sides known and angle required) Example: βIf area is 18cm2, work out π¦.β
Area is given, so use area formula:
1
2Γ π€ Γ 2π€ Γ sin 30Β° = 18
1
2π€2 = 18 β π€ = 6
Then using cosine rule to find π¦: π¦2 = 62 + 122 β 2 Γ 6 Γ 12 Γ cos 30Β° π¦ = 7.44ππ
Forgetting to square root at the end when using cosine rule to find a side.
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6.4 Use of Pythagorasβ Theorem in 2D and 3D.
We often have to form a 2D triangle βfloatingβ in 3D. e.g. βFind the length of the diagonal joining opposite corners of a unit cube.β We want the hypotenuse of the indicated shaded triangle. First use Pythagoras on base of cube
to get β2 bottom length of triangle. Then required length is
β2 + 12 = β3.
6.5 Find angle between a line and a plane, and the angle between two planes.
β(a) Work out the angle between line ππ΄ and plane π΄π΅πΆπ·.β I use what I call the βpen dropβ strategy. If a pen was the line VA and I dropped it onto the plane (ABCD), it would fall to AX. Thus the angle weβre after is between VA and
AX. By using simple trig on the triangle VAX (and using
Pythagoras on the square base to get π΄π = β34), we get
β ππ΄π = tanβ1 (5
β34)
(b) βWork out the angle between the planes πππ and πππ π.β
When the angle is between planes, our βpenβ this time must be perpendicular to the line formed by the intersection of the two planes. Thus we put our βpenβ between π and the midpoint of π π. We then drop the βpenβ onto the
plane ππππ . We get the pictured triangle.
In (b) in the example, we might accidentally find the angle between VQ and the plane (this angle will be too steep).
6.6 Graphs π¦ = sin π₯ , π¦ =cos π₯ , π¦ = tan π₯ for 0Β° β€π₯ β€ 360Β°
π¦ = sin π₯ π¦ = tan π₯
π¦ = cos π₯
(Note that π¦ = tan π₯ has asymptotes at π₯ = 90Β°, 270Β°, etc. The result is that tan is undefined at these values)
6.7 Be able to use the definitions sin π , cos π and tan π for any positive angle up to 360Β° (measured in degrees only)
The 4 rules of angles as I call them! 1. sin(π₯) = sin(180Β° β π₯) 2. cos(π₯) = cos(360Β° β π₯) 3. π ππ and πππ repeat every 360Β° 4. π‘ππ repeats every 180Β°
Weβll see this used in [6.10].
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6.8 Knowledge and use of 30Β°, 60Β°, 90Β° triangles and 45Β°, 45Β°, 90Β°.
We can use half a unit square (which has angles 45Β°, 45Β°, 90Β°) and half an equilateral triangle originally with sides 2 (angles 30Β°, 60Β°, 90Β°), as pictured below, to get exact values of sin 30Β° , sin 45Β°, etc. We use Pythagoras to obtain the remaining side length.
Then using simple trigonometry on these triangles:
sin 30Β° =1
2
sin 60Β° =β3
2
sin 45Β° =1
β2
Similarly cos 30Β° =β3
2, cos 60Β° =
1
2, cos 45Β° =
1
β2
tan 30Β° =1
β3, tan 60Β° = β3, tan 45Β° = 1
You donβt need to memorise all these, just the two triangles!
6.9 Trig identities tan π =sin π
cos π and sin2 π + cos2 π =
1
Remember that sin2 π just means (sin π)2 βProve that 1 β π‘ππ π π ππ π πππ π β‘ πππ 2 πβ
Generally a good idea to replace tan π with sin π
cos π.
1 βsin π
cos πsin π cos π β‘ cos2 π
1 βsin2 π cos π
cos πβ‘ cos2 π
1 β sin2 π β‘ cos2 π cos2 π β‘ cos2 π
βProve that π‘ππ π +1
π‘ππ πβ‘
1
π ππ π πππ πβ
Generally a good idea to combine any fractions into one. sin π
cos π+
cos π
sin πβ‘
1
sin π cos π
sin2 π + cos2 π
sin π cos πβ‘
1
sin π cos π
1
sin π cos πβ‘
1
sin π cos π
See my slides for more examples.
6.10 Solution of simple trigonometric equations in given intervals.
βSolve πππ(π) = βπ. π in the range πΒ° β€ π < πππΒ°β π₯ = sinβ1(β0.3) = β17.46Β°
At this point, we use the rules in [6.7] to get the solutions in the range provided. We usually get a pair of solutions for each 360Β° interval: 180 β β17.46 = 197.46Β° (since sin(π₯) = sin(180Β° β π₯)) β17.46Β° + 360Β° = 342.54Β° (since sin repeats every 360Β°) βSolve π πππ(π) = π in the range πΒ° β€ π < πππΒ°β
tan(π₯) =1
2
π₯ = tanβ1 (1
2) = 26.6Β°
26.6Β° + 180Β° = 206.6Β° (tan repeats every 180Β°) βSolve πππ π = π πππ π in the range πΒ° β€ π < πππΒ°β When you have a mix of π ππ and πππ (neither squared), divide both sides of the equation by πππ :
tan π₯ = 2 π₯ = tanβ1(2) = 63.4Β°, 243.4Β°
βSolve ππππ π½ + π πππ π½ = π in the range πΒ° β€ π < πππΒ°β Factorising: tan π (tan π + 3) = 0
tan π = 0 ππ tan π = β3 π = 0Β°, 180Β°, β 71.6Β°, 108.4Β°, 288.4Β°
(Cross out any solutions outside the range, i.e. β71.6Β°)
One of two main risks: (a) Missing out solutions, either because we havenβt used all the applicable rules in 6.7, or weβve forgotten the negative solution when square rooting both sides (where applicable). In tan2 π + 3 tan π = 0, it would be wrong to divide by tan π because we lose the solution where tan π = 0 (in general, never divide both sides of an equation by an expression involving a variable β always factorise!) (b) Mixing up the rules in 6.7, e.g. doing 180 β when you were supposed to 360 β
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βSolve ππππ π½ =π
π in the range πΒ° β€ π < πππΒ°β
You get both cos π =1
2 and cos π = β
1
2, so solve both!
βSolve π ππππ π½ β πππ π½ β π = π in the range πΒ° β€ π < πππΒ°β Again factorising: (2 sin π + 1)(sin π β 1) = 0
sin π = β1
2 ππ sin π = 1
β¦ βSolve πππππ π½ + π πππ π½ = π in the range πΒ° β€ π < πππΒ°β If you have a mix of sin and cos with one of them squared, use sin2 π₯ + cos2 π₯ = 1 to change the squared term.
2(1 β sin2 π) + 3 sin π = 0 2 β 2 sin2 π + 3 sin π = 0 2 sin2 π β 3 sin π β 2 = 0 (2 sin π β 1)(sin π β 1) = 0
sin π =1
2 ππ sin π = 1 β¦.