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Edexcel Mathematics: Core C1 6663 17th May 2017 [Exam Discussion] Watch

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    Welcome to the massive C1/C2 preperation thread for Edexcel, though other exam boards or students can use it for practice. The reason why I didn't include OCR or other exam boards is because they may have some different topics that are not tested in the Edexcel syllabus, however - for the exam boards, most of the topics are the same!

    This is the place to discuss, post problems, or ask any questions you may have regarding the exam!

    If your post or question gets missed or no one replies to it, you can ask these people individually who are willing to help: Cryptokyo Zacken crashMATHS RDKGames B_9710 xzh


    Edexcel C1 Exam Date - 17 MAY 2017
    Edexcel C2 Exam Date - 24 MAY 2017

    Revision Resources:

    Exam Solutions C1 Videos
    Exam Solutions C1 Papers
    crashMATHS Practice Papers (credit to crashMATHS )
    Edexcel A-level Maths Section (past papers, specification etc)
    Various C1 resources
    Various C2 resources
    M4ths
    Khan Academy (generalised, you have to search for topics)
    MyMaths
    MrBartonMaths C1
    MrBartonMaths C2

    Credit to Cryptokyo for this excellent resource, also attached in this post too.


    Top tips for the C1 Exam Tomorrow:
    (Courtesy of crashMATHS)

    • Take your time. Do not rush.
    • Read the questions carefully. Answer the questions that have been given to you, not the questions you hoped to see. If underlining key words helps you to focus your mind, then do this. For example, don't find the tangent if the question asks for the normal.
    • Differentiation is used to find the gradient at any point of a curve.
    • Integration is the opposite to differentiation. When you integrate, remember to include the constant of integration.
    • If you get stuck on a question, don't panic. Sit back, breathe and read the question again. Think: what information have I been given; what do I need to get to? For example, are you told the number of roots/intersection points for a function and need to show an inequality? The discriminant might be a good thing to consider because it involves an inequality itself.
    • Make links when reading a question. If you read the words 'gradient', 'tangent', 'normal', etc., when looking at a question involving curves, it probably will involve differentiation.
    • When working out the equation of a straight line, you need two bits of information: a point the line goes through and its gradient. Make sure you have both of these. If you don't have these, find them out!
    • If there is a tricky coordinate geometry question asking you to find the area of something, draw a diagram (if one isn't given) or study the diagram if one is given. Consider how you can perhaps deconstruct the shape into ones whose areas are easier to find. Find/write down any important lengths - these will usually gain you marks. Think of formulae you know and whether they may be relevant/useful.
    • For recursive relations, look out for patterns. If you have to calculate a sum that has a small upper limit, write out the terms and then do substitution. Writing out what the sum means shows understanding and will usually give you marks. If the limits seem odd, i.e. the lower limit is not 1 or the upper limit is a big number like 100, perhaps you'll need to think a bit more. For larger upper limits, you are not expected to list out all of the terms and add them. You may be expected to notice a pattern. Alternatively, if the quantity inside the sum is different to original sequence, perhaps you can manipulate your original sequence in some way to get your summand.
    • For sequences and series, those worded problems are usually quite simple. However, students seem to get lost and not isolate the key information. Ask yourself: what is the context, what is my first term, what is my common difference; am I working out a sum of terms or am I working at a specific term (and which formula should I thus be using?)?
    • Always use the information you are given. If you are asked to find something or show a result out of seemingly nowhere, convince yourself that all of the information you need is in front of you: read it carefully and, once again, try to isolate key words and make links.
    • Check your answers make sense. If a straight line is positive for  x>0 and you find that at  x = 2 , the line has  y coordinate  <0 , then something has gone wrong.
    • Check your working as you go along. After a few lines, stop, read the question again, read your working again, checking for any silly errors, and continue. Checking your working as you go along is more likely to help you identify errors than at the end, where you may be rushing or reading your work at a 'glance'.
    • Curve sketching: show all of the information they have asked in the question. Students forget to give intersection points, or equations of asymptotes, and lose unnecessary marks. Read the question and answer it. Remember a curve crosses the  y axis when  x = 0 and it crosses the  x axis when  y = 0 .
    • For asymptotes, make sure your curve looks like it is approaching the asymptote. To find asymptotes, consider what will make the equation 'blow up', so the speak. Division by 0 is a key one: what values of  x and  y will result in division by 0? These will be your asymptotes because your curve simply cannot take these values. For example  y = \frac{1}{x+2} . If  x = -2 , you would have division by 0, so there is an asymptote there. If  y = 0 , you'd require division by 0, so that must also be an asymptote.
    • When transforming graphs, I used to remember the transformations in two ways. If I have  y = f(x) and my transformation is inside the bracket, i.e.  y = f(x+a), y=f(ax) , then the transformation affects the  x coordinates and does the opposite to what you think it'd do, i.e.  y=f(x+3) takes 3 away from  x coordinates, rather than adding 3 to them and  y = f(3x) divides the  x coordinates by 3, rather than multiplying them by 3. If the transformation is outside the bracket, then it affects the  y and does exactly what you think it would do. This is a bit hand-waivey, but it is a nice way to remember it.
    • Understand function notation and its link to transformations. If you are given the curve  y = f(x) = x^3 + 2 and want to find the curve that results from translating the curve + 3 units parallel to the x-axis, then you work out  f(x-3) = (x-3)^3 + 2 = ... , as I'm sure you can expand and find.
    • Don't overdue it tonight. Just go over key ideas, make sure you really understand the basic ideas and keep asking 'why'.


    All the formula/rules for Edexcel C1:

    The formulae highlighted in red are the ones given to you in the formulae booklet, the rest you have to memorise - no excuses because I have noted down all the laws and necessary formulae that you are required to know in C1.

    Basic laws

     \displaystyle a^m \cdot a^n = a^{m+n}

    \displaystyle \frac{a^m}{a^n}=a^{m-n}

    \displaystyle (a^m)^n = a^{mn}

    \displaystyle a^{-m}=\frac{1}{a^m}

    \displaystyle a^{\frac{1}{m}}=\sqrt[m]{a}

    \displaystyle a^{\frac{n}{m}}=\sqrt[m]{a^n}

    \displaystyle a^0=1

    \displaystyle x^2-y^2=(x+y)(x-y) - difference of two squares

    \displaystyle \sqrt{ab}=\sqrt{a} \cdot \sqrt{b}

    \displaystyle \sqrt{ \frac{a}{b}}=\frac{ \sqrt{a} }{ \sqrt{b} }

    Rationalising the denominator involving surds:

    \displaystyle \sqrt{\frac{1}{a}}=\frac{1}{ \sqrt{a} }=\frac{1}{ \sqrt{a} }\cdot \frac{\sqrt{a}}{\sqrt{a}}=\frac{ \sqrt{a} }{a}

    \displaystyle \frac{1}{a\pm \sqrt{b}} = \frac{1}{a\pm \sqrt{b}} \cdot  \frac{a \mp \sqrt{b}}{a\mp \sqrt{b}}=\frac{a \mp \sqrt{b}}{a^2-b} - use of the conjugate whereby the conjugate of a+\sqrt{b} is a-\sqrt{b} (important to note that the conjugate of \sqrt{b}+a is -\sqrt{b}+a as some might get confused)

    Quadratics:

     \displaystyle x^2+bx=(x+\frac{b}{2})^2-(\frac{b}{2})^2 (completing the square)

    Solutions to ax^2+bx+c=0 are given by \displaystyle x=\frac{-b\pm \sqrt{b^2-4ac}}{2a} (quadratic formula)

    The discriminant of the quadratic formula is b^2-4ac = \Delta and if:-
    \Delta < 0 \Rightarrow no real roots
    \Delta = 0 \Rightarrow repeated real roots (or a single root)
    \Delta > 0 \Rightarrow 2 distinct real roots

    You should know the transformations of graphs.

    Equations of a straight line:

    A line passing through (x_1, y_1) and (x_2,y_2) has equation y-y_1=m(x-x_1) where \displaystyle m=\frac{y_2-y_1}{x_2-x_1}

    Different forms include y=mx+c, ax+by+c=0 and \displaystyle \frac{y-y_1}{y_2-y_1}=\frac{x-x_1}{x_2-x_1}

    Parallel lines have the same gradient. The product of the gradients of two lines which are perpendicular to one another is -1.
    Perpendicular lines have gradients -1/m.

    Gradient of a line: m = \frac {y2 - y1}{x2-x1} - difference in y coordinates / difference in x coordinates

    Sequences:

    The n^{th} term of an arithmetic sequence, u_n, with first term a and common difference d is given by u_n=a+(n-1)d

    The sum of the first n terms of an arithmetic sequence,S_n, is given by S_n=\frac{n}{2}[2a+(n-1)d]=\frac{n}{2}[a+L] where L is the final term.

    I cannot highlight this due to some latex problem, but: S_n=\frac{n}{2}[2a+(n-1)d] is given in the formulae booklet, whereas \frac{n}{2}[a+L][/tex] where L is the final term, is not.

    Basic Integration/Differentiation

    If y=x^n such that n is a constant, then \frac{dy}{dx}=n\cdot x^{n-1}

    If y=ax^n such that n is a constant, then \frac{dy}{dx}=n\cdot ax^{n-1}

    If \frac{dy}{dx}=x^n such that n=\text{const.} \neq -1 then y=\int \frac{dy}{dx} .dx = \int x^n .dx = \frac{x^{n+1}}{n+1}+c where c is a constant of integration.

    If \frac{dy}{dx}=kx^n such that n=\text{const.} \neq -1 and k is a constant, then y=\int \frac{dy}{dx} .dx = \int kx^n .dx = k\int x^n .dx = k\cdot \frac{x^{n+1}}{n+1}+c where c is a constant of integration.
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    Chittesh14 Sign me up as a helper pls. I will dig out some of the most difficult questions I had from last year some time later this week. Good idea setting up this thread!
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    (Original post by Cryptokyo)
    Chittesh14 Sign me up as a helper pls. I will dig out some of the most difficult questions I had from last year some time later this week. Good idea setting up this thread!
    No problem, I'll do that! Thanks for offering to be a helper and sharing some hard questions . Yeah, I want everyone to do well - like literally everyone to do amazing. This is going to be tons of practice - also for myself i am fully prepared, but anytime, anything could happen - I don't want to be challenged by a C1 question tbh lol.


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    Zacken crashMATHS physicsmaths Euclidean SeanFM notnek RDKGames B_9710
    Wondering if any of you also want to be helpers as you are all active in the Maths section.
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    (Original post by Chittesh14)
    Zacken crashMATHS physicsmaths Euclidean SeanFM notnek RDKGames B_9710
    Wondering if any of you also want to be helpers as you are all active in the Maths section.
    It's just C1, chill. But sure, add me to the helpers list.
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    (Original post by Chittesh14)
    Zacken crashMATHS physicsmaths Euclidean SeanFM notnek RDKGames B_9710
    Wondering if any of you also want to be helpers as you are all active in the Maths section.
    Nah sorry I forgot C1

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    jk I dont mind
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    (Original post by Chittesh14)
    Zacken crashMATHS physicsmaths Euclidean SeanFM notnek RDKGames B_9710
    Wondering if any of you also want to be helpers as you are all active in the Maths section.
    You can add me, sure.

    Just as a suggestion, it may be more useful and efficient if you turn this into the 2017 official C1 exam thread to avoid needing to create another one for the actual exam (unless it already exists).
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    (Original post by Chittesh14)
    Zacken crashMATHS physicsmaths Euclidean SeanFM notnek RDKGames B_9710
    Wondering if any of you also want to be helpers as you are all active in the Maths section.
    (Original post by Chittesh14)
    Zacken crashMATHS physicsmaths Euclidean SeanFM notnek RDKGames B_9710
    Wondering if any of you also want to be helpers as you are all active in the Maths section.
    Sure.
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    (Original post by Zacken)
    It's just C1, chill. But sure, add me to the helpers list.
    Loads of people struggle on C1 - well at least in my school. I'm going to make one for C2 too.. and maybe other modules :P


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    (Original post by RDKGames)
    Nah sorry I forgot C1

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    jk I dont mind
    Cool, I'll add you later .


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    (Original post by crashMATHS)
    You can add me, sure.

    Just as a suggestion, it may be more useful and efficient if you turn this into the 2017 official C1 exam thread to avoid needing to create another one for the actual exam (unless it already exists).
    Thanks, I'll add you later.
    Yes, I'll do that if the thread does not already exist. Thanks for the suggestion .


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    (Original post by B_9710)
    Sure.
    Thanks .


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    http://prnt.sc/e3ptsc

    need help with 9b

    Cryptokyo
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    (Original post by junayd1998)
    http://prnt.sc/e3ptsc

    need help with 9b

    Cryptokyo
    You substitution in is correct that when x=-\frac{1}{2}, \frac{\text{d}y}{\text{d}x}= \frac{3k}{4}+2. You have also correctly identified that the gradient of the tangent is \frac{7}{2}. \frac{\text{d}y}{\text{d}x} tells us the gradient of our curve at any given point and the tangent has the same gradient as the curve. Thus we get \frac{3k}{4}+2=\frac{7}{2}. You can find k from this.

    Hope this helps
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    (Original post by Cryptokyo)
    You substitution in is correct that when x=-\frac{1}{2}, \frac{\text{d}y}{\text{d}x}= \frac{3k}{4}+2. You have also correctly identified that the gradient of the tangent is \frac{7}{2}. \frac{\text{d}y}{\text{d}x} tells us the gradient of our curve at any given point and the tangent has the same gradient as the curve. Thus we get \frac{3k}{4}+2=\frac{7}{2}. You can find k from this.

    Hope this helps
    Okay also because the line and tangent are parallel that means they have equal gradients? So that's why we do 3k/4+2 = 7/2 ?
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    (Original post by junayd1998)
    Okay also because the line and tangent are parallel that means they have equal gradients? So that's why we do 3k/4+2 = 7/2 ?
    Yes, exactly! Two parallel lines have the same gradient. So the line parallel to our tangent has the same gradient as our tangent.
 
 
 
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