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

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    (Original post by Chittesh14)
    I've updated the thread guys, bloody hell - that Latex took me over 1 hours lmao :'(.
    Few things -

    (Original post by Chittesh14)
    Equations of a straight line: y = mx + c or ax + by + c = 0 (a, b and c are integers).
    a,b,c don't necessarily have to be integers, they can be in \mathbb{R}

    The gradient of perpendicular lines is equal to -1.
    Did you mean to say that the product of two gradients which are perpendicular to one another is -1? This does not make sense as it stands.


    or \frac{y-y1}{y2-y1} = \frac{x-x1}{x2-x1}
    What are x_1, x_2, y_1, y_2??

    If dy/dx = x^n, then y = \frac{1}{n+1}x^{n+1} + c

    \int {x^n}\ dx = \frac{x^{n+1}}{n+1} + c
    These two mean the same thing, you also need to comment that this is only true if n \neq -1


    Other than that I will fix some LaTex here and there

    Spoiler:
    Show




    Basic laws and whatnot

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

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

    \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

    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}

    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.

    (NOTE: You seem to have misused the word series for a sequence. A sequence is a list of elements, and a series is the summation of these elements. When you say "sum of an arithmetic series" it does not make sense as you are not summing up sums.)

    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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    (Original post by RDKGames)
    Few things -


    a,b,c don't necessarily have to be integers, they can be in \mathbb{R}



    Did you mean to say that the product of two gradients which are perpendicular to one another is -1? This does not make sense as it stands.




    What are x_1, x_2, y_1, y_2??

    These two mean the same thing, you also need to comment that this is only true if n \neq -1


    Other than that I will fix some LaTex here and there

    Spoiler:
    Show



    Basic laws and whatnot

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

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

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

    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}

    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.

    (NOTE: You seem to have misused the word series for a sequence. A sequence is a list of elements, and a series is the summation of these elements. When you say "sum of an arithmetic series" it does not make sense as you are not summing up sums.)

    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.


    I know, I didn't include what it's true for many things even a^0 = 1. I've made mistakes and I'm going to go over them tomorrow lol it's night that's why I done it in a rush. Mistakes happen lol obviously I know the difference between a sequence and series. Btw, Thanks for the latex.


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    Edexcel has a formula booklet...
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    (Original post by Chittesh14)
    ..
    (Original post by RDKGames)
    ..
    Typo: a^{-m} = \dfrac{1}{a^m}, not \dfrac{1}{m}
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    (Original post by Zacken)
    Edexcel has a formula booklet...
    I know. But not everyone knows what formulae they need to remember and I'm going to highlight the ones which are given. Half of the people forget these rules lol


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    (Original post by Chittesh14)
    Welcome to the massive C1 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 usually have Edexcel C2 content in their C1 content.

    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

    If anyone would like to be tagged in this thread (helpers) who are willing to help students with problems, can they please quote me and I'll tag them at the start of the post so people know who to tag if this thread ever becomes inactive.

    In essence, I would really like students to post the most difficult questions they've encountered or tricky ones (with confusing wording) or ones that are not straightforward because these are the ones which often make a student panic - simply because they don't understand the content well enough or are unsure about how to answer the question due to its style.


    CAN PEOPLE PLEASE POST LINKS TO REVISION RESOURCES OR ANY SPECIAL PAPERS THEY HAVE (EXCEPT EDEXCEL OR OTHER EXAM BOARDS C1 2016 FOR NOW) OR ANY MATERIALS THAT WOULD CONTRIBUTE TO THE THREAD. THANK YOU VERY MUCH!!

    Edexcel C1 Exam Date - 17 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)
    Daily revision sheets for C1 (complete one per day, each sheet dated)
    Maths videos by Jayates
    MyMaths
    MrBartonMaths C1
    MrBartonMaths C2

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

    All the formula/rules for Edexcel C1:

    a^m * a^n = a^{m+n}

 a^m / a^n = a^{m-n}

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

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

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

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

a^0 = 1

 x^2 - y^2 = (x+y)(x-y)

\sqrt{ab} = \sqrt a * \sqrt b

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

    For rationalising surds, fractions in the form:

    \sqrt \frac {1}{a} = \frac {1}{\sqrt a} = \frac {*\sqrt a}{*\sqrt a}

\dfrac {1}{a+ \sqrt{b}} = \dfrac {*a- \sqrt{b}}{*a-\sqrt{b}}

 \dfrac {1}{a- \sqrt{b}}\ = \dfrac {*a+ \sqrt{b}}{*a+\sqrt{b}}

     x^2 + bx = (x+b/2)^2 - (b/2)^2

x = \dfrac{-b  \pm \sqrt{b^2-4ac}}{2a}
    Discriminant (d) = b^2 - 4ac
    d < 0 = no real roots
    d = 0 = equal roots
    d > 0 = two real roots (distinct roots)

    You should know the transformations of graphs.

    Equations of a straight line: y = mx + c or ax + by + c = 0 (a, b and c are integers).
    Parallel lines have the same gradient. The gradient of perpendicular lines is equal to -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
    Equation of line formulae: y - y1 = m(x-x1)
    or \frac{y-y1}{y2-y1} = \frac{x-x1}{x2-x1}

    nth term of an arithmetic series = a + (n-1)d
    Sum of an arithmetic series = n/2[2a + (n-1)d] or n/2[a+L]
    where a is the first term, d is the common difference between the terms, n is the number of terms in the series and L is the last term in the series.

    If y = x^n, then dy/dx = nx^{n-1}
    If y = ax^n, then dy/dx = nax^{n-1}
    If dy/dx = x^n, then y = \frac{1}{n+1}x^{n+1} + c
    If dy/dx = kx^n, then y = \frac{kx^{n+1}}{n+1} + c
    \int {x^n}\ dx = \frac{x^{n+1}}{n+1} + c

    Difficult questions:

    Layout:-

    Question -
    Hint -
    Solution - (if you want to provide it) as many people will post solutions eventually and there may be different solutions to the same question.

    I'll be posting each question with a hint and solution in the spoiler so that students can attempt the question before looking at the answer. Just remember, if you directly look at the answer, you're only cheating yourself!
    Isn't C1 and C2 part of AS?
    Coz I'm studying them and it's AS. To be more appropriate, I'm studying IALs so that makes it as C12, but could I join in, if you don't mind?
    Coz the books are the same, aren't they? Only the exam dates differ.
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    (Original post by Reshyna)
    Isn't C1 and C2 part of AS?
    Coz I'm studying them and it's AS. To be more appropriate, I'm studying IALs so that makes it as C12, but could I join in, if you don't mind?
    Coz the books are the same, aren't they? Only the exam dates differ.
    That was my fault - I changed the forum title for the OP but made a mistake. I've fixed it now.

    And of course you're welcome to join in

    For non-IAL students I recommend trying the C12 past papers if you're aiming for a high grade since they often have harder questions.
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    (Original post by notnek)
    That was my fault - I changed the forum title for the OP but made a mistake. I've fixed it now.

    And of course you're welcome to join in

    For non-IAL students I recommend trying the C12 past papers if you're aiming for a high grade since they often have harder questions.
    Oh, I see
    Thanks a lot!
    Is practicing C1 and C2 papers beneficial for IAL students?
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    (Original post by Reshyna)
    Oh, I see
    Thanks a lot!
    Is practicing C1 and C2 papers beneficial for IAL students?
    It's all practice so yes, it is beneficial .


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    PhysicsandMathsTutor http://www.physicsandmathstutor.com/
    Provide Solomon Papers and Tiered Papers (Bronze, Silver and Gold - Being Hardest)

    Solomon Papers are good for those which find past papers repetitive. I certainly find them useful!

    Myself I'm resitting C1 and C2 as well!
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    (Original post by B_9710)
    Now this is probably an extension question since it uses limits that isn't really discussed anyway at AS. This also used C2 as well - binomial theorem. I think this is a decent question because it actually tells us what the derivative is by its definition.

     \text{Given that the definition of the derivative } f' \text{of a function } f \text{ is given as}
     \displaystyle \lim_{h\to 0} \frac{f(x+h)-f(x)}{h} ,
     \text{show that the derivative of } f  \text{ when } f(x)=x^n \text{ is } nx^{n-1} .
    First principles actually show up within the new specification, interestingly.
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    (Original post by _gcx)
    First principles actually show up within the new specification, interestingly.
    I think it's very weird to introduce the definition of the derivative without even talking about what a limit means or what it means for a function to be differentiable and/or continuous.
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    I'm in Year 13 and will be resitting C1 this year because I only got 78 UMS last year due to running out of time on the last two questions .

    I'll be doing the IYGB papers for practice, which I haven't seen mentioned here yet but I highly recommend them!
    http://www.madasmaths.com/archive_iy...ce_papers.html
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    I PROMISE TO UPDATE AND FIX EVERYTHING TOMORROW.
    Too tired today :/


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    (Original post by LeCroissant)
    I'm in Year 13 and will be resitting C1 this year because I only got 78 UMS last year due to running out of time on the last two questions .

    I'll be doing the IYGB papers for practice, which I haven't seen mentioned here yet but I highly recommend them!
    http://www.madasmaths.com/archive_iy...ce_papers.html
    Great decision , what a disappointment though 😒, I'm sure you'd have got 85+ if you'd have finished those 2 questions! Don't worry, this year you should get 100 .


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    (Original post by LeCroissant)
    I'm in Year 13 and will be resitting C1 this year because I only got 78 UMS last year due to running out of time on the last two questions .

    I'll be doing the IYGB papers for practice, which I haven't seen mentioned here yet but I highly recommend them!
    http://www.madasmaths.com/archive_iy...ce_papers.html
    I was trying to remember the website - that is TeeEm's website I think? I remember his papers but couldn't remember his website lol. Thanks a lot !!


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    lets do this!!!!!!!!
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    (Original post by Zacken)
    I think it's very weird to introduce the definition of the derivative without even talking about what a limit means or what it means for a function to be differentiable and/or continuous.
    Yeah but A level maths is not very rigorous at all. A lot of it is hand wavey and is certainly not the same sort of style as maths at university for sure. If you had to learn about all the stuff you mentioned it would probably seem very dull to some people - even though it would be for the greater good.
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    RDKGames notnek I've updated the thread. I will update again properly in the weekend with proper difficult questions and C2 formulae, but I've done everything I need to .
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    Who knows how where i can get the 2016 paper? (edexcel)
 
 
 
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