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    (Original post by ninuzu)
    I'd use partial fractions, but in C4 you are not expected to be able to do partial fractions where the order of the denominator is the same as the numerator...
    You are expected to be able to do partial fractions when the numerator is larger than or equal to the denominator. Check for example page 310 of the Core 4 book
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    (Original post by jassi1)
    You are expected to be able to do partial fractions when the numerator is larger than or equal to the denominator. Check for example page 310 of the Core 4 book
    Not equal to - that's FP2? Or if they do they will tell you how to structure it, since you have to have an 'A' term at the front with no denominator when the degrees are equal.
    Actually, that's not partial fractions entirely - if you look at the specification, they will only ever ask you partial fractions where the degree of the numerator < degree of the denominator (anything else is FP2). If they do have the degree greater than or equal to the denominator, you will need to do algebraic division first.
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    (Original post by Hello_ImJess)
    How would you integrate this?

    Attachment 227551


    Posted from TSR Mobile
    I'd use integration by parts of X(X+2)^-1
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    Hey guys, would we need to know the following integrations or are they given in the formula booklet? I can't find my copy of the FB

    sec^2(x) \ dx \ = tan(x)+c

cosec^2(x) \ dx \ = -cot(x) + c

cosec(x)cot(x) \ dx \ = -cosec(x)+c

    any help would be appreciated!
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    (Original post by Vip3rgt9)
    I'd use integration by parts of X(X+2)^-1
    No, integration by parts would take too long as you have to then integrate ln(x+2).

    Use substitution u=x+2 or turn the numerator into (x + 2) -2 and divide by (x+2).


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    (Original post by saberahmed786)
    Hey guys, would we need to know the following integrations or are they given in the formula booklet? I can't find my copy of the FB

    sec^2(x) \ dx \ = tan(x)+c

cosec^2(x) \ dx \ = -cot(x) + c

cosec(x)cot(x) \ dx \ = -cosec(x)+c

    any help would be appreciated!
    They are in the formula booklet. Look on ocr website and download a copy if you need to.


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    (Original post by Namige)
    No, integration by parts would take too long as you have to then integrate ln(x+2).

    Use substitution u=x+2 or turn the numerator into (x + 2) -2 and divide by (x+2).


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    Thanks. For OCR they always tell us which questions to do substitutions in though don't they?
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    Thanks everyone


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    (Original post by Vip3rgt9)
    Thanks. For OCR they always tell us which questions to do substitutions in though don't they?
    If they are deliberately looking for substitution, they will tell you in the question use the substitution u=blah.
    If the question is ambiguous i.e. they don't tell you what to do - you can use various different methods including substitution.
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    JAN 2009

    (1+2x)^0.5
    (1+x)^3

    Q3) iii) State the set of values of x for which the expansion is valid. [1]

    I know the answer is: -0.5 < x < 0.5

    but why is this?
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    (Original post by daniel_vincent)
    JAN 2009

    (1+2x)^0.5
    (1+x)^3

    Q3) iii) State the set of values of x for which the expansion is valid. [1]

    I know the answer is: -0.5 < x < 0.5

    but why is this?
    |2x| < 1

    so -0.5 < x < 0.5
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    (Original post by daniel_vincent)
    JAN 2009

    (1+2x)^0.5
    (1+x)^3

    Q3) iii) State the set of values of x for which the expansion is valid. [1]

    I know the answer is: -0.5 < x < 0.5

    but why is this?
    Because the formula (1+x)^n = 1 + nx + (n(n-1))/2! *x^2 + ... is only valid when |x|<1 so -1<x<1.
    In the previous part of the question, you have expanded (1+2x)^0.5 using this formula. Replacing x in the values for which the formula is valid, you get |2x|<1, so |x|<0.5 ---> -0.5<x<0.5
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    (Original post by Vip3rgt9)
    Thanks. For OCR they always tell us which questions to do substitutions in though don't they?
    Write x / (x+2) as (x + 2 - 2) / (x+2), then you have (x+2)/(x+2) - 2/(x+2).

    Which is just 1 - 2(x+2).
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    just finished the last paper Jan 13, I have to say that second part of the differential equation ****ed me over. Make sure u read the questions carefully guys they often give wayyy too much information to hide the very subtle hints, can very easily catch you out and unfortunately with DE if one part is wrong your pretty much screwed for the rest of it.
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    How to determine if vector lines are
    a) parallel, b) intersect, c) skew ??
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    (Original post by SexyAndIKnowIt.)
    How to determine if vector lines are
    a) parallel, b) intersect, c) skew ??
    a) same direction vector (usually the other one is a multiple).

    b) sub in and solve should be equal on both sides.

    c) sub in and solve and shouldn't be equal on both sides.
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    Can anyone suggest to me the hardest C4 paper so I can give it a go?
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    (Original post by Vip3rgt9)
    Can anyone suggest to me the hardest C4 paper so I can give it a go?
    Look at grade boundaries and the lowest will be the hardest
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    (Original post by Hello_ImJess)
    How would you integrate this?

    Attachment 227551


    Posted from TSR Mobile
    I would use long division. You would get 1 - 2/(x+2), integrate to get you x - 2ln(x+2).
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    June 11 lowest grades
 
 
 
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