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Edexcel A2 C4 Mathematics June 2016 - Official Thread Watch

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    https://a086a5a2f39bda93734c56a63fab...%20Edexcel.pdf

    Could someone please explain part 8b fully, including how you integrate 3^2x, what method was used as I don't get it at all, where did the ln3 come from etc., also how did they take the triangle area into consideration to minus it away from the total area under the curve, I'm co confused?
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    (Original post by ImNervous)
    I think so

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    Actually there would be no a.

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    If we have cotx = 0 are we supposed to know that tanx = 1/0 and we treat 1/0 as infinity? Just saw this in a solomon and it really threw me off
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    (Original post by ImNervous)
    Actually there would be no a.

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    And it would be (b-1). Found it on the integral calculator

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    (Original post by ParisInc.)
    https://a086a5a2f39bda93734c56a63fab...%20Edexcel.pdf

    Could someone please explain part 8b fully, including how you integrate 3^2x, what method was used as I don't get it at all, where did the ln3 come from etc., also how did they take the triangle area into consideration to minus it away from the total area under the curve, I'm co confused?
    Consider how you'd differentiate 3^x, you'd get 3^x.ln3
    The integral from part a uses this idea to give 3^x.1/ln3

    You could then solve 3^2x by using a substitution of u=2x and finding that with the part a answer.

    You then have the triangle (I've labelled this T), when the shape from 3^2x from 0 to 9 is rotated, it also rotates the triangle into a cone (with it's top at Q) with radius r and height h. You'll then need to take this area away, to get the volume of R rotated 360 degrees.
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    (Original post by Craig1998)
    Consider how you'd differentiate 3^x, you'd get 3^x.ln3
    The integral from part a uses this idea to give 3^x.1/ln3

    You could then solve 3^2x by using a substitution of u=2x and finding that with the part a answer.

    You then have the triangle (I've labelled this T), when the shape from 3^2x from 0 to 9 is rotated, it also rotates the triangle into a cone (with it's top at Q) with radius r and height h. You'll then need to take this area away, to get the volume of R rotated 360 degrees.
    I differentiated in part a to help find the gradient and equation of the tangent I didn't integrate anything in part a, which part a integral do you mean? Also how do you use the solution of differentiating 3^x to get the integral of 3^x, I get how to differentiate this function but I don't understand how it helps me find the integral of it? I'm just really confused could you please show me step by step?
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    (Original post by ParisInc.)
    I differentiated in part a to help find the gradient and equation of the tangent I didn't integrate anything in part a, which part a integral do you mean? Also how do you use the solution of differentiating 3^x to get the integral of 3^x, I get how to differentiate this function but I don't understand how it helps me find the integral of it? I'm just really confused could you please show me step by step?
    Sorry I misread the question, thought part a was integrating.

    Consider if you were asked to differentiate 3^x.1/ln3
    As the 1/ln3 is just a number, you do nothing to that. As for 3^x, the differential of that is 3^x.ln3
    This means that overall, the differential of 3^x.1/ln3 is 3^x, as the ln3's would cancel out with eachother.
    That means that, if you are asked to integrade 3^x, you would be able to do the reverse and get 3^x.1/ln3 + c.
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    (Original post by Craig1998)
    Sorry I misread the question, thought part a was integrating.

    Consider if you were asked to differentiate 3^x.1/ln3
    As the 1/ln3 is just a number, you do nothing to that. As for 3^x, the differential of that is 3^x.ln3
    This means that overall, the differential of 3^x.1/ln3 is 3^x, as the ln3's would cancel out with eachother.
    That means that, if you are asked to integrade 3^x, you would be able to do the reverse and get 3^x.1/ln3 + c.

    Ohhh okay I kind of understand it now, it was just that I'd never come across it before, thanks
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    (Original post by Ainsleyy)
    If we have cotx = 0 are we supposed to know that tanx = 1/0 and we treat 1/0 as infinity? Just saw this in a solomon and it really threw me off
    likely it wouldn't come up but now you know it anyways?
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    Guys why do we set 3/2root(2) = 543/256 for part c of this question?

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    https://a086a5a2f39bda93734c56a63fab...%20Edexcel.pdf
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    (Original post by Don Pedro K.)
    Guys why do we set 3/2root(2) = 543/256 for part c of this question?

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    Because that's what you get when you plug 1/10 into the approximation/expansion up to x^3
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    (Original post by 13 1 20 8 42)
    Because that's what you get when you plug 1/10 into the approximation/expansion up to x^3
    What does expanding up to X^3 have to do with it? We only went up to x^2?
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    (Original post by Don Pedro K.)
    What does expanding up to X^3 have to do with it? We only went up to x^2?
    Yeah I meant x^2 sorry
    The point is that it's what you plug into the expansion you already have, this serves as your approximation
    (well obvs you have to rearrange to get the root(2) approximation. I assume you can see that plugging 1/10 in to the actual expression gives you (3/2)root(2)
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    (Original post by yesyesyesno)
    likely it wouldn't come up but now you know it anyways?
    just scares me when I see something I didn't know this close to the exam
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    (Original post by Ainsleyy)
    If we have cotx = 0 are we supposed to know that tanx = 1/0 and we treat 1/0 as infinity? Just saw this in a solomon and it really threw me off
    Ive seen that a few times, like when solving a quadratic trig problem. Always tempting to flip the cot to get tan and leave zero as it is. But yes, 1/0 is infinity. If youre not sure, check it on your calculator and youll get maths error.
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    (Original post by 13 1 20 8 42)
    Yeah I meant x^2 sorry
    The point is that it's what you plug into the expansion you already have, this serves as your approximation
    (well obvs you have to rearrange to get the root(2) approximation. I assume you can see that plugging 1/10 in to the actual expression gives you (3/2)root(2)
    I'm sorry, I'm really confused as to what's going on... XD So in part A I did the expansion up to x^2. Got that.

    In part b, I subbed x = 1/10 into (4 + 5x)^0.5 to get the exact answer of (4 + 5x)^0.5.

    In part c, I subbed in (1/10) into my expansion to get 543/256.

    So because the exact answer in part b was 3/2root(2),

    3/2root(2) = 543/256. So to get an approximation for root(2), root(2) = 543/246 * 2/3.

    Is that reasoning correct?
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    (Original post by Don Pedro K.)
    I'm sorry, I'm really confused as to what's going on... XD So in part A I did the expansion up to x^2. Got that.

    In part b, I subbed x = 1/10 into (4 + 5x)^0.5 to get the exact answer of (4 + 5x)^0.5.

    In part c, I subbed in (1/10) into my expansion to get 543/256.

    So because the exact answer in part b was 3/2root(2),

    3/2root(2) = 543/256. So to get an approximation for root(2), root(2) = 543/246 * 2/3.

    Is that reasoning correct?
    Yeah that's exactly it (I presume you still mean 256 not 246 at the end there)
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    ok someone explain this
    (all in degrees)
    cot(90)= 1/tan(90) = Cos(90)/sin(90)

    but 1 over tan 90 gives error

    cos 90 over sine 90 gives 0

    i get that tan 90 is at an asymptote but surely this dosent work?
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    (Original post by 222gg222)
    ok someone explain this
    (all in degrees)
    cot(90)= 1/tan(90) = Cos(90)/sin(90)

    but 1 over tan 90 gives error

    cos 90 over sine 90 gives 0

    i get that tan 90 is at an asymptote but surely this dosent work?
    If you look at the limiting value of 1/(tan x) as x tends to 90 (in ugly degrees), you get 1/(massive huge infinite number), which is basically 0.
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    (Original post by Zacken)
    If you look at the limiting value of 1/(tan x) as x tends to 90 (in ugly degrees), you get 1/(massive huge infinite number), which is basically 0.
    ah that makes sense, so we assume a number so close to 0 is just 0, cheers
 
 
 
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