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    Hi, I've been doing a few questions from the text book and I'm not sure how to complete them. If someone could offer some help, that would be very much appreciated! In the images attached, I have written the question on the top of the page, and my working at the bottom part.

    With Q20, I have checked the expansion with the answers and it is correct. However I am not sure what to do to find dy/dx. I can't see the relationship between what I found with dy/dx.

    With Q27, the first part is correct. However to find the area R, you need the limits in terms of x right? However, we are given x = 4 and y= 1. I'm finding it difficult to convert this back into x? And I can't do part iii) as it relies on my finding the limits for the volume of revolution.



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    (Original post by Bazinga?)
    .
    For the first q, note that all but the second term, when differentiated term by term will be 0.

    The second, sub y = 1 into the equation given and solve for x
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    (Original post by joostan)
    For the first q, note that all but the second term, when differentiated term by term will be 0.

    The second, sub y = 1 into the equation given and solve for x
    For Q1, to clarify something, but the expansion (2 + 1/2x - 3/4x^2 + 17/8x^3) is essentially equal to y right? So you would just differentiate it as per usual?

    For 2, so that means I have to do 1 = 2/[(x-1)(x-3)] ? Wouldn't I get a quadratic equation? Which means there would be several answers?
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    (Original post by Bazinga?)
    For Q1, to clarify something, but the expansion (2 + 1/2x - 3/4x^2 + 17/8x^3) is essentially equal to y right? So you would just differentiate it as per usual?
    In essence, yues - sub in x=0 and all the terms of the expansion (except one) will be equal 0 too.
    (Original post by Bazinga?)
    For 2, so that means I have to do 1 = 2/[(x-1)(x-3)] ? Wouldn't I get a quadratic equation? Which means there would be several answers?
    It would help if I could see the region R - I'm assuming that there are some limitations imposed upon it?
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    (Original post by joostan)
    In essence, yues - sub in x=0 and all the terms of the expansion (except one) will be equal 0 too.

    It would help if I could see the region R - I'm assuming that there are some limitations imposed upon it?
    For the second question, I've written out all of the information that I am given. Unfortunately there isn't a graph to accompany it! So no limitations either.
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    (Original post by Bazinga?)
    For the second question, I've written out all of the information that I am given. Unfortunately there isn't a graph to accompany it! So no limitations either.
    There are two points where y=1. However there is an asymptote between one of the points where y =1, so we see that \ x =2+\sqrt{3} for there to be a defined integral.

    Also on your part c) you need to check your - signs
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    (Original post by joostan)
    There are two points where y=1. However there is an asymptote between one of the points where y =1, so we see that \ x =2+\sqrt{3} for there to be a defined integral.

    Also on your part c) you need to check your - signs
    When you do 1/4 = 2/(x-1)(x-3), don't you get x^2 - 4x - 5 = 0 when simplified? So when you factorise, you get x = 5 or x = -1? I'm assuming we take x = -1 as it is the lower bound, but why can't you use the other value?

    I'm a bit confused as I didn't get x = 2 square root 3 for either solution :/
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    (Original post by Bazinga?)
    When you do 1/4 = 2/(x-1)(x-3), don't you get x^2 - 4x - 5 = 0 when simplified? So when you factorise, you get x = 5 or x = -1? I'm assuming we take x = -1 as it is the lower bound, but why can't you use the other value?

    I'm a bit confused as I didn't get x = 2 square root 3 for either solution :/
    Where did the \frac{1}{4} come from?
    Allow me to refer you to wolfram.
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    (Original post by joostan)
    Where did the \frac{1}{4} come from?
    Allow me to refer you to wolfram.
    Oops, I just realised that I copied down the question wrong. The line is suppose to say, "The finite region bounded by the curve with equation y=... and the lines x=4 and y=1/4 is denoted by R". Sorry!

    In this case, x=5 or -1.
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    (Original post by Bazinga?)
    Oops, I just realised that I copied down the question wrong. The line is suppose to say, "The finite region bounded by the curve with equation y=... and the lines x=4 and y=1/4 is denoted by R". Sorry!

    In this case, x=5 or -1.
    In which case there are two asymptotes present between 4 and -1.
    So the solution x = 5 must be the other limit
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    (Original post by joostan)
    In which case there are two asymptotes present between 4 and -1.
    So the solution x = 5 must be the other limit
    How do you know that there is an asymptote present in the graph? And how did you deduce that there are two asymptotes between 4 and -1?
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    (Original post by Bazinga?)
    How do you know that there is an asymptote present in the graph? And how did you deduce that there are two asymptotes between 4 and -1?
    You cannot divide by zero.
    Inspect the function, and see where x causes you to divide by 0.
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    (Original post by joostan)
    You cannot divide by zero.
    Inspect the function, and see where x causes you to divide by 0.
    Thank you for your help, it's very much appreciated!

    Would you mind taking a look at these two questions? (They're the last two I promise!)

    With Q11, I'm really not sure what to do, I guess I see that when x = 10 then x^4 + 4 = 10004. But I don't really understand the question.

    With Q18, what does it mean by 'when x is large'? I've always been confused when they say things like this, like in binomial expansion questions, they sometimes say, 'when x is small find the expansion...'- what does this mean? So back to this question, I don't really understand what to do for part b



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    (Original post by Bazinga?)
    x
    No problem
    First q, you guessed corrctly, you just need to factorise the quartic, sub in x=10
    Second q, consider what happens to \dfrac{2x +3}{x^2 + x+1}
    when x gets very large
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    (Original post by joostan)
    No problem
    First q, you guessed corrctly, you just need to factorise the quartic, sub in x=10
    Second q, consider what happens to \dfrac{2x +3}{x^2 + x+1}
    when x gets very large
    Er back to 27 (I just tried the integration), so I integrated 1/(x-3) - 1/(x-1) with limits of 5 and 4. So I ended up with [ln|x-3| - ln|x-1|]5 and 4 so...

    ln|2| - ln|4| - (ln|1| - ln|3|) = ln1.5? Whereas in the question it says show that the area of R is ln3/2 - 1/4? Where does the 1/4 come from?

    For 11, when you say factorise the quartic, do you mean x^4 + 4 = (x^2 + 2x + 2)(x^2 - 2x + 2)? Surely that's the same as subbing in x=10 into the quartic?

    For 18. I think I understand it now, as when x is very large, then 2x-3 / x^2+x+1 tends to infinity. As a result, the graph is equal to x^2 - 3x + 2?

    One final thing, in binomial expansions, they sometimes say, "If x is small compared to a, expand ...", what does the "if x is small" bit mean?
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    (Original post by Bazinga?)
    Er back to 27 (I just tried the integration), so I integrated 1/(x-3) - 1/(x-1) with limits of 5 and 4. So I ended up with [ln|x-3| - ln|x-1|]5 and 4 so...

    ln|2| - ln|4| - (ln|1| - ln|3|) = ln1.5? Whereas in the question it says show that the area of R is ln3/2 - 1/4? Where does the 1/4 come from?

    For 11, when you say factorise the quartic, do you mean x^4 + 4 = (x^2 + 2x + 2)(x^2 - 2x + 2)? Surely that's the same as subbing in x=10 into the quartic?

    For 18. I think I understand it now, as when x is very large, then 2x-3 / x^2+x+1 tends to infinity. As a result, the graph is equal to x^2 - 3x + 2?

    One final thing, in binomial expansions, they sometimes say, "If x is small compared to a, expand ...", what does the "if x is small" bit mean?
    27) You need to subtract  1 \times \frac{1}{4} can you see why?
    Spoiler:
    Show
    y=1, yet you integrated the whole area down to y=0


    11) In essence yes, I hadn't checked to see if it factorised further.

    18) Not quite, the added bit tends to 0.

    For the expansion to be justified:
     |x| <1 I think that's basically what they're saying.
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    (Original post by joostan)
    27) You need to subtract  1 \times \frac{1}{4} can you see why?
    Spoiler:
    Show
    y=1, yet you integrated the whole area down to y=0


    11) In essence yes, I hadn't checked to see if it factorised further.

    18) Not quite, the added bit tends to 0.

    For the expansion to be justified:
     |x| <1 I think that's basically what they're saying.
    For the first one, I basically found R + 1/4 as the difference is 1/4 (in y) and 1 in x. So I subtract 1/4 to get R? ...This question is full of tricks! However, how do you know that the 'extra bit' is 1/4? What if the graph looks like this? :

    Uploaded with ImageShack.us Wouldn't it mean that the 'extra bit' is less than 1/4? I understand that the maximum area is 1/4...but it could be less?

    Okay thanks, I understand 11/18/binomial bit now.
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    (Original post by Bazinga?)
    For the first one, I basically found R + 1/4 as the difference is 1/4 (in y) and 1 in x. So I subtract 1/4 to get R? ...This question is full of tricks! However, how do you know that the 'extra bit' is 1/4? What if the graph looks like this? :
    It doesn't look like that, it looks like this.
    Personally, I think it's mean to set a question like this, but it should be obvious that it's underneath, as the function is positive here, so the area must be too.
    \ (5-4) \times \frac{1}{4}
    Spoiler:
    Show
    The area of a rectangle with height 1/4 as y=1/4 and width of the integral.
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    (Original post by joostan)
    It doesn't look like that, it looks like this.
    Personally, I think it's mean to set a question like this, but it should be obvious that it's underneath, as the function is positive here, so the area must be too.
    \ (5-4) \times \frac{1}{4}
    Spoiler:
    Show
    The area of a rectangle with height 1/4 as y=1/4 and width of the integral.
    (I was just drawing a random curve by the way!) That makes much more sense now, I suppose if I can do these harder questions, then I should be able to tackle the harder exam questions!

    I'm having a bit of trouble with part c. So with this, it is just a typical volume of revolution question, so you do pi multiplied by the integral of y^2? When it says rotated by 2pi radians about the x-axis, is this basically saying rotation by 360o? So I shouldn't need to factor that in my working?

    Here's my working, I'm not sure where I have gone wrong...as the answer is supposed to be pi(25/48 - ln3/2)


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    (Original post by Bazinga?)
    (I was just drawing a random curve by the way!) That makes much more sense now, I suppose if I can do these harder questions, then I should be able to tackle the harder exam questions!

    I'm having a bit of trouble with part c. So with this, it is just a typical volume of revolution question, so you do pi multiplied by the integral of y^2? When it says rotated by 2pi radians about the x-axis, is this basically saying rotation by 360o? So I shouldn't need to factor that in my working?

    Here's my working, I'm not sure where I have gone wrong...as the answer is supposed to be pi(25/48 - ln3/2)
    You need to subtract a volume - what do you think this is?
    and yes to the bold, you have to the underlined.
 
 
 
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