Year 13 Maths Help Thread

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    (Original post by medhelp)
    Q1.7
    How many zeros are there at the end of the number 100! ?
    I don't get what that exclamation mark means, and 2 isn't the answer. I remember using it as symbol in AS maths but not anything else.

    Q1.11
    A salesmen drives from L to B. The first half of the distance he drives at a constant speed of 80 mph. The second half he drives at a constant speed of 120 mph. What is his average speed for the total journey?

    I don't understand why the answer is 96 mph and not 100?
    100! just means 100 x 99 x 98 x 97 x ... x 1.

    A smaller example to show you in full; 5! is 5 x 4 x 3 x 2 x 1 = 120

    the second one is deceptive, it tricks you into just finding the average of the speeds but you have to remember that the first half of the journey took longer to complete than the second half. Work out the time taken for each half separately and go from there
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    How would you find the period in radians of y=cosec4x (does the period tell you the intervals where the graph repeats?)

    My book says y=cosecx repeats after 2pi intervals. How do you find this for transformed graphs?

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    cosec(x) is 1/sin(x).

    sin(x) repeats every 2pi, therefore so does 1/sin(x).

    regarding transformed graphs: consider x slowly increasing from 0 to 2*pi. Now consider what happens to 2x in the same time. It reaches twice the distance, from 0 to 4pi. So in the time it takes sin(x) to complete one cycle, sin(2x) will have completed two cycles. The period of the function decreases by a factor of two when the argument of the function is doubled. This is true of all functions, including sin(x) and cosec(x).
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    How using c4 methods do u integrate xe^x^2 in part b the 3rd line
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    (Original post by mik1a)
    cosec(x) is 1/sin(x).

    sin(x) repeats every 2pi, therefore so does 1/sin(x).

    regarding transformed graphs: consider x slowly increasing from 0 to 2*pi. Now consider what happens to 2x in the same time. It reaches twice the distance, from 0 to 4pi. So in the time it takes sin(x) to complete one cycle, sin(2x) will have completed two cycles. The period of the function decreases by a factor of two when the argument of the function is doubled. This is true of all functions, including sin(x) and cosec(x).
    Ohh so these questions get you to think about whats happening rather than using maths to find the period?

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    (Original post by kiiten)
    Ohh so these questions get you to think about whats happening rather than using maths to find the period?

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    Well, you can learn the rules, e.g. "f(a*x) squashes f(x) along the x-axis by a factor of a", it's a faster method but also more boring. I suggest you learn that rule for the exam, but also try to make sure you understand the method in case you forget the rule.
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    (Original post by mik1a)
    Well, you can learn the rules, e.g. "f(a*x) squashes f(x) along the x-axis by a factor of a", it's a faster method but also more boring. I suggest you learn that rule for the exam, but also try to make sure you understand the method in case you forget the rule.
    Is that graph transformation? Please could you show me using that method with the example?

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    (Original post by kiiten)
    Is that graph transformation? Please could you show me using that method with the example?

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    Sure, its the one on the far right. Squashes everything up towards the line x=0.

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    (Original post by asinghj)
    You know from C3:

     \frac{d}{dx} (e^(f(x))) = f`(x) e^(f(x))

    So you apply that in reverse, which means if you differentiate  e^\frac{-x^2}{2} you get the exactly what you are integrating...

    It's a weird way, I haven't done C4, so sorry if I waffled and stuff but watch this video:

    https://m.youtube.com/watch?v=t1EZyUxAkhg
    Thankyou
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    (Original post by mik1a)
    Sure, its the one on the far right. Squashes everything up towards the line x=0.

    Thank you

    So for my example would it be 2pi ÷ 4 ?

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    (Original post by youreanutter)
    How using c4 methods do u integrate xe^x^2 in part b the 3rd line
    Use the substitution  u = \frac{-x^2}{2} which transforms the integral to  \int -e^u\ du . More generally,  \int f'(x)e^{f(x)}\ dx = e^{f(x)} +c . To see why this works, consider the substitution  u = f(x) .
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    (Original post by medhelp)
    Q1.7
    How many zeros are there at the end of the number 100! ?
    I don't get what that exclamation mark means, and 2 isn't the answer. I remember using it as symbol in AS maths but not anything else.

    Q1.11
    A salesmen drives from L to B. The first half of the distance he drives at a constant speed of 80 mph. The second half he drives at a constant speed of 120 mph. What is his average speed for the total journey?

    I don't understand why the answer is 96 mph and not 100?
    Q1.7:

     100! = 100 \times 99 \times 98 ... \times 3 \times 2 \times 1
    For this question, consider how a zero at the end is formed - it means the number is multiplied by 10, so is formed from a 2 and a 5. Essentially, you have to find the number of pairs of 2s and 5s you have. I'll leave the answer in a spoiler, but try to work it out first. (Hint: if you express 100! in its prime factors, the number of 2s will be greater than the number of 5s)
    Spoiler:
    Show
    You only really need to calculate the number of 5s, as there will be at least a 2 for every 5, so the number of 5s = the number of pairs of 2s and 5s. How many 5s are there in 100!? There's one in 1 in 5, 1 in 10, 1 in 15 etc up to 100, which gives 20. However, 25, 50, 75 and 100 all have 2 lots of 5s in (as 25 is a factor), so there are overall 24 5s in 100!, meaning  100! can be expressed in the form  100! = 5 ^{24} \times k , where k is an integer, and so the number of zeroes at the end is 24. Generally, the number of zeroes at the end of  n! = \frac{n}{5} + \frac{n}{5^2} + \frac{n}{5^3} ... taking only the integer part of each term.
    Q1.11
     \mathrm{average\ speed} = \frac{\mathrm{total\ distance}}{ \mathrm{total\ time}}

     \mathrm{Let\ total\ distance} = 2x
    As distance travelled in both parts is the same, distance travelled in each half is x. Then find time taken for first half and the time taken for the second half (in terms of x), to find the total time taken, and then calculate the average speed, and the x s should cancel.
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    Name:  Complex transformation (region shading).png
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    I got the equation of the circle in part (a) as  (u-\frac{9}{8})^2 +v^2 =\frac{9}{64}

    For part (b) |z|<3 is the region inside the circle with centre (0,0) radius =  3 so why does the shaded region lie outside the circle after the transformation, as shown in the mark scheme?
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    (Original post by NotNotBatman)
    Name:  Complex transformation (region shading).png
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Size:  22.7 KB

    I got the equation of the circle in part (a) as  (u-\frac{9}{8})^2 +v^2 =\frac{9}{64}

    For part (b) |z|<3 is the region inside the circle with centre (0,0) radius =  3 so why does the shaded region lie outside the circle after the transformation, as shown in the mark scheme?
    for part b you should have \frac{|w|}{|1-w|}<3 so from there it should work out to give the region outside the circle
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    (Original post by solC)
    for part b you should have \frac{|w|}{|1-w|}<3 so from there it should work out to give the region outside the circle
    I have, \frac{|wi|}{|1-w|}<3, which leads me to the correct answer now. I have now realised how to work out the region from the Cartesian equation, just replacing the = sign with a > sign in later stages of working, so the distance from the centre to any point on a locus is greater than the radius, so all points outside the circle. Is there any way of noticing where to shade from the complex form?
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    (Original post by NotNotBatman)
    I have, \frac{|wi|}{|1-w|}<3, which leads me to the correct answer now. I have now realised how to work out the region from the Cartesian equation, just replacing the = sign with a > sign in later stages of working, so the distance from the centre to any point on a locus is greater than the radius, so all points outside the circle. Is there any way of noticing where to shade from the complex form?
    Hmm, I'm not too sure to be honest. I wouldn't think so because, unless you've already worked it out, you wouldn't even know what sort of locus z is mapped onto. For a question like this one though I think it would be the easiest way even if there were another method since you've already done all the calculations required, all you need to do is swap the = with an <.
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    (Original post by NotNotBatman)
    Is there any way of noticing where to shade from the complex form?
    Lol, yeah. Features of complex plane means that once you know where one point goes under the function, you can tell where all the others will (loosely speaking).

    So, in your case, take any z such that |z| &lt; 3, so say z = 1-i then w = \frac{1-i}{1-i + i} = 1 - i and this lies outside the circle in the w plane. So you can immediately tell that any point in |z| &lt; 3 will map to a point ouside the circle in the w-plane, so you shade outside the circle.
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    (Original post by Tau)
    Use the substitution  u = \frac{-x^2}{2} which transforms the integral to  \int -e^u\ du . More generally,  \int f'(x)e^{f(x)}\ dx = e^{f(x)} +c . To see why this works, consider the substitution  u = f(x) .
    Why on earth would you use u = -x^2/2? u=x^2 does the job fine and I can understand that u=x^2/2 helps remove ugly factors of 2^{-1} but why would you want to introduce an artificial negative sign for no good reason?
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    (Original post by Zacken)
    Why on earth would you use u = -x^2/2? u=x^2 does the job fine and I can understand that u=x^2/2 helps remove ugly factors of 2^{-1} but why would you want to introduce an artificial negative sign for no good reason?
    I agree it's unnecessary, it was to highlight that it was an example of the general case of  \int f'(x)e^{f(x)}\ dx = e^{f(x)} +c .
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    (Original post by Zacken)
    Why on earth would you use u = -x^2/2? u=x^2 does the job fine and I can understand that u=x^2/2 helps remove ugly factors of 2^{-1} but why would you want to introduce an artificial negative sign for no good reason?
    What is it to u, so racist


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