STEP Prep Thread 2017

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    (Original post by Zacken)
    You can go if you like, but in my experience, there's nothing that's going to help with STEP other than sitting down on your own and working through questions.
    Noted. In my current experience, this seems to be the case, too. But I am going to test whether or not it would be beneficial for me. And then see if I should continue. The previous cohort did very, very well. So the chances are high. However, I place emphasis on the fact that they worked hard to get those results themselves, so it's mostly just about guidance from the teachers.
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    (Original post by physicsmaths)
    Gna midnap him n take him to homerton


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    lol lim jeck is a joke fam
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    (Original post by Injective)
    lol lim jeck is a joke fam
    I wonder what Lim Jeck thinks of all these jokes about him lol (if he's seeing them).
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    (Original post by IrrationalRoot)
    I wonder what Lim Jeck thinks of all these jokes about him lol (if he's seeing them).
    Lol I doubt he even visits TSR. I spend more time on AoPS than this site, so wouldn't be surprised.
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    (Original post by IrrationalRoot)
    I wonder what Lim Jeck thinks of all these jokes about him lol (if he's seeing them).
    Looked him up, he's just spent two years in national service... what a waste
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    Hi guys,

    Thanks for your help with the second question of the year 2000 STEP 2 paper. Attached is my solution to the first question, any thoughts on my answer would be appreciated.

    Also enjoy your weekend
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    (Original post by Maths.)
    Hi guys,

    Thanks for your help with the second question of the year 2000 STEP 2 paper. Attached is my solution to the first question, any thoughts on my answer would be appreciated.

    Also enjoy your weekend
    Here is the attachment......
    Attached Images
  1. File Type: pdf STEP 2 Paper 2000 Q1.pdf (411.1 KB, 24 views)
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    (Original post by IrrationalRoot)
    I wonder what Lim Jeck thinks of all these jokes about him lol (if he's seeing them).
    Nah he doesn't see them thankfully :lol:


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    This isn't from a STEP paper (AFAIK) but can someone check my thinking for this question please and tell me any nicer ways to do this question.

    The question is:  \int^\frac{\pi}{2}_0 \log \sin \theta \ \mathrm{d}\theta

    Using the substitution  \theta \mapsto \frac{\pi}{2} - \theta transforms the integral into  \int^\frac{\pi}{2}_0 \log (\sin (\frac{\pi}{2} - \theta) ) \ \mathrm{d}\theta = \int^\frac{\pi}{2}_0 \log \cos \theta \ \mathrm{d}\theta

    If we call the integral  I , then  2I = \int^\frac{\pi}{2}_0 \log \sin \theta \ \mathrm{d}\theta + \int^\frac{\pi}{2}_0 \log \cos \theta \ \mathrm{d}\theta = \int^\frac{\pi}{2}_0 \log( \sin \theta \cos \theta ) \ \mathrm{d}\theta

    So  2I = \int^\frac{\pi}{2}_0 \log(\frac{1}{2} \sin 2\theta) \mathrm{d}\theta = \int^\frac{\pi}{2}_0 \log(\frac{1}{2}) \ \mathrm{d}\theta + \int^\frac{\pi}{2}_0 \log(\sin 2\theta) \ \mathrm{d}\theta

    We can rearrange this to give  \int^\frac{\pi}{2}_0 \log(\sin 2\theta) \ \mathrm{d}\theta  = 2I + \frac{\pi}{2} log2

    Using the substitution  x = 2\theta transforms this integral into
     \frac{1}{2} \int^\pi_0 \log \sin x \ \mathrm{d}x =\int^\frac{\pi}{2}_0 \log \sin x \ \mathrm{d} x = I

    So  I = 2I + \frac{\pi}{2} log2 , which means  I = -\frac{\pi}{2} log2

    Do I need to justify the use of  \frac{1}{2} \int^\pi_0 \log \sin x \ \mathrm{d}x =\int^\frac{\pi}{2}_0 \log \sin x \ \mathrm{d} x ?

    Thanks guys
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    (Original post by Tau)
    This isn't from a STEP paper (AFAIK) but can someone check my thinking for this question please and tell me any nicer ways to do this question.

    The question is:  \int^\frac{\pi}{2}_0 \log \sin \theta \ \mathrm{d}\theta

    Using the substitution  \theta \mapsto \frac{\pi}{2} - \theta transforms the integral into  \int^\frac{\pi}{2}_0 \log (\sin (\frac{\pi}{2} - \theta) ) \ \mathrm{d}\theta = \int^\frac{\pi}{2}_0 \log \cos \theta \ \mathrm{d}\theta

    If we call the integral  I , then  2I = \int^\frac{\pi}{2}_0 \log \sin \theta \ \mathrm{d}\theta + \int^\frac{\pi}{2}_0 \log \cos \theta \ \mathrm{d}\theta = \int^\frac{\pi}{2}_0 \log( \sin \theta \cos \theta ) \ \mathrm{d}\theta

    So  2I = \int^\frac{\pi}{2}_0 \log(\frac{1}{2} \sin 2\theta) \mathrm{d}\theta = \int^\frac{\pi}{2}_0 \log(\frac{1}{2}) \ \mathrm{d}\theta + \int^\frac{\pi}{2}_0 \log(\sin 2\theta) \ \mathrm{d}\theta

    We can rearrange this to give  \int^\frac{\pi}{2}_0 \log(\sin 2\theta) \ \mathrm{d}\theta  = 2I + \frac{\pi}{2} log2

    Using the substitution  x = 2\theta transforms this integral into
     \frac{1}{2} \int^\pi_0 \log \sin x \ \mathrm{d}x =\int^\frac{\pi}{2}_0 \log \sin x \ \mathrm{d} x = I

    So  I = 2I + \frac{\pi}{2} log2 , which means  I = -\frac{\pi}{2} log2

    Do I need to justify the use of  \frac{1}{2} \int^\pi_0 \log \sin x \ \mathrm{d}x =\int^\frac{\pi}{2}_0 \log \sin x \ \mathrm{d} x ?

    Thanks guys
    Yeh show that
    If f(x)=f(2a-x) then f is symmetric about a



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    (Original post by physicsmaths)
    Yeh show that
    If f(x)=f(2a-x) then f is symmetric about a



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    Ah alright, thanks.
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    (Original post by Tau)
    This isn't from a STEP paper (AFAIK) but can someone check my thinking for this question please and tell me any nicer ways to do this question.

    The question is:  \int^\frac{\pi}{2}_0 \log \sin \theta \ \mathrm{d}\theta

    Using the substitution  \theta \mapsto \frac{\pi}{2} - \theta transforms the integral into  \int^\frac{\pi}{2}_0 \log (\sin (\frac{\pi}{2} - \theta) ) \ \mathrm{d}\theta = \int^\frac{\pi}{2}_0 \log \cos \theta \ \mathrm{d}\theta

    If we call the integral  I , then  2I = \int^\frac{\pi}{2}_0 \log \sin \theta \ \mathrm{d}\theta + \int^\frac{\pi}{2}_0 \log \cos \theta \ \mathrm{d}\theta = \int^\frac{\pi}{2}_0 \log( \sin \theta \cos \theta ) \ \mathrm{d}\theta

    So  2I = \int^\frac{\pi}{2}_0 \log(\frac{1}{2} \sin 2\theta) \mathrm{d}\theta = \int^\frac{\pi}{2}_0 \log(\frac{1}{2}) \ \mathrm{d}\theta + \int^\frac{\pi}{2}_0 \log(\sin 2\theta) \ \mathrm{d}\theta

    We can rearrange this to give  \int^\frac{\pi}{2}_0 \log(\sin 2\theta) \ \mathrm{d}\theta  = 2I + \frac{\pi}{2} log2

    Using the substitution  x = 2\theta transforms this integral into
     \frac{1}{2} \int^\pi_0 \log \sin x \ \mathrm{d}x =\int^\frac{\pi}{2}_0 \log \sin x \ \mathrm{d} x = I

    So  I = 2I + \frac{\pi}{2} log2 , which means  I = -\frac{\pi}{2} log2

    Do I need to justify the use of  \frac{1}{2} \int^\pi_0 \log \sin x \ \mathrm{d}x =\int^\frac{\pi}{2}_0 \log \sin x \ \mathrm{d} x ?

    Thanks guys
    Depends on how pedantic you want to be; if this came up in STEP (which it sorta did this year in II) then you'd get away with stating that or at most a quick sketch.

    If you want to go for a more formal way, then what PM said works best, but may be overly much detail in STEP (imo).

    (BTW, that solution seems very familiar, would it happen to be mine/my question? Just curious)


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    (Original post by Zacken)
    Depends on how pedantic you want to be; if this came up in STEP (which it sorta did this year in II) then you'd get away with stating that or at most a quick sketch.

    If you want to go for a more formal way, then what PM said works best, but may be overly much detail in STEP (imo).

    (BTW, that solution seems very familiar, would it happen to be mine/my question? Just curious)


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    Okay, thanks. I saw the question in a post of yours about trigonometric symmetry but I don't think you ever posted the answer. Thanks for the question though
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    (Original post by physicsmaths)
    Yeh show that
    If f(x)=f(2a-x) then f is symmetric about a



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    I gave an explanation of this here:

    http://www.thestudentroom.co.uk/show...36&postcount=5

    I expect that if they expected any justification at all however, a sketch of the sine graph with the appropriate line of symmetry would be enough.
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    (Original post by Maths.)
    Here is the attachment......
    I guess they are still too busy to help you :2euk48l:


    Zacken
    jneill
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    (Original post by Tau)
    This isn't from a STEP paper (AFAIK) but can someone check my thinking for this question please and tell me any nicer ways to do this question.

    The question is:  \int^\frac{\pi}{2}_0 \log \sin \theta \ \mathrm{d}\theta

    Using the substitution  \theta \mapsto \frac{\pi}{2} - \theta transforms the integral into  \int^\frac{\pi}{2}_0 \log (\sin (\frac{\pi}{2} - \theta) ) \ \mathrm{d}\theta = \int^\frac{\pi}{2}_0 \log \cos \theta \ \mathrm{d}\theta

    If we call the integral  I , then  2I = \int^\frac{\pi}{2}_0 \log \sin \theta \ \mathrm{d}\theta + \int^\frac{\pi}{2}_0 \log \cos \theta \ \mathrm{d}\theta = \int^\frac{\pi}{2}_0 \log( \sin \theta \cos \theta ) \ \mathrm{d}\theta

    So  2I = \int^\frac{\pi}{2}_0 \log(\frac{1}{2} \sin 2\theta) \mathrm{d}\theta = \int^\frac{\pi}{2}_0 \log(\frac{1}{2}) \ \mathrm{d}\theta + \int^\frac{\pi}{2}_0 \log(\sin 2\theta) \ \mathrm{d}\theta

    We can rearrange this to give  \int^\frac{\pi}{2}_0 \log(\sin 2\theta) \ \mathrm{d}\theta  = 2I + \frac{\pi}{2} log2

    Using the substitution  x = 2\theta transforms this integral into
     \frac{1}{2} \int^\pi_0 \log \sin x \ \mathrm{d}x =\int^\frac{\pi}{2}_0 \log \sin x \ \mathrm{d} x = I

    So  I = 2I + \frac{\pi}{2} log2 , which means  I = -\frac{\pi}{2} log2

    Do I need to justify the use of  \frac{1}{2} \int^\pi_0 \log \sin x \ \mathrm{d}x =\int^\frac{\pi}{2}_0 \log \sin x \ \mathrm{d} x ?

    Thanks guys
    As physicsmaths pointed out about symmetry - I would at that point say: "So 2I +pi*ln(2)/2 = integral of ln(sin(2x)). Now we make substitution 2x = t to get integral of ln(sin(x))/2 from 0 to pi.

    However, since the graph of ln(sin(x)) is symmetric about x=pi/2, this is 2I/2=I so we get;

    2I+pi*ln(2)/2=I so I=-pi*ln(2)/2"

    Isn't needed, but worthwhile imo.
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    (Original post by atsruser)
    I gave an explanation of this here:

    http://www.thestudentroom.co.uk/show...36&postcount=5

    I expect that if they expected any justification at all however, a sketch of the sine graph with the appropriate line of symmetry would be enough.
    Actually, I note that the post in question dealt with reflections in a line x=a, which is of course related but not precisely the same.

    However if:

    for all x, f(x)=f(2a-x)

    then let x=a+u \Rightarrow 2a-x = 2a-a-u=a-u

    so that f(a+u)=f(a-u), for all u

    and so we have symmetry about the line x=a since the function takes the same value if we move a bit up the x-axis from a i.e. to a+u as when we move down the x-axis from a i.e. to a-u.

    This uses the trick of creating a new u-axis whose origin is at x=a
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    (Original post by Zacken)

    (BTW, that solution seems very familiar, would it happen to be mine/my question? Just curious)
    A similar question, \int_0^\pi 1+\cos x \ dx, came up on the hard integral thread, at least once. Someone or other posted this solution, whose elegance, efficiency, and panache may have stuck in your mind:

    http://www.thestudentroom.co.uk/show...&postcount=609
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    (Original post by Melanie Leconte)
    I guess they are still too busy to help you :2euk48l:


    :confused:

    :banghead:
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    (Original post by atsruser)
    A similar question, \int_0^\pi 1+\cos x \ dx, came up on the hard integral thread, at least once. Someone or other posted this solution, whose elegance, efficiency, and panache may have stuck in your mind.

    Heh, some self-plugging is always good. Good to see you around, helping the young 'uns.
 
 
 
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