The 2012 STEP Results Discussion Thread

Maths exam discussion - share revision tips in preparation for GCSE, A Level and other maths exams and discuss how they went afterwards.

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  1. Farhan.Hanif93's Avatar
    3481 13-06-2012  18:46
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    Re: The 2012 STEP Prep Thread!
    (Original post by shanban)
    this makes me feel much better- thank you!

    Thing is, I've developed a terrible habit of coming onto here to look for solutions when I reach a dead end...and altho everyone said not to look for hints after a few attempts, I did exactly that and now I'm regretting it..!

    Oh and I do about 3-4hrs of STEP before bed because then I don't lie in bed thinking I'm not prepared for it! Also I have 6 exams next week...STEP revision is alongside revision for 4 other modules!
    Whilst what I've said is true and probably encouraging, it doesn't mean that you should change your aim; the objective should still be to complete as many questions as you can in the given time. Only when you're completely stuck should you move on. At the same time, you need to keep in mind that STEP is a time pressured exam despite the fact that 3 hours seems like a lot; try not to get caught out by that.

    Good luck anyway, just keep the STEP stuff ticking over and make sure you leave enough time to get an A* in FMaths.
  2. shanban's Avatar
    3482 13-06-2012  19:19
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    Re: The 2012 STEP Prep Thread!
    (Original post by Farhan.Hanif93)
    Whilst what I've said is true and probably encouraging, it doesn't mean that you should change your aim; the objective should still be to complete as many questions as you can in the given time. Only when you're completely stuck should you move on. At the same time, you need to keep in mind that STEP is a time pressured exam despite the fact that 3 hours seems like a lot; try not to get caught out by that.

    Good luck anyway, just keep the STEP stuff ticking over and make sure you leave enough time to get an A* in FMaths.

    Aah really? I was thinking that usually I take about an hour and a half to two hours to get some decent solutions to ~4 questions done so 3hrs would give me plenty of time to check over things (which I never do when doing past papers) and try other qs...but will keep the time in mind

    Thanks!
  3. Anon 17's Avatar
    3483 13-06-2012  19:21
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    Re: The 2012 STEP Prep Thread!
    Hooray, finally did a question without looking at any hints/solutions or anything.

    ... Mucked up some simple maths in it (650 x 2 = 1300, not 1350 lmao) but I'm sure I would have spotted that checking it in the exam.

    That's three questions of STEP I 2009 done, it's time to get out of the house now.
    Post rating:
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  4. jack.hadamard's Avatar
    3484 13-06-2012  19:21
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    Re: The 2012 STEP Prep Thread!
    (Original post by TheJ0ker)
    On STEP 1 2001 Q2 for the second equality if you rearrange it to \sqrt{3x+10}-\sqrt{x+4} > 2 and then square, rearrange and square again why do you end up with a different quadratic to if you just do the process without rearranging it and consequently get the wrong answer?
    Why making your life difficult?
    Spoiler:
    Show

    \sqrt{3x + 10} > 2 + \sqrt{x + 4}


    Let v = x + 4.

    \sqrt{3v - 2} > 2 + \sqrt{v}

    \iff 3v - 2 > 4 + 4\sqrt{v} + v, \ \ \ \ \ \ \because v \geq \frac{2}{3}

    \iff (\sqrt{v} - 3)(\sqrt{v} + 1) > 0


    \therefore \sqrt{v} > 3 \ \ \Rightarrow \ \ x + 4 > 9.
    Last edited by jack.hadamard; 13-06-2012 at 22:10. Reason: Typo.
  5. jack.hadamard's Avatar
    3485 13-06-2012  19:25
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    Re: The 2012 STEP Prep Thread!
    (Original post by fruktas)
    Thank you! That was along the lines of what I was thinking, but for some reason it didn't sit right in my head. I see it now.
    As a rule of thumb, odd functions are equal to sums of odd functions, and even functions are sums of even function; always.
    A similar rule can also be deduced for products -- i.e. an even function can be a product that contains only an even number of odd functions.
  6. TheJ0ker's Avatar
    3486 13-06-2012  19:34
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    Re: The 2012 STEP Prep Thread!
    (Original post by jack.hadamard)
    Why making your life difficult?
    Spoiler:
    Show

    \sqrt{3x + 10} > 2 + \sqrt{x + 4}


    Let v = x + 4.

    \sqrt{3v - 2} > 2 + \sqrt{v}

    \iff 3v - 2 > 4 + 2\sqrt{v} + v, \ \ \ \ \ \ \because v \geq \frac{2}{3}

    \iff (\sqrt{v} - 3)(\sqrt{v} + 1) > 0


    \therefore \sqrt{v} > 3 \ \ \Rightarrow \ \ x + 4 > 9.
    I was just wondering why it didn't work. That method is indeed a lot easier, I didn't make the substitution when I first did it though

    Spoiler:
    Show
    Can I quickly ask for this line of working;

    \iff 3v - 2 > 4 + 2\sqrt{v} + v, \ \ \ \ \ \ \because v \geq \frac{2}{3}

    Why does that imply v \geq \frac{2}{3} I'm sure if v=2/3 the inequality doesn't hold.
  7. Lois:)'s Avatar
    3487 13-06-2012  19:43
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    Re: The 2012 STEP Prep Thread!
    (Original post by ThunderShade)
    Spoiler:
    Show
    {\bf p} = \lambda {\bf a} + (1 - \lambda ){\bf b}, so {\bf p}\cdot {\bf a} = (\lambda {\bf a} + (1 - \lambda ){\bf b})\cdot {\bf a} = (\lambda {\bf a})\cdot {\bf a} + ((1 - \lambda){\bf b})\cdot {\bf a} = \lambda a^2 + (1 - \lambda ){\bf a}\cdot {\bf b}. The guy made a typo the first time he writes that down, but he fixes it in the next line.

    Boldface vectors in LaTeX are quite big compared to normal characters. Anybody know how to fix that?
    Thanks! I didn't realise you could do that with dot products.. manipulate them like that. That's a lot clearer though
  8. Lois:)'s Avatar
    3488 13-06-2012  19:44
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    Re: The 2012 STEP Prep Thread!
    (Original post by mikelbird)
    It isnt really obvious...see document....
    Thanks that's really helpful
  9. TheMagicMan's Avatar
    3489 13-06-2012  21:58
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    Re: The 2012 STEP Prep Thread!
    (Original post by jack.hadamard)
    As a rule of thumb, odd functions are equal to sums of odd functions, and even functions are sums of even function; always.
    Not necessarily. You could have an odd function that was the sum of odd functions and even functions, as long as the even functions cancelled out when you performed the sums (although this is a technicality at best)
  10. jack.hadamard's Avatar
    3490 13-06-2012  21:59
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    Re: The 2012 STEP Prep Thread!
    (Original post by TheJ0ker)
    I was just wondering why it didn't work. That method is indeed a lot easier, I didn't make the substitution when I first did it though

    Spoiler:
    Show
    Can I quickly ask for this line of working;

    \iff 3v - 2 > 4 + 2\sqrt{v} + v, \ \ \ \ \ \ \because v \geq \frac{2}{3}

    Why does that imply v \geq \frac{2}{3} I'm sure if v=2/3 the inequality doesn't hold.
    It says because of this. (also a typo: should be 4\sqrt{v})
    Spoiler:
    Show

    If v could be less than that, one of the sides will be negative, and hence, it is not if and only if statement. This is just allowed values: v = x + 4 \geq -\frac{10}{3} + 4.
    Otherwise, if the second inequality is true, and you can take the square root to get the first one, so that it is valid both ways.

    One can justify this by monotonicity (preserving order), since y = x^2 and y = \sqrt{x} are monotonically increasing functions (for non-negative reals).
    Spoiler:
    Show

    When a function is monotonically increasing x_1 \geq x_2 \iff f(x_1) \geq f(x_2); i.e. preserves the order. Drop the equality for strictly increasing.

    So, for example, on the interval [-1, 1] the function y = x^2 is not monotonic, so you could not use similar logic for the implications.

  11. TheMagicMan's Avatar
    3491 13-06-2012  22:02
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    Re: The 2012 STEP Prep Thread!
    (Original post by TheJ0ker)
    I was just wondering why it didn't work. That method is indeed a lot easier, I didn't make the substitution when I first did it though

    Spoiler:
    Show
    Can I quickly ask for this line of working;

    \iff 3v - 2 > 4 + 2\sqrt{v} + v, \ \ \ \ \ \ \because v \geq \frac{2}{3}

    Why does that imply v \geq \frac{2}{3} I'm sure if v=2/3 the inequality doesn't hold.
    With regards to your spoiler...

    That condition is implicit in the initial question as it used \sqrt{3x+10} which implies that x \geq - \frac{10}{3}

    (At least I think that's what he talking about...i think he's trying to justify the validity of that inequality or something)
    Last edited by TheMagicMan; 13-06-2012 at 22:03.
  12. jack.hadamard's Avatar
    3492 13-06-2012  22:08
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    Re: The 2012 STEP Prep Thread!
    (Original post by TheMagicMan)
    Not necessarily. You could have an odd function that was the sum of odd functions and even functions, as long as the even functions cancelled out when you performed the sums (although this is a technicality at best)
    Technically you could, but these terms will not contribute anything to the actual function.
    Like, you could say f(x) = x can be expressed as f(x) = x + x^2 - x^2.

    EDIT: If you have an opinion, then share it. Don't just neg me.
    Last edited by jack.hadamard; 14-06-2012 at 18:40.
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  13. TheJ0ker's Avatar
    3493 14-06-2012  08:04
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    Re: The 2012 STEP Prep Thread!
    (Original post by TheMagicMan)
    With regards to your spoiler...

    That condition is implicit in the initial question as it used \sqrt{3x+10} which implies that x \geq - \frac{10}{3}

    (At least I think that's what he talking about...i think he's trying to justify the validity of that inequality or something)

    (Original post by jack.hadamard)
    It says because of this. (also a typo: should be 4\sqrt{v})
    Spoiler:
    Show

    If v could be less than that, one of the sides will be negative, and hence, it is not if and only if statement. This is just allowed values: v = x + 4 \geq -\frac{10}{3} + 4.
    Otherwise, if the second inequality is true, and you can take the square root to get the first one, so that it is valid both ways.

    One can justify this by monotonicity (preserving order), since y = x^2 and y = \sqrt{x} are monotonically increasing functions (for non-negative reals).
    Spoiler:
    Show

    When a function is monotonically increasing x_1 \geq x_2 \iff f(x_1) \geq f(x_2); i.e. preserves the order. Drop the equality for strictly increasing.

    So, for example, on the interval [-1, 1] the function y = x^2 is not monotonic, so you could not use similar logic for the implications.

    Ok thanks guys.
  14. ben-smith's Avatar
    3494 14-06-2012  11:56
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    Re: The 2012 STEP Prep Thread!
    going to do STEP II 2008 under timed conditions this afternoon. Wish me luck
  15. safmaster's Avatar
    3495 14-06-2012  12:01
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    Re: The 2012 STEP Prep Thread!
    (Original post by ben-smith)
    going to do STEP II 2008 under timed conditions this afternoon. Wish me luck
    Good Luck. How many more papers have you got left to do/is this your last one? The exam is next week !
  16. OmnipotentOmelette's Avatar
    3496 14-06-2012  14:19
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    Has anyone done the step 1 2000 paper? I did it timed and it seemed easier than others... Or did I just have a good day :-P

    This was posted from The Student Room's Android App on my GT-I9100
  17. Rahul.S's Avatar
    3497 14-06-2012  15:12
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    Re: The 2012 STEP Prep Thread!
    (Original post by OmnipotentOmelette)
    Has anyone done the step 1 2000 paper? I did it timed and it seemed easier than others... Or did I just have a good day :-P

    This was posted from The Student Room's Android App on my GT-I9100
    it was one of the first step papers i did during interview time. I would say relatively to recent papers it was nice - but it was 69 for a 1...so people did find it challenging :cool:
  18. jack.hadamard's Avatar
    3498 14-06-2012  18:42
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    Re: The 2012 STEP Prep Thread!
    Ready for Monday? Will we get the paper uploaded on here the next day?
  19. Ree69's Avatar
    3499 14-06-2012  19:13
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    Re: The 2012 STEP Prep Thread!
    Arg, exams haven't gone as well as planned. I'm even more pumped for STEP (I) now.
  20. Lord of the Flies's Avatar
    3500 14-06-2012  19:36
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    Re: The 2012 STEP Prep Thread!
    Gosh I feel so unprepared even though my life has literally been working through past STEP papers since God knows when... Ach!!
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