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    STEP I (paper here)
    1: Solution by Zacken
    2: Solution by Zacken
    3: Solution by Rishab_01
    4: Solution by Zacken
    5: Solution by Rishab_01
    6: Solution by Zacken
    7: Solution by harlem_basket
    8: Solution by Zacken
    9: Solution by FractalSteinway
    10: Solution by FractalSteinway
    11: Solution by JohnCraig
    12: Solution by Farhan.Hanif93
    13: Solution by 13 1 20 8 42

    STEP II (paper here)
    1: Solution by Zacken
    2: Solution by Farhan.Hanif13
    3: Solution by Slice of Pi
    4: Solution by Slice of Pi and SolC
    5: Solution by Rishab_01
    6: Solution by IrrationalRoot
    7: Solution by Zacken
    8: Solution by Farhan.Hanif13
    9: Solution by atsruser
    10: Solution by Forecast
    11: Solution by DFranklin
    12: Solution by DFranklin
    13: Solution by 13 1 20 8 42


    STEP III (paper here):
    1: Solution by Zacken
    2: Solution by SliceofPi
    3: Solution by Jamvicious
    4: Solution by DFranklin
    5: Solution by ThatPerson
    6: Solution by DFranklin
    7: Solution by DFranklin
    8: Solution by IrrationalRoot
    9: Solution by Stelios98
    10: Solution by DFranklin
    11: Solution by DFranklin
    12: Solution by Zacken
    13: Solution by Zacken
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    STEP I, Question 1:

    Let u=x\sin x+\cos x so that \frac{\mathrm{d}u}{\mathrm{d}x}= x\cos x . Hence the integral is

    \displaystyle

\begin{equation*}\int \frac{x\cos x}{x\sin x + \cos x} \, \mathrm{d}x = \int \frac{\mathrm{d}u}{u} = \log u +\mathcal{C} = \log (x\sin x + \cos x) + \mathcal{C} \end{equation*}

    Let u = x\cos x - \sin x so that \frac{\mathrm{d}u}{\mathrm{d}x} = -x\sin x. Hence the integral is

    \displaystyle

\begin{equation*}\int \frac{x\sin x}{x\cos x - \sin x} \, \mathrm{d}x = -\int\frac{\mathrm{d}u}{u} = -\log u + \mathcal{C}= -\log (x\cos x - \sin x) + \mathcal{C}\end{equation*}

    Let u = x\sec^2 x - \tan x so that \frac{\mathrm{d}u}{\mathrm{d}x} = 2x\tan x \sec^2 x. Hence the integral is

    \displaystyle

\begin{equation*}\int \frac{x\sec^2 x \tan x}{x\sec^2 x - \tan x} \, \mathrm{d}x = \frac{1}{2}\int \frac{\mathrm{d}u}{u} = \frac{1}{2}\log u + \mathcal{C} = \frac{\log |x\sec^2 x - \tan x|}{2} + \mathcal{C}\end{equation*}

    Let u = x\sec^2  x - \tan x so that \frac{\mathrm{d}u}{\mathrm{d}x} = 2x\tan x \sec^2 x. Hence the integral is

    \displaystyle

\begin{equation*}\int \frac{x\sin x \cos x}{(x-\sin x \cos x)^2} \, \mathrm{d}x = \int \frac{x\sec^2 x\tan x}{(x\sec^2  x - \tan x)^2} \, \mathrm{d}x = \frac{1}{2}\int \frac{\mathrm{d}u}{u^2} = -\frac{1}{2u} + \mathcal{C} = \frac{1}{2(\tan x - x\sec^2 x)} + \mathcal{C}\end{equation*}

    This was embarrassingly easy for a STEP question, so it's hard to predict the spread of marks but something like: 3/4/5/8 or 3/4/6/7 is what I'd guess.
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    1. STEP IV (paper here):
      1:
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    (Original post by Zacken)
    Question 1:

    Let u=x\sin x+\cos x so that \frac{\mathrm{d}u}{\mathrm{d}x}= x\cos x . Hence the integral is

    \displaystyle

\begin{equation*}\int \frac{x\cos x}{x\sin x + \cos x} \, \mathrm{d}x = \int \frac{\mathrm{d}u}{u} = \log u +\mathcal{C} = \log (x\sin x + \cos x) + \mathcal{C} \end{equation*}

    Let u = x\cos x - \sin x so that \frac{\mathrm{d}u}{\mathrm{d}x} = -x\sin x. Hence the integral is

    \displaystyle

\begin{equation*}\int \frac{x\sin x}{x\cos x - \sin x} \, \mathrm{d}x = -\int\frac{\mathrm{d}u}{u} = -\log u + \mathcal{C}= -\log (x\cos x - \sin x) + \mathcal{C}\end{equation*}

    Let u = x\sec^2 x - \tan x so that \frac{\mathrm{d}u}{\mathrm{d}x} = 2x\tan x \sec^2 x. Hence the integral is

    \displaystyle

\begin{equation*}\int \frac{x\sec^2 x \tan x}{x\sec^2 x - \tan x} \, \mathrm{d}x = \frac{1}{2}\int \frac{\mathrm{d}u}{u} = \frac{1}{2}\log u + \mathcal{C} = \frac{\log |x\sec^2 x - \tan x|}{2} + \mathcal{C}\end{equation*}

    Let u = x\sec^2  x - \tan x so that \frac{\mathrm{d}u}{\mathrm{d}x} = 2x\tan x \sec^2 x. Hence the integral is

    \displaystyle

\begin{equation*}\int \frac{x\sin x \cos x}{(x-\sin x \cos x)^2} \, \mathrm{d}x = \int \frac{x\sec^2 x\tan x}{(x\sec^2  x - \tan x)^2} \, \mathrm{d}x = \frac{1}{2}\int \frac{\mathrm{d}u}{u^2} = -\frac{1}{2u} + \mathcal{C} = \frac{1}{2(\tan x - x\sec^2 x)} + \mathcal{C}\end{equation*}

    This was embarrassingly easy for a STEP question, so it's hard to predict the spread of marks but something like: 3/4/5/8 or 3/4/6/7 is what I'd guess.
    zain set me link for good copy for the paper
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    Question 3
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    Do you have the paper?
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    (Original post by A Slice of Pi)
    Do you have the paper?
    No but i remember this question, it was pretty standard. If you happened to study parabola then its a piece of cake else I would rate it moderate.
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    Question 2:

    Note that, since integration preserves rank, we have \int_1^x \frac{\mathrm{d}t}{t^2} \leqslant \int_1^x \frac{\mathrm{d}t}{t} \leqslant \int_1^x \,\mathrm{d}t for all x\geqslant 1. This gives

    \displaystyle

\begin{equation*}1 - \frac{1}{x} \leqslant \log x \leqslant x-1\end{equation*}

    for all x \geq 1. If 0 < x \leqslant 1 then replace the above with x\mapsto \frac{1}{x}\geqslant 1 to get

    \displaystyle

\begin{equation*}1-x\leqslant \log (1/x) \leqslant \frac{1}{x}-1 \implies 1-\frac{1}{x} \leqslant \log x \leqslant x-1\end{equation*}

    by multiplying both sides by -1. This completes (i) and (ii). Again. Integration preserves rank so for all y > 1 we have

    \displaystyle 

\begin{equation*}\int_1^y 1-\frac{1}{x} \, \mathrm{d}x \leqslant \int_1^y \log x \, \mathrm{d}x \leqslant \int_1^y x-1 \, \mathrm{d}x \end{equation*}

    Hence

    \displaystyle

\begin{equation*}y- 1- \log y \leqslant y \log y - y + 1 \leqslant \frac{1}{2}(y-1)^2\end{equation*}

    so that y \log y \leqslant \frac{(y-1)(y+1)}{2} from above and \logy \geqslant 2(y-1)/(y+1) from below by some trivial rearranging. Hence we get

    \displaystyle

\begin{equation*}\frac{2}{y+1} \leqslant \frac{\log y}{y-1}\leqslant \frac{y+1}{2y}\end{equation*}

    for all y > 1. For 0<y<1 simply invert y \mapsto \frac{1}{y} and you get the same thing after some trivial rearranging, completing (iii).

    I'd predict a split of about 5/7/8.
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    (Original post by solC)
    Question 11: (Credit to JohnCraig)
    The attachment doesn't seem to be working for me - could you try uploading it so something like imgur and then inject the image here using the "insert image via URL"?

    (Original post by A Slice of Pi)
    Do you have the paper?
    Yep, but just a bunch of badly taken pictures - I'll rotate/crop and pdf them at some point today.

    (Original post by Rishabh_01)
    ...
    If anybody is willing to LaTeX this up, please do!
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    (Original post by Zacken)
    The attachment doesn't seem to be working for me - could you try uploading it so something like imgur and then inject the image here using the "insert image via URL"?



    Yep, but just a bunch of badly taken pictures - I'll rotate/crop and pdf them at some point today.



    If anybody is willing to LaTeX this up, please do!
    dished why u blanking me
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    (Original post by Zacken)
    The attachment doesn't seem to be working for me - could you try uploading it so something like imgur and then inject the image here using the "insert image via URL"?
    Oh right, sorry. I'll try to fix it now.

    What are your opinions on the paper thus far, from the questions you've seen?
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    Question 5
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    What do you thing you be the marks boundries for grade 1 and grade S
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    Question 11: (credit to JohnCraig)

    Name:  2017 Paper I Q11.jpg
Views: 848
Size:  15.0 KB

    If this doesn't work I give up
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    STEP I, Question 8:

    Base case is trivial. Assume a_n^2 + 2a_n b_n - b_n^2 = 1 then

    \displaystyle 

\begin{align*}a^2_{n+1} + 2a_{n+1}b_{n+1}- b^2_{n+1} &= a_n^2 + 4a_n b_n + 4b_n^2 + 2(2a_n^2 + 9a_nb_n+10b_n^2) - 4a_n^2 - 20a_nb_n - 25b_n^2 \\ &=a_n^2 + 2a_nb_n -b_n^2 = 1.\end{align*}

    (i) Assume that a_n \geqslant 0 and b_n \geqslant 2\cdot 5^{n-1}. Then a_{n+1} = a_n + 4\cdot 5^{n-1} \geqslant a_n \geqslant 0 and b_{n+1} = 2a_n + 5b_n \geqslant 5b_{n} \geqslant 2 \cdot 5^n.

    Now let c_n = \frac{a_n}{b_n} so that c_n^2 + 2c_n - 1 = 1/b_n^2, so that as n \to \infty we have b_n unbounded above so 1/b_n^2 \to 0 and assuming that c_n \to c gives c^2+ 2c - 1 = 0, i.e c=-1+\sqrt{2} since c_n \geqslant 0.

    (ii) Assume that c_n > \sqrt{2}-1, then since c_{n+1} = \frac{a_{n+1}}{b_{n+1}} = \frac{a_n + 2b_n}{2a_n + 5b_{n}} = \frac{c_n + 2}{2c_n + 5} = \frac{1}{2}\left(1 - \frac{1}{2c_n + 5}\right) we have \frac{1}{2c_n + 5} < \frac{1}{2\sqrt{2} + 3} = 3 - 2\sqrt{2} so c_{n+1} > \frac{1}{2}(1-(3-2\sqrt{2})) = \sqrt{2} - 1.

    Hence since c_n + 1 > \sqrt{2} we have \frac{2}{c_n + 1} < \frac{2}{\sqrt{2}} = \sqrt{2} giving \frac{2}{c_n + 1} < \sqrt{2} < c_n + 1 as required.

    Now note that from the recurrence for c_n we can compute c_3 = \frac{29}{70} so \frac{140}{99} < \sqrt{2} < \frac{99}{70}.
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    If anyone has a copy of the paper can you send it to me? I could write up a few solutions.
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    Paper uploaded.
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    STEP I, Question 6:

    (i) Assume that f(x) takes only (wlog) non-negative values in the interval [0,1]. Let x_0 be such that f(x_0) > 0. Then by continuity we have an interval [0,1] \cap (x_0 - \delta, x_0 + \delta) where |f(x)- f(x_0)| < f(x_0)/2 \Rightarrow f(x) > 0. But then \int_0^1 f > \int_{x_0 - \delta}^{x_0 + \delta} f > 0. Contradiction. (really contraposition)

    (ii) We have \int_0^1 (x-\alpha)^2 g = \int_0^1 x^2 g - 2\alpha \int_0^1 xg + \alpha^2 \int_0^1 g = 0. Hence (x-\alpha)^2 g must take both positive and negative values on [0,1] (or g identically 0\, and we are done anyway). But since squares are non-negative this means that g must take positive and negative values on [0,1] hence must hit 0\, at some point.

    We get 2a+b = 1, 3a + 2b = 6\alpha and 4a + 3b = 12\alpha^2 so two possible solutions are (1 \pm \sqrt{3}, \mp 2\sqrt{3}, \frac{1}{2}\mp \frac{1}{2\sqrt{3}}). Both of which hit 0 at some point in (0,1) which you can check using obvious inequalities. e.g x = \frac{1+ \sqrt{3}}{2\sqrt{3}} = \frac{1}{2} + \frac{1}{2\sqrt{3}} < 1.

    (iii) Set g = h' then \int_0^1 h' = h(1) - h(0) = 1. Also \int_0^1 xh' = xh]_0^1 - \int_0^1 h = 1 - \beta. And finally \int_0^1 x^2 h' = x^2h]_0^1 - 2\int xh = 1 - \beta(2-\beta) = 1 - 2\beta + \beta^2 = (1-\beta)^2.

    So apply (ii) with g = h' and \alpha = 1-\beta get g = h' = 0 somewhere in [0,1].
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    Could you add my solution of question 5 to pinned thread
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    (Original post by Rishabh_01)
    Could you add my solution of question 5 to pinned thread
    I already have.
 
 
 
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