STEP Maths I, II, III 1995 Solutions

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  1. klausw's Avatar
    141 18-05-2012  18:54
    • Junior Member
    • Posts: 55
    Re: STEP Maths I, II, III 1995 Solutions
    (Original post by insparato)
    Writing out STEP III Question 2 now.

    \displaystyle I_n = \int_0^a x^{n+\frac{1}{2}}(a-x)^{\frac{1}{2}}

    \displaystyle I_0 = \int_0^a x^{\frac{1}{2}}(a-x)^{\frac{1}{2}}

    x = asin^2u

    \frac{dx}{du} = 2asinucosu

    dx = 2asinucosu. du

    a = asin^2u

    u = \frac{\pi}{2}

    0 = asin^2u

    u = 0

    \displaystyle I_0 = \int_0^{\frac{\pi}{2}} a^{\frac{1}{2}}sinu(a^{\frac{1}{ 2}}cosu).2asinucosu. du

    \displaystyle = 2a^2\int_0^{\frac{\pi}{2}} sin^2ucos^2u

    \displaystyle = 2a^2\int_0^{\frac{\pi}{2}} \frac{1}{4}(1-cos2u)(1+cos2u)

    \displaystyle = \frac{a^2}{2} \int_0^{\frac{\pi}{2}} 1-cos^22u

    \displaystyle = \frac{a^2}{2} \int_0^{\frac{\pi}{2}} 1-\frac{1}{2}(1-cos4u)

    \displaystyle = \frac{a^2}{2} \int_0^{\frac{\pi}{2}} \frac{1}{2}-\frac{1}{2}cos4u

    = \frac{a^2}{2} [\frac{1}{2}u - \frac{1}{8}sin4u]_0^{\frac{\pi}{2}}

    = \frac{a^2}{2} ([\frac{\pi}{4} - 0]-[0-0])

    = \frac{\pi a^2}{8}

    ii)
    \displaystyle I_n = \int_0^a x^{n+\frac{1}{2}}(a-x)^{\frac{1}{2}}

    u = x^{n+\frac{1}{2}}

    \frac{du}{dx} = (n+\frac{1}{2})x^{n-\frac{1}{2}}

    \frac{dv}{dx} = (a-x)^{\frac{1}{2}}

    v = -\frac{2}{3}(a-x)^{\frac{3}{2}}

    \displaystyle I_n = [-\frac{2}{3}x^{n+\frac{1}{2}}}(a-x)^{\frac{3}{2}}]_0^a - \int_0^a -\frac{2}{3}(a-x)^{\frac{3}{2}.(n+\frac{1}{2}x^ {n-\frac{1}{2}}

    \displaystyle I_n = \frac{2}{3}(n+\frac{1}{2})\int (a-x)^{\frac{3}{2}}x^{n-\frac{1}{2}}

    Unparseable or potentially dangerous latex formula. Error 4: no dvi output from LaTeX. It is likely that your formula contains syntax errors or worse.
    \displaystyle I_n = \frac{2}{3}(n+\frac{1}{2})\int (a-x)(a-x)^{\frac{1}{2}x^{n-\frac{1}{2}}


    Unparseable or potentially dangerous latex formula. Error 4: no dvi output from LaTeX. It is likely that your formula contains syntax errors or worse.
    \displaystyle I_n = \frac{2}{3}(n+\frac{1}{2})\int a(a-x)^{\frac{1}{2}x^{n-\frac{1}{2}}-(a-x)^{\frac{1}{2}}x^{n+\frac{1}{2} }


    I_n = \frac{2}{3}(n+\frac{1}{2})[aI_{n-1} - I_n]

    3I_n = (2n+1)(aI_{n-1}-I_n)

    3I_n = 2anI_{n-1} - 2nI_n -I_n + aI_{n-1}

    (2n+4)I_n = a(2n+1)I_{n-1}

    I_n = \frac{a(2n+1)}{2n+4}I_{n-1}

    I think this answer is not finished. Here we are asked to evaluate I_n, but I think not in terms of I_n-1 but in terms of a.

    I use the reduction formula to expand I_n as

    (a^n+2)(Pi)(2n+1)!!/(2n+4)!!

    Also if I am allowed to use double factorial notation here?
  2. brianeverit's Avatar
    142 31-05-2012  12:04
    • Exalted Member
    • Location: Lincoln
    • Posts: 293
    Re: STEP Maths I, II, III 1995 Solutions
    Last part of III/2
    \text{From }(2n+4)I_n=a(2n+1)I_{n-1} \text{ we have }
    I_n=\frac{a(2n+1)}{2n+4}I_{n-1}= \frac{a(2n+1)}{2n+4}\times\frac{ a(2n-1)}{2n+2}I_{n-2}\cdots\frac{a.3}{6}I_0
    = \frac{a^n(2n+1)!}{2^n n!}\times \frac{1}{2^{n-1} (n+2)!}=\frac{a^n(2n+1)!}{2^{2n-1} n!(n+2)!}\frac{\pi a^2}{8}
  3. brianeverit's Avatar
    143 17-06-2012  10:33
    • Exalted Member
    • Location: Lincoln
    • Posts: 293
    Re: STEP Maths I, II, III 1995 Solutions
    (Original post by insparato)
    Writing out STEP III Question 2 now.

    \displaystyle I_n = \int_0^a x^{n+\frac{1}{2}}(a-x)^{\frac{1}{2}}

    \displaystyle I_0 = \int_0^a x^{\frac{1}{2}}(a-x)^{\frac{1}{2}}

    x = asin^2u

    \frac{dx}{du} = 2asinucosu

    dx = 2asinucosu. du

    a = asin^2u

    u = \frac{\pi}{2}

    0 = asin^2u

    u = 0

    \displaystyle I_0 = \int_0^{\frac{\pi}{2}} a^{\frac{1}{2}}sinu(a^{\frac{1}{ 2}}cosu).2asinucosu. du

    \displaystyle = 2a^2\int_0^{\frac{\pi}{2}} sin^2ucos^2u

    \displaystyle = 2a^2\int_0^{\frac{\pi}{2}} \frac{1}{4}(1-cos2u)(1+cos2u)

    \displaystyle = \frac{a^2}{2} \int_0^{\frac{\pi}{2}} 1-cos^22u

    \displaystyle = \frac{a^2}{2} \int_0^{\frac{\pi}{2}} 1-\frac{1}{2}(1-cos4u)

    \displaystyle = \frac{a^2}{2} \int_0^{\frac{\pi}{2}} \frac{1}{2}-\frac{1}{2}cos4u

    = \frac{a^2}{2} [\frac{1}{2}u - \frac{1}{8}sin4u]_0^{\frac{\pi}{2}}

    = \frac{a^2}{2} ([\frac{\pi}{4} - 0]-[0-0])

    = \frac{\pi a^2}{8}

    ii)
    \displaystyle I_n = \int_0^a x^{n+\frac{1}{2}}(a-x)^{\frac{1}{2}}

    u = x^{n+\frac{1}{2}}

    \frac{du}{dx} = (n+\frac{1}{2})x^{n-\frac{1}{2}}

    \frac{dv}{dx} = (a-x)^{\frac{1}{2}}

    v = -\frac{2}{3}(a-x)^{\frac{3}{2}}

    \displaystyle I_n = [-\frac{2}{3}x^{n+\frac{1}{2}}}(a-x)^{\frac{3}{2}}]_0^a - \int_0^a -\frac{2}{3}(a-x)^{\frac{3}{2}.(n+\frac{1}{2}x^ {n-\frac{1}{2}}

    \displaystyle I_n = \frac{2}{3}(n+\frac{1}{2})\int (a-x)^{\frac{3}{2}}x^{n-\frac{1}{2}}

    Unparseable or potentially dangerous latex formula. Error 4: no dvi output from LaTeX. It is likely that your formula contains syntax errors or worse.
    \displaystyle I_n = \frac{2}{3}(n+\frac{1}{2})\int (a-x)(a-x)^{\frac{1}{2}x^{n-\frac{1}{2}}


    Unparseable or potentially dangerous latex formula. Error 4: no dvi output from LaTeX. It is likely that your formula contains syntax errors or worse.
    \displaystyle I_n = \frac{2}{3}(n+\frac{1}{2})\int a(a-x)^{\frac{1}{2}x^{n-\frac{1}{2}}-(a-x)^{\frac{1}{2}}x^{n+\frac{1}{2} }


    I_n = \frac{2}{3}(n+\frac{1}{2})[aI_{n-1} - I_n]

    3I_n = (2n+1)(aI_{n-1}-I_n)

    3I_n = 2anI_{n-1} - 2nI_n -I_n + aI_{n-1}

    (2n+4)I_n = a(2n+1)I_{n-1}

    I_n = \frac{a(2n+1)}{2n+4}I_{n-1}
    \text{But the question asked you to evaluate } I_n
  4. und's Avatar
    144 02-10-2012  21:30
    • TSR Demigod
    Re: STEP Maths I, II, III 1995 Solutions
    (Original post by DFranklin)
    Of course, there's nothing special about painting the first post red (as opposed to blue or white), so the number of ways of painting n+1 posts so no two posts are the same colour and the first and last posts are coloured the same is just 3 times what we've just calculated. But then by identifying the 1st and last posts (which are the same colour), we are left with the same situation as n posts in a circle. (e.g. if we consider n=13, then we can think of having posts 1,2,...,12 as the hours on a clock, and we consider post 13 to be the same as post 1).

    So the required answer is just 3r_{n+1} = 2^n+2(-1)^n.

    Edit: Mistake spotted by DVS.
    Sorry for bringing this post back from five years ago, but I just did this question and decided that the answer to the last part is given by \frac{3r_{n+1}}{n} because the arrangement can be rotated n times. If you try with the case n=3, there are two ways of doing it (RBW or RWB in any order). However, having thought about it further, I'm going to assume the question implies that this does not need to be considered, and my assumption about dividing by n is wrong anyway due to other concerns of a similar nature that would need to be factored in. I wonder how many marks I'd lose for putting it over n...
    Last edited by und; 02-10-2012 at 22:13.
  5. Nayim's Avatar
    145 07-10-2012  20:26
    • Respected Member
    • Posts: 169
    Re: STEP Maths I, II, III 1995 Solutions
    STEP 1 QUESTION 9 ALTERNATE SOLUTION
    (up) y = tsin(theta)sqrt(2gh) - 1/2(gt^2)
    (side) x = tcos(theta)sqrt(2gh) so x^2 = t^2cos^2(theta)2gh
    using pythagorean identity, x^2 = 2ght^2 - (y + 1/2(gt^2))^2
    putting y into x^2 give quadratic in t^2
    i.e. (1/4)(g^2)t^4 + (yg -2gh)t^2 + (y^2 + x^2) = 0
    for t = real then (yg -2gh)^2 > (g^2)(y^2 + x^2)
    which, after expanding and simplifying, gives us x^2 < 4h(h-y)
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