Results are out! Find what you need...fast. Get quick advice or join the chat
Hey! Sign in to get help with your study questionsNew here? Join for free to post

STEP Maths I, II, III 1988 solutions

Announcements Posted on
Will you get the grades you need for uni? Get prepared today and de-stress, sign up to email alerts for course places! 02-06-2015
Waiting on IB results? Our IB results hub explains everything you need to know 01-06-2015
  1. Offline

    ReputationRep:
    (Updated as far as #68) SimonM - 15.06.2009

    Seeing as people seem to have asked for a thread for this for quite a while I thought I might as well make one.

    I guess you're all familiar what these threads are for by now; post your solutions to the older STEP papers as there are no solutions available freely on the net - plus it's good revision Feel free to post alternative solutions, and please point out when you see a weakness or mistake in a solution.

    If you look in the STEP megathread you will find a link to the papers - if you still have problems accessing them, PM me or someone else who has them.

    STEP I:
    1: Solution by nota bene
    2: Solution by nota bene
    3: Solution by Glutamic Acid
    4: Solution by brianeverit
    5: Solution by Swayum
    6: Solution by brianeverit
    7: Solution by Glutamic Acid
    8: Solution by brianeverit
    9: Solution by nota bene
    10: Solution by brianeverit
    11: Solution by brianeverit
    12: Solution by brianeverit
    13: Solution by brianeverit
    14: Solution by brianeverit
    15: Solution by nota bene
    16: Solution by brianeverit


    STEP II:
    1: Solution by kabbers
    2: Solution by SimonM
    3: Solution by Glutamic Acid
    4: Solution by brianeverit
    5: Solution by Squeezebox
    6: Solution by SimonM
    7: Solution by Square
    8: Solution by brianeverit
    9: Solution by SimonM
    10: Solution by brianeverit
    11: Solution by brianeverit
    12: Solution by brianeverit
    13: Solution by brianeverit
    14: Solution by brianeverit
    15: Solution by brianeverit
    16: Solution by brianeverit


    STEP III
    1: Solution by kabbers
    2: Solution by Squeezebox
    3: Solution by mikelbird and brianeverit
    4: Solution by mikelbird
    5: Solution by kabbers
    6: Solution by kabbers
    7: Solution by squeezebox
    8: Solution by SimonM
    9: Solution by SimonM
    10: Solution by ben-smith
    11: Solution by Jkn
    12. Solution by ben-smith
    13: Solution by ben-smith
    14: Solution by ben-smith
    15:
    16: Solution by ben-smith


    Solutions written by TSR members:
    1987 - 1988 - 1989 - 1990 - 1991 - 1992 - 1993 - 1994 - 1995 - 1996 - 1997 - 1998 - 1999 - 2000 - 2001 - 2002 - 2003 - 2004 - 2005 - 2006 - 2007
  2. Offline

    ReputationRep:
    STEP I Q1

    h(x)=\frac{\log x}{x}
    h'(x)=-\frac{\log x}{x^2}+\frac{1}{x^2}=\frac{1}{x  ^2}(1-\log x)
    For max/min set h'(x)=0 i.e. 0=\frac{1}{x^2}(1-\log x) \Rightarrow x=e
    To find the nature of the stationary point, consider e.g. h'(e-0.2)=positive and h'(e+0.2)=negative hence a (global) maximum. For sketching purposes might be worth noting that \displaystyle\lim_{x\to0^+}h(x)=-\infty and \displastyle\lim_{x\to\infty}h(x  )=0. Graph see attached (Mathematica because of lack of scanner:p:).

    Solving n^m=m^n is equivalent of solving \log n^m=\log m^n\Leftrightarrow mlog n=n\log m \Leftrightarrow \frac{\log n}{n}=\frac{\log m}{m} (last step valid as neither m nor n can be 0).
    m=n is the trivial solution.
    Consider the line f=c (i.e. a constant). For 0< c< \frac{1}{e} there are two solutions. There's one solution when c=\frac{1}{e} or c\le0. And for c>\frac{1}{e} there are no solutions.

    Now if m\not=n then \frac{n}{m}=k (k\not=1).

    \frac{\log n}{n}=\frac{\log \frac{n}{k}}{\frac{n}{k}}
    \log n=\frac{nk(\log n-\log k)}{n}=k(\log n - \log k)
    \log n=\frac{k}{k-1}\log k
    Thus n=k^{\frac{k}{k-1}} and m=k^{\frac{                  1}{k-1}}.

    (What am I missing here, this looks trivial for a STEP question...)

    ----

    STEP I Q2

    g(x)=f(x)+\frac{1}{f(x)} (f(x)\not=0)
    g'(x)=f'(x)-\frac{f'(x)}{(f(x))^2}=f'(x)(1-\frac{1}{(f(x))^2})
    g''(x)=f''(x)(1-\frac{1}{(f(x))^2})+f'(x)(\frac{  2f'(x)}{(f(x))^3})=f''(x)(1-\frac{1}{(f(x))^2})+\frac{2(f'(x  ))^2}{(f(x))^3}

    Let f(x)=4+\cos(2x)+2\sin(x)
    f'(x)=-2\sin(2x)+2\cos(x)=2\cos(x)(-2\sin(x)+1)
    Now WLOG g'(x)=0\Leftrightarrow f'(x)=0 \text{or} (f(x))^2=1

    f'(x)=0 \Leftrightarrow 2\cos(x)(1-2\sin(x))=0 Has solutions cos(x)=0 and sin(x)=1/2
    i.e. x=\frac{\pi}{2},\frac{3\pi}{2},\  frac{\pi}{6},\frac{5\pi}{6}

    f(x)=1\Leftrightarrow 1=4+1-2\sin^2(x)+2\sin(x) \Leftrightarrow 0=2(2-u^2+u) where u=\sin(x)
    By quadratic formula the solutions are u=\frac{1\pm\sqrt{1+4\times2}}{2  } i.e. u=-1, 2
    \sin(x)=2 has no real solutions and \sin(x)=-1 has the root x=\frac{3\pi}{2} (which we have already found).

    f(x)=-1\Leftrightarrow -1=4+1-2\sin^2(x)+2\sin(x)\Leftrightarr  ow 0=2(3-u^2+u) where u=\sin(x)
    Solutions to this are u=\frac{1\pm\sqrt{1+4\times3}}{2  } and as \frac{1\pm\sqrt{1+4\times3}}{2}>  1 sin(x)=u will have no real solutions.

    f'(\frac{\pi}{2}-\frac{\pi}{16})<0, f'(\frac{\pi}{2}+\frac{\pi}{16})  >0 so a (local)min f(\frac{\pi}{2})=4-1+2=5 \Rightarrow g(\frac{\pi}{2})=5+\frac{1}{5}
    f'(\frac{3\pi}{2}-\frac{\pi}{16})<0, f'(\frac{3\pi}{2}+\frac{\pi}{16}  )>0 so a (global) min f(\frac{3\pi}{2})=4-1-2=1 \Rightarrow g(\frac{3\pi}{2})=2
    f'(\frac{\pi}{6}+\frac{\pi}{18})  <0, f'(\frac{\pi}{6}-\frac{\pi}{18})>0 so a (global) max f(\frac{\pi}{6})=4+\frac{\sqrt{2  }}{2}+1=5+\frac{\sqrt{2}}{2} \Rightarrow g(\frac{\pi}{6})=5+\frac{\sqrt{2  }}{2}+\frac{1}{5+ \frac{\sqrt{2}}{2}}
    f'(\frac{5\pi}{6}+\frac{\pi}{18}  )<0,f'(\frac{5\pi}{6}-\frac{\pi}{18})>0 so a (local) max f(\frac{5\pi}{6})=4+\frac{1}{2}+  1=\frac{11}{2} \Rightarrow g(\frac{5\pi}{6})=\frac{11}{2}+\  frac{2}{11}
    Attached Thumbnails
    Click image for larger version. 

Name:	I 88q1.JPG 
Views:	273 
Size:	20.3 KB 
ID:	51013  
  3. Online

    ReputationRep:
    I haven't bothered to download the paper, so this may not be a valid objection, but I feel I need to point out that 2^4 = 4^2...
  4. Offline

    ReputationRep:
    Yeah, I certainly need to! Had forgotten that one, which is by inspection.
  5. Offline

    ReputationRep:
    High, quick question: What programme are you using to write out the Mathematics? I've seen such quite alot here but have no idea to do so myself. :|

    Thanks.
  6. Online

    ReputationRep:
    Nota: I think you would be expected to give at least some justification why (2,4) and (4,2) are the only 'non-trivial' solutions. (Without giving you a hard time for the slipup: if you missed (2,4), how do you know you didn't miss anything else?)
  7. Offline

    ReputationRep:
    (Original post by The Lyceum)
    High, quick question: What programme are you using to write out the Mathematics? I've seen such quite alot here but have no idea to do so myself. :|

    Thanks.
    LaTeX. It's installed on the forum software - see the thread in the maths forum called "PLEASE READ BEFORE POSTING !!" for an explanation of how to use it.
  8. Offline

    ReputationRep:
    (Original post by generalebriety)
    LaTeX. It's installed on the forum software - see the thread in the maths forum called "PLEASE READ BEFORE POSTING !!" for an explanation of how to use it.
    Thanks.
  9. Offline

    ReputationRep:
    Im about half way through II/7.

    So if someone could refrain from spoiling my fun and posting the solution before I wake up tomorrow and finish the question please!
  10. Offline

    ReputationRep:
    Good idea Nota . I'll have a look at the paper(s) tomorrow and see if I get anything out.

    Also....1989 has many unsolved questions. [/shameless plug]

    generalebriety - could you link this in the Big fat STEP mega thread, pretty please :puppyeyes:?
  11. Offline

    ReputationRep:
    (Original post by DFranklin)
    Nota: I think you would be expected to give at least some justification why (2,4) and (4,2) are the only 'non-trivial' solutions. (Without giving you a hard time for the slipup: if you missed (2,4), how do you know you didn't miss anything else?)
    Ah, yes. *feels stupid*

    Consider the line f=c (i.e. a constant). For 0< c< \frac{1}{e} there are two solutions. There's one solution when c=\frac{1}{e} or c\le0. And for c>\frac{1}{e} there are no solutions.

    Now if m\not=n then \frac{n}{m}=k (k\not=1).

    \frac{\log n}{n}=\frac{\log \frac{n}{k}}{\frac{n}{k}}
    \log n=\frac{nk(\log n-\log k)}{n}=k(\log n - \log k)
    \log n=\frac{k}{k-1}\log k
    Thus n=k^{\frac{k}{k-1}} and m=k^{\frac{                  1}{k-1}}.
  12. Online

    ReputationRep:
    Sorry, but I'm not seeing how the above shows there's only two solutions.

    (I would give a very simple solution based on the graph and the location of the turning point).
  13. Offline

    ReputationRep:
    III/1:

    Sketch y = \frac{x^2 e^{-x}}{x+1}

    I'm not going to bother posting most of the differentiating legwork, there arent too many tricks, its just an algebra bash.

    Differentiating (quotient rule is your friend), we get

    \frac{dy}{dx} = e^{-x}\frac{x(2-x^2)}{(x+1)^2}

    So we have turning points at x = 0, and x = +\sqrt{2}, -\sqrt{2}


    Differentiating again, we get

    \frac{d^2 y}{dx^2} = e^{-x}\frac{(2-3x^2)(x+1)^2 - 2(x+1)(2x-x^3) - (2x - x^3)(x+1)^2}{(x+1)^4}

    Substituting in values of x for the turning points, we find that x = 0 is a minimum, and the other two maximums.


    Noticing the denominator of y, we find that there is an asymptote at x = -1.

    And because of the exponential properties of e^-x, y tends to zero as x \to +\infty

    y tends to -\infty as x \to -\infty since the numerator remains positive and given the exponenential properties of e^x much greater than the denominator, while the denominator becomes negative.


    So the graph will look http://www.thestudentroom.co.uk/show...8&postcount=14



    Prove \displaystyle 0 < \int^{\infty}_0 \frac{x^2 e^{-x}}{1+x} dx < 1

    First note that \frac{x^2}{x+1} = \frac{x^2+x}{x+1} - \frac{x}{x+1} = x - \frac{x}{x+1} = x - (\frac{x+1}{x+1} - \frac{1}{x+1}) = x - 1 + \frac{1}{x+1}

    Hence we may split the integral into \displaystyle\int^{\infty}_0 (x - 1)e^{-x} dx + \int^{\infty}_0 \frac{1}{x+1}e^{-x} dx

    Consider \displaystyle\int^{\infty}_0 (x - 1)e^{-x} dx

    = \displaystyle\int^{\infty}_0 xe^{-x} dx - \int^{\infty}_0 e^{-x} dx

    = \displaystyle([-xe^{-x}]^{\infty}_0 + \int^{\infty}_0 e^{-x} dx - \int^{\infty}_0 e^{-x} dx

    = 0


    Now consider \displaystyle\int^{\infty}_0 \frac{1}{x+1}e^{-x} dx

    I posit the inequality \frac{1}{x+1}e^{-x} < e^{-x} for x > 0

    \frac{1}{x+1} < 1
    1 < (x+1)
    0 < x

    So our inequality holds.

    So:

    \displaystyle\int^{\infty}_0 \frac{1}{x+1}e^{-x} dx < \int^{\infty}_0 e^{-x} dx

    \displaystyle\int^{\infty}_0 \frac{1}{x+1}e^{-x} dx < [-e^{-x}]^{\infty}_0

    \displaystyle\int^{\infty}_0 \frac{1}{x+1}e^{-x} dx < 1

    Thus \displaystyle\int^{\infty}_0 (x - 1)e^{-x} dx + \int^{\infty}_0 \frac{1}{x+1}e^{-x} dx < 1

    And therefore, \int^{\infty}_0 \frac{x^2 e^{-x}}{1+x} dx < 1


    Now notice that the graph of y = \frac{x^2 e^{-x}}{1+x} is greater than 0 for x > 0, and hence so is the infinite integral.

    So, \displaystyle 0 < \int^{\infty}_0 \frac{x^2 e^{-x}}{1+x} dx < 1

    please point out any mistakes
  14. Offline

    ReputationRep:
    k here's the graph from the above q. if i've made any mistakes, please feel free to point them out
    Attached Thumbnails
    Click image for larger version. 

Name:	graph.jpg 
Views:	229 
Size:	80.5 KB 
ID:	51025  
  15. Online

    ReputationRep:
    You've made a mistake integrating (x-1)e^{-x}. (Still right answer, but some of what you've written is wrong).
  16. Offline

    ReputationRep:
    (Original post by DFranklin)
    You've made a mistake integrating (x-1)e^{-x}. (Still right answer, but some of what you've written is wrong).
    ok thanks, corrected
  17. Offline

    ReputationRep:
    I/9

    i) \displaystyle\int_1^e \frac{\log x}{x^2} dx Recall that \int \log x dx= x(\log (x)-1)+c, now we can use this to integrate by parts.
    I=\displaystyle\int \frac{\log x}{x^2}dx=\frac{1}{x^2}x(\log(x)-1)-\displaystyle\int-2\frac{x(\log(x)-1)}{x^3}dx
    -I=\frac{1}{x}(\log(x)-1)+2\frac{1}{x}=\frac{\log x +1}{x}
    Putting in the limits gives -I=\frac{2}{e}-1= So I=1-\frac{2}{e}

    ii) \displaystyle\int \frac{\cos(x)}{\sin(x)\sqrt{1+\s  in(x)}}dx Let u^2=1+\sin(x)\Rightarrow 2u\frac{du}{dx}=\cos(x)\Leftrigh  tarrow dx=\frac{2u}{\cos(x)}du
    So the integral becomes \displastyle\int \frac{2}{\sin x}du=\displaystyle\int\frac{2}{u  ^2-1}du=\displaystyle\int\frac{1}{u-1}du-\displaystyle\int\frac{1}{u+1}du  =\log|u-1|-\log|u+1|+c=\log|\frac{u-1}{u+1}|+c
    Going back to x we have \log|\frac{u-1}{u+1}|+c=\log|\frac{\sqrt{1+\s  in x}-1}{\sqrt{1+\sin x}+1}|+c
  18. Offline

    ReputationRep:
    I've got III/3 out. Will post later. Suprisingly easy, its more like a FP3 question to be fair. Might need a check over because my complex transformations are more than a little rusty.
  19. Offline

    ReputationRep:
    Done rough "workings out" for I/9 again two surprisingly easy integrals.
  20. Offline

    ReputationRep:
    (Original post by insparato)
    Done rough "workings out" for I/9 again two surprisingly easy integrals.
    Agreed, took me 10 minutes on the tube this morning when on my way to the library:p:

Reply

Submit reply

Register

Thanks for posting! You just need to create an account in order to submit the post
  1. this can't be left blank
    that username has been taken, please choose another Forgotten your password?
  2. this can't be left blank
    this email is already registered. Forgotten your password?
  3. this can't be left blank

    6 characters or longer with both numbers and letters is safer

  4. this can't be left empty
    your full birthday is required
  1. By joining you agree to our Ts and Cs, privacy policy and site rules

  2. Slide to join now Processing…

Updated: February 5, 2015
TSR Support Team

We have a brilliant team of more than 60 Support Team members looking after discussions on The Student Room, helping to make it a fun, safe and useful place to hang out.

New on TSR

Improving your uni offer

Why now is the time to think about Adjustment

Study resources
x

Think you'll be in clearing or adjustment?

Hear direct from unis that want to talk to you

Get email alerts for university course places that match your subjects and grades. Just let us know what you're studying.

Quick reply
Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.