STEP Maths I, II, III 1988 solutions Watch

SimonM
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(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
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nota bene
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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}
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DFranklin
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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...
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nota bene
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Yeah, I certainly need to! Had forgotten that one, which is by inspection.
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The Lyceum
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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.
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DFranklin
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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?)
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generalebriety
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(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.
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The Lyceum
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(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.
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Square
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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!
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squeezebox
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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:?
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nota bene
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(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}}.
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DFranklin
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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).
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kabbers
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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
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kabbers
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k here's the graph from the above q. if i've made any mistakes, please feel free to point them out
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DFranklin
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You've made a mistake integrating (x-1)e^{-x}. (Still right answer, but some of what you've written is wrong).
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kabbers
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(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
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nota bene
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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
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insparato
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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.
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insparato
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Done rough "workings out" for I/9 again two surprisingly easy integrals.
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nota bene
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(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:
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