The Student Room Group

STEP Maths I, II, III 1988 solutions

(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:wink: 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
(edited 10 years ago)

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Reply 1
STEP I Q1

h(x)=logxxh(x)=\frac{\log x}{x}
h(x)=logxx2+1x2=1x2(1logx)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=1x2(1logx)x=e0=\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 limx0+h(x)=\displaystyle\lim_{x\to0^+}h(x)=-\infty and
Unparseable latex formula:

\displastyle\lim_{x\to\infty}h(x)=0

. Graph see attached (Mathematica because of lack of scanner:p:).

Solving nm=mnn^m=m^n is equivalent of solving lognm=logmnmlogn=nlogmlognn=logmm\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<1e0< c< \frac{1}{e} there are two solutions. There's one solution when c=1ec=\frac{1}{e} or c0c\le0. And for c>1ec>\frac{1}{e} there are no solutions.

Now if mnm\not=n then nm=k\frac{n}{m}=k (k1k\not=1).

lognn=lognknk\frac{\log n}{n}=\frac{\log \frac{n}{k}}{\frac{n}{k}}
logn=nk(lognlogk)n=k(lognlogk)\log n=\frac{nk(\log n-\log k)}{n}=k(\log n - \log k)
logn=kk1logk\log n=\frac{k}{k-1}\log k
Thus n=kkk1n=k^{\frac{k}{k-1}} and m=k1k1m=k^{\frac{ 1}{k-1}}.

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

----

STEP I Q2

g(x)=f(x)+1f(x)g(x)=f(x)+\frac{1}{f(x)} (f(x)0f(x)\not=0)
g(x)=f(x)f(x)(f(x))2=f(x)(11(f(x))2)g'(x)=f'(x)-\frac{f'(x)}{(f(x))^2}=f'(x)(1-\frac{1}{(f(x))^2})
g(x)=f(x)(11(f(x))2)+f(x)(2f(x)(f(x))3)=f(x)(11(f(x))2)+2(f(x))2(f(x))3g''(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)+2sin(x)f(x)=4+\cos(2x)+2\sin(x)
f(x)=2sin(2x)+2cos(x)=2cos(x)(2sin(x)+1)f'(x)=-2\sin(2x)+2\cos(x)=2\cos(x)(-2\sin(x)+1)
Now WLOG g(x)=0f(x)=0or(f(x))2=1g'(x)=0\Leftrightarrow f'(x)=0 \text{or} (f(x))^2=1

f(x)=02cos(x)(12sin(x))=0f'(x)=0 \Leftrightarrow 2\cos(x)(1-2\sin(x))=0 Has solutions cos(x)=0 and sin(x)=1/2
i.e. x=π2,3π2,π6,5π6x=\frac{\pi}{2},\frac{3\pi}{2},\frac{\pi}{6},\frac{5\pi}{6}

f(x)=11=4+12sin2(x)+2sin(x)0=2(2u2+u)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)u=\sin(x)
By quadratic formula the solutions are u=1±1+4×22u=\frac{1\pm\sqrt{1+4\times2}}{2} i.e. u=-1, 2
sin(x)=2\sin(x)=2 has no real solutions and sin(x)=1\sin(x)=-1 has the root x=3π2x=\frac{3\pi}{2} (which we have already found).

f(x)=11=4+12sin2(x)+2sin(x)0=2(3u2+u)f(x)=-1\Leftrightarrow -1=4+1-2\sin^2(x)+2\sin(x)\Leftrightarrow 0=2(3-u^2+u) where u=sin(x)u=\sin(x)
Solutions to this are u=1±1+4×32u=\frac{1\pm\sqrt{1+4\times3}}{2} and as 1±1+4×32>1\frac{1\pm\sqrt{1+4\times3}}{2}>1 sin(x)=u will have no real solutions.

f(π2π16)<0f'(\frac{\pi}{2}-\frac{\pi}{16})<0, f(π2+π16)>0f'(\frac{\pi}{2}+\frac{\pi}{16})>0 so a (local)min f(π2)=41+2=5g(π2)=5+15f(\frac{\pi}{2})=4-1+2=5 \Rightarrow g(\frac{\pi}{2})=5+\frac{1}{5}
f(3π2π16)<0f'(\frac{3\pi}{2}-\frac{\pi}{16})<0, f(3π2+π16)>0f'(\frac{3\pi}{2}+\frac{\pi}{16})>0 so a (global) min f(3π2)=412=1g(3π2)=2f(\frac{3\pi}{2})=4-1-2=1 \Rightarrow g(\frac{3\pi}{2})=2
f(π6+π18)<0f'(\frac{\pi}{6}+\frac{\pi}{18})<0, f(π6π18)>0f'(\frac{\pi}{6}-\frac{\pi}{18})>0 so a (global) max f(π6)=4+22+1=5+22g(π6)=5+22+15+22f(\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(5π6+π18)<0f'(\frac{5\pi}{6}+\frac{\pi}{18})<0,f(5π6π18)>0f'(\frac{5\pi}{6}-\frac{\pi}{18})>0 so a (local) max f(5π6)=4+12+1=112g(5π6)=112+211f(\frac{5\pi}{6})=4+\frac{1}{2}+1=\frac{11}{2} \Rightarrow g(\frac{5\pi}{6})=\frac{11}{2}+\frac{2}{11}
Reply 2
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...
Reply 3
Yeah, I certainly need to! Had forgotten that one, which is by inspection.
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.
Reply 5
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?)
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. :smile:
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. :smile:


Thanks. :smile:
Reply 8
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!
Good idea Nota :smile:. I'll have a look at the paper(s) tomorrow and see if I get anything out.

Also....1989 has many unsolved questions.

generalebriety - could you link this in the Big fat STEP mega thread, pretty please :puppyeyes:?
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<1e0< c< \frac{1}{e} there are two solutions. There's one solution when c=1ec=\frac{1}{e} or c0c\le0. And for c>1ec>\frac{1}{e} there are no solutions.

Now if mnm\not=n then nm=k\frac{n}{m}=k (k1k\not=1).

lognn=lognknk\frac{\log n}{n}=\frac{\log \frac{n}{k}}{\frac{n}{k}}
logn=nk(lognlogk)n=k(lognlogk)\log n=\frac{nk(\log n-\log k)}{n}=k(\log n - \log k)
logn=kk1logk\log n=\frac{k}{k-1}\log k
Thus n=kkk1n=k^{\frac{k}{k-1}} and m=k1k1m=k^{\frac{ 1}{k-1}}.
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).
Reply 12
III/1:

Sketch y=x2exx+1y = \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

dydx=exx(2x2)(x+1)2\frac{dy}{dx} = e^{-x}\frac{x(2-x^2)}{(x+1)^2}

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


Differentiating again, we get

d2ydx2=ex(23x2)(x+1)22(x+1)(2xx3)(2xx3)(x+1)2(x+1)4\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+x \to +\infty

y tends to -\infty as xx \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/showpost.php?p=11883448&postcount=14



Prove 0<0x2ex1+xdx<1\displaystyle 0 < \int^{\infty}_0 \frac{x^2 e^{-x}}{1+x} dx < 1

First note that x2x+1=x2+xx+1xx+1=xxx+1=x(x+1x+11x+1)=x1+1x+1\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 0(x1)exdx+01x+1exdx\displaystyle\int^{\infty}_0 (x - 1)e^{-x} dx + \int^{\infty}_0 \frac{1}{x+1}e^{-x} dx

Consider 0(x1)exdx\displaystyle\int^{\infty}_0 (x - 1)e^{-x} dx

=0xexdx0exdx= \displaystyle\int^{\infty}_0 xe^{-x} dx - \int^{\infty}_0 e^{-x} dx

=([xex]0+0exdx0exdx= \displaystyle([-xe^{-x}]^{\infty}_0 + \int^{\infty}_0 e^{-x} dx - \int^{\infty}_0 e^{-x} dx

=0= 0


Now consider 01x+1exdx\displaystyle\int^{\infty}_0 \frac{1}{x+1}e^{-x} dx

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

1x+1<1\frac{1}{x+1} < 1
1<(x+1)1 < (x+1)
0<x0 < x

So our inequality holds.

So:

01x+1exdx<0exdx\displaystyle\int^{\infty}_0 \frac{1}{x+1}e^{-x} dx < \int^{\infty}_0 e^{-x} dx

01x+1exdx<[ex]0\displaystyle\int^{\infty}_0 \frac{1}{x+1}e^{-x} dx < [-e^{-x}]^{\infty}_0

01x+1exdx<1\displaystyle\int^{\infty}_0 \frac{1}{x+1}e^{-x} dx < 1

Thus 0(x1)exdx+01x+1exdx<1\displaystyle\int^{\infty}_0 (x - 1)e^{-x} dx + \int^{\infty}_0 \frac{1}{x+1}e^{-x} dx < 1

And therefore, 0x2ex1+xdx<1\int^{\infty}_0 \frac{x^2 e^{-x}}{1+x} dx < 1


Now notice that the graph of y=x2ex1+xy = \frac{x^2 e^{-x}}{1+x} is greater than 0 for x > 0, and hence so is the infinite integral.

So, 0<0x2ex1+xdx<1\displaystyle 0 < \int^{\infty}_0 \frac{x^2 e^{-x}}{1+x} dx < 1

please point out any mistakes :smile:
Reply 13
k here's the graph from the above q. if i've made any mistakes, please feel free to point them out :smile:
You've made a mistake integrating (x-1)e^{-x}. (Still right answer, but some of what you've written is wrong).
Reply 15
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 :smile:
I/9

i) 1elogxx2dx\displaystyle\int_1^e \frac{\log x}{x^2} dx Recall that logxdx=x(log(x)1)+c\int \log x dx= x(\log (x)-1)+c, now we can use this to integrate by parts.
I=logxx2dx=1x2x(log(x)1)2x(log(x)1)x3dxI=\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=1x(log(x)1)+21x=logx+1x-I=\frac{1}{x}(\log(x)-1)+2\frac{1}{x}=\frac{\log x +1}{x}
Putting in the limits gives I=2e1=-I=\frac{2}{e}-1= So I=12eI=1-\frac{2}{e}

ii) cos(x)sin(x)1+sin(x)dx\displaystyle\int \frac{\cos(x)}{\sin(x)\sqrt{1+\sin(x)}}dx Let u2=1+sin(x)2ududx=cos(x)dx=2ucos(x)duu^2=1+\sin(x)\Rightarrow 2u\frac{du}{dx}=\cos(x)\Leftrightarrow dx=\frac{2u}{\cos(x)}du
So the integral becomes
Unparseable latex formula:

\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 logu1u+1+c=log1+sinx11+sinx+1+c\log|\frac{u-1}{u+1}|+c=\log|\frac{\sqrt{1+\sin x}-1}{\sqrt{1+\sin x}+1}|+c
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.
Done rough "workings out" for I/9 again two surprisingly easy integrals.
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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