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Reply 1
I just tried a few questions from paper III and it seemed relatively easy - I managed to complete 1, 3, 4, 5 and 7 as well as attempt (badly) 2 in about 2 hours. I was under the impression that the older papers were harder? Hmm...

I will post my solutions for 1, 3, 4, 5 and 7 shortly. In the meantime - does anybody have a hint for 2? There's obviously at least one root (x = 0), but I'm not sure I can see a way to count the non-trivial roots...
Err... which paper?

Edit: ah, you edited. I'll have a look.
In STEP questions like that, you can be almost certain that the second part relates to the first part in some way. In this case, put y = e^x - 1 - k tan^-1 x, and find dy/dx...
Reply 4
Aaah. I see. I will try that now.

In the meantime, these are my solutions for questions 1, 3, 4, 5, and 7 of paper III. Incidentally, one of my interview questions was to prove that there were only five regular polyhedra (last part of question 4), but by considering the sum of angles at each vertex instead... Much less long-winded, I think, even though it would be necessary to assume that all regular polyhedra are convex.

Also, I think this post needs updating... (There's also a thread for 2000 now.)
Reply 5
Those all look fine to me. Funnily, I don't think I ever looked at this paper last year - I don't remember any of the (pure) questions at any rate! My feeling is those questions are relatively short, but also require a bit more familiarity with "non A-level maths" than the average. This is probably right up your street, so it worked out as a particularly easy paper for you.
I think I've got 8 and 6 (paper III) scribbled somewhere...I'll post them if I can find.
Reply 7
Those all look fine to me. Funnily, I don't think I ever looked at this paper last year - I don't remember any of the (pure) questions at any rate! My feeling is those questions are relatively short, but also require a bit more familiarity with "non A-level maths" than the average. This is probably right up your street, so it worked out as a particularly easy paper for you.
STEP III - Question 8

i)

0≀x≀2Ο€β‡’n=1 0 \leq x \leq 2\pi \Rightarrow n = 1 .

So the differential equation becomes:

d2ydx2+y=0 \frac{d^{2}y}{dx^{2}} + y = 0, whose general solution is: y(x)=Acos⁑(x)+Bsin⁑(x) y(x) = A\cos(x) + B\sin(x) .
Now, since when x=0, y=0 and dydx=1\frac{dy}{dx} = 1 , the particular solution is y(x)=sin⁑(x)y(x) =\sin(x) (*).

ii) Since 2π≀x≀4Ο€,β‡’n=2 2\pi \leq x \leq 4\pi, \Rightarrow n = 2 .

So this time the differential equation is:

d2ydx2+4y=0\frac{d^{2}y}{dx^{2}} + 4y = 0

general solution: y(x)=Ccos⁑(2x)+Dsin⁑(2x)y(x) = C\cos(2x) + D\sin(2x) (**).

Now, since y and dydx\frac{dy}{dx} are continuous at x=2Ο€x=2\pi, (**) has the same value as (*) at x=2Ο€x = 2\pi and also their derivatives take the same value. From this we can work out the values of C and D, (C=0 and D= 12\frac{1}{2}). Which gives us: y(x)=12sin⁑(2x)y(x) = \frac{1}{2}\sin(2x).
Using the same method in the previous part, we can deduce that y(x)=1nsin⁑(nx)y(x) = \frac{1}{n}\sin(nx).

iii)

∫0∞y2 dx=∫02Ο€y2 dx+∫2Ο€4Ο€y2 dx+∫4Ο€6Ο€y2 dx+...........+∫2(rβˆ’1)Ο€2rΟ€y2 dx+.... \displaystyle\int^\infty_0 y^{2} \, dx = \displaystyle\int^{2\pi}_{0} y^{2} \, dx + \displaystyle\int^{4\pi}_{2\pi} y^{2} \, dx + \displaystyle\int^{6\pi}_{4\pi} y^{2} \, dx + ...........+ \displaystyle\int^{2r\pi}_{2(r-1)\pi} y^{2} \, dx + ....




= ∫02Ο€sin⁑2(x) dx+∫2Ο€4Ο€122sin⁑2(x) dx+∫4Ο€6Ο€132sin⁑2(x) dx+..... \displaystyle\int^{2\pi}_{0} \sin^{2}(x) \, dx + \displaystyle\int^{4\pi}_{2\pi} \frac{1}{2^{2}}\sin^{2}(x) \, dx + \displaystyle\int^{6\pi}_{4\pi} \frac{1}{3^{2}}\sin^{2}(x) \, dx + .....



= ∫02Ο€(1βˆ’cos⁑(2x)) dx+∫2Ο€4Ο€122(1βˆ’cos⁑(4x)) dx+∫4Ο€6Ο€132(1βˆ’cos⁑(6x)) dx+..... \displaystyle\int^{2\pi}_{0} (1-\cos(2x)) \, dx + \displaystyle\int^{4\pi}_{2\pi} \frac{1}{2^{2}}(1-\cos(4x)) \, dx + \displaystyle\int^{6\pi}_{4\pi} \frac{1}{3^{2}}(1-\cos(6x))\, dx + .....

(using cos⁑(2x)=1βˆ’2sin⁑2(x) \cos(2x) = 1 - 2\sin^{2}(x) )

= 12[2Ο€βˆ’0]+122[12(4Ο€βˆ’2Ο€)]+132[12(6Ο€βˆ’4Ο€)]+.... \frac{1}{2}[ 2\pi - 0] + \frac{1}{2^{2}}[\frac{1}{2}( 4\pi - 2\pi )] + \frac{1}{3^{2}}[\frac{1}{2}( 6\pi - 4\pi)] + ....


= Ο€+Ο€22+Ο€32+.....=Ο€βˆ‘n=1∞1n2 \pi + \frac{\pi}{2^{2}} + \frac{\pi}{3^{2}} + ..... = \pi \displaystyle\sum_{n=1}^\infty \frac{1}{n^{2}}
Zhen Lin
I just tried a few questions from paper III and it seemed relatively easy - I managed to complete 1, 3, 4, 5 and 7 as well as attempt (badly) 2 in about 2 hours. I was under the impression that the older papers were harder? Hmm...

I will post my solutions for 1, 3, 4, 5 and 7 shortly. In the meantime - does anybody have a hint for 2? There's obviously at least one root (x = 0), but I'm not sure I can see a way to count the non-trivial roots...


I would find the stationary points by differentiating, then see where there is a +ve/-ve sign change between stationary points in the original equation, if that makes sense.
Zhen, for question 2, I found sketching the graph of f(x) makes it alot eaiser and clearer. (same for part ii) ).
Reply 11
Well, for the first part it's very easy to see the answer, even without sketching the graph - we know it's monotonic increasing from calculating the derivative. The second part is not so clear-cut because I can't tell whether (exβˆ’1)\left( e^x - 1 \right) is increasing/decreasing faster than ktanβ‘βˆ’1xk \tan^{-1} x. I tried generalebriety's suggestion and found a derivative that is similar to the first part, but then I got stuck because turning points are not necessarily the same as roots...
I dunno if what I did was correct, but I considered; g(x) = exβˆ’1tanβˆ’1(x) \frac{e^{x} - 1}{tan^{-1}(x)}. By playing around with it, I noticed that as x βŸΆβˆ’βˆž\longrightarrow -\infty, g(x) ⟢2Ο€\longrightarrow \frac{2}{\pi}. If you want me to into anyfurther detail , I will.
I'm currently struggling my way through II/3, will type up when done.

under construction

Zhen Lin
Well, for the first part it's very easy to see the answer, even without sketching the graph - we know it's monotonic increasing from calculating the derivative. The second part is not so clear-cut because I can't tell whether (exβˆ’1)\left( e^x - 1 \right) is increasing/decreasing faster than ktanβ‘βˆ’1xk \tan^{-1} x. I tried generalebriety's suggestion and found a derivative that is similar to the first part, but then I got stuck because turning points are not necessarily the same as roots...


You know the roots must be between turning points, before the first turning point, or after the last one, so if there is a sign change between two turning points there must be one root between the turning points. It is quite easy to work out if there is another root before or after the last turning point.
II/2

solution

For (ii), I don't see why you say that s is positive. Given the conditions they've given, it's pretty obvious you're supposed to use "B^2-4AC<0" on n^2(p-r)-ns+q.
I'll fix that - I thought I'd take a shortcut but turns out I didn't check carefully as obviously Β±(4q(pβˆ’r))>s\pm(\sqrt{4q(p-r)})>s
II/6

solution

II/7

This question seems too easy, so I've probably made a mistake:

y=xx2βˆ’2x+ay=\frac{x}{\sqrt{x^2-2x+a}} Then dydx=βˆ’xxβˆ’1x2βˆ’2x+aβˆ’x2βˆ’2x+ax2βˆ’2x+a=βˆ’x(xβˆ’1)βˆ’(x2βˆ’2x+a)(x2βˆ’2x+a)32=βˆ’xβˆ’a(x2βˆ’2x+a)32=aβˆ’x(x2βˆ’2x+a)32\frac{dy}{dx}= - \frac{x\frac{x-1}{\sqrt{x^2-2x+a}}-\sqrt{x^2-2x+a}}{x^2-2x+a}= - \frac{x(x-1)-(x^2-2x+a)}{(x^2-2x+a)^{\frac{3}{2}}}= - \frac{x-a}{(x^2-2x+a)^{\frac{3}{2}}} = \frac{a-x}{(x^2-2x+a)^{\frac{3}{2}}}
Stationary points can occur when dy/dx=0 i.e. the numerator needs to be zero <=> 0=a-x <=>x=a Hence there is at most one stationary point. The derivative is undefined if x2βˆ’2x+a=0x^2-2x+a=0 i.e. x=2Β±4βˆ’4a2x=\frac{2\pm\sqrt{4-4a}}{2} and for a>1 this has no (real) solutions and thus the function has more than zero stationary points. I.e. there is one stationary point, as desired.

i) a=2 f(0)=0 f(2)=2f(2)=\sqrt{2} so stationary point at (2,2)(2,\sqrt{2}) fβ€²(1.9)<0f'(1.9)<0and fβ€²(2.1)>0f'(2.1)>0 and thus the point is a maximum.
Also note lim⁑xβ†’Β±βˆž[f(x)]=Β±1\displaystyle\lim_{x\to\pm\infty}[f(x)]=\pm1

ii) a=1 f(0)=0 From above we have that there are no stationary points because the denominator of the function (and it's derivative) goes to 0 as a goes to 1 i.e. vertical asymptote at x=1 Same as in i) note that lim⁑xβ†’Β±βˆž[f(x)]=Β±1\displaystyle\lim_{x\to\pm\infty}[f(x)]=\pm1


Sketches see attached. (Very rough, would put much more effort in during a real exam)

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