STEP I, II, III 2000 solutions

I think the word "always" comes into play here; i.e. for any values of p,q,r, rather than just the ones in this instance that depend upon a,b,c.

The question isn't brilliantly worded, but I think the way it's supposed to flow is:

Suppose this quartic identity holds. Using the various relationships between p,q,r and a,b,c, show that a^2 is a root of {cubic}.

Now, suppose you have a cubic of the form {cubic}. Show that this cubic always has a non-negative root.

In particular, in the 2nd case you should not assume that we can find a, b, c such that the given quartic identity holds.

I doubt it was many marks though - I think the main intent is to get you to look at a certain cubic that may help for the last bit (and to explciitly point out a root of that cubic).

I think what Brian did diverges from the examiners plan for the next part, incidentally.

Is it me or is this a particularly tricky III paper?

Is it me or is this a particularly tricky III paper?

Judging by the grade boundaries, it seems like it's a bit easier than the average paper.

I meant more in the vein of what you thought.

I would expect my thoughts to be largely irrelevant given that we have the grade boundaries (and thus practically the general consensus on the difficulty of the paper).

Personally, I agree with the grade boundaries. At first glance, I'm liking the look of Q1, 2, 3, 4, 6 and 8. I recall doing most of those when I worked the paper, albeit not in exam conditions. (Q2 on this paper was the first ever STEP question that I completed. )

your first ever Q was a STEP III Q? respect.

Gah, just did STEP III Q12 and was all happy with myself at what I thought was a really clever way of doing it and then I saw glutamic acid's solution which took about two lines
Whilst his solution is a teensy bit hand wavey in that he doesn't explain his really clever method a great deal, I'm pretty sure it's correct.
My solution is different, it's more from 1st principles. Should I post it?

Square: I'm pretty sure you've misunderstood a (small) part of the question.

When it says "explain why this equation always has a non-negative root", I think you're supposed to be looking at the equation by itself, without assuming that you know about a^2 being a root. In which case:



I don't think this could cost you many marks, and I'm not even sure I'm right in my interpretation here.

For the rest of it, it's very hard to comment when you've included so little working; assuming your working is right, you should be OK.

For what it's worth, this really doesn't feel like it should be a "3 sides of A4" question, but that's easy for me to say.

II / 1

Guess a solution to $\displaystyle \frac{1}{N} = \frac{1}{a} + \frac{1}{b}$ you are provided with $\displaystyle \frac{1}{2} = \frac{1}{3} + \frac{1}{6}$ and also $\displaystyle \frac{1}{3} = \frac{1}{4} + \frac{1}{12}$

The solution is clear - $\displaystyle \frac{1}{N} = \frac{1}{N + 1} + \frac{1}{N^2 + N}$

To prove this is so; $\displaystyle \frac{1}{N} \equiv \frac{1}{N + 1} + \frac{1}{N^2 + N} = \frac{N+1}{N(N+1)} = \frac{1}{N}$

Looking at $\displaystyle (a-N)(b-N) = N^2$ for N is a prime it is apparent that the only factors of N^2 are $\displaystyle N^2 = N \times N = N^2 \times 1$

This would imply that if distinct natural numbers (a,b) are sought for the equation $\frac{1}{N} = \frac{1}{a} + \frac{1}{b}$ for N prime then there is only one pair of solutions.

wip for $\displaystyle \frac{2}{N}$

I/2

Second part:

I couldn't see a systematic way of dealing with the trinomial and making sure I got all the terms correct other than just expanding it out. I can write in the coefficient given by the initial expansion as they'll be the same as in the first expansion.

$(x^4 + (x^2+1))^5 = x^{20} + 5(x^{16})(x^2+1) + 10(x^{12})(x^2+1)^2 + 10(x^8)(x^2+3)^3 + 5(x^4)(x^2+1)^4 + (x^2+1)^5$
$= x^{20} + 5x^{16}(x^2+1) + 10x^{12}(x^4+2x^2+1) + 10x^8(x^6 +3x^4+3x^2+1) + 5x^4(x^8+4x^6+6x^4+4x^2+1) + (x^{10}+5x^8+10x^6+10x^4+5x^2+1)$

...

Did anyonr prove it for $\frac{2}{N}$???

I've shown that for there to be two distinct unit fractions where N is prime, i.e $\frac{2}{N} = \frac{1}{a} + \frac{1}{b}$? a and b must either both be odd or both be even, and have proved it for if they are both even then there is only one way that $\frac{2}{N}$ is a sum of two distinct unit fractions.. but cant prove that there are none for if a and b are both odd...

II / 8

$\displaystyle - \int \frac{dy}{(y+3)^\frac{1}{2}} = -2 \int -2xe^{-x^2} dx \implies -2(y+3)^{\frac{1}{2}} = -2e^{-x^2} + C \implies y = e^{-2x^2} + 2ke^{-x^2} + (k^2 - 2)$

After subbing (0,6) ; $\displaystyle (k+4)(k-2) = 0$

$\displaystyle \therefore y = e^{-2x^2} - 8e^{-x^2} + 13$ and case 2 $\displaystyle y = e^{-2x^2} + 4e^{-x^2} + 1$

There is a vlaue of k such that as x tends to $\displaystyle \infty$ y tends to $\displaystyle 1$

(ii) wip. Got
$\displaystyle y^2e^{-3x^2}$ but the answer requires $\displaystyle ye^{-3x^2}$

I/2:
I think my answers are right, but the method is rather tedious so I've most likely missed a trick somewhere along the line.

Considering the expansion of (x^4-1/x^2)^5 =
$(x^4)^5 + \dbinom{5}{1}(x^4)^4(-1/x^2) + \dbinom{5}{2}(x^4)^3(-1/x^2)^2 + \dbinom{5}{3}(x^4)^2(-1/x^2)^3 + \dbinom{5}{1}(x^4)(-1/x^2)^4 + (-1/x^2)^5$
$= x^{20} - 5x^{14} + 10x^8 - 10x^2 + 5x^{-4} - x^{-10}$

Considering the expansion of (x-1/x)^6 =
$x^6 + \dbinom{6}{1}(x^5)(-1/x) + \dbinom{6}{2}(x^4)(-1/x)^2 + \dbinom{6}{3}(x^3)(-1/x)^3 + \dbinom{6}{4}(x^2)(-1/x)^2 + \dbinom{6}{5}(x)(-1/x)^5 + (-1/x)$

$= x^6 - 6x^4 + 15x^2 - 20 + 15x^{-2} - 6x^{-4} + x^{-6}$

Consider the multiplication of these, the exponents of x that will add to give -12 are -x^-10 and 15x^-2. Therefore the coefficient of x^-12 is -1 x 15 = -15.

The exponents of x that will add to give 2, are 10x^8 and x^-6, -10x^2 and -20, and 5x^-4 and x^6. So the coefficient of x^2 is (10 x 1) + (-10 x -20) + (5 x 1) = 10 + 200 + 5 = 215.

Second part:
I couldn't see a systematic way of dealing with the trinomial and making sure I got all the terms correct other than just expanding it out. I can write in the coefficient given by the initial expansion as they'll be the same as in the first expansion.
$(x^4 + (x^2+1))^5 = x^{20} + 5(x^{16})(x^2+1) + 10(x^{12})(x^2+1)^2 + 10(x^8)(x^2+3)^3 + 5(x^4)(x^2+1)^4 + (x^2+1)^5$
$= x^{20} + 5x^{16}(x^2+1) + 10x^{12}(x^4+2x^2+1) + 10x^8(x^6 +3x^4+3x^2+1) + 5x^4(x^8+4x^6+6x^4+4x^2+1) + (x^{10}+5x^8+10x^6+10x^4+5x^2+1)$

$= x^{20} + 5x^{18} + 5x^{16} + 10x^{16} + 20x^{14} + 10x^{12} + 10x^{14} + 30x^{12} + 30x^{10} + 10x^8 + 5x^{12} + 20x^{10} + 30x^8 + 20x^6 + 5x^4 + x^{10} + 5x^8 + 10x^6 + 10x^4 + 5x^2 + 1$

$= x^{20} + 5x^{18} + 15x^{16} + 30x^{14} + 45x^{12} + 51x^{10} + 45x^8 + 30x^6 + 15x^4 + 5x^2 + 1$.

Now, the expansion of (x^2-1)^11 will have exponents of 22,20,18 .... 4,2,0. These which add to 4 are 4 and 0, 2 and 2 and 0 and 4. So this is given by
$(1 \times \dbinom{11}{9} \times (-1)^9) + (5 \times \dbinom{11}{10} \times (-1)^{10}) + (15 \times (-1)^{11}) = -55 + 55 -15 = -15.$

The exponents adding to 38 are 20 and 18, 18 and 20, 16 and 22. So this is given by $(1 \times \dbinom{11}{2} \times (-1)^2) + (5 \times \dbinom{11}{1} \times (-1)) + (15 \times 1) = 55 - 55 + 15 = 15.$

II / 8
...

(ii) wip. Got
$\displaystyle y^2e^{-3x^2}$ but the answer requires $\displaystyle ye^{-3x^2}$