Here are my thoughts on STEP 1:

1) Quite straightforward: write 5 as (10.3)/2 and 6 as 2.3 for the decimal answers. Then directly evaluate the logs of the 3 numbers and see the inequalities. So obviously the first digit is 5.

Next do the same, see what the number is in decimal form, and take away the integer part to get the index of 10. The answers I think were 1, 1 and 9.

2) I have to confess I just expanded it out, and picked the only term. Similarly the x^2 term is 215. The trick in the next bit is to combine 5 of the left brackets with the right, and multiply out to get (x² -1)⁶ (x⁶ -1)⁵ . Then take out x⁶ from the left and x¹⁰ from the left, to get the original expansion time x¹⁶ . For the first, the answer is the result for -12 before (15) and get the coefficient of x²² from the original series (+15).

3) The graph is a step stone, and the integral is easily obtained by getting the area underneath (10). Next the graph is identical, only replacing 1,2,3 etc on the x-axis with ln2, ln3 ...
To get the area, say it is contained in a box of area n.ln(n), and then take away all the individual rectangles. The sum of the area of thse is ln(1) + ln(2) + ln(3) ... ln(n) which is ln(n!) and so that gives the answer.

4) i) show that there are no SPs in the range, and that the gradient is positive, and deduce that the maximum is then at the end of the range, ie, when x=1.
ii) I did this by direct differentiation, and then multiply by (x² +1)⁴ . Equate coefficients to get A=C=D=1, and B=5/3.

iii) Then integrate the expression in ii), and leave the bit with D as an integral. You get the integrated part as 11/24, and you already know that the integral of the part with D is < = 1/16, so the answer follows.

I ignored 5 because it had too many words, and completely missed 6!

7) I thought it looked too short so didn't bother.

8) This is just a case of working out where the modulus bit is negative and positive. The integral is 2(1-e^-1 ). i) Factorise the cubic, and according to the factors, split into 3 separate integrals (the middle one negative I think) and the answer is 16 + 5/6. ii) Draw a sketch graph of each [alternatively write as root2[sin(x+pi/4)] and again split up into separate integrals depending where it is negative/positive. The answer I got was 4root2.

9/10.. didn't fancy them much

11) Take moments and equate the wall normal with the ground friction. If both limiting, then both your 'inequalities' will be equalities, and sub one into the other to show that that's a contradiction. Next, do the same again with the moments and the equalities but this time solve for AP. This was astoundingly easy I thought. I got AP=0.438.

12/13/14) I almost changed my mind about avoiding these, but decided that I would go home instead of doing #13. That's dedication for you.

Overall) The ones I could do I did perfectly. I am annoyed at not seeing Q6, because it looked quite good.

So, STEP 3. I managed Q1 and 2 only at ASDA today, which was an achievement in itself, but was too busy to do any more.

1) Bread and butter stuff.. the reciprocal and a simple hyperbola. Obviously orthogonal intersections. Sketch can take 1-5 mins depending how many coloured crayons you have... Q= (-a, -b)

Next I worked out the equations of the tangents (regarding M) and subbed them into each other, using the relationships of a and b to evaluate the resultant expression. Basically you needed to know that (1-3b⁴ )/b⁶ +² ) = 1, for the x coord, and that a^-3 + 2^-1 = a. And then from the sketch, clearly N = (b,-a).

Next, you have perpendicular lines, so it is a rectangle. If a pair of the perpendicular lines are of equal length, you have a square. A quick pythagoras shows that they are.

2) Aaah! The good old sqrt[(1-cost)/(1+cost)]. How I waited for this in STEP 2. Your integral should a few steps after the substitution become intg{1-cost}dt between pi/3 and pi/2. This is (1/2)root3 - 1 + pi/6, which can be checked by looking at the next bit.

The key is to make the right substitution.. x = (a+b)/2 + (a-b)cost/2. It is then delightfully simple, and you get (b-a)/2 times your original integral. Since you already know this, you are two steps away from showing the required result.

Impression so far:

I can't believe that was STEP 3! That was easier than STEP 2 surely. The one thing I would say is that they took a bit longer to do, but were easier to get. I should have done STEP 1 and STEP 3. But I might guff the next 4 required questions of course.

Neil M

PS: what do you think of my answers?

Neil - Did you simplify your equation of tangents (using the fact that (a,b) lies on both hyperbolae)? It looks as if you haven't been bothered about it and so obtaining a slightly harder version for the points of intersection.

Question 2 looks alright, have you attempted any more in the weekend?

Kerwin.

Yes, this is what I meant by using the relationships of A and B. I did Q6 this morning at work (was working over the weekend too).

Q6) quite simple, equate coefficients. Then expand out in the cubic. a² is always going to be positive. Then for the last bit, I missed the key point, but it said hence or otherwise, so I quated it to (x² +ax+b)(x² +cx+d). Then bd =39, so b=13, d=3 or b=1 d=39. So then you try the first one, and it works. Took me about 15 minutes, most of that time spent trying to figure not the key that I missed. Surely not a STEP III question.

A few points about this.

Firstly saying a² > = 0 will not answer the question: "why does this equation always have a non-negative root". The reason is you haven't shown that there exists real a,b,c which solve the relationships for p,q,r. a could end up complex in which case a² is not necessarily positive. What you have to do is notice that the quartic is negative when u = 0, and will tend to infinity as u -> infinity. Therefore as polynomials are continuous functions, there must exist a value x such the quartic is zero - therefore positive root.

For the next part you shouldn't really assume that b,c are integral. You have got the right answer though. What they wanted you to do was to use the previous results.

We have to express:

y⁴ - 8y³ + 23y² - 34y + 39 = 0

as the product of two quadratic factors.

First of all the first part deals with a quartic without a x³ term. So eliminate the y³ term in the quartic at the bottom:

Let u = y - 2:

(u+2)⁴ - 8(u+2)³ + 23(u+2)² - 34(u+2) + 39 = 0

Expand out:

u⁴ - u² - 6u + 15 = 0

So set p = -1, q = -6, r = 15.

From the first part we therefore know that if we factor it into (x² -ax+b)(x² +ax+c) a² must be a root of:

u³ + 2pu² + (p² -4r)u - q² = 0

and we're given that one root is u = 9. So why not try a² = 9? Take a = 3. Now equate some coefficients in the equation:

bc = 15
b + c = 8

b(8-b) = 15

b² - 8b + 15 = 0
(b-4)² = 1

b = 5, c = 3 will do.

(u² -3u+5)(u² +3u+3)

Actually we've got b,c the wrong way round here as it happens because if you expand it you get a 6u term. So replace u with -u:

(u² + 3u + 5)(u² - 3u + 3)

And finally substitute y = u + 2:

(y² -y+3)(y² -7y+13)

This seems a ridiculously long way round compared to your way (I didn't quite have time to finish it in the exam) but at least you don't have to assume b,c are integers.

Yours,

Michael

Michael-

I knew removing the cube term was vital, but I didn't want to make a substitution since the answer was so easy. I thought about adding multiples of the other equation but I just didn't bother. I guessed that if the last part involved using the first part, the constants were going to be integers, but I agree its not the best way to do it in general.

Neil M

Neil - have a go at Q8 next. This looks an extremely easy question, but one I wasted loads of time over.

Yours,

Michael

OK, here goes Q8 STEP III 2000:

Firstly. Hmmm. Assume for n=k and n=k+1, sub in, and rearrange on of your assumed ones to get rid of a_k-2 . Then the answer should be forthcoming. Next, I used induction again, rather than the alternative of repeated substitution:

Show it works for n=1, and sub in the results. Remembering at all times it might be useful that a is the golden ratio, so a ² = a +1. Then multiply numerator and denomnator by a for simplicity. Next comes the key: to realise (or show by working backwards) that a ⁵ + a ^-3 = 7a . Once you've done this, the numerator will factorise, with one of them being the same as the denominator, so you get the answer. That was a bit harder I'll grant you!

Neil M

The answer is immediately forthcoming

Neil - I don't quite follow the first part. The second part I think I get (but I did it by assuming a specific solution of the form ${\lambda }^n$, which is fine also).

Michael