Someone else said this was what you got if you worked out one of the terms (2, 8, 24 etc.)* instead of the sum (2, 10, 34 etc.).

*or something like that - not sure any more what the sequence actually was

Reply 1523

RuairiMorrissey

5

Original post by ShatnersBassoon

Someone else said this was what you got if you worked out one of the terms (2, 8, 24 etc.)* instead of the sum (2, 10, 34 etc.).

*or something like that - not sure any more what the sequence actually was

I thought that was part V?

Reply 1524

louisforrest

2

Original post by ShatnersBassoon

I haven't seen much discussion about question 5 yet, other than the first two parts:

(i) and (ii) A = C = 2; B = -2.
(iii) This was induction, or at least, induction without the wordy conclusion. Basically you found

s_{k+1}

=

s_k

+ [the (k+1)th term] and showed it was equal to f(k+1).

(iv) I split

t_k

up into two series: the first, the coefficients of n, was a GP - I think with powers of 2? - so it was easy to find a formula for that. The second was just

s_k

so sub in f(k), add the two together and you've got

t_k

.

u_k

was the sum of fractions in the form n/(2^n), so you put them over a common denominator (2^n), with 2^n + 2^(n-1) + ... 1 as the numerator, use the GP formula to find an expression for the numerator and then simplify the fraction.

(v) This was meta. The sum of sums. I split it into the sum of three terms from 1 to n: the first GP I found in

t_k

; another weird GP; -2 [which summed to -2n]. My final answer was

(something \times n - somethingelse)^{k-1} + 4

.

But I felt like I was doing part (v) in particular in at best an inefficient, and at worst a completely wrong, way. What did everyone else do?

What do you think the first two parts of this will be in terms of the marks?

Reply 1525

ShatnersBassoon

4

Original post by RuairiMorrissey

I thought that was part V?

Part (v) was the sum of the sums; that is:

\sum_{k=1}^{n} s_k = 2 + (2+8) + (2+8+24) + ... + s_n

Unless this is the third question I've read wrong.

Original post by louisforrest

What do you think the first two parts of this will be in terms of the marks?

Based on the time I spent on each part, I would guess 4 marks. But if I was doing things inefficiently, as I suspect I was, then it could have been worth a little bit more. I think the main point of the question was to get onto the induction and the "recognise that this sequence relates to the one we've just shown you"-type stuff, so it was really just an introduction. Sorry if that's bad news for you - I suspect a lot of people will have just stopped after part (ii).

(edited 7 years ago)

Reply 1526

louisforrest

2

did anyone else find a<1 for the one about area of the graphs enclosed by x axis. I dont see how it is (6/5)^4

Reply 1527

RuairiMorrissey

5

Original post by ShatnersBassoon

Part (v) was the sum of the sums; that is:

\sum_{k=1}^{n} s_k = 2 + (2+8) + (2+8+24) + ... + s_n

Unless this is the third question I've read wrong.

I've read loads wrong I wouldn't worry, we'll have to wait until the paper is out.

Reply 1528

peted0herty

1

anyone got any ideas about the weightings of the marks for each part of question 3? (or any of the long questions, for that matter)

Reply 1529

RuairiMorrissey

5

Original post by ShatnersBassoon

Part (v) was the sum of the sums; that is:

\sum_{k=1}^{n} s_k = 2 + (2+8) + (2+8+24) + ... + s_n

Unless this is the third question I've read wrong.

Based on the time I spent on each part, I would guess 4 marks. But if I was doing things inefficiently, as I suspect I was, then it could have been worth a little bit more. I think the main point of the question was to get onto the induction and the "recognise that this sequence relates to the one we've just shown you"-type stuff, so it was really just an introduction. Sorry if that's bad news for you - I suspect a lot of people will have just stopped after part (ii).

Nevermind you were right.

Please end me, i've miss-read SO many : (

Reply 1530

anony-mousse

10

Original post by peted0herty

anyone got any ideas about the weightings of the marks for each part of question 3? (or any of the long questions, for that matter)

I suspect that the last parts of q3 will be 2-3 marks each but I can't remember exactly what the question structure was

Reply 1531

louisforrest

2

**** already down to 5 in the multiple choice

Reply 1532

m0.4444

7

Original post by boombox111

for what values of a is the area between y = a^2 - x^2 and the x-axis greater than the area between y = x^4 - a and the x-axis?
something like this.

Tbh i got that one wrong. But i worked it out after. The x^4 graph has x-intercepts wider than the x^2 for x>1. Therefore the x^4 has a greater area for a>1. However, for x<1 the x^4 graph has a smaller gap between the x-intercepts and so has a smaller area.

Reply 1533

m0.4444

7

Original post by louisforrest

did anyone else find a<1 for the one about area of the graphs enclosed by x axis. I dont see how it is (6/5)^4

I think it was one of the a<1 or a>1. I cant remember the details. I got it wrong but in hindsight that was the right answer i think.

Reply 1534

louisforrest

2

Original post by m0.4444

I think it was one of the a<1 or a>1. I cant remember the details. I got it wrong but in hindsight that was the right answer i think.

No I've realised I'm wrong its (6/5)^4

Reply 1535

Palette

14

Predicted grade boundaries anyone?

Reply 1536

m0.4444

7

Original post by louisforrest

No I've realised I'm wrong its (6/5)^4

Why? I know i got it wrong i put all values of a.

Reply 1537

ShatnersBassoon

4

Original post by peted0herty

anyone got any ideas about the weightings of the marks for each part of question 3? (or any of the long questions, for that matter)

There were 7 parts to this question, right? I don't think I've ever seen that many before in a MAT question. I'd expect the marks to have been fairly homogeneous - i.e. 2 marks for each of them, with a third mark for whichever part they thought was trickiest.

It was basically all "show that" and "explain that", from what I can remember, so I don't think there's much to talk about, but:

Can't remember the first parts.

For

\int_a^b x^n dx = -\int_b^a x^n dx

, I think you just needed to evaluate them and show they were both equal to

\frac{x^{b+1}-x^{a+1}}{n+1}

.

The subsequent "show that" seemed to me more like an "explain that", but I said

p(x) = a_0 + a_1x + a_2x^2 + ... + a_kx^k

Then I set its indefinite integral,

a_0x + \frac{a_1x^2}{2} + \frac{a_2x^3}{3} + ... + \frac{a_kx^k}{k}

equal to a new function, F(x).
Then the LHS of their equation was F(b) - F(a), and the RHS was -(F(a) - F(b)) = F(b) - F(a).

Then we had an "explain that" with 2α-t and so on as the limits of the integral. This was because the function was bilateral; I don't think I explained it very well but basically the value of f(t) at any point from t to α corresponded 1-to-1 to a point from 2α-t to 2α, so the symmetry meant the areas were equal.

G(x) - there was some kind of show that I can't quite recall requiring the