STEP 2 in 2024: Sharing Your Story! [PLUS WITH SOME SOLUTIONS AND PREDICTION]

IIRC, There was a premise in the question that showed that a sequence tends to a limit L if it bounded by some value b, so L < b. So i just manipulated the sequence to find a value of b that would act as the bound value, which I found as x_0, thus M < x_0 But yea, your methods probably right anyway.

I'm just gonna list my thoughts overall, i'd really appreciate if anyone had anything to add.
Q1) Consecutive Numbers
P sure I got this one. Iirc, the final solutions were (35, 49) and (204, 288) for (n, k). I just remember you had to group one set of k^2 + k as 2n^2 to get the 2N'^2 = K'^2 + K equivalence, before applying it to the (6, 8) solution that you had to figure out for the first part. As long as i explained the stuff about changing k and c properly, i think i should get 20 on this.
Q2) Series Integration
Okay, so part i was basically just a bunch of manipulation. The next part was just an application of the first part, and you get the integral under question, which should be integrated to ln(3/2) i think.This next part, I think i got the right methodology, applying partial fractions to the summation and getting 24 x ([integral] + [integral]). Then you have to write (x+1)/(x^3 + 8) [the integrand to be found] as (x+1)/(x + 2)(x^2 - 2x + 4). I majorly ****ed up and mistakenly went through the partial fractions to solve 1/(x + 2)(x^2 - 2x + 4) [i missed the x, **** my life]. Anyway, you end up getting the coefficients as multiples of 1/12, i don't remember exactly what I got. Then you get an integral of 1/(x+2) [which we found out as ln(3/2) + integral of (bx + c)/(x^2 - 2x + 4) which you break into integral of (k(2x-2)(x^2 - 2x + 4) + m/((x-1)^2 + 4). then you get 2 ln integrals and an arctan integral, which should get you the solution.i'm pretty positive this is right, but how many marks do you think i lost because i calculated the wrong partial fraction? i only noticed in the last 2 seconds of the exam, its too bad. i hope i'll get at least 17, but who knows.
Q5) ranges
i don't know why the hell i attempted this question, im terrible with domains and ranges, i didn't know what i was doing. i majorly regret trying this, i don't even remember what i was doing. i just completed a few squares and called it a day. i think i maybe got 2 or 3 marks. im annoyed there was no matrix question, legit my favourite types, and i'd be a lot more confident if i hadn't had to do this one, but i didn't have a ruler so i couldn't attempt the graph question and 'm not the best at vectors, so i had no choice.
Q6) standardising summations of a_n and b_n
i think i kinda mucked this all up. i got all the answers, but i basically never did what the question asked me to. i just wrote out the expansion and managed to get all the coefficients of the forms that you required. i didn't use t_n to get a_n, and i didn't use (b_n/a_n) to find b_n. i think that's gonna dock me at least 6 marks. the final part was using the fact that (1-x)^(-1) x (1-x)^(-1/2) = (1-x)^(-3/2), which leads to the summation of a_r from r = 1 to n = b_n, which can then be manipulated to the final result.i should have just manipulated my answers to fit the method they required us to use. ahh.
q8) a(n) and g(n) sequences integration
this was just inequality manipulation, i found it pretty nice. i got that y_n is bounded by m < x_0? hopefully that's right. then as n tends to infinity, x_n - y_n tends to 0, so x_n tends to y_n which tends to m. then the t substitution was fairly mechanical but i think it was just a standard substitution. then you apply the i(p,q) = i((a(p,q), (g(p,q)) result to i(x_0, y_0) = i(a(x_0, y_0), g(x_0, y_0)) = i(x_1, y_1) = ... = i(x_n, y_n) as n tends to infinity. then, as x_n and y_n tend to m, the final integral is of 1/(x^2 + m^2) from 0 to infinity, which is 1/marctan(x/m) [infty, 0] which I got as pi/2M. Did anyone else get that?
Q9) Ditch mechanics
I think one felt a bit light? I hope I didn't miss a part. The first part was setting it so the time was when the x component reached d, and then s < 0 in the suvat at that time, it leads you straight to the result. Then you have to get tan alpha > 1 so alpha is greater than 45 degree. The next part was something similar, set the time so the x component reached 2d and then s < -2d (as it should be below the length of the wall by the time it would theoretically hit it for it to reach the bottom of the pit) which you then manipulate and you get the requisite result.
I think the final part was just showing that lambda < tan alpha < lambda + some term (I can't remember exactly) for the ball to hit the bottom of the ditch (basically just using the stuff we solved in the previous part), and as tan alpha can take all values from 1 to infinity, a suitable value of alpha can be used to satisfy the equation for all values of lambda.I think I should have mentioned that lambda > 1 in the last part but it was proven in an earlier part (and hopefully it's implied by ' tan alpha can take valued from 1 to infinity'). Who knows.
Anyway, that's how my paper went. I think I got between 90-95 overall, I'm still annoyed that there were no matrices, they're my favourite type of question. Im so annoyed man . But oh well. What do you think the S grade boundary will be? I think I'll be borderline.
NOTE - I think my earlier comment deleted, so I've just recopied it.

I found Question 5 on ranges pretty straightforward. I started by completing the square to transform the function into vertex form, which helped me identify the minimum or maximum value for the range. For the domain, I considered any restrictions on the input, like non-negative square roots or avoiding zero denominators. I also checked the function's behavior at the domain boundaries to confirm the range. Using interval notation, I clearly expressed both the domain and the range. It was manageable overall, though I wish there had been a matrix question instead!

Do you notice that n is in Z, meaning that it is an integer?

yuh

I found Question 5 on ranges pretty straightforward. I started by completing the square to transform the function into vertex form, which helped me identify the minimum or maximum value for the range. For the domain, I considered any restrictions on the input, like non-negative square roots or avoiding zero denominators. I also checked the function's behavior at the domain boundaries to confirm the range. Using interval notation, I clearly expressed both the domain and the range. It was manageable overall, though I wish there had been a matrix question instead!

yuh

Do you remember your solutions for the first few parts of 5? I just wanna see if I at least scraped a few. I didn't realise that it was in the integer realm, guess that makes more sense. ****k

What do you think the boundary for S will be?

which one? vector?

nah, domains and ranges. It's the one that's gonna cost me my S unless I'm lucky and the boundary is like, idk, high 80s (rather than mid 90s)

Did 1,2,4,5,6 and 12
Thought the number theory for 1 and 5 was relatively straight forward (just difference of two squares and squares mod 4).
Vector question was really nice especially the last part spamming the dot product.
Stat p12 was not too bad but the last part was slightly strange.
6 looked horrible at first sight but ended up being really quick and the integration for 2 again with the binomial expansion looked a lot harder than it was.
Overall very average paper (unfortunately no matrices tho! hopefully will come up in S3 tho...) so grade boundary for S probably around 90-95 and 1 about 60-65 I think

for another series question, about T_r, a_r, b_r, you can finish the first several questions step by step. For the last part, notice that:
the coefficient of xⁿ in (1-x)^-3/2
= the sum of the product of 'the coefficient of x^r in (1-x)^-1' times 'the coefficient of x^n-r in (1-x)^-1/2'
= the sum of the coefficient of x^r in (1-x)^-1, where r from 0 to n
since after binomial expansion of (1-x)^-1, the coefficient of x^r is 1 for any r greater or equal to 0
By using this, we can get the answer required.
If you like, you can see similar question at:
STEP2 1999 4, STEP2 1999 8, STEP3 2007 2, STEP2 2016 5

How many marks do you think u got? I'm rly bad at number theory so I completely messed up Q5, it's gonna cost me an S i think.

How many marks do you think u got? I'm rly bad at number theory so I completely messed up Q5, it's gonna cost me an S i think.

I think you should still be able to make an S with 5 out of 6 comfortably

I would not be able to comment on qs 8 and 9 since i didn't attempt them but I think I got very similar results for 1,2 and 6. Also I wouldn't worry about using a different method to what the markscheme wants because they will give you marks for rigorous maths. If your logic is correct and coherent you will get the marks.
As for q5 the way I did it was just completing the square and noticing they were the same function just shifted. so like f1(x) = (x-1)^2 + 2 or smthg like that and so plugging in x = y+1 you get f1(x) = y^2 + 2 and since x is in Z y is also in Z and also any y exists since y-1 would also be in Z so the range of f1 is a subset of F1 and vice versa so they must be identical. Second part was something like m^2 - n^2 = 2 for some m and n and then you show there are no solutions. Third part was basically the same and last part was the same trick of 'shifting' but with a cubic instead.
As for the rest of the people in my school I think they all thought the paper was just average not particularly hard or easy. Questions weren't horrible tho!