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

Hi There
Im in year 12 and one of my friends did step 2 this year in year 12
how was it?
he also got distinction in BMO2

Hard to say🤣 I think I could do a better job if I can find out the method of drawing graphs immediately

lmao no calculator is tough

how did you do the series question

how did you do the series question

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

thanks a lot. I don't think i got it, but i'll try to find the answer, for reference, do you remember the question

that paper i found to be so easy compared to the other years, i needed a one and i think i definately met the mark requirement but i feel like the grade boundaries will be much higher this year as all the people who sat it in my school also found it really easy. so im praying 80 should be a grade 1 hopefully they dont raise it by too much.

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. But oh well. What do you think the S grade boundary will be? I think I'll be borderline.

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 values 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. I think this might have been a slightly easier paper as well. Damn, why were there no matrices 👿 ☹️ . But oh well. What do you think the S grade boundary will be? I think I'll be borderline.

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. But oh well. What do you think the S grade boundary will be? I think I'll be borderline.

yeah i aggree with what you wrote mostly im so glad i found the mechanics quite easy, im failry sure i got 40 on those questions i know some people found them tricky, but i dont know i just figured out how to do them in the test. like it just came to me xD

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.

same answer in the final part for Q8. By the way, I think, for the boundary of y_n, I prove it by showing that, as n approximates to infinite, (sqrt(x_n)-sqrt(y_n)) then to be zero, so sqrt(x_n) tends to equal to sqrt(y_n), thus y_(n+1) tends to equal to y_n.