STEP 2
Took STEP 2. I did the questions in the order of 12,11,1,6,5,2. Basically, the strat was to farm section C and then pick 4 reasonable questions from section A. As much as possible, do not touch physics, diff eq and any sketching.
11 and 12 were standard run of the mill section C questions which are easy to do. The only thing unusual was that in 11, you had to sketch the graph of $x^{\frac{1}{x}}$ which I managed to screw up magnificently since I managed to calculate $\frac{d}{dx} x^{\frac{1}{x}}$ wrongly and didn’t think twice about what $\lim\limits_{x \to \infty} x^{\frac{1}{x}}$ is supposed to be so I just assumed $0$. So I’m hoping that it is worth very little marks…
1 was some number theory problem where you had to find triplets $(c,n,k)$ such that $(c)+(c+1)+\ldots+(c+n+k-1) = (c+n+k) + (c+n+k+1) + \ldots + (c+2n+k-1)$, which is just bashing algebra.
2 was slightly interesting, where you swap the order of integration and summation to evaluate integrals as infinite sums and vice versa. For example, $\int_0^1 \frac{1}{8-x^3} \, dx = \frac{1}{8} \int_0^1 \sum\limits_{k\geq 0} (\frac{1}{2}x)^{3k} \, dx = \frac{1}{8} \sum\limits_{k\geq 0} \int_0^1 (\frac{1}{2}x)^{3k} \, dx = \frac{1}{8} \sum\limits_{k\geq 0} \frac{1}{2^{3k}(3k+1)}$. The final step of the problem required evaluating $\int_0^1 \frac{1+x}{8-x^3} \, dx$ which I took super long to do because I am absolutely terrible at doing algebra. Somehow managed to get the answer a few minutes before the test ended. I managed to split $\frac{1+x}{8-x^3}$ into partial fractions wrongly twice. The question is posted in this reddit post.
5 was about finding interections of polynomial sets, that is sets generated by $\{ax^3 + bx^2 + cx + d \mid x \in \mathbb{Z} \}$. For all but the last subproblem, $a=0$, so the problem just became mostly bashing equations of the form $x^2 - y^2 = c$ where $c$ is given. The basic idea is that if $\lvert x \rvert$ is large enough, the $\lvert x \rvert^2 - (\lvert x \rvert -1)^2 = 2\lvert x \rvert-1$ will be bigger than $c$.
6 was kind of generating functions. Basically, we know $(1-x)^{a} = \sum_{k \geq 0} (-1)^k \binom{a}{k} x^{ak}$ from extended binomial theorem. Then we can write this $(-1)^{k+1} \binom{a}{k+1}$ in terms of $(-1)^{k} \binom{a}{k}$ and evantually write $(-1)^{i} \binom{a}{i}$ in a neat closed form (containing positive integer binomials) for $a = -\frac{1}{2}$ and $a = -\frac{3}{2}$. Then, in the last part, we have a nice binomial equation by comparing coefficients of $(1-x)^{-\frac{1}{2}}(1-x)^{-1} = (1-x)^{-\frac{3}{2}}$, where we note that in generating functions, $(1-x)^{-1}$ is the prefix sum power series for ogfs.