It is hard to give anything more than general advice, which is to go through practice papers and calculus type questions. Doing calculus type questions refreshes you on the techniques and doing enough of them gets you faster, while doing practice papers also gets you used to using in-exam methods, taking advantage of the nature of these multiple choice questions.
For example, rather than finding the actual equation of the plane and plugging them in, I would have computed the gradient vector, which is $$ ( -2z^2+6xy, 3x^2+2zy, y^2-4xz)\mid_{(1,1,1)}= (4,5,-3).$$ Any point $P$ that lines on the plane in question must satisfy $(P- (1,1,1))\cdot (4,5,-3)=0$ so subtract 1 from each coordinate and just compute dot products until you find one that is $0.$ You wouldn't even have to compute (a) or (d) properly, you'll notice by signs that the dot product will be positive. It is quite similar to finding the equation of the plane first and plugging them in, but seems to save at least a little bit of time.
General Tips:
- Don't spend more than 2 minutes on any question, after your two minutes if you haven't got the answer, you should at least be able to rule out 2 answers. Mark off those 2 answers for now.
- Dimensional analysis is a good trick. There are some calculus type rates questions (A car uses this much fuel per mile when travelling at some speed...). The options might be some integrals with various integrands, and only one of them even has the correct units.
- If you can rule out 2 options, good, and if you can rule out 3 even better, guessing is benefical. Don't guess if you can't rule anything out, because there is only 1/5 chance of getting the right answer, and an incorrect answer subtracts 0.25.
- Some things they seem to like testing, so worth revising are: Green's theorem, and using it to find areas, Volumes of surfaces of revolution, finding volumes by setting up double integrals, Decompositions of finite abelian groups, basic statistics and probability was something I had to revise, Residue theorem, Cauchy-Riemann equations. Also, many of the incorrect answer choices for questions on Lebesgue theory seem to come from incorrectly thinking uncountable implies 0 measure. Keep in mind the Cantor set for these questions.
Good luck!
Well, there are iterative algorithms. There are two beautiful ones by the Borwein brothers, based on work by Ramanujan. Algorithm 1 involves the silver ratio, and Algorithm 2 involves the cube of the golden ratio.
I. Algorithm 1. Start with seed values:
$$y_0 = -1+\sqrt{2}$$
$$a_0 = 2(-1+\sqrt{2})^2$$
and two iterative rules,
$$y_{n+1} = \frac{1-(1-{y_n}^4)^{1/4}}{1+(1-{y_n}^4)^{1/4}}\tag1$$
$$a_{n+1} = a_n(y_{n+1}+1)^4-2^{2n+3}\,y_{n+1}\big(y_{n+1}^2+y_{n+1}+1\big)\tag2$$
Then,
$$\quad\quad\quad\lim_{n\to\infty} \frac{1}{a_n} = \pi\quad\text{(very fast)}$$
The difference grows quartically,
$$\quad\quad\quad\quad\frac{1}{a_n} - \pi \approx 4^{n+2} q^{4^n},\quad \text{where}\;q = e^{-2\pi}$$
Thus for $n=1,2,3$, the difference is about $10^{-10},\,10^{-42},\,10^{-172},$ or more than the fourth power of the previous. It's that fast.
Best Answer
Perhaps not quite the way you are looking for, but:
You can derive Taylor's theorem with the integral form of the remainder by repeated integration by parts: $$ f(x)-f(a) = \int_a^x f'(t) \, dt = \left[-(x-t)f'(t) \right]_a^x + \int_a^x (x-t) f''(t) \, dt \\ = (x-a)f'(a) + \int_a^x (x-t) f''(t) \, dt, $$ and so on, integrating the $(x-t)$ and differentiating the $f$ each time, to arrive at $$ R_N = \int_a^x \frac{(x-t)^N}{N!} f^{(N+1)}(t) \, dt. \tag{1} $$ Interpretation for this is simply that integrating by parts in the other direction will give you back precisely $f(x) - f(a) - \dotsb - \frac{1}{N!}(x-a)^N f^{(N)}(a)$.
Now, we can get from (1) to the Lagrange and Cauchy forms of the remainder by using the Mean Value Theorem for Integrals, in the form:
(this is easy if you think about weighted averages and the usual Mean Value Theorem).
Applying this to (1) with $h=f$, $g(t)=(x-t)^N/N!$ gives $$ R_N = f^{(N+1)}(c)\frac{(x-a)^{N+1}}{(N+1)!}, $$ which is the Lagrange form of the remainder; using $h(t)=f(t)(x-t)^N/N!$, $g(t)=1$ gives $$ R_N = f^{(N+1)}(c')\frac{(x-c')^N}{N!}(x-a), $$ which is the Cauchy form of the remainder.
The weighted averages mentioned above are a way to think about what we did here: we take an average of $\frac{(x-t)^N}{N!} f^{(N+1)}(t)$ over $[a,x]$, and use the MVT to equate this to a value of the function at a specific point: how much of the function we count as in the weighting affects what answer we obtain.