I attempted the Oxford Math Admission Test (MAT) in Nov 4th, 2020 and got 82 (did not get interview invite unfortunately), now I am preparing for the Cambridge math admission test (STEP) just for extra challenge.

In this blog I would give some reflections on STEP and share my personal experience, the MAT blog would be posted later. Hope this blog would help you :)


STEP takes place in June every year, aiming at shortlisting the most talented candidates during the university application process. So unlike other university entrance exams like Alevel, HKDSE etc., STEP problems are much more difficult and challenging.

There are three papers in the examination, namely STEP1, STEP2 and STEP3, ranked according to difficulty (means 1 is the easiest and 3 is the most difficult). In recent years STEP1 is cancelled and two other exams still run as normal.

As I attempt STEP2 this year, I would mainly talk about STEP2 in this blog.

The syllabus of STEP2 can be found in this website: STEP 2021 Specifications. In general the pure math section of the exam would cover Calculus, Matrices, Polynomials, Number theory and many other topics.

My Preparation

I finished my last DSE exam - Math Extended Part 2 (Calculus and Algebra) in May 17th, and STEP2 would be in June 14th, so there was not much time left for preparing the exam.

Since I studied under the HKDSE curriculum, some content in STEP’s curriculum is not covered. Therefore, I bought Alevel Further Math’s textbook and started self-study. I learnt topics like relationship between roots in polynomial, linear transformation, and first-order differential equation on my own.

After that, I began practising real STEP problems by year, after several days I’ve done five past papers (2016 ~ 2020). Then I started to do the exercises provided in the book Advanced Problems in mathamatics. Most of the problems there are selected from STEP1 questions so they are not very difficult, but by doing that you could be familar with various topics, and could employ relevant knowledge to solve the problems in that topic during the actual exam.

In the last several days, I began doing STEP problems by topic, I classified topics that I has the most confidence, and selected relevant problems on STEP database for practising purposes.

Problem Characteristics

In one STEP problem, there are always many subproblems, starting from trivial ones to challenging ones, and guiding you through the whole question by giving hints. So the problems would not be very difficult if you can notice them, it is highly possible that you could use the results of previous subproblems to help solve the current one. Therefore, when you are attempting the question, make sure that you check the previous subproblems and see if they could give some hints and guidence.

Reflections by topic


If PQP \Rightarrow Q, then Q is necessary condition of P, P is sufficient condition of Q. If PQP \Leftrightarrow Q, then P is necessary and sufficient condition of Q.


The calculus questions in STEP mainly test basic calculus manipulation skills such as integration by substitution, by parts etc. And sometimes would include reduction formula as well.

For by substitution questions, the first subproblem would always tell you what to substitute, for example:

By using the substitution u=1/xu = 1/x, show that for b>0b > 0:

1/bbxlnx(a2+x2)(a2x2+1)dx=0\int_{1/b}^b{\frac{x \ln x}{(a^2+x^2)(a^2x^2+1)}}\, dx = 0

( STEP2 2014 Question 4 )

And in the last subproblem would ask you to choose suitable substitution to prove or evaluate something.

By using a substituion of the form u=k/xu = k/x, for suitable k, show that:

01(a2+x2)2dx=π4a3\int_{0}^{\infty}{\frac{1}{(a^2+x^2)^2}}\, dx = \frac{\pi}{4a^3}

In such problem, you would have two hints: the first is the hint from the previous subproblems, and the second is the upper and lower limit of the integral, they give you idea about what to substitute.

Example 2 ( STEP2 1998 Q7 ):

Prove that, if

I=01(1+x2)k(1+x)k+1dxI = \int_{0}^{1}{\frac{(1+x^2)^k}{(1+x)^{k+1}}}\, dx


I=014πdθ[2cosθcos(14πθ)]k+1dxI = \int_{0}^{\frac{1}{4}\pi}{\frac{d\theta}{[\sqrt{2} \cos \theta \cos (\frac{1}{4}\pi - \theta)]^{k+1}}} \, dx

Show further that:

I=2018πdθ[2cosθcos(14πθ)]k+1dx=2021(1+x2)k(1+x)k+1dxI = 2\int_{0}^{\frac{1}{8}\pi}{\frac{d\theta}{[\sqrt2 \cos \theta \cos (\frac{1}{4}\pi - \theta)]^{k+1}}} \, dx = 2\int_{0}^{\sqrt{2} - 1}{\frac{(1+x^2)^k}{(1+x)^{k+1}}}\, dx

For integration by parts problem, a lot of calculations are required so make sure that you do not make careless mistakes, STEP2 1994 Question 2 is one of the very challenging questions. Lower and upper limit of the integral is still very useful in the problem, 2003 STEP2 Question 7 is a case in point:

Show that if n>0n > 0 then:

e1/nlnxxn+1dx=2n2e\int_{e^{1/n}}^{\infty}{\frac{\ln x}{x^{n+1}}} \, dx = \frac{2}{n^2 e}

From the problem we could discover that the lower limit of the integral is e1/ne^{1/n}, but the result contains 1e\frac{1}{e}, that means the result of the integral should contain xnx^{-n}, it could only be obtained by integrating x(n+1)x^{-(n+1)}, therefore we could do the following procedure:

e1/nlnxxn+1dx=e1/nlnxd(1nxn)dxdx=...=2n2e\int_{e^{1/n}}^{\infty}{\frac{\ln x}{x^{n+1}}} \, dx = \int_{e^{1/n}}^{\infty}{\ln x \frac{d(-\frac{1}{n}x^{-n})}{dx}} \, dx = ... = \frac{2}{n^2 e}

Trapezium rule is in syllabus too, STEP2 1987 Question 7 is worth attempting, which tests approximation of the integral.

Sometimes the problem needs to evaluate integral which upper or lower limit contains infinity, for example:

0xnexdx=[xnex]0+n0xn1exdx\int_{0}^{\infty} x^n e^{-x}\, dx = [-x^n e^{-x}]^{\infty}_{0} + n\int_{0}^{\infty}{x^{n-1} e^{-x}} \, dx

For xnex-x^n e^{-x}, it equals to xnex\frac{-x^n}{e^x},as exe^x grows much faster than xnx^n, when xx \to \infty, it approaches zero.


Linear transformation is an important concept in STEP, including rotation, shearing, reflection etc.

STEP2 1998 Question 5, STEP2 1993 Question 10, STEP2 1989 Question 7 are good practice materials.

Besides, STEP problems may test matrices in an algebaric way, for example:

If M is a 2×22\times 2 matrix, prove that:

Tr(M2)=Tr(M)22Det(M)\mathrm{Tr}(M^2) = \mathrm{Tr}(M)^2 - 2\mathrm{Det}(M)