STEP Preparation

The Sixth Term Examination Paper (STEP) is a university admissions test used primarily by the University Of Cambridge for undergraduate mathematics courses. It is also accepted or required by several other UK Universities as part of their conditional offers.

There are three papers:

Each paper is 3 hours long. Candidates choose 6 questions from a total of 12 (8 pure, 2 mechanics, 2 Probability/statistics). Each question is worth 20 marks. There is no penalty for incorrect working: Marks are awarded for correct progress towards the solution.

Grades are S (Outstanding), 1 (Very Good), 2 (Good), 3 (Satisfactory), and U (Unclassified). Typical Cambridge offers require Grade 1 in two STEP papers. The distribution is not linear: the gap between Grade 2 and Grade 1 is substantial, reflecting the exam”s emphasis on sustained, complete solutions rather than partial credit.

• Questions are multi-part and require sustained reasoning. A single question may involve three or four distinct ideas chained together.
• Proofs are expected, not just computations. Candidates must justify steps, not merely state results.
• There is no module-specific focus: questions freely combine topics (e.g., trigonometry and integration, or algebra and geometry).
• The difficulty lies in the depth of each question rather than the breadth of the syllabus.

STEP frequently requires manipulation of inequalities, often combining algebraic techniques with Calculus or induction.

Cauchy-Schwarz Inequality. For real numbers $a_1, \ldots, a_n$ and $b_1, \ldots, b_n$:

\left(\sum_{i=1}^{n} a_i b_i\right)^2 \leq \left(\sum_{i=1}^{n} a_i^2\right)\left(\sum_{i=1}^{n} b_i^2\right)

With equality if and only if $a_i$ and $b_i$ are proportional.

AM-GM Inequality. For non-negative real numbers $x_1, \ldots, x_n$:

\frac{x_1 + x_2 + \cdots + x_n}{n} \geq \sqrt[n]{x_1 x_2 \cdots x_n}

With equality if and only if all $x_i$ are equal.

Technique: completing the square. Many STEP inequalities can be reduced to completing the square. For example, to show that $x^2 + y^2 + z^2 \geq xy + yz + zx$ for all real $x, y, z$:

2(x^2 + y^2 + z^2 - xy - yz - zx) = (x-y)^2 + (y-z)^2 + (z-x)^2 \geq 0

Technique: substitution to standardise. When an inequality involves a constraint, a well-chosen Substitution can reduce it to a known form. If $a + b + c = 1$ with $a, b, c > 0$ Setting $a = x/(x+y+z)$ Etc., can convert a constrained problem into an unconstrained one.

Partial fraction decomposition is a routine technique that appears in integration and series questions.

Repeated linear factors. For a denominator with $(x-a)^k$:

\frac{P(x)}{(x-a)^k} = \frac{A_1}{x-a} + \frac{A_2}{(x-a)^2} + \cdots + \frac{A_k}{(x-a)^k}

Irreducible quadratic factors. For a denominator with $(x^2 + bx + c)^m$ where $b^2 < 4c$:

\frac{P(x)}{(x^2 + bx + c)^m} = \frac{B_1 x + C_1}{x^2 + bx + c} + \cdots + \frac{B_m x + C_m}{(x^2 + bx + c)^m}

Cover-up rule. For distinct linear factors, the coefficient of $1/(x - a_i)$ is obtained by Evaluating the remaining expression at $x = a_i$.

Remainder theorem. When $P(x)$ is divided by $(x - a)$ The remainder is $P(a)$.

Factor theorem. $(x - a)$ is a factor of $P(x)$ if and only if $P(a) = 0$.

Vieta’s formulae. For a cubic $x^3 + px^2 + qx + r = 0$ with roots $\alpha, \beta, \gamma$:

\alpha + \beta + \gamma = -p, \quad \alpha\beta + \beta\gamma + \gamma\alpha = q, \quad \alpha\beta\gamma = -r

Resultants and symmetric functions. STEP often asks for expressions in the roots of a polynomial Without finding the roots explicitly. For example, to find $\alpha^2 + \beta^2 + \gamma^2$:

\alpha^2 + \beta^2 + \gamma^2 = (\alpha + \beta + \gamma)^2 - 2(\alpha\beta + \beta\gamma + \gamma\alpha) = p^2 - 2q

Integration is the single most tested topic on STEP. Candidates must be fluent with the following Techniques.

Substitution. The standard $u = g(x)$ substitution. The key is recognising when the integrand Contains $g'(x)$ alongside a function of $g(x)$.

Integration by parts. For $\int u \, dv = uv - \int v \, du$. Strategy: apply repeatedly when the Integrand involves $x^n$ multiplied by $\sin x$, $\cos x$, $e^x$ Or $\ln x$.

Reduction formulae. Many STEP questions establish a recurrence relation. For example, setting $I_n = \int_0^{\pi/2} \sin^n x \, dx$ and integrating by parts yields:

I_n = \frac{n-1}{n} I_{n-2}

With $I_0 = \pi/2$ and $I_1 = 1$.

Trigonometric substitutions. For integrands involving $\sqrt{a^2 - x^2}$ Use $x = a\sin\theta$. For $\sqrt{a^2 + x^2}$ Use $x = a\tan\theta$. For $\sqrt{x^2 - a^2}$ Use $x = a\sec\theta$.

Rational functions. Decompose via partial fractions, then integrate term by term. A common STEP Trick is to write $\int \frac{1}{x^2 + a^2} \, dx = \frac{1}{a}\arctan\frac{x}{a}$.

Improper integrals. Evaluate limits at points of discontinuity or at infinity. Always check Convergence before computing the value.

First order: separable. $\frac{dy}{dx} = f(x)g(y)$ rearranges to $\int \frac{1}{g(y)}\,dy = \int f(x)\,dx$.

First order: integrating factor. For $\frac{dy}{dx} + P(x)y = Q(x)$ The integrating factor is $\mu(x) = e^{\int P(x)\,dx}$ Giving $\frac{d}{dx}(\mu y) = \mu Q$.

Second order: constant coefficients. For $a\frac{d^2y}{dx^2} + b\frac{dy}{dx} + cy = f(x)$:

1. Solve the complementary function from $a m^2 + bm + c = 0$.
2. Find a particular integral appropriate to $f(x)$.
3. Combine: $y = y_c + y_p$.

Taylor and Maclaurin series. For $f(x)$ infinitely differentiable at $x = a$:

f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(a)}{n!}(x - a)^n

Standard series to know:

e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!}, \quad \sin x = \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n+1}}{(2n+1)!}, \quad \cos x = \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n}}{(2n)!}

\ln(1+x) = \sum_{n=1}^{\infty} \frac{(-1)^{n+1} x^n}{n} \quad (|x| \leq 1, , x \neq -1)

\frac{1}{1-x} = \sum_{n=0}^{\infty} x^n \quad (|x| < 1)

Convergence tests. Ratio test, comparison test, integral test. STEP often asks candidates to prove Convergence or divergence of a given series using these methods.

Summation by parts. Analogous to integration by parts:

\sum_{k=m}^{n} a_k b_k = A_n b_n - \sum_{k=m}^{n-1} A_k(b_{k+1} - b_k)

Where $A_k = a_1 + \cdots + a_k$.

Scalar product. $\mathbf{a} \cdot \mathbf{b} = |\mathbf{a}||\mathbf{b}|\cos\theta = a_1b_1 + a_2b_2 + a_3b_3$.

Vector product. $\mathbf{a} \times \mathbf{b}$ is perpendicular to both $\mathbf{a}$ and $\mathbf{b}$ With magnitude $|\mathbf{a}||\mathbf{b}|\sin\theta$.

Triple scalar product. $[\mathbf{a}, \mathbf{b}, \mathbf{c}] = \mathbf{a} \cdot (\mathbf{b} \times \mathbf{c})$ Equals the signed volume of the parallelepiped spanned by $\mathbf{a}, \mathbf{b}, \mathbf{c}$.

Vector equation of a line. $\mathbf{r} = \mathbf{a} + t\mathbf{d}$.

Vector equation of a plane. $\mathbf{r} \cdot \mathbf{n} = d$ Or $\mathbf{r} = \mathbf{a} + s\mathbf{b} + t\mathbf{c}$.

Matrix multiplication. $(AB)_{ij} = \sum_k A_{ik}B_{kj}$. Matrix multiplication is associative but Not commutative.

Determinant (3x3). Expand by cofactors along any row or column:

\det A = a_{11}(a_{22}a_{33} - a_{23}a_{32}) - a_{12}(a_{21}a_{33} - a_{23}a_{31}) + a_{13}(a_{21}a_{32} - a_{22}a_{31})

Inverse. A^{-1} = \frac{1}{\det A}\text{adj(A). A matrix is invertible if and only if $\det A \neq 0$.

Eigenvalues and eigenvectors. Solve $\det(A - \lambda I) = 0$ for eigenvalues $\lambda$ Then solve $(A - \lambda I)\mathbf{v} = \mathbf{0}$ for eigenvectors $\mathbf{v}$.

Transformation matrices. Reflection in the line $y = x\tan\theta$ Rotation by angle $\theta$ And other geometric transformations can be represented as $2 \times 2$ or $3 \times 3$ matrices.

Addition principle. If task $A$ can be done in $m$ ways and task $B$ in $n$ ways, and the tasks are Mutually exclusive, then $A$ or $B$ can be done in $m + n$ ways.

Multiplication principle. If task $A$ can be done in $m$ ways and task $B$ in $n$ ways Independently, then $A$ and $B$ together can be done in $mn$ ways.

Permutations and combinations. The number of permutations of $r$ objects from $n$ is $P(n,r) = \frac{n!}{(n-r)!}$. The number of combinations is $C(n,r) = \binom{n}{r} = \frac{n!}{r!(n-r)!}$.

Inclusion-exclusion principle. For sets $A$ and $B$:

|A \cup B| = |A| + |B| - |A \cap B|

Conditional probability. $P(A \mid B) = \frac{P(A \cap B)}{P(B)}$.

Bayes’ theorem.

P(A \mid B) = \frac{P(B \mid A) , P(A)}{P(B)}

Expected value and variance. For a discrete random variable $X$:

\mathbb{E}(X) = \sum_i x_i p_i, \quad \text{Var(X) = \mathbb{E}(X^2) - [\mathbb{E}(X)]^2

Linearity of expectation. $\mathbb{E}(X + Y) = \mathbb{E}(X) + \mathbb{E}(Y)$ always, regardless of Independence.

Stars and bars. The number of solutions to $x_1 + x_2 + \cdots + x_k = n$ in non-negative integers Is $\binom{n + k - 1}{k - 1}$.

Bijection arguments. Show that the objects being counted are in one-to-one correspondence with a Simpler set.

Generating functions. The coefficient of $x^n$ in $(1 + x)^N$ is $\binom{N}{n}$ And in $(1 + x + x^2 + \cdots)^k$ is $\binom{n + k - 1}{k - 1}$.

Distance. $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$.

Midpoint. $M = \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right)$.

Line equations. Gradient-intercept: $y = mx + c$. Two-point form: $\frac{y - y_1}{x - x_1} = \frac{y_2 - y_1}{x_2 - x_1}$.

Perpendicular distance from a point to a line. For line $ax + by + c = 0$ and point $(x_0, y_0)$:

d = \frac{|ax_0 + by_0 + c|}{\sqrt{a^2 + b^2}}

Circle. $(x - a)^2 + (y - b)^2 = r^2$. Centre $(a, b)$ Radius $r$.

Ellipse. $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ with $a > b$. Foci at $(\pm c, 0)$ where $c^2 = a^2 - b^2$.

Hyperbola. $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$. Foci at $(\pm c, 0)$ where $c^2 = a^2 + b^2$. Asymptotes: $y = \pm \frac{b}{a}x$.

Parabola. $y^2 = 4ax$ with focus $(a, 0)$ and directrix $x = -a$. Parametric form: $(at^2, 2at)$.

Tangent to a parametric curve. For $(x(t), y(t))$:

\frac{dy}{dx} = \frac{dy/dt}{dx/dt}

Envelopes. Given a family of curves $F(x, y, t) = 0$ parameterised by $t$ The envelope satisfies $F(x, y, t) = 0$ and $\frac{\partial F}{\partial t} = 0$ simultaneously.

Prove that for all positive real numbers $a, b, c$:

$$$$

Solution. We use the substitution $x = b + c$, $y = c + a$, $z = a + b$. Then $a = \frac{y + z - x}{2}$, $b = \frac{z + x - y}{2}$, $c = \frac{x + y - z}{2}$.

By the triangle inequalities on $a, b, c > 0$ We have $y + z > x$, $z + x > y$, $x + y > z$ So $x, y, z$ are the side lengths of a (non-degenerate) triangle.

The inequality becomes:

\frac{y + z - x}{2x} + \frac{z + x - y}{2y} + \frac{x + y - z}{2z} \geq \frac{3}{2}

Which simplifies to:

\frac{y}{2x} + \frac{z}{2x} + \frac{z}{2y} + \frac{x}{2y} + \frac{x}{2z} + \frac{y}{2z} - \frac{3}{2} \geq \frac{3}{2}

\frac{y}{2x} + \frac{z}{2x} + \frac{z}{2y} + \frac{x}{2y} + \frac{x}{2z} + \frac{y}{2z} \geq 3

By AM-GM on each pair:

\frac{y}{2x} + \frac{x}{2y} \geq 2\sqrt{\frac{y}{2x} \cdot \frac{x}{2y}} = 1

\frac{z}{2x} + \frac{x}{2z} \geq 1, \quad \frac{z}{2y} + \frac{y}{2z} \geq 1

Summing these three inequalities gives the result. Equality holds when $x = y = z$ I.e., $a = b = c$.

Evaluate:

and deduce that $I_n = \frac{(2n)!}{4^n(n!)^2} \cdot \frac{\pi}{2}$.

Solution. Write $I_n = \int_0^{\pi/2} \cos^{2n}\theta \, d\theta$.

Integration by parts with $u = \cos^{2n-1}\theta$ and $dv = \cos\theta\, d\theta$:

du = -(2n-1)\cos^{2n-2}\theta \sin\theta , d\theta, \quad v = \sin\theta

I_n = \left[\cos^{2n-1}\theta \sin\theta\right]_0^{\pi/2} + (2n-1)\int_0^{\pi/2} \cos^{2n-2}\theta \sin^2\theta , d\theta

The boundary term vanishes. Using $\sin^2\theta = 1 - \cos^2\theta$:

\frac{u}{x} = -x + C