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 $a, b, c$:

$\frac{a}{b + c} + \frac{b}{c + a} + \frac{c}{a + b} \geq \frac{3}{2}$

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:

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

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 $\sin^2\theta = 1 - \cos^2\theta$:

$I_n = (2n-1)\int_0^{\pi/2} \cos^{2n-2}\theta(1 - \cos^2\theta)\,d\theta = (2n-1)(I_{n-1} - I_n)$

Solving for $I_n$:

$I_n + (2n-1)I_n = (2n-1)I_{n-1}$

$2n I_n = (2n-1)I_{n-1}$

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

Since $I_0 = \pi/2$:

$I_n = \frac{2n-1}{2n} \cdot \frac{2n-3}{2n-2} \cdots \frac{3}{4} \cdot \frac{1}{2} \cdot \frac{\pi}{2} = \frac{(2n)!}{4^n(n!)^2} \cdot \frac{\pi}{2}$

Where the last equality follows by writing the product of odd terms as $(2n)!/(2^n \cdot n!)$ and the Product of even terms as $2^n \cdot n!$So:

$\frac{(2n)!}{2^n n! \cdot 2^n n!} = \frac{(2n)!}{4^n (n!)^2}$

Solve the differential equation:

$x\frac{dy}{dx} + y = x^2 y^2$

subject to $y(1) = 1$. Find the range of $x$ for which the solution is defined.

Solution. This is a Bernoulli equation. Rewrite as:

$\frac{dy}{dx} + \frac{y}{x} = xy^2$

Divide by $y^2$ (valid where $y \neq 0$):

$y^{-2}\frac{dy}{dx} + \frac{y^{-1}}{x} = x$

Let $u = y^{-1}$So $\frac{du}{dx} = -y^{-2}\frac{dy}{dx}$:

$-\frac{du}{dx} + \frac{u}{x} = x$

$\frac{du}{dx} - \frac{u}{x} = -x$

This is a linear first-order ODE. Integrating factor: $\mu = e^{\int -1/x\,dx} = e^{-\ln x} = \frac{1}{x}$.

$\frac{d}{dx}\left(\frac{u}{x}\right) = -1$

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

$u = -x^2 + Cx = x(C - x)$

Since $u = 1/y$:

$y = \frac{1}{x(C - x)}$

Applying $y(1) = 1$: $1 = \frac{1}{1 \cdot (C - 1)}$So $C = 2$.

Therefore:

$y = \frac{1}{x(2 - x)} = \frac{1}{2x - x^2}$

The denominator $2x - x^2 = x(2 - x)$ must be non-zero, so $x \neq 0$ and $x \neq 2$. For the Solution through $(1, 1)$We have $y > 0$ at $x = 1$And the solution is defined on $(0, 2)$.

Let $a_1 = 1$ and $a_{n+1} = \frac{1}{2}\left(a_n + \frac{2}{a_n}\right)$ for $n \geq 1$.

(i) Show that $a_n \geq \sqrt{2}$ for all $n \geq 2$. (ii) Show that $(a_n)$ is decreasing for $n \geq 2$. (iii) Deduce that $(a_n)$ converges and find its limit.

Solution.

(i) For $n \geq 1$By AM-GM:

$a_{n+1} = \frac{1}{2}\left(a_n + \frac{2}{a_n}\right) \geq \sqrt{a_n \cdot \frac{2}{a_n}} = \sqrt{2}$

Since $a_2 = \frac{1}{2}(1 + 2) = \frac{3}{2} \geq \sqrt{2}$By induction $a_n \geq \sqrt{2}$ for all $n \geq 2$.

(ii) For $n \geq 2$:

$a_{n+1} - a_n = \frac{1}{2}\left(a_n + \frac{2}{a_n}\right) - a_n = \frac{2 - a_n^2}{2a_n}$

Since $a_n \geq \sqrt{2}$We have $a_n^2 \geq 2$So $2 - a_n^2 \leq 0$ and $2a_n > 0$Giving $a_{n+1} - a_n \leq 0$. Hence $(a_n)$ is decreasing for $n \geq 2$.

(iii) The sequence $(a_n)_{n \geq 2}$ is decreasing and bounded below by $\sqrt{2}$So by the Monotone convergence theorem it converges to some limit $L \geq \sqrt{2}$.

Taking limits in the recurrence:

$L = \frac{1}{2}\left(L + \frac{2}{L}\right)$

$2L = L + \frac{2}{L}$

$L = \frac{2}{L}$

$L^2 = 2$

Since $L \geq \sqrt{2} > 0$We have $L = \sqrt{2}$.

A bag contains $n$ balls, of which $r$ are red and $n - r$ are blue. Balls are drawn one at a time without replacement until a red ball is drawn. Let $X$ be the number of draws required.

(i) Find $\mathbb{E}(X)$. (ii) Find \text{Var(X). (iii) Show that $\mathbb{E}(X) = \frac{n+1}{r+1}$.

Solution. This is a negative hypergeometric distribution.

(i) Use indicator random variables. Let $I_j$ be the indicator that the $j$-th draw is required, I.e., $I_j = 1$ if the first $j-1$ draws are all blue and the $j$-th is red.

The probability that the first $j-1$ draws are all blue is:

P(\text{first j-1 \text{ are blue) = \frac{\binom{n-r}{j-1}}{\binom{n}{j-1}}

Provided $j - 1 \leq n - r$. Then:

$P(X = j) = \frac{\binom{n-r}{j-1}}{\binom{n}{j-1}} \cdot \frac{r}{n - (j-1)}$

For $j = 1, 2, \ldots, n - r + 1$.

For the expectation, we use a different approach. Consider the blue balls in the bag. There are $r$ Red balls creating $r + 1$ “gaps” (before the first red, between consecutive reds, after the last red). The blue balls are distributed uniformly at random among these $r + 1$ gaps.

Let $B_i$ be the number of blue balls in gap $i$. Then $B_1 + B_2 + \cdots + B_{r+1} = n - r$And by Symmetry $\mathbb{E}(B_i) = \frac{n - r}{r + 1}$ for each $i$.

The number of draws $X$ equals $B_1 + 1$ (the blue balls before the first red, plus one for the first Red ball itself). Therefore:

$\mathbb{E}(X) = \mathbb{E}(B_1) + 1 = \frac{n - r}{r + 1} + 1 = \frac{n - r + r + 1}{r + 1} = \frac{n + 1}{r + 1}$

(ii) For the variance, use a similar symmetry argument. We need $\mathbb{E}(B_1^2)$. The blue balls are distributed multinomially among $r + 1$ gaps with equal Probabilities $\frac{1}{r+1}$ each. For a multinomial distribution:

\text{Var(B_1) = (n-r)\cdot\frac{1}{r+1}\cdot\frac{r}{r+1} = \frac{r(n-r)}{(r+1)^2}

Since \text{Var(B_1) = \mathbb{E}(B_1^2) - [\mathbb{E}(B_1)]^2:

$\mathbb{E}(B_1^2) = \frac{r(n-r)}{(r+1)^2} + \frac{(n-r)^2}{(r+1)^2} = \frac{(n-r)(r + n - r)}{(r+1)^2} = \frac{(n-r)n}{(r+1)^2}$

Since $X = B_1 + 1$:

\text{Var(X) = \text{Var(B_1) = \frac{r(n-r)}{(r+1)^2}

(iii) This was established in part (i): $\mathbb{E}(X) = \frac{n+1}{r+1}$.

Arithmetic errors in long calculations. STEP questions involve sustained algebra. A single sign Error early in a solution can invalidate an entire question. Always check intermediate results and Verify that special cases are consistent.

Neglecting to state conditions. When dividing by an expression, you must state that it is non-zero. When taking square roots, you must consider both signs. When applying a convergence test, you must Verify its hypotheses.

Confusing necessary and sufficient conditions. Showing that a result holds for specific values does Not prove it . Showing that a condition is necessary does not show it is sufficient.

Incomplete integration by parts. When applying integration by parts repeatedly, track the signs Carefully. The alternating sign pattern is a frequent source of error.

Missing edge cases in combinatorics. When counting, verify that no configuration is counted twice And none is omitted. Check boundary cases (e.g., empty sets, all elements identical).

Unjustified interchange of limits. You cannot always differentiate under an integral sign or Interchange the order of summation. In a proof-based exam, such interchanges must be justified.

Failing to check the answer. Substitute your answer back into the original equation or verify Against known special cases. A quick check can catch errors that are otherwise invisible.

With 12 questions and only 6 to attempt, selection is critical. Spend the first 10—15 minutes reading All 12 questions and identifying the 6 you are most confident about. Prioritise questions where you Can see a clear path to a complete solution.

• State what you are doing at each stage. A sequence of unexplained equations is not a proof.
• Draw diagrams where appropriate, especially for geometry and mechanics.
• Label all variables . If a question introduces notation, use it consistently.
• If you cannot complete a question, write down what you have done. Partial marks are available for correct progress.

At 3 hours for 6 questions, you have approximately 30 minutes per question. If a question is taking Longer than 35 minutes, consider moving on. It is better to attempt 6 questions partially than 3 Questions fully and 0 partially.

Reserve the final 10 minutes to review your work. Check that:

• Every equation follows from the previous one.
• Boundary conditions have been applied.
• The final answer is in the form requested by the question.
• No algebraic errors have been introduced in the final steps.

1. Work through past papers systematically, starting with the earliest available and progressing to the most recent.
2. For each question, write a full solution before checking the mark scheme. The act of writing the solution is where the learning happens.
3. Maintain a list of techniques that you find difficult and revisit them regularly.
4. Time yourself on full papers under exam conditions at least once a week in the months leading up to the exam.
5. Study the mark schemes to understand what earns full marks. STEP rewards completeness and rigour, not just correct answers.

Example 1: Applying key concepts

When working with step preparation, follow a structured approach:

1. Identify the key concepts and definitions relevant to the question
2. Apply the appropriate methods, equations, or frameworks
3. Support your answer with evidence, examples, or calculations
4. Evaluate your answer critically, considering limitations and alternative perspectives

• STEP (Sixth Term Examination Paper) consists of STEP 2 and STEP 3, each a 3-hour paper with 12 questions (answer 6); used by Cambridge for maths offers.
• STEP 2 covers A-Level Maths and Further Maths (AS content); STEP 3 covers full A-Level Further Maths.
• Questions require sustained reasoning, multi-step proofs, and the ability to connect different areas of mathematics.
• Effective preparation: start early (January of Year 13), work through STEP Foundation modules, then full past papers under timed conditions.
• Marking rewards clear explanation and correct reasoning, not just final answers; partial credit is generous for good working.