STEP2 2025 -- Pure Mathematics

Exam: STEP2  |  Year: 2025  |  Questions: Q1—Q8  |  Total marks per question: 20

All questions below are pure mathematics. Problem statements are transcribed from the STEP Support Programme 2025 worked paper.

Overview

Q	Topic	Difficulty	Key Techniques
1	Functions and graph sketching	Standard	Piecewise definitions, intersections, local extrema
2	Complex numbers	Challenging	Modulus equations, real/imaginary cases, cubic equations
3	Logarithmic comparison	Challenging	Graph of $\ln x/x$, exponent comparison, monotonicity
4	Floor functions	Challenging	Periodicity, floor identities, telescoping floor sums
5	Improper integrals	Standard	Substitution, symmetry, parameter matching
6	Coordinate geometry	Hard	Tangency, circles, parabolas, simultaneous constraints
7	Differential equations	Challenging	Reduction of order, separation of variables, qualitative motion
8	Balanced partitions	Hard	Equal sums, equal squares, cubic identities, construction

---

Question 1

Topic: Functions and graph sketching  |  Difficulty: Standard  |  Marks: 20

Problem

1 The function $\operatorname{Min}$ is defined as

$$\operatorname{Min}(a,b)= \begin{cases} a, & a\leq b,\\ b, & a>b. \end{cases}$$

(i) Sketch the graph $y=\operatorname{Min}(x^2,2x)$.

(ii) Solve the equation

$2\operatorname{Min}(x^2,2x)=5x-3.$

(iii) Solve the equation

$\operatorname{Min}(x^2,2x)+\operatorname{Min}(x^3,4x)=mx$

in the cases $m=2$ and $m=6$.

(iv) Show that $(1,-3)$ is a local maximum point on the curve

$y=2\operatorname{Min}(x^2,x^3)-5x$

and find the other three local maxima and minima on this curve. Sketch the curve.

Hint

First find the switch points where the two expressions inside each $\operatorname{Min}$ are equal. In part (iv), compare the quadratic and cubic pieces on either side of each switch point.

Model Solution

See the official worked solution in the STEP 2 2025 Worked Paper, Question 1.

Examiner Notes

This was a popular question. Common issues included missing range checks after solving piecewise equations, smoothing corners where the selected function changes, and not marking significant points clearly on the sketch.

---

Question 2

Topic: Complex numbers  |  Difficulty: Challenging  |  Marks: 20

Problem

2 (i) (a) Show that if the complex number $z$ satisfies the equation

$z^2+|z+b|=a,$

where $a$ and $b$ are real numbers, then $z$ must be either purely real or purely imaginary.

(b) Show that the equation

$z^2+\left|z+\frac52\right|=\frac72$

has no purely imaginary roots.

(c) Show that the equation

$z^2+\left|z+\frac72\right|=\frac52$

has no purely real roots.

(d) Show that, when $\frac12<b<\frac34$, the equation

$z^2+|z+b|=\frac12$

will have at least one purely imaginary root and at least one purely real root.

(ii) Solve the equation

$z^3+|z+2|^2=4.$

Hint

Write $z=x+iy$ and compare imaginary parts. The first part tells you exactly which cases need to be checked.

Model Solution

See the official worked solution in the STEP 2 2025 Worked Paper, Question 2.

Examiner Notes

Many candidates handled part (i)(a) well. The main errors were missing one of the real-case subcases, stopping too early after finding a positive discriminant, and not explaining why all solutions had been found in part (ii).

---

Question 3

Topic: Logarithmic comparison  |  Difficulty: Challenging  |  Marks: 20

Problem

3 (i) Sketch a graph of

$y=\frac{\ln x}{x}\qquad (x>0).$

(ii) Use your graph to show the following.

(a)

$3^\pi>\pi^3.$

(b)

$\left(\frac94\right)^{\sqrt5}>(\sqrt5)^{9/4}.$

(iii) Given that $1<x<2$, decide, with justification, which is the larger of $x^{x+2}$ or $(x+2)^x$.

(iv) Show that the inequalities

$9^{\sqrt2}>(\sqrt2)^9$

and

$3^{2\sqrt2}>(2\sqrt2)^3$

are equivalent. Given that $e^2<8$, decide, with justification, which is the larger of $9^{\sqrt2}$ and $(\sqrt2)^9$.

(v) Decide, with justification, which is the larger of

$8^{\sqrt[3]{3}}\quad\text{and}\quad(\sqrt[3]{3})^8.$

Hint

For positive bases, compare $a^b$ and $b^a$ by comparing $\ln a/a$ and $\ln b/b$. The turning point of $\ln x/x$ is at $x=e$.

Model Solution

See the official worked solution in the STEP 2 2025 Worked Paper, Question 3.

Examiner Notes

The graph in part (i) was usually done well. Later parts rewarded careful direction of implication: candidates needed to show how the ordering on the graph implies the required exponential inequality.

---

Question 4

Topic: Floor functions  |  Difficulty: Challenging  |  Marks: 20

Problem

4 Let $\lfloor x\rfloor$ denote the largest integer that satisfies $\lfloor x\rfloor\leq x$. For example, if $x=-4.2$, then $\lfloor x\rfloor=-5$.

(i) Show that, if $n$ is an integer, then

$\lfloor x+n\rfloor=\lfloor x\rfloor+n.$

(ii) Let $n$ be a positive integer and define the function $f_n$ by

$$f_n(x)=\lfloor x\rfloor+\left\lfloor x+\frac1n\right\rfloor +\left\lfloor x+\frac2n\right\rfloor+\cdots +\left\lfloor x+\frac{n-1}{n}\right\rfloor-\lfloor nx\rfloor.$$

(a) Show that

$f_n\left(x+\frac1n\right)=f_n(x).$

(b) Evaluate $f_n(t)$ for $0\leq t<\frac1n$.

(c) Hence show that $f_n(x)\equiv0$.

(iii) (a) Show that

$\left\lfloor\frac{x}{2}\right\rfloor+\left\lfloor\frac{x+1}{2}\right\rfloor=\lfloor x\rfloor.$

(b) Hence, or otherwise, simplify

$$\left\lfloor\frac{x+1}{2}\right\rfloor+ \left\lfloor\frac{x+2}{2^2}\right\rfloor+\cdots+ \left\lfloor\frac{x+2^k}{2^{k+1}}\right\rfloor+\cdots .$$

Hint

For part (ii), use the period found in (a) to reduce every input to $0\leq t<1/n$. For part (iii), look for a telescoping pattern after applying part (a) with $n=2$.

Model Solution

See the official worked solution in the STEP 2 2025 Worked Paper, Question 4.

Examiner Notes

The floor notation was the main source of difficulty. Strong solutions used the periodicity in part (ii) cleanly and avoided treating floor brackets as ordinary linear brackets.

---

Question 5

Topic: Improper integrals  |  Difficulty: Standard  |  Marks: 20

Problem

5 You need not consider the convergence of the improper integrals in this question.

(i) Use the substitution $x=u^{-1}$ to show that

$$\int_0^\infty \frac{\sqrt{x}-1}{\sqrt{x(x^3+1)}}\,dx=0.$$

(ii) Use the substitution $x=u^{-2}$ to show that

$$\int_0^\infty \frac{1}{\sqrt{x^3+1}}\,dx =2\int_0^\infty \frac{1}{\sqrt{x^6+1}}\,dx.$$

(iii) Find, in terms of $p$ and $s$, a value of $r$ for which

$$\int_0^\infty \frac{x^r-1}{\sqrt{x^s(x^p+1)}}\,dx=0,$$

given that $p$ and $s$ are fixed values for which the required integrals converge.

(iv) Show that, for any positive value of $k$, it is possible to find values of $p$ and $q$ for which

$$\int_0^\infty \frac{1}{\sqrt{x^p+1}}\,dx =k\int_0^\infty \frac{1}{\sqrt{x^q+1}}\,dx.$$

Hint

Track both the reversed limits and the power of $u$ introduced by $dx/du$. In part (iv), try $x=u^{-k}$ and match the resulting denominator to the required form.

Model Solution

See the official worked solution in the STEP 2 2025 Worked Paper, Question 5.

Examiner Notes

The first two substitutions were generally successful. The more common later errors were index mistakes and insufficient explanation that suitable parameters really exist.

---

Question 6

Topic: Coordinate geometry  |  Difficulty: Hard  |  Marks: 20

Problem

6 (i) The circle

$x^2+(y-a)^2=r^2$

touches the parabola $2ky=x^2$, where $k>0$, tangentially at two points. Show that

$r^2=k(2a-k).$

Show further that if $r^2=k(2a-k)$ and $a>k>0$, then the circle $x^2+(y-a)^2=r^2$ touches the parabola $2ky=x^2$ tangentially at two points.

(ii) The lines $y=c\pm x$ are tangents to the circle $x^2+(y-a)^2=r^2$. Find $r^2$, and the coordinates of the points of contact, in terms of $a$ and $c$.

(iii) $C_1$ and $C_2$ are circles with equations

$x^2+(y-a_1)^2=r_1^2$

and

$x^2+(y-a_2)^2=r_2^2$

respectively, where $a_1\ne a_2$ and $r_1\ne r_2$. Each circle touches the parabola $2ky=x^2$ tangentially at two points and the lines $y=c\pm x$ are tangents to both circles.

(a) Show that

$a_1+a_2=2c+4k$

and

$a_1^2+a_2^2=2c^2+16kc+12k^2.$

(b) The circle $x^2+(y-d)^2=p^2$ passes through the four points of tangency of the lines $y=c\pm x$ to the two circles $C_1$ and $C_2$. Find $d$ and $p^2$ in terms of $k$ and $c$.

(c) Show that the circle $x^2+(y-d)^2=p^2$ also touches the parabola $2ky=x^2$ tangentially at two points.

Hint

Use the parabola equation to substitute for $y$ at a tangency point, then force the resulting equation in $x$ to have a repeated root. For the line tangencies, the perpendicular distance from the centre to the line is useful.

Model Solution

See the official worked solution in the STEP 2 2025 Worked Paper, Question 6.

Examiner Notes

This was one of the more demanding pure questions. Candidates who kept the tangency conditions algebraic and avoided unnecessary geometric assumptions made the best progress.

---

Question 7

Topic: Differential equations  |  Difficulty: Challenging  |  Marks: 20

Problem

7 The differential equation

$$\frac{d^2x}{dt^2}=2x\frac{dx}{dt}$$

describes the motion of a particle with position $x(t)$ at time $t$. At $t=0$, $x=a$, where $a>0$.

(i) Solve the differential equation in the case where

$\frac{dx}{dt}=a^2$

when $t=0$. What happens to the particle as $t$ increases from $0$?

(ii) Solve the differential equation in the case where

$\frac{dx}{dt}=a^2+p^2$

when $t=0$, where $p>0$. What happens to the particle as $t$ increases from $0$?

(iii) Solve the differential equation in the case where

$\frac{dx}{dt}=a^2-q^2$

when $t=0$, where $q>0$. What happens to the particle as $t$ increases from $0$? Give conditions on $a$ and $q$ for the different cases which arise.

Hint

Let $v=dx/dt$ and use $d^2x/dt^2=(dv/dx)(dx/dt)$ where appropriate. This reduces the differential equation to $dv/dx=2x$.

Model Solution

See the official worked solution in the STEP 2 2025 Worked Paper, Question 7.

Examiner Notes

Successful solutions reduced the order before separating variables. The qualitative final comments needed to distinguish the different signs in the three initial-velocity cases.

---

Question 8

Topic: Balanced partitions  |  Difficulty: Hard  |  Marks: 20

Problem

8 If we split a set $S$ of integers into two subsets $A$ and $B$ whose intersection is empty and whose union is the whole of $S$, and such that

• the sum of the elements of $A$ is equal to the sum of the elements of $B$;
• the sum of the squares of the elements of $A$ is equal to the sum of the squares of the elements of $B$,

then we say that we have found a balanced partition of $S$ into two subsets.

(i) Find a balanced partition of the set $\{1,2,3,4,5,6,7,8\}$ into two subsets $A$ and $B$, each of size 4.

(ii) Given that $a_1,a_2,\ldots,a_m$ and $b_1,b_2,\ldots,b_m$ are sequences with

$$\sum_{k=1}^m a_k=\sum_{k=1}^m b_k \qquad\text{and}\qquad \sum_{k=1}^m a_k^2=\sum_{k=1}^m b_k^2,$$

show that

$$\sum_{k=1}^m a_k^3+\sum_{k=1}^m(c+b_k)^3 =\sum_{k=1}^m b_k^3+\sum_{k=1}^m(c+a_k)^3$$

for any real number $c$.

(iii) Find, with justification, a balanced partition of the set $\{1,2,3,\ldots,16\}$ into two subsets $A$ and $B$, each of size 8, which also has the property that the sum of the cubes of the elements of $A$ is equal to the sum of the cubes of the elements of $B$.

(iv) You are given that the sets $A=\{1,3,4,5,9,11\}$ and $B=\{2,6,7,8,10\}$ form a balanced partition of the set $\{1,2,3,\ldots,11\}$.

Let

$S=\{n^2,(n+1)^2,(n+2)^2,\ldots,(n+11)^2\},$

where $n$ is any positive integer. Find, with justification, two subsets $C$ and $D$ of $S$ whose intersection is empty and whose union is the whole of $S$, and such that the sum of the elements of $C$ is equal to the sum of the elements of $D$.

Hint

Part (ii) is the engine of the question: expand the cubes and use the equal-sum and equal-square assumptions. Then use shifted copies of an earlier balanced partition.

Model Solution

See the official worked solution in the STEP 2 2025 Worked Paper, Question 8.

Examiner Notes

The proof in part (ii) was the key gateway. Candidates who used the provided balanced partition structurally, rather than searching blindly, were better placed for the construction questions.