STEP3 2025 -- Pure Mathematics

Exam: STEP3  |  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	Beta-type integrals	Standard	Substitution, symmetry, trigonometric forms
2	Iteration and periodicity	Challenging	Piecewise linear maps, function iteration, cycles
3	Means and comparison	Challenging	Secants, monotonic ratios, AM-GM-HM
4	Transformations and implicit curves	Hard	Rotation matrices, reflection, implicit differentiation
5	Vectors in 3D geometry	Challenging	Vector parametrisation, planes, barycentric coordinates
6	Complex numbers	Hard	Symmetric identities, moduli, roots of polynomials
7	Inverse trigonometric functions	Challenging	Asymptotics, graph sketching, inverse tan identities
8	De Moivre’s theorem	Challenging	Induction, trigonometric sums, identities

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Question 1

Topic: Beta-type integrals  |  Difficulty: Standard  |  Marks: 20

Problem

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

For $p,q>0$, define

$$b(p,q)=\int_0^1 x^{p-1}(1-x)^{q-1}\,dx.$$

(i) Show that $b(p,q)=b(q,p)$.

(ii) Show that

$b(p+1,q)=b(p,q)-b(p,q+1)$

and hence that

$b(p+1,p)=\frac12 b(p,p).$

(iii) Show that

$$b(p,q)=2\int_0^{\pi/2}(\sin\theta)^{2p-1}(\cos\theta)^{2q-1}\,d\theta.$$

Hence show that

$$b(p,p)=\frac{1}{2^{2p-1}}b\left(p,\frac12\right).$$

(iv) Show that

$$b(p,q)=\int_0^\infty \frac{t^{p-1}}{(1+t)^{p+q}}\,dt.$$

(v) Evaluate

$$\int_0^\infty \frac{t^{3/2}}{(1+t)^6}\,dt.$$

Hint

Use $x=1-u$ for symmetry, $x=\sin^2\theta$ for the trigonometric form, and $x=1/(1+t)$ or an equivalent fractional substitution for the improper integral form.

Model Solution

See the official worked solution in the STEP 3 2025 Worked Paper, Question 1.

Examiner Notes

This was the most popular STEP 3 question. Most candidates handled the early substitutions well, but full credit required careful justification when changing limits and reusing the earlier beta-function identities.

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Question 2

Topic: Iteration and periodicity  |  Difficulty: Challenging  |  Marks: 20

Problem

2 Let

$f(x)=7-2|x|.$

A sequence $u_0,u_1,u_2,\ldots$ is defined by $u_0=a$ and

$u_n=f(u_{n-1})\qquad(n>0).$

(i) (a) Sketch, on the same axes, the graphs with equations $y=f(x)$ and $y=f(f(x))$.

(b) Find all solutions of the equation

$f(f(x))=x.$

(c) Find the values of $a$ for which the sequence $u_0,u_1,u_2,\ldots$ has period 2.

(d) Show that, if $a=\frac{28}{5}$, then the sequence $u_2,u_3,u_4,\ldots$ has period 2, but neither $u_0$ nor $u_1$ is equal to either of $u_2$ or $u_3$.

(ii) (a) Sketch, on the same axes, the graphs with equations $y=f(x)$ and $y=f(f(f(x)))$.

(b) Consider the sequence $u_0,u_1,u_2,\ldots$ in the cases $a=1$ and $a=-\frac79$. Hence find all the solutions of the equation

$f(f(f(x)))=x.$

(c) Find a value of $a$ such that the sequence $u_3,u_4,u_5,\ldots$ has period 3, but where none of $u_0,u_1$ or $u_2$ is equal to any of $u_3,u_4$ or $u_5$.

Hint

Write the iterates piecewise and keep track of which interval the input lies in after each application of $f$. Periodic sequences can include preperiodic starting values.

Model Solution

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

Examiner Notes

Graph accuracy was important: small piecewise errors in $f(f(x))$ or $f(f(f(x)))$ changed the period analysis. The later parts tested the distinction between a periodic tail and an initially periodic sequence.

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Question 3

Topic: Means and comparison  |  Difficulty: Challenging  |  Marks: 20

Problem

3 Let $f(x)$ be defined and positive for $x>0$. Let $a$ and $b$ be real numbers with $0<a<b$ and define the points

$A=(a,f(a)),\qquad B=(b,-f(b)).$

Let $X=(m,0)$ be the point of intersection of line $AB$ with the $x$-axis.

(i) Find an expression for $m$ in terms of $a,b,f(a)$ and $f(b)$.

(ii) Show that, if $f(x)=\sqrt{x}$, then

$m=\sqrt{ab}.$

Find, in terms of $n$, a function $f(x)$ such that

$$m=\frac{a^{n+1}+b^{n+1}}{a^n+b^n}.$$

(iii) Let $g_1(x)$ and $g_2(x)$ be defined and positive for $x>0$. Let $m=M_1$ when $f(x)=g_1(x)$ and let $m=M_2$ when $f(x)=g_2(x)$.

Show that if

$\frac{g_1(x)}{g_2(x)}$

is a decreasing function then $M_1>M_2$.

Hence show that

$$\frac{a+b}{2}>\sqrt{ab}>\frac{2ab}{a+b}.$$

(iv) Let $p$ and $c$ be chosen so that the curve

$y=p(c-x)^3$

passes through both $A$ and $B$. Show that

$$\frac{c-a}{b-c}=\left(\frac{f(a)}{f(b)}\right)^{1/3}$$

and hence determine $c$ in terms of $a,b,f(a)$ and $f(b)$.

Show that if $f$ is a decreasing function, then $c<m$.

Hint

For the secant intersection, start with the two-point equation of the line. In the comparison part, compare two choices of $f$ through the ratio $g_1/g_2$ rather than through the final formula alone.

Model Solution

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

Examiner Notes

Many candidates could derive $m$, but the inequality part required careful use of monotonicity. The final comparison $c<m$ was sensitive to the direction of the decreasing-function argument.

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Question 4

Topic: Transformations and implicit curves  |  Difficulty: Hard  |  Marks: 20

Problem

4 (i) $x_2$ and $y_2$ are defined in terms of $x_1$ and $y_1$ by the equation

$$\begin{pmatrix}x_2\\y_2\end{pmatrix} = \begin{pmatrix} \frac1{\sqrt2} & -\frac1{\sqrt2}\\ \frac1{\sqrt2} & \frac1{\sqrt2} \end{pmatrix} \begin{pmatrix}x_1\\y_1\end{pmatrix}.$$

$G_1$ is the graph with equation

$$\frac{x^2}{9}+\frac{y^2}{4}=1$$

and $G_2$ is the graph with equation

$$\frac{\left(\frac{x}{\sqrt2}+\frac{y}{\sqrt2}\right)^2}{9} +\frac{\left(-\frac{x}{\sqrt2}+\frac{y}{\sqrt2}\right)^2}{4}=1.$$

Show that, if $(x_1,y_1)$ is a point on $G_1$, then $(x_2,y_2)$ is a point on $G_2$.

Show that $G_2$ is an anti-clockwise rotation of $G_1$ through $45^\circ$ about the origin.

(ii) (a) The matrix

$$\begin{pmatrix} -0.6 & 0.8\\ 0.8 & 0.6 \end{pmatrix}$$

represents a reflection. Find the line of invariant points of this matrix.

(b) Sketch, on the same axes, the graphs with equations

$y=2x$

and

$0.8x+0.6y=2^{-0.6x+0.8y}.$

(iii) Sketch, on the same axes, for $0\leq x\leq 2\pi$, the graphs with equations

$y=\sin x$

and

$y=\sin(x-2y).$

You should determine the exact coordinates of the points on the graph with equation $y=\sin(x-2y)$ where the tangent is horizontal and those where it is vertical.

Hint

Part (i) is a coordinate transformation: substitute the inverse rotated coordinates into $G_1$. For part (iii), differentiate implicitly and find where $dy/dx$ is $0$ or undefined.

Model Solution

See the official worked solution in the STEP 3 2025 Worked Paper, Question 4.

Examiner Notes

This was one of the less popular pure questions. The exam report highlighted insufficient detail in the transformation proof and a need for care with exact coordinates in the implicit sketch.

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Question 5

Topic: Vectors in 3D geometry  |  Difficulty: Challenging  |  Marks: 20

Problem

5 Three points, $A,B$ and $C$, lie in a horizontal plane, but are not collinear. The point $O$ lies above the plane.

Let

$$\overrightarrow{OA}=\mathbf a,\qquad \overrightarrow{OB}=\mathbf b,\qquad \overrightarrow{OC}=\mathbf c.$$

$P$ is a point with

$$\overrightarrow{OP}=\alpha\mathbf a+\beta\mathbf b+\gamma\mathbf c,$$

where $\alpha,\beta$ and $\gamma$ are all positive and $\alpha+\beta+\gamma<1$.

Let

$k=1-(\alpha+\beta+\gamma).$

(i) The point $L$ is on $OA$, the point $X$ is on $BC$ and $LX$ passes through $P$. Determine $\overrightarrow{OX}$ in terms of $\beta,\gamma,\mathbf b$ and $\mathbf c$ and show that

$$\overrightarrow{OL}=\frac{\alpha}{k+\alpha}\mathbf a.$$

(ii) Let $M$ and $Y$ be the unique pair of points on $OB$ and $CA$ respectively such that $MY$ passes through $P$, and let $N$ and $Z$ be the unique pair of points on $OC$ and $AB$ respectively such that $NZ$ passes through $P$.

Show that the plane $LMN$ is also horizontal if and only if $OP$ intersects plane $ABC$ at the point $G$, where

$$\overrightarrow{OG}=\frac13(\mathbf a+\mathbf b+\mathbf c).$$

Where do points $X,Y$ and $Z$ lie in this case?

(iii) State what the condition $\alpha+\beta+\gamma<1$ tells you about the position of $P$ relative to the tetrahedron $OABC$.

Hint

Parametrise the point $X$ on $BC$ and the point $L$ on $OA$, then equate coefficients in the vector equation for $P$. Symmetry gives the corresponding formulae for $M,N,Y,Z$.

Model Solution

See the official worked solution in the STEP 3 2025 Worked Paper, Question 5.

Examiner Notes

This question rewarded clean vector parametrisation. The if-and-only-if argument needed both directions, with the positivity condition ensuring the relevant point lies inside the tetrahedron.

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Question 6

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

Problem

6 (i) Let $a,b$ and $c$ be three non-zero complex numbers with the properties

$a+b+c=0$

and

$a^2+b^2+c^2=0.$

Show that $a,b$ and $c$ cannot all be real. Show further that $a,b$ and $c$ all have the same modulus.

(ii) Show that it is not possible to find three non-zero complex numbers $a,b$ and $c$ with the properties

$a+b+c=0$

and

$a^3+b^3+c^3=0.$

(iii) Show that if any four non-zero complex numbers $a,b,c$ and $d$ have the properties

$a+b+c+d=0$

and

$a^3+b^3+c^3+d^3=0,$

then at least two of them must have the same modulus.

(iv) Show, by taking $c=1,d=-2$ and $e=3$, that it is possible to find five real numbers $a,b,c,d$ and $e$ with distinct magnitudes and with the properties

$a+b+c+d+e=0$

and

$a^3+b^3+c^3+d^3+e^3=0.$

Hint

Keep the algebra symmetric. For three variables, use $a=-(b+c)$ and compare with the sum of squares or cubes; for four variables, think of the numbers as roots of a polynomial.

Model Solution

See the official worked solution in the STEP 3 2025 Worked Paper, Question 6.

Examiner Notes

This question was challenging. Splitting every complex number into real and imaginary parts tended to create too many equations; symmetric identities and polynomial-root arguments were more effective.

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Question 7

Topic: Inverse trigonometric functions  |  Difficulty: Challenging  |  Marks: 20

Problem

7 Let

$$f(x)=\sqrt{x^2+1}-x.$$

(i) Using a binomial series, or otherwise, show that, for large $|x|$,

$$\sqrt{x^2+1}\sim |x|+\frac{1}{2|x|}.$$

Sketch the graph $y=f(x)$.

(ii) Let

$g(x)=\tan^{-1}f(x)$

and, for $x\ne0$, let

$$k(x)=\frac12\tan^{-1}\left(\frac1x\right).$$

(a) Show that

$g(x)+g(-x)=\frac{\pi}{2}.$

(b) Show that

$k(x)+k(-x)=0.$

(c) Show that

$\tan k(x)=\tan g(x)$

for $x>0$.

(d) Sketch the graphs $y=g(x)$ and $y=k(x)$ on the same axes.

(e) Evaluate

$$\int_0^1 k(x)\,dx$$

and hence write down the value of

$$\int_{-1}^0 g(x)\,dx.$$

Hint

Rationalise $f(x)$ to understand its behaviour for large positive and negative $x$. For part (c), use the double-angle formula for tangent.

Model Solution

See the official worked solution in the STEP 3 2025 Worked Paper, Question 7.

Examiner Notes

The inverse-trigonometric identities required attention to domains. The sketch in part (d) depended on combining symmetry from earlier parts with the positive-$x$ equality.

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Question 8

Topic: De Moivre’s theorem  |  Difficulty: Challenging  |  Marks: 20

Problem

8 (i) Show that

$$z^{m+1}-\frac{1}{z^{m+1}} =\left(z-\frac1z\right)\left(z^m+\frac1{z^m}\right) +\left(z^{m-1}-\frac1{z^{m-1}}\right).$$

Hence prove by induction that, for $n\geq1$,

$$z^{2n}-\frac{1}{z^{2n}} =\left(z-\frac1z\right) \sum_{r=1}^n\left(z^{2r-1}+\frac1{z^{2r-1}}\right).$$

Find similarly

$z^{2n}-\frac1{z^{2n}}$

as a product of $\left(z+\frac1z\right)$ and a sum.

(ii) (a) By choosing $z=e^{i\theta}$, show that

$$\sin 2n\theta =2\sin\theta\sum_{r=1}^n\cos(2r-1)\theta.$$

(b) Use this result, with $n=2$, to show that

$$\cos\frac{2\pi}{5}=\cos\frac{\pi}{5}-\frac12.$$

(c) Use this result, with $n=7$, to show that

$$\cos\frac{2\pi}{15}+\cos\frac{4\pi}{15}+\cos\frac{8\pi}{15}+\cos\frac{16\pi}{15} =\frac12.$$

(iii) Show that

$$\sin\frac{\pi}{14}-\sin\frac{3\pi}{14}+\sin\frac{5\pi}{14}=\frac12.$$

Hint

The first recurrence peels off the highest odd power. For the similar formula with $z+1/z$, the alternating signs in the sum matter.

Model Solution

See the official worked solution in the STEP 3 2025 Worked Paper, Question 8.

Examiner Notes

The induction was often clear, but the alternating-sign summation in the similar identity was frequently mishandled. Later parts needed explicit trigonometric identities rather than several compressed manipulation steps.