past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2026/ib/paper-2.bigb

= Paper 2
{scope}

https://www.maths.cam.ac.uk/undergrad/pastpapers/files/2026/Paperib_2_2026.pdf

= 1E
{parent=Paper 2}
{scope}
{title2=Groups, Rings and Modules}

= a
{parent=1e}
{scope}

= Solution
{parent=a}

Use the norm $N(a+b\sqrt{-n})=a^2+nb^2$. If $n$ is composite, compare $n=(\sqrt{-n})(-\sqrt{-n})$ with an integer factorisation; if $n$ is prime, $n$ is odd and $n+1=(1+\sqrt{-n})(1-\sqrt{-n})$ has a nontrivial integer factorisation. Norm considerations show the relevant factors are irreducible and nonassociate, so unique factorisation fails.

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= b
{parent=1e}
{scope}

= Solution
{parent=b}

At the maximal <ideal> $\mathfrak m=(2,X)$, the images of $4,2X,X^2$ form a three-dimensional <basis> of $I/\mathfrak mI$ over $\mathbb F_2$, so two generators are impossible. More generally $(2^{n-1},2^{n-2}X,\ldots,2X^{n-2},X^{n-1})$ needs exactly $n$ generators by the same argument.

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= 2F
{parent=Paper 2}
{scope}
{title2=Analysis II}

= Solution
{parent=2f}

Differentiability at $p$ means $f(p+h)=f(p)+Lh+o(|h|)$ for a <linear map> $L=Df(p)$; <partial derivatives> are its values on coordinate <vectors>. Integrating the <partial derivatives> along the two coordinate segments and using continuity at $p$ proves the stated sufficient condition. Here $f(x,y)=g(x)+g(y)$ with $g(t)=t^2\sin(1/t)$ and $g(0)=0$. Since $g\prime(0)=0$, $f$ is <differentiable> everywhere. But $g\prime(t)=2t\sin(1/t)-\cos(1/t)$ does not approach $0$ at zero, so $f$ is $C^1$ exactly where $xy\ne0$.

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= 3D
{parent=Paper 2}
{scope}
{title2=Methods}

= i
{parent=3d}
{scope}

= Solution
{parent=i}

Square the Fourier <series>, integrate, and use orthogonality; cross terms vanish. This gives $\int_{-1}^1|f|^2=a_0^2/2+\sum_{n\ge1}(a_n^2+b_n^2)$.

Solved by gpt-5.6-sol high.

= ii
{parent=3d}
{scope}

= Solution
{parent=ii}

For $f=x^2$, $\int_{-1}^1x^4dx=2/5$, $a_0=2/3$, and $a_n=4(-1)^n/(\pi^2n^2)$. Parseval yields $2/5=2/9+16\pi^{-4}\sum n^{-4}$, hence $\sum n^{-4}=\pi^4/90$.

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= 4D
{parent=Paper 2}
{scope}
{title2=Electromagnetism}

= Solution
{parent=4d}

For $z\gt 0$, $\phi=q(4\pi\varepsilon_0)^{-1}(|x-d\hat z|^{-1}-|x+d\hat z|^{-1})$ and $E=-\nabla\phi$. The normal field just above the plane is $-qd/[2\pi\varepsilon_0(r^2+d^2)^{3/2}]$, so $\sigma=\varepsilon_0E_z$ is the stated expression. Integrating $2\pi r\sigma(r)dr$ from zero to infinity gives total induced charge $-q$.

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= 5A
{parent=Paper 2}
{scope}
{title2=Fluid Dynamics}

= i
{parent=5a}
{scope}

= Solution
{parent=i}

With $u=\psi_y$, $v=-\psi_x$, one may take $\psi=(\alpha+t)y^2/2-x$.

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= ii
{parent=5a}
{scope}

= Solution
{parent=ii}

At fixed $t$, <streamlines> satisfy $(\alpha+t)y^2/2-x=C$: right-opening parabolas, more tightly curved at $t=1$ than at $t=0$.

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= iii
{parent=5a}
{scope}

= Solution
{parent=iii}

The relative coefficient change is $t/\alpha$, so for $t\ll\alpha$ the flow is approximately steady and its <streamlines> are $\alpha y^2/2-x=C$.

Solved by gpt-5.6-sol high.

= 6H
{parent=Paper 2}
{scope}
{title2=Statistics}

= Solution
{parent=6h}

The MLE is $\hat\mu=\bar X$ with MSE $1/n$. Combining likelihood and the $N(0,\tau^{-2})$ prior gives posterior $N(n\bar X/(n+\tau^2),1/(n+\tau^2))$. The frequentist MSE of its mean is $[n+\tau^4\mu^2]/(n+\tau^2)^2$, which is below $1/n$ exactly when $\mu^2\lt 2\tau^{-2}+1/n$.

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= 7H
{parent=Paper 2}
{scope}
{title2=Optimisation}

= Solution
{parent=7h}

The maximum-flow problem maximises source-to-sink flow subject to capacity and conservation constraints. Ford–Fulkerson repeatedly finds an augmenting path in the residual network and augments by its bottleneck capacity, stopping when none exists. Multiplying rational capacities by a common denominator makes them integers; every augmentation then raises the flow by at least one while the value is bounded by the finite source capacity, so termination is finite.

Solved by gpt-5.6-sol high.

= 8E
{parent=Paper 2}
{scope}
{title2=Linear Algebra}

= a
{parent=8e}
{scope}

= Solution
{parent=a}

If $p=m_{\alpha_W}$ and $q=m_{\alpha_{V/W}}$, then $m_\alpha$ is divisible by both. Also $q(\alpha)V\subseteq W$, so $p(\alpha)q(\alpha)=0$; hence $\operatorname{lcm}(p,q)\mid m_\alpha\mid pq$. A single size-two nilpotent Jordan block with its invariant eigenline makes the lcm too small; a diagonal operator with the same <eigenvalue> on $W$ and $V/W$ makes the product too large. Square-free $m_\alpha$ implies both induced minimal <polynomials> square-free, so diagonalizability descends. The nilpotent-block example disproves the converse.

Solved by gpt-5.6-sol high.

= b
{parent=8e}
{scope}

= Solution
{parent=b}

Each eigenspace of $\alpha_1$ is stable under every commuting $\alpha_i$. Restrict the remaining diagonalizable maps to each eigenspace and induct; concatenating their simultaneous eigenbases proves the result.

Solved by gpt-5.6-sol high.

= c
{parent=8e}
{scope}

= Solution
{parent=c}

Yes. In finite dimension, construct a common-eigenspace decomposition by repeatedly refining with members of the family. Every proper refinement increases the finite number of summands, so finitely many maps already produce a decomposition stable under all remaining maps; diagonalize their restrictions as above.

Solved by gpt-5.6-sol high.

= 9E
{parent=Paper 2}
{scope}
{title2=Groups, Rings and Modules}

= a
{parent=9e}
{scope}

= Solution
{parent=a}

Sylow I gives a <subgroup> of order equal to the full $p$-part of $|G|$. Sylow II says every $p$-subgroup lies in a conjugate of any Sylow $p$-subgroup; act by the <subgroup> on the Sylow cosets and use a fixed orbit because the number of cosets is prime to $p$.

Solved by gpt-5.6-sol high.

= b
{parent=9e}
{scope}

= Solution
{parent=b}

Let an order-$p^{d-1}$ <subgroup> $H$ act on $G/H$. Since the number of cosets is divisible by $p$, orbit counting and Cauchy produce a fixed coset with stabiliser extending $H$ by index $p$. Induction yields <subgroups> of every $p$-power order. Applying the same orbit count to the action of $H$ on $G/H$ leaves fixed cosets indexed by $N(H)/H$, giving $|G:H|\equiv|N(H):H|\pmod p$.

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= c
{parent=9e}
{scope}

= Solution
{parent=c}

No. In an abelian <group> conjugacy is equality; for example distinct order-$p$ <subgroups> of $C_p\times C_p\times C_p$ are non-maximal and not conjugate.

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= 10F
{parent=Paper 2}
{scope}
{title2=Analysis II}

= Solution
{parent=10f}

The inverse <function> theorem says that a $C^1$ map with invertible <derivative> at $x$ restricts to a $C^1$ diffeomorphism between neighbourhoods of $x$ and $f(x)$. It makes every zero isolated. Compactness of $\bar\Omega$, together with separation from zero on the boundary, then makes the zero set finite and interior. Continuity of $Df$ gives the displayed radii. Choose disjoint balls around the zeros and a positive lower bound for $|f|$ on their complement and boundary. A sufficiently small $C^1$ perturbation has no zeros outside, while the quoted local lemma gives exactly one zero in each ball with the same <determinant> sign. Summing proves $N(g)=N(f)$.

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= 11F
{parent=Paper 2}
{scope}
{title2=Topological Spaces}

= a
{parent=11f}
{scope}

= Solution
{parent=a}

Connected means having no separation into two nonempty disjoint open subsets. Images of intervals under $f$ are intervals. Thus for $a\lt b\lt c$, injectivity forces $f(b)$ to lie between $f(a)$ and $f(c)$, so the quoted criterion makes $f$ monotone. A monotone bijection $\mathbb R\to\mathbb R$ has no jumps, since a jump would omit an interval; hence it is continuous.

Solved by gpt-5.6-sol high.

= b
{parent=11f}
{scope}

= Solution
{parent=b}

Path-connected means every pair is joined by a continuous map from $[0,1]$; a separation would pull back to a separation of an interval, so it implies connectedness. One valid boundary pair follows the bottom then right edges and the left then top edges. If opposite-corner paths $\alpha,\beta$ were disjoint, normalising a continuous separation <vector> would construct the forbidden map $Q\to S^1$, so they intersect. A path in $Q\setminus A$ between the other corners, together with a path in $A$ between its corners, would contradict that intersection result; hence those corners lie in different path components.

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= 12
{parent=Paper 2}
{scope}
{title2=Complex Analysis OR Complex Methods}

= 12.1G
{parent=12}
{scope}

= Solution
{parent=12.1g}

Cauchy’s formula is $f(w)=(2\pi i)^{-1}\int_\gamma f(z)/(z-w)\,dz$, proved by removing the removable singularity of $[f(z)-f(w)]/(z-w)$ and applying Cauchy’s theorem; similarly $f\prime(w)=(2\pi i)^{-1}\int f(z)/(z-w)^2dz$. The resulting estimates prove Liouville’s theorem. Finally $\cos^2x=(1+\cos2x)/2$ and residues in the upper half-plane give $\int_{-\infty}^\infty\cos^2x/(x^2+1)dx=\frac\pi2(1+e^{-2})$.

Solved by gpt-5.6-sol high.

= 12.2C
{parent=12}
{scope}

= Solution
{parent=12.2c}

Choose $\sqrt{z^2-1}\sim z$ at infinity. On $-1\lt x\lt 1$ its upper/lower values are $\pm i\sqrt{1-x^2}$, divided by $x+c$. Under $z=1/\zeta$ the transformed integrand is the stated $g$; inversion reverses orientation, maps the slit exterior/interior accordingly, and encloses $0,-1/c$ negatively. Their residues give $J=-2\pi i(c-\sqrt{c^2-1})$. Comparing the two slit banks gives $J=-2iI$, hence $I=\pi(c-\sqrt{c^2-1})$.

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= 13B
{parent=Paper 2}
{scope}
{title2=Variational Principles}

= a
{parent=13b}
{scope}

= Solution
{parent=a}

With $V=\pi a^2l$ and $A=2\pi a^2+2\pi al$, eliminating $l$ and differentiating gives $a^3=V/(2\pi)$ and $A=3(2\pi)^{1/3}V^{2/3}$, so $B=3$. Under $a^2+l^2/4=R^2$, maximising gives $l/a=\sqrt5-1$, hence $A=2\pi\sqrt5,a^2$ and $C=5$.

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= b
{parent=13b}
{scope}

= Solution
{parent=b}

Substitute $y=-x-z$. The stationary equations are $-6x^2+2(x+z)=0$ and $2x+8z=0$, giving $(x,y,z)=(0,0,0)$ and $(1/4,-3/16,-1/16)$.

Solved by gpt-5.6-sol high.

= 14D
{parent=Paper 2}
{scope}
{title2=Methods}

= Solution
{parent=14d}

Separation gives angular modes $e^{im\theta}$ and $-[rR\prime]\prime+m^2R/r=\lambda rR$. Frobenius gives $p=m$ and $a_k=-a_{k-2}/[k(k+2m)]$, producing $J_m$; termwise <differentiation> gives the two stated recurrences. A radial solution is $u=\sum_n[A_n\cos(cj_{0n}t)+B_n\sin(cj_{0n}t)]J_0(j_{0n}r)$. Here $B_n=0$ and
$$A_n=\frac{2}{J_0\prime(j_{0n})^2}\int_0^1r(1-r)J_0(j_{0n}r)dr,$$
equivalently $A_n=2[j_{0n}J_0\prime(j_{0n})^2]^{-1}\int_0^1rJ_1(j_{0n}r)dr$ after integration by parts.

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= 15B
{parent=Paper 2}
{scope}
{title2=Quantum Mechanics}

= Solution
{parent=15b}

With the proton fixed, $[-\hbar^2\nabla^2/(2m_\mu)-e^2/(4\pi\varepsilon_0r)]\psi=E\psi$: <kinetic energy> plus Coulomb attraction, assuming nonrelativistic point particles and neglecting spin, radiation, and proton motion (or replacing $m_\mu$ by the reduced mass). Substitution of the $2s$ form gives $E=-\hbar^2\alpha^2/(8m_\mu)$ and $\alpha=m_\mu e^2/(4\pi\varepsilon_0\hbar^2)$. Direct <differentiation> verifies it. The curve starts positive, crosses at $r=2/\alpha$, then approaches zero from below; ordinary hydrogen has the same shape but the smaller electron mass gives a larger radial scale.

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= 16D
{parent=Paper 2}
{scope}
{title2=Electromagnetism}

= Solution
{parent=16d}

<Faraday's law> is $\mathcal E=\oint_C(E+v\times B)\cdot dl=-d\Phi_B/dt$. Here $I=-L(B\dot x+x\dot B)/R$ (sign relative to the chosen circuit orientation), and the bar force is $ILB$. For $B=B_0/(1+t)$ this gives the displayed equation with $\alpha=B_0^2L^2/(mR)$. At $\alpha=1$, the independent solutions are $u$ and $ue^{1/u}$, $u=1+t$; the data give $x=x_0ue^{1/u-1}$. Thus $\dot x\to x_0/e$ and total mechanical work is $mx_0^2/(2e^2)$. The current circulates to reinforce the decaying $+z$ flux, and its force drives increasing area, both opposing the flux decrease as <Lenz's law> requires.

Solved by gpt-5.6-sol high.

= 17C
{parent=Paper 2}
{scope}
{title2=Numerical Analysis}

= Solution
{parent=17c}

Odd terms vanish and direct evaluation gives $L(p)=(8/45)c_4$, so the rule is exact through degree $3$. With $n=3$, the Peano kernel is
$$K(t)=\frac16L[(x-t)_+^3]=\frac16\left[\frac{(1-t)^4}{4}-(-s-t)_+^3-(s-t)_+^3\right].$$
Its nonnegativity and $\int K=L(x^4)/4!=1/135$ give $|L(f)|\le\|f^{(4)}\|_\infty/135$, so $N=135$. Equality holds for a quartic with fourth <derivative> of constant sign.

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= 18H
{parent=Paper 2}
{scope}
{title2=Markov Chains}

= Solution
{parent=18h}

Invariant means $\pi P=\pi$. In a finite irreducible chain, $E_iT_i^+=1/\pi_i$ and the expected visits to $i$ between visits to $j$ are $\pi_i/\pi_j$. The first burglar has uniform invariant law on $64$ PINs, so (a) is $64$ and (b) is $16$. First-step equations by Hamming distance give the adjacent-state hitting time (c) as $63$. For the modified one-coordinate chain, destination weights are $1$ for digits $0$–$5$ and $2$ for $6,7$; its invariant weights are proportional to $w_a(10-w_a)$, namely $9$ and $16$. The product invariant law and the occupation formula give (d) $\pi_{(6,7)}/\pi_{(0,0)}=256/81$.

Solved by gpt-5.6-sol high.