past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2022/ib/paper-2.bigb

= Paper 2
{scope}

https://www.maths.cam.ac.uk/undergrad/pastpapers/files/2022/paperib_2_2022.pdf

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

= a
{parent=1e}
{scope}

= Solution
{parent=a}

The zero element is torsion. If $m,n\in T$, choose nonzero $r,s\in R$ with $rm=0$ and $sn=0$. Since $R$ is an <integral domain>, $rs\ne0$, and
$$
rs(m+n)=s(rm)+r(sn)=0.
$$
Thus $m+n\in T$. If $a\in R$, then $r(am)=a(rm)=0$, so $am\in T$. Hence $T$ is closed under addition and scalar multiplication and is an <R-module>[submodule] of $M$.

Solved by gpt-5.6-sol high.

= b
{parent=1e}
{scope}

= Solution
{parent=b}

If $m\in T_1$, then $rm=0$ for some nonzero $r$, and
$$
r\phi(m)=\phi(rm)=0.
$$
Thus $\phi(T_1)\subseteq T_2$. The formula
$$
\boxed{\Phi(m+T_1)=\phi(m)+T_2}
$$
is consequently independent of the representative and defines an $R$-module homomorphism on the <quotient module>.

If $\phi$ is injective and $\Phi(m+T_1)=T_2$, then $\phi(m)$ is torsion, so $r\phi(m)=0$ for some nonzero $r$. Injectivity gives $rm=0$, hence $m\in T_1$. Therefore
$$
\boxed{\phi\text{ injective }\Longrightarrow\Phi\text{ injective}}.
$$

The converse fails. For $R=\mathbb Z$, take
$$
M_1=\mathbb Z\oplus\mathbb Z/2\mathbb Z,\qquad
M_2=\mathbb Z,
\qquad
\phi(n,\bar a)=n.
$$
Then $\phi$ is not injective, but quotienting $M_1$ by its torsion submodule leaves $\mathbb Z$, and $\Phi:\mathbb Z\to\mathbb Z$ is the identity.

Solved by gpt-5.6-sol high.

= 2G
{parent=Paper 2}
{scope}
{title2=Analysis and Topology}

= Solution
{parent=2G}

Because $f$ is a bijection and $e$ is a metric,
$$
d'(x,y)=e(f(x),f(y))
$$
is nonnegative, symmetric, obeys the triangle inequality, and vanishes exactly when $x=y$. Hence it is a metric. Its open sets are precisely the inverse images under $f$ of $e$-open sets in $N$. Since $f:(M,d)\to(N,e)$ is a <homeomorphism>, these are exactly the $d$-open sets. Thus $d'$ and $d$ are <equivalent metrics>.

Let
$$
h(x)=\frac12+\frac1\pi\arctan x,
$$
a homeomorphism $\mathbb R\to(0,1)$, and define
$$
\boxed{d'(x,y)=|h(x)-h(y)|}.
$$
This metric induces the standard topology on $\mathbb R$. However, the sequence $x_n=n$ is $d'$-Cauchy because $h(n)\to1$. If it converged to $x\in\mathbb R$, then continuity of $h$ would give $h(x)=1$, impossible. Therefore $(\mathbb R,d')$ is not complete.

Solved by gpt-5.6-sol high.

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

= Solution
{parent=3B}

The <method of characteristics> gives
$$
\frac{dx}{ds}=-y,\qquad
\frac{dy}{ds}=x,\qquad
\frac{du}{ds}=0.
$$
Along a characteristic,
$$
\frac d{ds}(x^2+y^2)=2x(-y)+2yx=0.
$$
Thus the characteristics are circles $x^2+y^2=C$, traversed counterclockwise, and $u$ is constant on each circle. Their intersections with the $x$-axis have $x^2=C$, where the boundary value is $f(C)$. Hence
$$
\boxed{u(x,y)=f(x^2+y^2)}.
$$
The characteristic sketch is the family of concentric circles centred at the origin.

Solved by gpt-5.6-sol high.

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

= Solution
{parent=4D}

The uniform charge density is
$$
\rho_q=\frac{3Q}{4\pi R^3}.
$$
Spherical symmetry and <Gauss's law> give
$$
\boxed{
\mathbf E(\mathbf r)=
\begin{cases}
\displaystyle
\frac{Q}{4\pi\varepsilon_0R^3}\,\mathbf r,
&|\mathbf r|<R,\\[6pt]
\displaystyle
\frac{Q}{4\pi\varepsilon_0}
\frac{\mathbf r}{|\mathbf r|^3},
&|\mathbf r|>R.
\end{cases}}
$$

Take the first centre as the origin, so the second centre is at $\mathbf d$. In the overlap, each point lies inside both spheres. Superposition gives
$$
\mathbf E_{\rm total}
=\frac{Q}{4\pi\varepsilon_0R^3}\mathbf r
-\frac{Q}{4\pi\varepsilon_0R^3}(\mathbf r-\mathbf d)
=\boxed{\frac{Q}{4\pi\varepsilon_0R^3}\mathbf d}.
$$
The field is therefore constant throughout the overlap region.

Solved by gpt-5.6-sol high.

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

= a
{parent=5c}
{scope}

= Solution
{parent=a}

The <divergence> is
$$
\nabla\cdot\mathbf u
=\frac{\partial(2t)}{\partial x}
+\frac{\partial(xt)}{\partial y}=0.
$$
Thus the flow is <incompressible flow>[incompressible].

Solved by gpt-5.6-sol high.

= b
{parent=5c}
{scope}

= Solution
{parent=b}

A particle released at the origin at time $s$ satisfies
$$
\dot x=2t,\qquad \dot y=xt,
\qquad x(s)=y(s)=0.
$$
Integration gives
$$
x(t)=t^2-s^2,
\qquad
y(t)=\frac{(t^2-s^2)^2}{4}.
$$
Its distance from the origin is therefore
$$
\boxed{
\sqrt{x^2+y^2}
=(t^2-s^2)
\sqrt{1+\frac{(t^2-s^2)^2}{16}}}.
$$

Solved by gpt-5.6-sol high.

= c
{parent=5c}
{scope}

= Solution
{parent=c}

At a fixed observation time $t$, varying the release time $0<s<t$ gives
$$
x=t^2-s^2,\qquad y=\frac{x^2}{4}.
$$
Hence the dye <streakline> is the parabolic arc
$$
\boxed{y=\frac{x^2}{4},\qquad0<x<t^2}.
$$

Solved by gpt-5.6-sol high.

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

= a
{parent=6h}
{scope}

= Solution
{parent=a}

Let
$$
N(\theta)=\#\{i:X_i<\theta\}.
$$
Ignoring the immaterial choice of density values at the sample points, the <likelihood function> is
$$
\boxed{
L(\theta)
=2^{-n}\left(1+\frac1\theta\right)^{N(\theta)}}.
$$
Between consecutive <order statistics>, $N(\theta)$ is constant while $(1+1/\theta)^{N(\theta)}$ decreases with $\theta$. A maximum can therefore be moved to the left endpoint of one of these intervals. With an equivalent version using $\mathbf1_{\{x\leq\theta\}}$, the maximum is attained at a sample order statistic. Thus the <maximum-likelihood estimator> coincides with one of $X_1,\ldots,X_n$.

Solved by gpt-5.6-sol high.

= b
{parent=6h}
{scope}

= Solution
{parent=b}

The density is an equal mixture of the uniform distributions on $[0,1]$ and $[0,\theta]$. Hence
$$
\mathbb E_\theta X
=\frac12\cdot\frac12+\frac12\cdot\frac\theta2
=\frac{1+\theta}{4},
$$
so
$$
\boxed{\mathbb E_\theta(4\overline X-1)=\theta}.
$$
Thus $\widetilde\theta$ is unbiased.

Also,
$$
\mathbb E_\theta X^2=\frac{1+\theta^2}{6},
\qquad
\operatorname{Var}_\theta X
=\frac{5\theta^2-6\theta+5}{48}.
$$
The <central limit theorem> gives
$$
\sqrt n(\widetilde\theta-\theta)
\xrightarrow{d}
N\left(0,\frac{5\theta^2-6\theta+5}{3}\right).
$$
Replacing $\theta$ in the asymptotic standard error by the consistent estimator $\widetilde\theta$ gives the <Wald confidence interval>
$$
\boxed{
\widetilde\theta
\mathbin{\pm}
z_{1-\alpha/2}
\sqrt{\frac{5\widetilde\theta^2-6\widetilde\theta+5}{3n}}}.
$$
It may be intersected with the parameter space $(0,1)$.

Solved by gpt-5.6-sol high.

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

= Solution
{parent=7H}

The <Lagrange sufficiency theorem> says that for a convex differentiable objective and convex differentiable inequality constraints, any feasible point satisfying the <Karush-Kuhn-Tucker conditions> with nonnegative multipliers is a global minimizer.

Write
$$
f=-x_1-3x_2,\qquad
g_1=x_1^2+x_2^2-25,\qquad
g_2=-x_1+2x_2-5.
$$
The unconstrained maximizer of $x_1+3x_2$ on the disc violates $g_2\leq0$, so both boundaries are active at the optimum. Their intersections are
$$
(-5,0)\quad\text{and}\quad(3,4),
$$
and $(3,4)$ gives the smaller objective.

To certify it, at $x_*=(3,4)$ choose
$$
\lambda=\frac14,\qquad\mu=\frac12.
$$
Then
$$
\nabla f(x_*)+\lambda\nabla g_1(x_*)
+\mu\nabla g_2(x_*)=0,
$$
both multipliers are nonnegative, and complementary slackness holds because both constraints are active. The sufficiency theorem therefore gives
$$
\boxed{x_1=3,\qquad x_2=4,\qquad f_{\min}=-15}.
$$

Solved by gpt-5.6-sol high.

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

= Solution
{parent=8F}

Let $f_1,f_2\in V^*$ satisfy $f_1(v)f_2(v)=0$ for every <vector> $v$. If $f_1=0$ there is nothing to prove. Otherwise choose $u$ with $f_1(u)\ne0$. For every $v\in\ker f_1$ and every <real number> $t$,
$$
f_1(u+tv)=f_1(u)\ne0,
$$
so the hypothesis gives $f_2(u+tv)=0$. Taking two values of $t$ shows that $f_2(u)=f_2(v)=0$. Since
$$
V=\mathbb Ru+\ker f_1,
$$
we obtain $f_2=0$. Thus one of the two <linear functionals> is zero.

<Sylvester's law of inertia> says that, in a suitable <basis>, every real <quadratic form> is
$$
q=x_1^2+\cdots+x_p^2-x_{p+1}^2-\cdots-x_{p+m}^2.
$$
The <rank> is $r=p+m$, and the <signature> is $\sigma=p-m$; both are independent of the chosen basis.

If $q=f_1f_2$, its <polarization identity>[polar form] has image contained in $\operatorname{span}\{f_1,f_2\}$, so $r\leq2$. If $r=2$, the two functionals must be independent and, after taking their sum and difference, $q$ is a difference of two squares. Hence $\sigma=0$. If $r=1$, then $|\sigma|=1$. In every case,
$$
r+|\sigma|\leq2.
$$
Conversely, this inequality leaves only
$$
q=0,\qquad q=\pm x_1^2,qquad q=x_1^2-x_2^2,
$$
up to a <change of basis>[linear change of coordinates]. These factor respectively as $0$, $(\pm x_1)x_1$, and $(x_1-x_2)(x_1+x_2)$. Therefore
$$
\boxed{q=f_1f_2\quad\Longleftrightarrow\quad r+|\sigma|\leq2}.
$$

Finally suppose that $q$ takes both positive and negative values. Sylvester's law supplies normalized basis vectors
$$
e_1,\ldots,e_p,\quad f_1,\ldots,f_m,\quad z_1,\ldots,z_k
$$
with $q(e_i)=1$, $q(f_j)=-1$, and the $z_l$ spanning the <radical of a bilinear form>[radical]. The $p+m+k$ vectors
$$
e_1+f_1,\ e_1-f_1,\ e_i+f_1\ (i\geq2),\ e_1+f_j\ (j\geq2),\ z_l
$$
are linearly independent and all satisfy $q(v)=0$. They form the required basis of <isotropic vector>[isotropic vectors].

Solved by gpt-5.6-sol high.

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

= Solution
{parent=9E}

A <Sylow subgroup> of a <finite group> $G$ is a subgroup whose order is the largest <prime power>[power of a prime] $p$ dividing $|G|$. The <Sylow theorems> say that such subgroups exist, that every $p$-subgroup lies in one, that all Sylow $p$-subgroups are conjugate, and that their number $n_p$ divides $|G|/p^a$ and obeys $n_p\equiv1\pmod p$.

For the last congruence, let one Sylow $p$-subgroup $P$ act by conjugation on the set of all Sylow $p$-subgroups. Every <orbit of a group action>[orbit] other than a fixed point has size divisible by $p$. If $Q$ is fixed, then $P,Q\leq N_G(Q)$. They are Sylow subgroups of this <normalizer>, so they are conjugate within $N_G(Q)$; because every element of $N_G(Q)$ fixes $Q$ under conjugation, this forces $P=Q$. Thus there is exactly one fixed point and $n_p\equiv1\pmod p$.

Now let $H<A_n$ have <index of a subgroup>[index] $m$ with $1<m<n$. The <coset action> gives a homomorphism
$$
A_n\longrightarrow S_m.
$$
Its <kernel of a group homomorphism>[kernel] is a <normal subgroup> of the <simple group> $A_n$. It cannot be all of $A_n$, since the action is transitive and nontrivial, so it is trivial. This would embed $A_n$ into $S_m$, contrary to $|A_n|=n!/2>m!=|S_m|$. Hence no such subgroup exists.

Suppose finally that a group $G$ of order $90$ were simple. The Sylow count satisfies
$$
n_5\mid18,\qquad n_5\equiv1\pmod5,
$$
and simplicity excludes $n_5=1$, so $n_5=6$. Conjugation on these six subgroups gives a nontrivial homomorphism $G\to S_6$, which simplicity makes injective. Its image lies in $A_6$, because the composite with the <sign homomorphism> $S_6\to\{\pm1\}$ must be trivial. It would therefore be a subgroup of $A_6$ of index $360/90=4$, contradicting the result just proved. Thus
$$
\boxed{\text{no group of order }90\text{ is simple}.}
$$

Solved by gpt-5.6-sol high.

= 10G
{parent=Paper 2}
{scope}
{title2=Analysis and Topology}

= Solution
{parent=10G}

The <implicit function theorem> states that if $F(x_0,y_0)=0$ and the partial derivative $D_yF(x_0,y_0)$ is an <invertible linear map>, then near $(x_0,y_0)$ the zero set of $F$ is uniquely the graph $y=g(x)$ of a continuously differentiable function.

If $f$ is a <differentiable map>[differentiable] bijection with differentiable inverse $g$, the <chain rule> applied to $g\circ f$ and $f\circ g$ gives
$$
Dg_{f(x)}Df_x=I,
\qquad
Df_xDg_{f(x)}=I.
$$
Thus $Df_x$ is an isomorphism with inverse $Dg_{f(x)}$.

If a continuously differentiable map $F:\mathbb R^n\to\mathbb R^n$ has invertible derivative everywhere, the <inverse function theorem> makes it a local diffeomorphism. In particular it is an <open map>, so its image is open. Its image need not be closed: $x\mapsto\arctan x$ has nonzero derivative everywhere and image $(-\pi/2,\pi/2)$.

For the given map of <elementary symmetric polynomial>[elementary symmetric polynomials],
$$
DF=
\begin{pmatrix}
1&1&1\\
y+z&x+z&x+y\\
yz&xz&xy
\end{pmatrix},
$$
and direct evaluation of the <determinant> gives
$$
\det DF=(x-y)(y-z)(z-x).
$$
Hence the <critical set> is
$$
\boxed{C=\{x=y\}\cup\{y=z\}\cup\{z=x\}}.
$$
Its complement is the <Zariski-open set> on which the three coordinates are pairwise distinct. Each point has one of the six possible strict coordinate orderings, and each ordering defines a nonempty convex open region. A <continuous path> cannot change an ordering without crossing $C$. Therefore $\mathbb R^3\setminus C$ has exactly
$$
\boxed{3!=6}
$$
<connected component>[connected components].

Solved by gpt-5.6-sol high.

= 11F
{parent=Paper 2}
{scope}
{title2=Geometry}

= a
{parent=11f}
{scope}

= Solution
{parent=a}

With
$$
X(u,v)=(\sinh u\cos v,\sinh u\sin v,v),
$$
the coordinate tangent vectors satisfy
$$
X_u\cdot X_u=\cosh^2u,
\qquad X_u\cdot X_v=0,
\qquad X_v\cdot X_v=\cosh^2u.
$$
Thus the <first fundamental form> is
$$
\boxed{ds^2=\cosh^2u\,(du^2+dv^2)}.
$$
Geometrically, $S$ is the half of a <helicoid> with positive radius, winding upwards indefinitely.

Consider the upper half of the <catenoid>, parametrized as the <surface of revolution>
$$
Y(u,v)=(\cosh u\cos v,\cosh u\sin v,u),
\qquad u,v>0.
$$
It has the same first fundamental form, so $(u,v)\mapsto Y(u,v)$ defines a <local isometry>. By the invariance of <Gaussian curvature> under local isometry, the curvatures agree; directly, the conformal metric above gives
$$
\boxed{K=-\operatorname{sech}^4u}.
$$

Solved by gpt-5.6-sol high.

= b
{parent=11f}
{scope}

= Solution
{parent=b}

For an oriented regular surface with unit normal $N$, the <Gauss map> sends each point $p$ to $N(p)\in S^2$. The outward normal to the catenoid is
$$
N(u,v)=(\operatorname{sech}u\cos v,
\operatorname{sech}u\sin v,-\tanh u).
$$
Its image is the open southern hemisphere, apart from its limiting equator and south pole. The Gauss map is one-to-one after taking $v$ modulo $2\pi$ and reverses orientation. Since the Jacobian of the Gauss map is the <Gaussian curvature>, its signed spherical area is the <total Gaussian curvature>:
$$
\boxed{\int_{S'}K\,dA=-\operatorname{area}(\text{hemisphere})=-2\pi}.
$$
Equivalently, $K\,dA=-\operatorname{sech}^2u\,du\,dv$, whose integral over $u>0$ and $0\leq v<2\pi$ is $-2\pi$.

Solved by gpt-5.6-sol high.

= c
{parent=11f}
{scope}

= Solution
{parent=c}

The helicoid has no angular identification: its height coordinate is $v$. Consequently
$$
\int_S K\,dA
=-\int_0^\infty\int_0^\infty
\operatorname{sech}^2u\,dv\,du
=-\infty.
$$
The catenoid has total curvature $-2\pi$. A global <Riemannian isometry> preserves both the Gaussian curvature and the area element, hence preserves total curvature. Therefore
$$
\boxed{S\text{ and }S'\text{ are not globally isometric}.}
$$
The same obstruction is visible in the parameter map: increasing $v$ by $2\pi$ gives a different point of the helicoid but the same point of the catenoid.

Solved by gpt-5.6-sol high.

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

= a
{parent=12a}
{scope}

= Solution
{parent=a}

Close the contour in the upper half-plane. The degree assumption and <Jordan lemma> make the integral over the large semicircle tend to zero, while the absence of real zeros avoids indentations of the contour. The <residue theorem> therefore gives
$$
\boxed{
\int_{-\infty}^{\infty}R(x)e^{ix}\,dx
=2\pi i\sum_{\substack{Q(z_k)=0\\ \operatorname{Im}z_k>0}}
\operatorname{Res}_{z=z_k}\bigl(R(z)e^{iz}\bigr)}.
$$
Repeated roots are handled by the usual higher-order <residue> formula.

Solved by gpt-5.6-sol high.

= b
{parent=12a}
{scope}

= Solution
{parent=b}

Apply the formula to
$$
R(z)=\frac{z}{1+z^4}.
$$
The upper-half-plane poles are
$$
z_1=e^{i\pi/4},\qquad z_2=e^{3i\pi/4},
$$
and their residues are $e^{iz_k}/(4z_k^2)$. Writing $c=1/\sqrt2$, their sum is
$$
\frac{i}{4}\left(e^{iz_2}-e^{iz_1}\right)
=\frac12e^{-c}\sin c.
$$
Since the cosine part of $x e^{ix}/(1+x^4)$ is <odd function>[odd], its integral vanishes, while the sine part is <even function>[even]. Hence
$$
i\int_{-\infty}^{\infty}\frac{x\sin x}{1+x^4}\,dx
=2\pi i\left(\frac12e^{-c}\sin c\right),
$$
so
$$
\boxed{
\int_{-\infty}^{\infty}\frac{x\sin x}{1+x^4}\,dx
=\pi e^{-1/\sqrt2}\sin\!\left(\frac1{\sqrt2}\right)}.
$$

Solved by gpt-5.6-sol high.

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

= a
{parent=13d}
{scope}

= Solution
{parent=a}

For fixed endpoint values, the first variation of the <functional>
$$
I[z]=\int_a^b f(z,z';x)\,dx
$$
vanishes exactly when the <Euler-Lagrange equation>
$$
\boxed{\frac{d}{dx}\frac{\partial f}{\partial z'}
-\frac{\partial f}{\partial z}=0}
$$
holds. If $f$ has no explicit $x$ dependence, differentiation along an extremal gives the <Beltrami identity>
$$
\boxed{f-z'\frac{\partial f}{\partial z'}=\text{constant}}.
$$
For a further constraint $J[z]=J_0$, introduce a constant <Lagrange multiplier> $\lambda$ and apply the Euler-Lagrange equation to $I+\lambda J$; $\lambda$ is then chosen so that the constraint holds.

Solved by gpt-5.6-sol high.

= b
{parent=13d}
{scope}

= i
{parent=b}
{scope}

= Solution
{parent=i}

Enforce the fixed <arc length> with a multiplier. Apart from an irrelevant common factor, the augmented integrand is
$$
f=(gz+\lambda)\sqrt{1+z'^2}.
$$
The <Beltrami identity> gives
$$
\frac{gz+\lambda}{\sqrt{1+z'^2}}=\text{constant}.
$$
After absorbing constants into $z_0$ and a positive scale $B$, integration yields the <catenary>
$$
\boxed{z-z_0=-B\cosh(x/B)}.
$$
The endpoint conditions and prescribed length give
$$
\boxed{z_0=B\cosh(a/B),
\qquad L=2B\sinh(a/B)}.
$$
The second equation determines $B>0$ implicitly because $L>2a$, and the first then determines $z_0$.

Solved by gpt-5.6-sol high.

= ii
{parent=b}
{scope}

= Solution
{parent=ii}

For the lower semicircle
$$
z(x)=-\sqrt{a^2-x^2},
\qquad
\sqrt{1+z'^2}=\frac{a}{-z}.
$$
For a general potential, the Beltrami first integral is
$$
\frac{\rho\Phi(z)+\lambda}{\sqrt{1+z'^2}}=C.
$$
Substitution of the semicircle therefore requires
$$
-\frac{z}{a}\bigl(\rho\Phi(z)+\lambda\bigr)=C.
$$
Thus, up to an arbitrary additive constant and a multiplicative strength,
$$
\boxed{\Phi(z)=\Phi_0+\frac{A}{z}}.
$$
The additive constant is absorbed by the length multiplier. The semicircle has length $\pi a$, as required.

Solved by gpt-5.6-sol high.

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

= a
{parent=14a}
{scope}

= Solution
{parent=a}

Substitution of $y=e^{-x}$ gives
$$
(x+\lambda+1)e^{-x}-(x+\lambda)e^{-x}-e^{-x}=0.
$$
For a second solution $y=ax+b$, the <ordinary differential equation> reduces to $a\lambda-b=0$. Taking $a=1$ gives the linearly independent solution
$$
\boxed{y_1=x+\lambda}.
$$

Solved by gpt-5.6-sol high.

= b
{parent=14a}
{scope}

= Solution
{parent=b}

The solution $x+\lambda$ satisfies the left <boundary condition>, while $e^{-x}$ satisfies decay at infinity. The proposed expression for $x<\xi$ is therefore the correct left homogeneous solution. For $x>\xi$, write $G=Ce^{-x}$. Continuity at $x=\xi$ gives
$$
Ce^{-\xi}=-\frac{\xi+\lambda}{\xi+\lambda+1}.
$$
Hence the <Green function> is
$$
\boxed{
G(x;\xi)=
\begin{cases}
-\dfrac{x+\lambda}{\xi+\lambda+1},&0\leq x<\xi,\\[6pt]
-\dfrac{\xi+\lambda}{\xi+\lambda+1}e^{\xi-x},&x>\xi.
\end{cases}}
$$
Indeed, its derivative has the required unit jump,
$$
G_x(\xi^+;\xi)-G_x(\xi^-;\xi)=1,
$$
which produces the <Dirac delta function>[Dirac delta distribution] in $L[G]$.

Solved by gpt-5.6-sol high.

= c
{parent=14a}
{scope}

= Solution
{parent=c}

For $\lambda=2$, the Green representation of the <boundary value problem> is
$$
y(x)=\int_0^\infty G(x;\xi)\bigl[-(\xi+3)e^{-\xi}\bigr],d\xi.
$$
Splitting the integral at $\xi=x$ gives
$$
y(x)=e^{-x}\int_0^x(\xi+2),d\xi
 +(x+2)\int_x^\infty e^{-\xi},d\xi.
$$
Therefore
$$
\boxed{y(x)=e^{-x}\left(\frac{x^2}{2}+3x+2\right)}.
$$
It satisfies $y(0)=2y'(0)$ and tends to zero as $x\to\infty$.

Solved by gpt-5.6-sol high.

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

= a
{parent=15b}
{scope}

= Solution
{parent=a}

Inside the <infinite square well>, the time-independent <Schrödinger equation> has <Dirichlet boundary conditions> at $0$ and $a$. Its normalized <eigenstate>[energy eigenstates] and <energy eigenvalue>[energy eigenvalues] are
$$
\boxed{
\chi_n(x)=\sqrt{\frac2a}\sin\frac{n\pi x}{a},
\qquad
E_n=\frac{n^2\pi^2\hbar^2}{2ma^2},
\qquad n=1,2,\ldots .}
$$

Solved by gpt-5.6-sol high.

= b
{parent=15b}
{scope}

= i
{parent=b}
{scope}

= Solution
{parent=i}

Expand the normalized initial <wavefunction> in the orthonormal energy basis:
$$
c_n=\int_0^a\chi_n(x)^*f(x),dx.
$$
The <time-dependent Schrödinger equation> then gives the <unitary time evolution>
$$
\boxed{
\psi(x,t)=\sum_{n=1}^\infty
c_n\chi_n(x)e^{-iE_nt/\hbar}}.
$$

Solved by gpt-5.6-sol high.

= ii
{parent=b}
{scope}

= Solution
{parent=ii}

For
$$
T=\frac{2ma^2}{\pi\hbar},
$$
the phase of the $n$th stationary state is
$$
e^{-iE_nT/\hbar}=e^{-in^2\pi}=(-1)^n.
$$
Reflection in the midpoint gives
$$
\chi_n(a-x)=(-1)^{n+1}\chi_n(x),
$$
so
$$
\boxed{\psi(x,T)=-f(a-x)}.
$$
The minus sign is a global <quantum phase> and therefore does not change the physical state. The relative even--odd phases first acquire this reflection pattern at $T$. Applying it twice gives
$$
\boxed{\psi(x,2T)=f(x)}.
$$
If $f$ vanishes on the right half of the well, then $\psi(x,T)$ vanishes on the left half. The <Born rule> consequently gives zero probability of finding the particle in $0\leq x\leq a/2$ at time $T$.

Solved by gpt-5.6-sol high.

= iii
{parent=b}
{scope}

= Solution
{parent=iii}

The specified initial state is normalized. The requested energy is $E_2$, and its <probability amplitude> is
$$
c_2=\int_0^{a/2}
\sqrt{\frac2a}\sin\frac{2\pi x}{a}
\frac2{\sqrt a}\sin\frac{2\pi x}{a},dx
=\frac1{\sqrt2}.
$$
Thus the <Born rule> gives
$$
\boxed{\mathbb P(E=E_2)=|c_2|^2=\frac12}.
$$
Unitary evolution changes each energy coefficient only by a phase, so this probability remains $1/2$ at both $T$ and $2T$. If an <energy measurement> returns $E_2$, the state collapses to the nondegenerate eigenstate $\chi_2$; a subsequent energy measurement therefore returns $E_2$ with probability
$$
\boxed{1}.
$$

Solved by gpt-5.6-sol high.

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

= a
{parent=16d}
{scope}

= Solution
{parent=a}

The <Maxwell equations>[Maxwell equation] $\nabla\times\mathbf E=-\partial\mathbf B/\partial t$, integrated over a fixed spanning surface $S$, gives by <Stokes theorem>
$$
\oint_{\partial S}\mathbf E\cdot d\mathbf l
=-\int_S\frac{\partial\mathbf B}{\partial t}\cdot d\mathbf S
=-\frac{d}{dt}\int_S\mathbf B\cdot d\mathbf S.
$$
Thus the <electromotive force> and <magnetic flux> obey <Faraday's law>
$$
\boxed{\mathcal E=-\frac{d\Phi_B}{dt}}.
$$

Solved by gpt-5.6-sol high.

= b
{parent=16d}
{scope}

= i
{parent=b}
{scope}

= Solution
{parent=i}

By cylindrical symmetry the field is azimuthal and constant on a circle of radius $r$. The <Ampère-Maxwell equation>[Ampère's law] for a steady current gives
$$
2\pi r B=\mu_0I,
$$
so
$$
\boxed{\mathbf B=\frac{\mu_0I}{2\pi r}\,\mathbf e_\phi}.
$$
In the plane $x=0$ with $y>0$, it points in the $-x$ direction.

Solved by gpt-5.6-sol high.

= ii
{parent=b}
{scope}

= Solution
{parent=ii}

At time $t$, let the centre of the loop have coordinate $Y=d+vt$. With normal $+\mathbf e_x$, the <magnetic flux> is
$$
\Phi_B
=-\int_{Y-a}^{Y+a}\int_{-a}^{a}
\frac{\mu_0I}{2\pi y},dz,dy
=-\frac{\mu_0Ia}{\pi}\log\frac{Y+a}{Y-a}.
$$
<Faraday's law> and <Ohm's law> therefore give a current of magnitude
$$
\boxed{
|I_{\rm loop}(t)|
=\frac{2\mu_0Ia^2v}
{\pi R\bigl((d+vt)^2-a^2\bigr)}}.
$$
The original flux points into the plane and decreases in magnitude as the loop recedes. By <Lenz's law>, the induced current reinforces the into-plane field: it is clockwise when viewed from the $+x$ side.

Solved by gpt-5.6-sol high.

= iii
{parent=b}
{scope}

= Solution
{parent=iii}

Translation parallel to the wire leaves every point of the loop at the same distance from the wire, so its <magnetic flux> is constant. Equivalently, the line integrals of the motional field $\mathbf u\times\mathbf B$ cancel on opposite sides. Hence
$$
\boxed{\mathcal E=0,
\qquad I_{\rm loop}=0}.
$$

Solved by gpt-5.6-sol high.

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

= a
{parent=17c}
{scope}

= Solution
{parent=a}

Writing $f,f',f''$ at $y_n$, the implicit stage has the expansion
$$
k_2=f+h f'f+h^2\left(a(f')^2f+\frac12f''f^2\right)+O(h^3).
$$
Thus one step is
$$
y_{n+1}=y_n+hf+\frac{h^2}{2}f'f
+h^3\left(\frac a2(f')^2f+\frac14f''f^2\right)+O(h^4).
$$
The exact <Taylor theorem>[Taylor expansion] has third-order term
$$
\frac{h^3}{6}\bigl((f')^2f+f''f^2\bigr).
$$
The coefficients agree through order two for every real $a$, while the $f''f^2$ coefficient prevents order three for any $a$. Hence the <order of a Runge-Kutta method> is
$$
\boxed{2\quad\text{for every }a\in\mathbb R}.
$$
(The linear special case has one extra matched term when $a=1/3$, but the order for general nonlinear equations remains two.)

Solved by gpt-5.6-sol high.

= b
{parent=17c}
{scope}

= Solution
{parent=b}

Apply the method to the <linear stability domain>[linear test equation] $y'=\lambda y$ and put $z=h\lambda$. Solving the implicit stage gives the <stability function>
$$
\boxed{
R(z)=\frac{1+(1-a)z+\tfrac12(1-2a)z^2}{1-az}}.
$$
If $a\ne1/2$, the quadratic numerator makes $|R(z)|$ unbounded as $z\to-\infty$, so the method cannot be <A-stable>. For $a=1/2$,
$$
R(z)=\frac{1+z/2}{1-z/2},
$$
which is the <trapezoidal rule> stability function and satisfies $|R(z)|\leq1$ whenever $\operatorname{Re}z\leq0$. Therefore
$$
\boxed{\text{the method is A-stable exactly when }a=\frac12}.
$$

Solved by gpt-5.6-sol high.

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

= a
{parent=18h}
{scope}

= Solution
{parent=a}

The chain is an irreducible <birth-death process>. The <detailed balance> equations are
$$
\pi_i\frac13=\pi_{i+1}\frac23,
$$
so $\pi_{i+1}=\pi_i/2$. Normalization gives the <stationary distribution>
$$
\boxed{\pi_i=2^{-(i+1)},\qquad i\geq0}.
$$
An irreducible countable-state Markov chain that possesses a stationary probability distribution is <positive recurrent state>[positive recurrent]. Hence $X$ is positive recurrent.

Solved by gpt-5.6-sol high.

= b
{parent=18h}
{scope}

= Solution
{parent=b}

The self-loop at zero makes the irreducible chain <aperiodic Markov chain>[aperiodic]. The convergence theorem for irreducible, aperiodic, positive recurrent countable-state chains gives
$$
\mathbb P(X_n=i)\longrightarrow\pi_i.
$$
Because $X$ and $Y$ are <independent random variables>[independent],
$$
\mathbb P(X_n=0,Y_n=1)
=\mathbb P(X_n=0)\mathbb P(Y_n=1)
\longrightarrow\pi_0\pi_1.
$$
Therefore
$$
\boxed{\lim_{n\to\infty}\mathbb P(X_n=0,Y_n=1)
=\frac12\cdot\frac14=\frac18}.
$$

Solved by gpt-5.6-sol high.

= c
{parent=18h}
{scope}

= Solution
{parent=c}

The pair $Z_n=(X_n,Y_n)$ is an irreducible positive recurrent <product Markov chain> with stationary distribution
$$
\Pi_{ij}=\pi_i\pi_j.
$$
The <stationary cycle occupation formula> says that, during one return cycle to a state $z$, the expected number of visits to a set $A$ is $\Pi(A)/\Pi(z)$. Take
$$
z=(0,0),
\qquad A=\{(i,1):i\geq0\}.
$$
Then
$$
\Pi(A)=\pi_1=\frac14,
\qquad
\Pi(0,0)=\pi_0^2=\frac14.
$$
The initial and terminal states both have $Y=0$, so either convention for including the endpoints gives the same count. Thus
$$
\boxed{\mathbb E\!\left[\#\{0\leq n\leq T:Y_n=1\}\right]=1}.
$$

Solved by gpt-5.6-sol high.