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

= Paper 2
{scope}

https://www.maths.cam.ac.uk/undergrad/pastpapers/files/2026/Paperia_2_2026.pdf

= 1C
{parent=Paper 2}
{scope}
{title2=Differential Equations}

= Solution
{parent=1c}

<Differentiation> under the <integral> gives $I\prime\prime=\int_0^1y^4e^{xy}dy$. Integrating the <derivative> of $e^{xy}(xy^3-3y^2)$ and then once more yields $x^2\int_0^1y^4e^{xy}dy-12\int_0^1y^2e^{xy}dy=e^x(x-4)$.

Solved by gpt-5.6-sol high.

= 2C
{parent=Paper 2}
{scope}
{title2=Differential Equations}

= a
{parent=2c}
{scope}

= Solution
{parent=a}

With $y=z/x^2$ the equation becomes $z\prime\prime+z=0$. Thus $y_1=\cos x/x^2$, $y_2=\sin x/x^2$, and $W(y_1,y_2)=x^{-4}\ne0$.

Solved by gpt-5.6-sol high.

= b
{parent=2c}
{scope}

= Solution
{parent=b}

The transformed inhomogeneous equation is $z\prime\prime+z=x^2$. Taking $z=x^2-2$ gives the particular <integral> $y_p=1-2/x^2$.

Solved by gpt-5.6-sol high.

= 3F
{parent=Paper 2}
{scope}
{title2=Probability}

= i
{parent=3f}
{scope}

= Solution
{parent=i}

Completing the square gives $X,Y\sim N(0,4/3)$, so each marginal is $\sqrt3(2\sqrt{2\pi})^{-1}e^{-3x^2/8}$. Their covariance is $2/3$, so they are not independent.

Solved by gpt-5.6-sol high.

= ii
{parent=3f}
{scope}

= Solution
{parent=ii}

From part (i),
$$
\operatorname{var}(X)=\operatorname{var}(Y)=\frac43,
\qquad \operatorname{cov}(X,Y)=\frac23.
$$
The conditional-normal formula therefore gives
$$
Y\mid X=x\sim
N\left(\frac{\operatorname{cov}(X,Y)}{\operatorname{var}(X)}x,
\operatorname{var}(Y)-
\frac{\operatorname{cov}(X,Y)^2}{\operatorname{var}(X)}\right)
=N\left(\frac x2,1\right).
$$
Consequently the <Gaussian conditional expectation> is
$$
\mathbb E[Y\mid X]=\frac X2.
$$

Solved by gpt-5.6-sol high.

= 4F
{parent=Paper 2}
{scope}
{title2=Probability}

= i
{parent=4f}
{scope}

= Solution
{parent=i}

Writing $S_n=\sum I_i$, each $\mathbb EI_i=1/n$. Distinct indicators have joint probability $1/[n(n-1)]$ and zero covariance, so $\mathbb ES_n=1$ and $\operatorname{var}S_n=n(1/n)(1-1/n)=1-1/n$.

Solved by gpt-5.6-sol high.

= ii
{parent=4f}
{scope}

= Solution
{parent=ii}

<Cauchy-Schwarz inequality> applied to $X1_{X\gt 0}$ gives $(\mathbb EX)^2\le\mathbb E(X^2)\mathbb P(X\gt 0)$.

Solved by gpt-5.6-sol high.

= iii
{parent=4f}
{scope}

= Solution
{parent=iii}

Apply part (ii): $\mathbb P(S_n\gt 0)\ge1/\mathbb E(S_n^2)=1/(2-1/n)\ge1/2$.

Solved by gpt-5.6-sol high.

= 5C
{parent=Paper 2}
{scope}
{title2=Differential Equations}

= i
{parent=5c}
{scope}

= Solution
{parent=i}

$y_0\prime\prime=(x^2-1)y_0$, so $\alpha=1$.

Solved by gpt-5.6-sol high.

= ii
{parent=5c}
{scope}

= Solution
{parent=ii}

Substitution and cancellation give $f\prime\prime-2xf\prime+(\alpha-1)f=0$.

Solved by gpt-5.6-sol high.

= iii
{parent=5c}
{scope}

= Solution
{parent=iii}

Equating powers gives $(n+2)(n+1)a_{n+2}+(\alpha-1-2n)a_n=0$.

Solved by gpt-5.6-sol high.

= iv
{parent=5c}
{scope}

= Solution
{parent=iv}

$y\prime(0)=0$ forces the odd <series> to vanish. The even recurrence terminates at degree $2N$ precisely when $\alpha=4N+1$.

Solved by gpt-5.6-sol high.

= v
{parent=5c}
{scope}

= Solution
{parent=v}

After division by $e^{-x^2/2}$, $f\prime\prime-2xf\prime+8f=2$. The initial data and evenness give $f=1-3x^2+x^4$, hence $y=e^{-x^2/2}(1-3x^2+x^4)$.

Solved by gpt-5.6-sol high.

= 6C
{parent=Paper 2}
{scope}
{title2=Differential Equations}

= i
{parent=6c}
{scope}

= Solution
{parent=i}

$\nabla U=(3x^2-3+y^2,,2y(x+1))$, so the stationary points are $(\pm1,0)$. The Hessian at $(1,0)$ is $\operatorname{diag}(6,4)$, making it the local minimum; $(-1,0)$ is degenerate and not a minimum.

Solved by gpt-5.6-sol high.

= ii
{parent=6c}
{scope}

= Solution
{parent=ii}

$dU/dt=\nabla U\cdot(-\nabla U)=-|\nabla U|^2\le0$. The descending trajectory from $(2,1)$ stays in its bounded contour basin and tends to the only critical point there, $(1,0)$.

Solved by gpt-5.6-sol high.

= iii
{parent=6c}
{scope}

= Solution
{parent=iii}

Writing $\xi=x-1$, $\eta=y$, the linearisation is $\dot\xi=-6\xi$, $\dot\eta=-4\eta$. Eliminating $t$ gives $\eta=C\xi^{2/3}$.

Solved by gpt-5.6-sol high.

= iv
{parent=6c}
{scope}

= Solution
{parent=iv}

Now $\dot U=-|\nabla U|^2+U_y\cos\omega t$, whose second term has either sign. Near the minimum, $\xi=C e^{-6t}$ and $\dot\eta+4\eta=\cos\omega t$, so
$$\eta=C_2e^{-4t}+\frac{4\cos\omega t+\omega\sin\omega t}{16+\omega^2}.$$

Solved by gpt-5.6-sol high.

= 7C
{parent=Paper 2}
{scope}
{title2=Differential Equations}

= a
{parent=7c}
{scope}

= i
{parent=a}
{scope}

= Solution
{parent=i}

Spatial <derivatives> vanish for $u=y(t)$, giving $\ddot y+\gamma\dot y+\omega^2y=f(t)$.

Solved by gpt-5.6-sol high.

= ii
{parent=a}
{scope}

= Solution
{parent=ii}

The roots are $r=(-\gamma\pm\sqrt{\gamma^2-4\omega^2})/2$. They give oscillatory decay for $\gamma\lt 2\omega$, two exponential decays for $\gamma\gt 2\omega$, and $(A+Bt)e^{-\gamma t/2}$ at equality. All tend to zero.

Solved by gpt-5.6-sol high.

= iii
{parent=a}
{scope}

= Solution
{parent=iii}

Put $T=\pi/\omega$, $t_0=T/2$, and $\Omega=\omega/2$. A solution satisfying both endpoint conditions is
$$y=Ae^{-\gamma t/2}\sin(\Omega t)+\frac I\Omega e^{-\gamma(t-t_0)/2}\sin(\Omega(t-t_0))H(t-t_0),$$
where $A=-\sqrt2I\omega^{-1}e^{\sqrt3\pi/4}$. The <derivative> jumps by $I$ at $t_0$, as required.

Solved by gpt-5.6-sol high.

= b
{parent=7c}
{scope}

= Solution
{parent=b}

Substitution gives $(\dot X^2-c^2)\phi\prime\prime-(\ddot X+\gamma\dot X)\phi\prime+\omega^2\phi=0$. Multiply by $\phi\prime$ and integrate over $\mathbb R$; decay kills the first and last <integrals>, leaving $(\ddot X+\gamma\dot X)\int(\phi\prime)^2=0$. Thus a nonconstant profile requires $X=X_\infty+Ce^{-\gamma t}$ and approaches a fixed position.

Solved by gpt-5.6-sol high.

= 8C
{parent=Paper 2}
{scope}
{title2=Differential Equations}

= i
{parent=8c}
{scope}

= Solution
{parent=i}

During warming, $\dot\theta=\alpha(\theta_1-\theta)$ and $\theta=\theta_1-(\theta_1-\theta_0)e^{-\alpha t}$. During cooling, with $\tau=t-T$, $\dot\theta=-\alpha(\theta-\theta_0)$ and $\theta=\theta_0+(\theta_1-\theta_0)(1-e^{-\alpha T})e^{-\alpha\tau}$.

Solved by gpt-5.6-sol high.

= ii
{parent=8c}
{scope}

= Solution
{parent=ii}

$N(2T)/N(0)=\exp[-\int_0^{2T}\beta(\theta(t))dt]$. Evaluation gives the implicit equation
$$\beta_{\max}\left[T-\frac{e^{-\alpha T}(1-e^{-\alpha T})}{\alpha}\right]=\log100.$$

Solved by gpt-5.6-sol high.

= iii
{parent=8c}
{scope}

= Solution
{parent=iii}

For $x=\alpha T\ll1$, $e^{-x}(1-e^{-x})=x-\tfrac32x^2+O(x^3)$. Hence $\tfrac32\beta_{\max}\alpha T^2\sim\log100$ and
$$T\sim\sqrt{\frac{2\log100}{3\beta_{\max}\alpha}}.$$

Solved by gpt-5.6-sol high.

= 9F
{parent=Paper 2}
{scope}
{title2=Probability}

= i
{parent=9f}
{scope}

= Solution
{parent=i}

Each of the $\binom n3$ triples is a triangle with probability $p^3$, so $\mathbb ET=\binom n3p^3$.

Solved by gpt-5.6-sol high.

= ii
{parent=9f}
{scope}

= Solution
{parent=ii}

<Markov inequality> gives $\mathbb P(T\gt 0)\le\mathbb ET=O(n^{3-3\alpha})\to0$ when $\alpha\gt 1$.

Solved by gpt-5.6-sol high.

= iii
{parent=9f}
{scope}

= Solution
{parent=iii}

<Chebyshev inequality> gives $\mathbb P(T=0)\le\mathbb P(|T-\mathbb ET|\ge\mathbb ET)\le\operatorname{var}(T)/(\mathbb ET)^2\to0$.

Solved by gpt-5.6-sol high.

= iv
{parent=9f}
{scope}

= Solution
{parent=iv}

Only triangle pairs sharing an edge have nonzero covariance, so
$$\operatorname{var}T=\binom n3(p^3-p^6)+2\binom n2\binom{n-2}2(p^5-p^6).$$
After division by $(\mathbb ET)^2$ this is $O(n^{-3}p^{-3}+n^{-2}p^{-1})\to0$ for $p=n^{-\alpha}$, $0\lt \alpha\lt 1$; use part (iii).

Solved by gpt-5.6-sol high.

= 10F
{parent=Paper 2}
{scope}
{title2=Probability}

= i
{parent=10f}
{scope}

= Solution
{parent=i}

The waiting time is geometric with mean $1/p$.

Solved by gpt-5.6-sol high.

= ii
{parent=10f}
{scope}

= Solution
{parent=ii}

First-step equations for states “no trailing H” and “one trailing H” give $E_0=1+qE_0+pE_1$, $E_1=1+qE_0$, hence $E_0=(1+p)/p^2$.

Solved by gpt-5.6-sol high.

= iii
{parent=10f}
{scope}

= Solution
{parent=iii}

The word $HT$ has no proper self-overlap, so its mean waiting time is the reciprocal of its probability: $1/[p(1-p)]$. This also follows from two first-step equations.

Solved by gpt-5.6-sol high.

= iv
{parent=10f}
{scope}

= Solution
{parent=iv}

The standard run recursion gives $E_n=1+p^{-1}+\cdots+p^{-n}=(1-p^n)/[(1-p)p^n]$.

Solved by gpt-5.6-sol high.

= v
{parent=10f}
{scope}

= Solution
{parent=v}

For $k\ge n$, $p_k(n)=\binom{k-1}{n-1}p^n(1-p)^{k-n}$. It is the sum of $n$ independent geometric variables, so the <central limit theorem> applies with mean $n/p$ and standard deviation $\sqrt{n(1-p)}/p$. Take
$k_a(n)=\lceil n/p+a\sqrt{n(1-p)}/p\rceil$ and $k_b(n)=\lfloor n/p+b\sqrt{n(1-p)}/p\rfloor$.

Solved by gpt-5.6-sol high.

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

= a
{parent=11f}
{scope}

= Solution
{parent=a}

Shift the absorbing interval to $\{0,\ldots,a+1\}$ and start at $1$. The gambler’s-ruin harmonic <functions> give $\mathbb P(\tau_a\lt \tau_{-1})=1/(a+1)$ and $\mathbb ET=1(a+1-1)=a$.

Solved by gpt-5.6-sol high.

= b
{parent=11f}
{scope}

= i
{parent=b}
{scope}

= Solution
{parent=i}

At time $T_k$ the visited vertices form a contiguous arc and the walk is at an endpoint. Exiting an interval of $k$ visited vertices, starting one step from an absorbing endpoint, has mean $k$. Therefore $\mathbb E(T_{k+1}-T_k)=k$ and $\mathbb ET_n=\sum_{k=1}^{n-1}k=n(n-1)/2$.

Solved by gpt-5.6-sol high.

= ii
{parent=b}
{scope}

= Solution
{parent=ii}

For each $j\ne0$, cut the cycle at $j$. The event that $j$ is last is the event that the lifted walk covers the other $n-1$ residues before crossing that cut. Gambler’s ruin (or cyclic symmetry of the two expanding endpoints) gives $\mathbb P(Z=j)=1/(n-1)$.

Solved by gpt-5.6-sol high.

= 12F
{parent=Paper 2}
{scope}
{title2=Probability}

= i
{parent=12f}
{scope}

= Solution
{parent=i}

Conditioning on generation $n$ and using independence gives $F_{n+1}(s)=F_n(F_1(s))$.

Solved by gpt-5.6-sol high.

= ii
{parent=12f}
{scope}

= Solution
{parent=ii}

Differentiating at $s=1$ gives $\mathbb EX_{n+1}=\mu\mathbb EX_n$, hence $\mathbb EX_n=\mu^n$.

Solved by gpt-5.6-sol high.

= iii
{parent=12f}
{scope}

= Solution
{parent=iii}

The extinction probability is the smallest fixed point of the convex generating <function> $F_1$ in $[0,1]$. Since $F_1\prime(1)=\mu\le1$ and $p_0,p_2\gt 0$ make strict convexity relevant, the only fixed point is $1$.

Solved by gpt-5.6-sol high.

= iv
{parent=12f}
{scope}

= Solution
{parent=iv}

Here $F_1(s)=1/(2-s)$ and $\mu=1$, so extinction is certain. Iteration gives
$$F_n(s)=\frac{n-(n-1)s}{n+1-ns},$$
and $\mathbb P(X_n=0)=F_n(0)=n/(n+1)$.

Solved by gpt-5.6-sol high.