past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2025/ia/paper-4.bigb

= Paper 4
{scope}

https://www.maths.cam.ac.uk/undergrad/pastpapers/files/2025/paperia_4_2025.pdf

= 1F
{parent=Paper 4}
{scope}
{title2=Numbers and Sets}

= Solution
{parent=1f}

The <series> $\sum a_n$ converges when its partial sums tend to a finite real <limit>. Put $m=n+1$. Then
$$a_n=\frac{2\sqrt m}{\sqrt{m-1}+\sqrt{m+1}}-1>0$$
by strict concavity of the square root. Taylor expansion at $m$ gives
$$\sqrt{m-1}+\sqrt{m+1}=2\sqrt m\left(1-\frac1{8m^2}+O(m^{-4})\right),$$
so $m^2a_n\to1/8$. <Limit> comparison with $\sum n^{-2}$ proves convergence.

Solved by gpt-5.6-sol high.

= 2E
{parent=Paper 4}
{scope}
{title2=Numbers and Sets}

= a
{parent=2e}
{scope}

= Solution
{parent=a}

Every integer $n>1$ is a product of primes, uniquely up to order. Existence follows by <strong induction>: if $n$ is not prime, factor it into smaller positive integers and apply induction. For uniqueness, if
$$p_1\cdots p_r=q_1\cdots q_s,$$
Euclid's lemma makes $p_1$ divide some $q_j$, hence equal it. Cancel that prime and induct on the number of factors.

Solved by gpt-5.6-sol high.

= b
{parent=2e}
{scope}

= Solution
{parent=b}

Prime exponents must obey simultaneous congruences modulo $2,3,5$. The choices
$$v_2(n)=15,\qquad v_3(n)=10,\qquad v_5(n)=6$$
satisfy the required congruences. Thus one example is
$$n=2^{15}3^{10}5^6.$$

Solved by gpt-5.6-sol high.

= 3B
{parent=Paper 4}
{scope}
{title2=Dynamics and Relativity}

= Solution
{parent=3b}

Newton's equation and its energy <integral> are
$$m\ddot x=-V'(x),\qquad \frac12m\dot x^2+V(x)=E.$$
Thus
$$T=\sqrt{2m}\int_{x_-}^{x_+}\frac{dx}{\sqrt{E-V(x)}}.$$
Let $a=-V''(x_+^*)>0$. Near the maximum,
$$V(x)=E^*-\frac a2(x-x_+^*)^2+O(|x-x_+^*|^3),$$
and the nearby turning point has distance $\epsilon\sim\sqrt{2E^*\delta/a}$ from $x_+^*$. The singular part of the period is an $\operatorname{arcosh}$ <integral>:
$$\sqrt{2m}\sqrt{\frac2a}\int_\epsilon^c\frac{dy}{\sqrt{y^2-\epsilon^2}}
=\sqrt{\frac ma}\log(1/\delta)+O(1).$$
Hence
$$T=\sqrt{\frac m{-V''(x_+^*)}}\{\log(1/\delta)+O(1)\}.$$
Physically, the limiting orbit approaches the <unstable equilibrium> at the barrier top with vanishing speed and spends arbitrarily long there.

Solved by gpt-5.6-sol high.

= 4B
{parent=Paper 4}
{scope}
{title2=Dynamics and Relativity}

= Solution
{parent=4b}

Put $R=R_A+R_B$, $c=\sqrt{R^2-b^2}/R$, and $s=b/R$. At impact the unit <vector> from A to B is $n=(-c,s)$. The <impulse> is normal, so tangential <velocity> is unchanged, while the normal components undergo a one-dimensional <elastic collision>. Therefore
$$v_A=\frac{2M_B}{M_A+M_B}(u e_x\cdot n)n
=-\frac{2M_Buc}{M_A+M_B}n,$$
$$v_B=ue_x-\frac{2M_A}{M_A+M_B}(ue_x\cdot n)n
=ue_x+\frac{2M_Auc}{M_A+M_B}n.$$

Solved by gpt-5.6-sol high.

= 5F
{parent=Paper 4}
{scope}
{title2=Numbers and Sets}

= Solution
{parent=5f}

If $\sqrt2+\sqrt3$ were rational, squaring would make $\sqrt6$ rational, a contradiction. Also
$$\tan(\pi/8)=\frac{1}{\sqrt2+1}=\sqrt2-1,$$
so it is irrational but algebraic, satisfying $x^2+2x-1=0$.

For lattice vertices $u,v,w$, the triangle area is
$$\frac12|\det(v-u,w-u)|,$$
a half-integer. Triangulating a convex <lattice polygon> from one vertex makes its area rational.

A regular octagon of side length $s$ has area $2(1+\sqrt2)s^2$. Lattice endpoints make $s^2$ a positive integer, so this area would be irrational, contradicting the preceding result. Hence no such octagon exists.

Solved by gpt-5.6-sol high.

= 6E
{parent=Paper 4}
{scope}
{title2=Numbers and Sets}

= a
{parent=6e}
{scope}

= Solution
{parent=a}

For pairwise coprime positive integers $n_i$, every system $x\equiv a_i\pmod{n_i}$ has a unique solution modulo $\prod_i n_i$.

Solved by gpt-5.6-sol high.

= b
{parent=6e}
{scope}

= Solution
{parent=b}

Euler's totient is $\phi(n)=|\{1\le a\le n:(a,n)=1\}|$, equivalently the number of units modulo $n$.

Solved by gpt-5.6-sol high.

= c
{parent=6e}
{scope}

= Solution
{parent=c}

If $(a,n)=1$, then $a^{\phi(n)}\equiv1\pmod n$.

Solved by gpt-5.6-sol high.

= d
{parent=6e}
{scope}

= Solution
{parent=d}

The Chinese remainder isomorphism $\mathbb Z/(nm)\cong\mathbb Z/n\times\mathbb Z/m$ restricts to a bijection on units. Counting those units gives $\phi(nm)=\phi(n)\phi(m)$.

Solved by gpt-5.6-sol high.

= e
{parent=6e}
{scope}

= Solution
{parent=e}

Write squarefree $n=\prod p$. For each $p\mid n$, either $p\mid a$, when both sides vanish modulo $p$, or Fermat's theorem applies. Since $p-1\mid\phi(n)$ and $k\equiv1\pmod{\phi(n)}$, it gives $a^k\equiv a\pmod p$. The <Chinese remainder theorem> combines these congruences to give $a^k\equiv a\pmod n$.

Solved by gpt-5.6-sol high.

= f
{parent=6e}
{scope}

= Solution
{parent=f}

RSA chooses $n=pq$ and exponents $e,d$ with $ed\equiv1\pmod{\phi(n)}$. The public key is $(n,e)$, encryption is $c\equiv m^e\pmod n$, and the secret exponent decrypts via $c^d\equiv m\pmod n$. Security relies on the difficulty of recovering $\phi(n)$ from a large unfactored $n$.

Solved by gpt-5.6-sol high.

= g
{parent=6e}
{scope}

= Solution
{parent=g}

Choose a prime with $p^2\mid n$, and set $a=0$, $b=n/p$. Then $b\not\equiv0\pmod n$, but $b^2=n^2/p^2$ is divisible by $n$. Therefore $a^k\equiv b^k\equiv0\pmod n$ for every $k>1$.

Solved by gpt-5.6-sol high.

= 7D
{parent=Paper 4}
{scope}
{title2=Numbers and Sets}

= a
{parent=7d}
{scope}

= Solution
{parent=a}

Injective means $f(x)=f(y)$ implies $x=y$. It need not imply surjective on an infinite set: $f(n)=n+1$ is injective from $\mathbb N$ to itself but misses its least element.

Solved by gpt-5.6-sol high.

= b
{parent=7d}
{scope}

= Solution
{parent=b}

An element of the left side is $f(s)$ for some $s\in S$ with $f(s)\in T$, exactly an element of $T\cap f(S)$. The second equality need not hold, even for a surjection: map $\{1,2\}$ to a singleton and take $S=\{1\}$, $S'=\{2\}$. Then the image of the intersection is empty but the intersection of the images is not.

Solved by gpt-5.6-sol high.

= c
{parent=7d}
{scope}

= i
{parent=c}
{scope}

= Solution
{parent=i}

True. If $p(a,a')=p(b,b')$, then $a'=b'$, so $f(a)=g(a')=g(b')=f(b)$. Injectivity of $f$ gives $a=b$.

Solved by gpt-5.6-sol high.

= ii
{parent=c}
{scope}

= Solution
{parent=ii}

True. Given $a'\in A'$, surjectivity supplies $a\in A$ with $f(a)=g(a')$. Then $(a,a')$ lies in the fibre product and maps to $a'$.

Solved by gpt-5.6-sol high.

= 8D
{parent=Paper 4}
{scope}
{title2=Numbers and Sets}

= Solution
{parent=8d}

A set is countable when it is finite or admits a bijection with a subset of $\mathbb N$. For each degree, the coefficient tuples for <polynomials> in $A[x]$ form a finite Cartesian power of the <countable set> $A$; a countable union over degrees is countable. In particular the algebraic numbers, being roots of countably many integer <polynomials> with finitely many roots each, are countable. Since $\mathbb R$ is uncountable, uncountably many real numbers are transcendental.

The set $Q$ is uncountable. Partition $N$ into four-element blocks and, independently on each block, choose either of two fixed-point-free permutations. Binary <sequences> then inject into $Q$.

Finally choose a line $L$ distinct from and nonparallel to every $L_i$, possible because only countably many directions are excluded. Each $L\cap L_i$ has at most one point, so the union covers only countably many points of the uncountable line $L$. It cannot cover the plane.

Solved by gpt-5.6-sol high.

= 9B
{parent=Paper 4}
{scope}
{title2=Dynamics and Relativity}

= Solution
{parent=9b}

With $\omega=B/m$, magnetic force does no work, so $|\dot x|$ is constant; its zero $z$ component also makes $\dot z$ constant. For the stated data,
$$\dot x=u\cos\omega t,\qquad \dot y=-u\sin\omega t,$$
$$x=\frac u\omega\sin\omega t,\qquad y=\frac u\omega(\cos\omega t-1).$$
This is a circle of radius $u/\omega$ centred at $(0,-u/\omega)$.

With $\alpha=\mu/m$, the complex <velocity> is $\dot x+i\dot y=u e^{-(\alpha+i\omega)t}$. Hence
$$x=\frac{u[\alpha(1-e^{-\alpha t}\cos\omega t)+\omega e^{-\alpha t}\sin\omega t]}{\alpha^2+\omega^2},$$
$$y=\frac{u[\alpha e^{-\alpha t}\sin\omega t-\omega(1-e^{-\alpha t}\cos\omega t)]}{\alpha^2+\omega^2}.$$
The path spirals into $u(\alpha,-\omega)/(\alpha^2+\omega^2)$. As $u$ varies, this lies on $y=-(\omega/\alpha)x$, at distance
$$\frac u{\sqrt{\alpha^2+\omega^2}}=\frac{mu}{\sqrt{\mu^2+B^2}}$$
from the origin.

Solved by gpt-5.6-sol high.

= 10B
{parent=Paper 4}
{scope}
{title2=Dynamics and Relativity}

= Solution
{parent=10b}

Binet's equation gives
$$r=\frac{r_0}{1+e\cos\theta},\qquad r_0=\frac{l^2}{k}.$$
Consequently
$$l=\sqrt{kr_0},\qquad E=\frac{mk(e^2-1)}{2r_0}.$$

Solved by gpt-5.6-sol high.

= a
{parent=10b}
{scope}

= Solution
{parent=a}

At infinity the specific energy is $v^2/2$ and $l=bv$, so
$$e=\sqrt{1+\frac{b^2v^4}{k^2}}.$$
The angle through which the <velocity> is deflected between the incoming and outgoing asymptotes is
$$\chi=2\tan^{-1}\left(\frac{k}{bv^2}\right).$$

Solved by gpt-5.6-sol high.

= b
{parent=10b}
{scope}

= Solution
{parent=b}

Zero energy gives $e=1$. Since the periapsis is $d$, $r_0=2d$ and $l=\sqrt{2kd}$. Put $T=\tan(\theta/2)$; then
$$r=d(1+T^2),\qquad \frac{dt}{dT}=\sqrt{\frac{2d^3}{k}}(1+T^2).$$
Taking $t=0$ at periapsis yields
$$\theta(t)=2\tan^{-1}T(t),\qquad T^3+3T=3t\sqrt{\frac{k}{2d^3}}.$$

Solved by gpt-5.6-sol high.

= 11B
{parent=Paper 4}
{scope}
{title2=Dynamics and Relativity}

= Solution
{parent=11b}

For axis unit <vector> $\hat\omega$,
$$T=\frac12\sum_i m_i|\omega\times x_i|^2=\frac12I\omega^2,\qquad I=\sum_i m_i|x_i-(x_i\cdot\hat\omega)\hat\omega|^2.$$
For a body, replace the sum by $I=\int\rho(x)r_\perp^2\,dV$.

For the hollow ball,
$$I=\frac25M\frac{R_+^5-R_-^5}{R_+^3-R_-^3}
=\beta MR_+^2,\qquad \beta=\frac25\frac{1-\mu^5}{1-\mu^3}.$$
Energy conservation and $v=R_+\Omega$ give
$$v=\sqrt{\frac{2gh}{1+\beta}}.$$
The factor $\beta$ increases with $\mu$, so if $\mu_A>\mu_B$, ball B accelerates more and arrives first. Moving the cavity off centre removes material with a larger lever arm and lowers the <moment of inertia> about the centre of mass; despite the resulting wobble, ball C therefore rolls down faster than A.

Solved by gpt-5.6-sol high.

= 12B
{parent=Paper 4}
{scope}
{title2=Dynamics and Relativity}

= Solution
{parent=12b}

With $\gamma=(1-v^2/c^2)^{-1/2}$,
$$U^\mu=\gamma(c,v),\qquad U_\mu U^\mu=\gamma^2(c^2-v^2)=c^2.$$
<Differentiation> with respect to <proper time> gives
$$A^\mu=\left(\frac{\gamma^4v\cdot a}{c},\ \gamma^2a+\frac{\gamma^4(v\cdot a)v}{c^2}\right).$$
Orthogonality $U\cdot A=0$ implies that in the instantaneous rest frame $A'^\mu=(0,a')$. For collinear motion a Lorentz boost gives
$$a'=\gamma^3a=\frac{a}{(1-v^2/c^2)^{3/2}}.$$
For constant positive <proper acceleration> $a'$,
$$v(t)=\frac{a't}{\sqrt{1+(a't/c)^2}},\qquad
x(t)=\frac{c^2}{a'}\left(\sqrt{1+(a't/c)^2}-1\right).$$
This hyperbola has future null asymptote $x=ct-c^2/a'$. A right-moving signal emitted from $x_0$ at $t=0$ follows $x=ct+x_0$; if $x_0<-c^2/a'$, it lies beyond that <acceleration> horizon and never intersects the worldline.

Solved by gpt-5.6-sol high.