past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2026/ia/paper-1.bigb

= Paper 1
{scope}

https://www.maths.cam.ac.uk/undergrad/pastpapers/files/2026/Paperia_1_2026.pdf

= 1A
{parent=Paper 1}
{scope}
{title2=Vectors and Matrices}

= a
{parent=1a}
{scope}

= Solution
{parent=a}

The principal value is $\log z=\log|z|+i\operatorname{Arg}z$, where $-\pi\lt \operatorname{Arg}z\leq\pi$ and $z\ne0$.

Solved by gpt-5.6-sol high.

= b
{parent=1a}
{scope}

= Solution
{parent=b}

Since $1+i=\sqrt2e^{i\pi/4}$, the principal value is $\tfrac12\log2+i\pi/4$. All values are $\tfrac12\log2+i(\pi/4+2\pi k)$, $k\in\mathbb Z$.

Solved by gpt-5.6-sol high.

= c
{parent=1a}
{scope}

= Solution
{parent=c}

Write $2=2e^{2\pi i k}$. Then $z=i+2^{1/5}e^{2\pi i k/5}$ for $k=0,\ldots,4$: the vertices of a regular pentagon of radius $2^{1/5}$ centred at $i$.

Solved by gpt-5.6-sol high.

= 2C
{parent=Paper 1}
{scope}
{title2=Vectors and Matrices}

= Solution
{parent=2c}

Let $P$ have the old domain <basis> as columns, $C$ the old codomain <basis>, and $D$ the new codomain <basis>. Change of coordinates gives $A\prime=D^{-1}CAP^{-1}$, hence
$$A\prime=\begin{pmatrix}1&4&3\\0&-3&-1\end{pmatrix}.$$

Solved by gpt-5.6-sol high.

= 3E
{parent=Paper 1}
{scope}
{title2=Analysis I}

= a
{parent=3e}
{scope}

= Solution
{parent=a}

The radius is $R=\sup\{r\geq0:\sum a_nz^n\text{ converges whenever }|z|\lt r\}$; equivalently $R^{-1}=\limsup|a_n|^{1/n}$. The same definition applies over $\mathbb R$ and $\mathbb C$.

Solved by gpt-5.6-sol high.

= b
{parent=3e}
{scope}

= i
{parent=b}
{scope}

= Solution
{parent=i}

Here $a_{2n}=(-1)^n$ and $a_{2n+1}=0$, so $\sum a_nz^n=\sum(-z^2)^n=(1+z^2)^{-1}=((z-i)(z+i))^{-1}$ for $|z|\lt 1$.

Solved by gpt-5.6-sol high.

= ii
{parent=b}
{scope}

= Solution
{parent=ii}

Along real $z=x\uparrow1$, the <modulus> is $1/(1+x^2)$ and stays bounded, but along $z=iy$ with $y\uparrow1$ it is $1/(1-y^2)\to\infty$. Since $\limsup|a_n|^{1/n}=1$, $R=1$.

Solved by gpt-5.6-sol high.

= c
{parent=3e}
{scope}

= Solution
{parent=c}

No. For example $\sum_{n\ge0}(-1)^nx^n=(1+x)^{-1}$ has radius $1$ and is bounded by $1$ on $(0,1)$.

Solved by gpt-5.6-sol high.

= 4E
{parent=Paper 1}
{scope}
{title2=Analysis I}

= a
{parent=4e}
{scope}

= Solution
{parent=a}

Every bounded <sequence> in $\mathbb R$ has a convergent subsequence.

Solved by gpt-5.6-sol high.

= b
{parent=4e}
{scope}

= Solution
{parent=b}

$a_n\to a$ means that for every $\varepsilon\gt 0$ there is $N$ such that $n\ge N$ implies $|a_n-a|\lt \varepsilon$.

Solved by gpt-5.6-sol high.

= c
{parent=4e}
{scope}

= Solution
{parent=c}

If $(a_n)$ is bounded, apply Bolzano–Weierstrass to $(\Re a_n,\Im a_n)$ (or successively to both coordinates). If it is unbounded, recursively choose $n_k\gt n_{k-1}$ with $|a_{n_k}|\ge k^2$. The alternatives are not exclusive: $0,1,0,4,0,9,\ldots$ has both kinds of subsequence.

Solved by gpt-5.6-sol high.

= 5A
{parent=Paper 1}
{scope}
{title2=Vectors and Matrices}

= a
{parent=5a}
{scope}

= Solution
{parent=a}

$x\cdot y=\sum_i x_iy_i$ and $\|x\|=(x\cdot x)^{1/2}$.

Solved by gpt-5.6-sol high.

= b
{parent=5a}
{scope}

= Solution
{parent=b}

<Cauchy-Schwarz inequality> states $|x\cdot y|\le\|x\|\|y\|$. For $y\ne0$, positivity of $\|x-ty\|^2$ at $t=(x\cdot y)/\|y\|^2$ gives $\|x\|^2-(x\cdot y)^2/\|y\|^2\ge0$; $y=0$ is immediate.

Solved by gpt-5.6-sol high.

= c
{parent=5a}
{scope}

= Solution
{parent=c}

$(b-a)\times(c-a)=(-1,0,0)\times(-1,-1,1)=(0,1,1)$. Thus one choice is $\hat n=(0,1,1)/\sqrt2$ and $\beta=\sqrt2$.

Solved by gpt-5.6-sol high.

= d
{parent=5a}
{scope}

= Solution
{parent=d}

The distance is $|d\cdot\hat n-\beta|=|3/\sqrt2-\sqrt2|=1/\sqrt2$.

Solved by gpt-5.6-sol high.

= e
{parent=5a}
{scope}

= Solution
{parent=e}

The centre-to-plane distance is $|g-2|/\sqrt2$. Intersection occurs iff this is at most $2\sqrt2$, i.e. $-2\le g\le6$. Equality gives one tangent point; strict inequality gives a circle of radius $\sqrt{8-(g-2)^2/2}$.

Solved by gpt-5.6-sol high.

= 6C
{parent=Paper 1}
{scope}
{title2=Vectors and Matrices}

= a
{parent=6c}
{scope}

= i
{parent=a}
{scope}

= Solution
{parent=i}

$A(A^2-4A+5I)=2I$, so $A$ has inverse.

Solved by gpt-5.6-sol high.

= ii
{parent=a}
{scope}

= Solution
{parent=ii}

$A^{-1}=\tfrac12(A^2-4A+5I)$.

Solved by gpt-5.6-sol high.

= iii
{parent=a}
{scope}

= Solution
{parent=iii}

The relation $A^3=4A^2-5A+2I$ reduces $A^k$ inductively to a linear combination of $I,A,A^2$.

Solved by gpt-5.6-sol high.

= b
{parent=6c}
{scope}

= i
{parent=b}
{scope}

= Solution
{parent=i}

$Bx=0\iff ASx=0\iff x\in S^{-1}(\ker A)$; invertibility of $S$ preserves dimension.

Solved by gpt-5.6-sol high.

= ii
{parent=b}
{scope}

= Solution
{parent=ii}

<Rank-nullity theorem> gives $\operatorname{rank}T+\dim\ker T=n$. Equal nullities therefore give equal ranks.

Solved by gpt-5.6-sol high.

= iii
{parent=b}
{scope}

= Solution
{parent=iii}

$B^k=S^{-1}A^kS$, so the preceding argument applied to $A^k,B^k$ gives equal ranks.

Solved by gpt-5.6-sol high.

= c
{parent=6c}
{scope}

= Solution
{parent=c}

<Cayley-Hamilton theorem> says $\chi_B(B)=0$. Since $B$ is nilpotent, all <eigenvalues> and hence every non-leading coefficient of $\chi_B(t)$ vanish, so $\chi_B(t)=t^n$ and $B^n=0$.

Solved by gpt-5.6-sol high.

= 7B
{parent=Paper 1}
{scope}
{title2=Vectors and Matrices}

= a
{parent=7b}
{scope}

= Solution
{parent=a}

From $Hv=\mu v$, $v^\dagger Hv=\mu\|v\|^2$ is real, so $\mu\in\mathbb R$. For distinct <eigenvalues>, $\mu_i v_i^\dagger v_j=v_i^\dagger Hv_j=\mu_jv_i^\dagger v_j$, hence orthogonality. Normalising the <eigenvectors> and taking them as the rows of $U$ gives $H=U^\dagger\Lambda U$.

Solved by gpt-5.6-sol high.

= b
{parent=7b}
{scope}

= Solution
{parent=b}

$\chi_H(\lambda)=\det(H-\lambda I)=\prod_i(\mu_i-\lambda)$ for this sign convention.

Solved by gpt-5.6-sol high.

= c
{parent=7b}
{scope}

= Solution
{parent=c}

The subspace $y^\perp$ is fixed by $M=I-xy^\dagger$, while the remaining <determinant> factor is $1-y^\dagger x$. Thus $\det M=1-y^*\cdot x$.

Solved by gpt-5.6-sol high.

= d
{parent=7b}
{scope}

= Solution
{parent=d}

Apply the <matrix determinant lemma> to $HP-\lambda I=(H-\lambda I)-Hu,u^T$: its <determinant> is $\chi_H(\lambda)[1-u^T(H-\lambda I)^{-1}Hu]$.

Solved by gpt-5.6-sol high.

= e
{parent=7b}
{scope}

= Solution
{parent=e}

$Pu=0$, hence $HPu=0$. In the diagonal <basis>, $f(\lambda)=1-\sum_j u_j^2\mu_j/(\mu_j-\lambda)=-\lambda\sum_j u_j^2/(\mu_j-\lambda)$. Between consecutive $\mu_j$ its nonzero factor is strictly monotone from one infinite sign to the other, giving one root; zero weights give equality at endpoints. Hence $\mu_1\le\sigma_1\le\mu_2\le\cdots\le\sigma_{n-1}\le\mu_n$.

Solved by gpt-5.6-sol high.

= 8B
{parent=Paper 1}
{scope}
{title2=Vectors and Matrices}

= a
{parent=8b}
{scope}

= Solution
{parent=a}

All three Pauli <matrices> are Hermitian. Direct multiplication gives $\sigma_a^2=I$ and, for $a\ne b$, $\sigma_a\sigma_b=i\varepsilon_{abc}\sigma_c$, which combines as the displayed identity.

Solved by gpt-5.6-sol high.

= b
{parent=8b}
{scope}

= Solution
{parent=b}

Because $(n\cdot\sigma)^2=I$, even and odd powers sum separately: $R=I\cos\theta+i(n\cdot\sigma)\sin\theta$. Then $RR^\dagger=I$.

Solved by gpt-5.6-sol high.

= c
{parent=8b}
{scope}

= Solution
{parent=c}

$A=\begin{pmatrix}x_3&x_1-ix_2\\x_1+ix_2&-x_3\end{pmatrix}$, so $\operatorname{tr}A=0$ and $\det A=-|x|^2$. Unitary conjugation preserves Hermiticity, trace, and <determinant>; every traceless Hermitian $2\times2$ <matrix> is $x\prime\cdot\sigma$ with real $x\prime$, and $|x\prime|=|x|$.

Solved by gpt-5.6-sol high.

= d
{parent=8b}
{scope}

= Solution
{parent=d}

Using $(a\cdot\sigma)(b\cdot\sigma)=(a\cdot b)I+i(a\times b)\cdot\sigma$ gives
$$x\prime=x\cos(2\theta)-(n\times x)\sin(2\theta)+n(n\cdot x)(1-\cos(2\theta)).$$
The sign follows from the convention $R=e^{i\theta n\cdot\sigma}$.

Solved by gpt-5.6-sol high.

= 9E
{parent=Paper 1}
{scope}
{title2=Analysis I}

= a
{parent=9e}
{scope}

= Solution
{parent=a}

For a <polynomial> of degree $d$, Taylor expansion about $x_0$ terminates: $P(x_0+h)=\sum_{n=0}^dP^{(n)}(x_0)h^n/n!$.

Solved by gpt-5.6-sol high.

= b
{parent=9e}
{scope}

= Solution
{parent=b}

If $f$ has $n+1$ <derivatives> near $x_0$, then $f(x)=\sum_{k=0}^nf^{(k)}(x_0)(x-x_0)^k/k!+f^{(n+1)}(\xi)(x-x_0)^{n+1}/(n+1)!$ for some intermediate $\xi$.

Solved by gpt-5.6-sol high.

= c
{parent=9e}
{scope}

= Solution
{parent=c}

Choose a compact interval around $x_0$ and its constant $C$. For $|h|\lt 1/C$, the Lagrange remainder after order $n$ is at most $(C|h|)^{n+1}\to0$. Thus the <taylor series> with $a_n=f^{(n)}(x_0)/n!$ equals $f(x_0+h)$.

Solved by gpt-5.6-sol high.

= d
{parent=9e}
{scope}

= Solution
{parent=d}

$e^x$ is real analytic because $e^{x_0+h}=e^{x_0}\sum_{n\ge0}h^n/n!$; it is not a <polynomial> since all its <derivatives> are nonzero.

Solved by gpt-5.6-sol high.

= 10E
{parent=Paper 1}
{scope}
{title2=Analysis I}

= a
{parent=10e}
{scope}

= Solution
{parent=a}

If $f$ is continuous on $[a,b]$, <differentiable> on $(a,b)$, and $f(a)=f(b)$, then some $c\in(a,b)$ has $f\prime(c)=0$. A maximum or minimum is attained; unless $f$ is constant, one extremum lies inside, where the <derivative> vanishes.

Solved by gpt-5.6-sol high.

= b
{parent=10e}
{scope}

= Solution
{parent=b}

The <mean value theorem> gives $f(b)-f(a)=(b-a)f\prime(c)\gt 0$.

Solved by gpt-5.6-sol high.

= c
{parent=10e}
{scope}

= Solution
{parent=c}

Since $f\prime\prime\prime\lt 0$, $f\prime\prime$ decreases from $0$, so $f\prime$ strictly decreases from $1$ and $f(x)\lt x$. If $f\prime$ never vanished, its decreasing positive <limit> would make $f$ stay positive, but $f\prime\prime\lt 0$ bounded away from zero after any fixed point forces $f\prime$ negative eventually. Thus $f$ rises to one maximum and then decreases strictly to $-\infty$, crossing zero exactly once; it is positive before that crossing.

Solved by gpt-5.6-sol high.

= d
{parent=10e}
{scope}

= Solution
{parent=d}

$\sin x=\sum_{n\ge0}(-1)^nx^{2n+1}/(2n+1)!$ and $\cos x=\sum_{n\ge0}(-1)^nx^{2n}/(2n)!$. Here $\sin\prime\prime\prime=-\cos$ is negative while $\cos\gt 0$ near zero; applying the preceding shape argument until the first zero of $\cos$ and then the addition identities establishes a first positive zero $x_0$ with positivity before it.

Solved by gpt-5.6-sol high.

= e
{parent=10e}
{scope}

= Solution
{parent=e}

From the addition formula and $\sin x_0=0$, $\cos x_0=-1$, so $2x_0$ is a period. Repeatedly subtract $2x_0$ from a positive period $2x_1$; the remainder in $[0,2x_0)$ is also a period, and positivity plus the addition identities force it to be $0$. Thus $x_1=nx_0$; the converse follows from periodicity and oddness.

Solved by gpt-5.6-sol high.

= 11E
{parent=Paper 1}
{scope}
{title2=Analysis I}

= a
{parent=11e}
{scope}

= Solution
{parent=a}

For continuous $f$, $F(x)=\int_a^xf(t)dt$ is <differentiable> with $F\prime=f$. Conversely, if $F\prime$ is continuous, $\int_a^bF\prime=F(b)-F(a)$.

Solved by gpt-5.6-sol high.

= b
{parent=11e}
{scope}

= Solution
{parent=b}

Put $h(x)=|f(0)|+\int_0^xg(t)|f(t)|dt$. Then $|f|\le h$ and $h\prime\le gh$. If $h\gt 0$, $(\log h)\prime\le g$; integration gives $h(x)\le h(0)e^{\int_0^xg}$. For $f(0)=0$, apply the same argument after adding $\varepsilon$ and let $\varepsilon\downarrow0$.

Solved by gpt-5.6-sol high.

= c
{parent=11e}
{scope}

= Solution
{parent=c}

For $h=f_1-f_2$, $h\prime=\sqrt{1+x}(f_1+f_2)h$ and $h(0)=0$. On every compact subinterval the coefficient is continuous, so <Gronwall inequality> gives $h=0$; hence equality holds throughout $[0,a)$.

Solved by gpt-5.6-sol high.

= 12E
{parent=Paper 1}
{scope}
{title2=Analysis I}

= a
{parent=12e}
{scope}

= Solution
{parent=a}

False. Every interval contains rationals and irrationals, so every lower Darboux sum is $0$ and every upper sum is $1$.

Solved by gpt-5.6-sol high.

= b
{parent=12e}
{scope}

= Solution
{parent=b}

False. Thomae’s <function> is Riemann integrable and discontinuous at every rational, an infinite countable set.

Solved by gpt-5.6-sol high.

= c
{parent=12e}
{scope}

= Solution
{parent=c}

False. <Continuous functions> can converge pointwise to the non-Riemann-integrable Dirichlet <function>; for example enumerate the rationals and use continuous narrow bumps whose first $n$ peaks cover the first $n$ rationals while widths shrink suitably pointwise off $\mathbb Q$.

Solved by gpt-5.6-sol high.

= d
{parent=12e}
{scope}

= Solution
{parent=d}

False without continuity: the <function> equal to $1$ at one point and $0$ elsewhere is nonnegative, Riemann integrable with <integral> $0$, but is not identically zero.

Solved by gpt-5.6-sol high.

= e
{parent=12e}
{scope}

= Solution
{parent=e}

False. The <Cantor function> has <derivative> $0$ outside the Cantor set, hence its <derivative> extends to the Riemann-integrable zero <function>, while $F(1)-F(0)=1$.

Solved by gpt-5.6-sol high.

= f
{parent=12e}
{scope}

= Solution
{parent=f}

True. If $|f(x_0)|\gt C$, continuity gives an interval of positive length on which $|f|\gt C+\delta$. Its $L^p$ norm exceeds $(C+\delta)\ell^{1/p}$, which tends to $C+\delta\gt C$ as $p\to\infty$, a contradiction.

Solved by gpt-5.6-sol high.