past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2025/ia/paper-1.bigb

= Paper 1
{scope}

https://www.maths.cam.ac.uk/undergrad/pastpapers/files/2025/paperia_1_2025.pdf

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

= a
{parent=1b}
{scope}

= Solution
{parent=a}

Translate and scale the circle to $|z|=1$. For distinct points $u,v,w$ on it, the angle at $w$ is right exactly when
$$R=\frac{u-w}{v-w}$$
is purely imaginary. Since $\bar u=u^{-1}$ and similarly for $v,w$, direct simplification gives $\bar R=(v/u)R$. Thus $\bar R=-R$ exactly when $v=-u$, which says that the opposite side joins antipodal points and is a diameter. The converse is the same calculation in reverse.

Solved by gpt-5.6-sol high.

= b
{parent=1b}
{scope}

= Solution
{parent=b}

The vertices satisfy
$$(z+1)^N-1=0.$$
After removing the root $z=0$, the product of the other roots has <modulus> equal to the constant term of $((z+1)^N-1)/z$, namely $N$. Hence the product of the $N-1$ chord lengths from one vertex of a unit <regular polygon> is $N$. Multiplying this identity over all $N$ vertices counts every chord twice, so the product of all chord lengths is $N^{N/2}$. Scaling the circle by $R$ scales each of its $N(N-1)/2$ chords by $R$, giving
$$N^{N/2}R^{N(N-1)/2}.$$

Solved by gpt-5.6-sol high.

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

= a
{parent=2c}
{scope}

= Solution
{parent=a}

Expansion gives $\det A=(a-3)(a+2)$. Thus uniqueness fails precisely for $a=-2$ and $a=3$.

Solved by gpt-5.6-sol high.

= b
{parent=2c}
{scope}

= Solution
{parent=b}

Solvability requires $b$ to be orthogonal to the left nullspace. For $a=-2$ one may take $n=(-3,-2,2)^T$; for $a=3$ one may take $n=(1,-1,1)^T$.

Solved by gpt-5.6-sol high.

= c
{parent=2c}
{scope}

= Solution
{parent=c}

For $b=(2,b,0)^T$, compatibility gives $b=-3$ when $a=-2$ and $b=2$ when $a=3$. The respective solution families are
$$x=(1,-1,0)^T+t(1,1,1)^T$$
and
$$x=(1,-1,0)^T+t(-3/2,7/2,1)^T,$$
where $t\in\mathbb R$.

Solved by gpt-5.6-sol high.

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

= a
{parent=3e}
{scope}

= Solution
{parent=a}

For every $z\in\mathbb C$,
$$e^z=\sum_{n=0}^{\infty}\frac{z^n}{n!},\qquad
\sin z=\sum_{n=0}^{\infty}(-1)^n\frac{z^{2n+1}}{(2n+1)!}.$$

Solved by gpt-5.6-sol high.

= b
{parent=3e}
{scope}

= Solution
{parent=b}

Termwise <differentiation> gives
$$f\prime(z)=\sum_{n=1}^{\infty}n a_nz^{n-1}.$$
The differentiated <series> has the same radius of convergence $R$.

Solved by gpt-5.6-sol high.

= c
{parent=3e}
{scope}

= Solution
{parent=c}

Fix $a$ and define $g(b)=e^{a+b}e^{-b}$ using only the power <series>. Termwise <differentiation> gives $(e^z)\prime=e^z$, and the product rule gives $g\prime(b)=0$. Hence $g$ is constant, so $g(b)=g(0)=e^a$. Multiplication by $e^b$ yields $e^{a+b}=e^ae^b$.

Solved by gpt-5.6-sol high.

= d
{parent=3e}
{scope}

= Solution
{parent=d}

The removable definition gives the entire <series>
$$f(z)=\sum_{n=0}^{\infty}(-1)^n\frac{z^{2n}}{(2n+1)!}.$$
Therefore
$$f^{(k)}(0)=\begin{cases}0,&k\text{ odd},\\(-1)^{k/2}/(k+1),&k\text{ even}.
\end{cases}$$

Solved by gpt-5.6-sol high.

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

= a
{parent=4e}
{scope}

= Solution
{parent=a}

The Darboux–Riemann criterion says that a <bounded function> on $[a,b]$ is Riemann integrable exactly when, for every $\varepsilon>0$, some partition $P$ satisfies $U(f,P)-L(f,P)<\varepsilon$. Indeed, every lower sum is at most every upper sum. Taking the supremum of lower sums and infimum of upper sums, the criterion makes their difference smaller than every positive $\varepsilon$, so they are equal; this common value is the Riemann <integral>.

Solved by gpt-5.6-sol high.

= b
{parent=4e}
{scope}

= Solution
{parent=b}

Given $\varepsilon>0$, choose a partition of $[a,d]$ whose upper-minus-lower sum is below $\varepsilon$, and refine it by inserting $b,c$. The contribution from subintervals lying in $[b,c]$ is nonnegative and no larger than the total difference. Restricting the refined partition to $[b,c]$ therefore gives upper-minus-lower sum below $\varepsilon$, so the criterion proves integrability there.

Solved by gpt-5.6-sol high.

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

= Solution
{parent=5b}

Taking the parallelogram spanned by $e_2,e_3$ as base gives base area $|e_2\times e_3|$ and height $|e_1\cdot(e_2\times e_3)|/|e_2\times e_3|$. A tetrahedron has one third of the corresponding pyramid volume and the triangle has half the parallelogram base, hence
$$V=\frac16|\Delta|,\qquad \Delta=e_1\cdot(e_2\times e_3).$$
The reciprocal <vectors> are
$$f_1=\frac{e_2\times e_3}{\Delta},\quad f_2=\frac{e_3\times e_1}{\Delta},\quad f_3=\frac{e_1\times e_2}{\Delta},$$
which directly satisfy $f_i\cdot e_j=\delta_{ij}$.

Solved by gpt-5.6-sol high.

= i
{parent=5b}
{scope}

= Solution
{parent=i}

The final face contains $v_i=v+e_i$, so $c\cdot(v+e_i)+d=0$. Therefore $e_i\cdot c=-(c\cdot v+d)$.

Solved by gpt-5.6-sol high.

= ii
{parent=5b}
{scope}

= Solution
{parent=ii}

The $i$th face through $v$ contains the two edge directions $e_j$ with $j\ne i$, so $a_i\cdot e_j=0$ for those $j$. The one-dimensional space with these two orthogonality conditions is spanned by $f_i$, hence $a_i=\lambda_i f_i$ for some nonzero real $\lambda_i$.

Solved by gpt-5.6-sol high.

= iii
{parent=5b}
{scope}

= Solution
{parent=iii}

Dotting $c=\sum_j\gamma_ja_j$ with $e_i$ gives $e_i\cdot c=\gamma_i\lambda_i$, proving $\gamma_i=(e_i\cdot c)/\lambda_i$. Also
$$a_1\cdot(a_2\times a_3)=\frac{\lambda_1\lambda_2\lambda_3}{\Delta},\qquad
\gamma_1\gamma_2\gamma_3=-\frac{(c\cdot v+d)^3}{\lambda_1\lambda_2\lambda_3}.$$
Combining these with $V=|\Delta|/6$ gives the displayed face formula, with the orientation chosen so its signed right-hand side is positive.

Solved by gpt-5.6-sol high.

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

= a
{parent=6c}
{scope}

= Solution
{parent=a}

The distinguished unit <vector> is $n=(1,1,1)/\sqrt3$. Comparing diagonal, symmetric off-diagonal, and antisymmetric parts gives
$$\alpha=a-\frac{b+c}{2},\qquad \beta=\frac32(b+c),\qquad \gamma=\frac{\sqrt3}{2}(b-c),$$
so $A_{ij}=\alpha\delta_{ij}+\beta n_in_j+\gamma\varepsilon_{ijk}n_k$.

Solved by gpt-5.6-sol high.

= b
{parent=6c}
{scope}

= Solution
{parent=b}

On the line spanned by $n$, $A$ multiplies by $L=\alpha+\beta=a+b+c$. On $n^\perp$, it is the composition of a rotation with a dilation by $r=\sqrt{\alpha^2+\gamma^2}$. Consequently every area in that plane is multiplied by $r^2=\alpha^2+\gamma^2$.

Solved by gpt-5.6-sol high.

= c
{parent=6c}
{scope}

= i
{parent=c}
{scope}

= Solution
{parent=i}

For a plane reflection, the $n$ direction is reversed while $n^\perp$ is fixed. Thus $L=-1$, $\alpha=1$, and $\gamma=0$, equivalently
$$a=\frac13,\qquad b=c=-\frac23.$$

Solved by gpt-5.6-sol high.

= ii
{parent=c}
{scope}

= Solution
{parent=ii}

The map is a rotation about the $n$ axis exactly when
$$a+b+c=1,\qquad \alpha^2+\gamma^2=1,$$
or explicitly
$$(2a-b-c)^2+3(b-c)^2=4.$$
These conditions make the axis fixed and the perpendicular-plane action length preserving; they are also necessary.

Solved by gpt-5.6-sol high.

= d
{parent=6c}
{scope}

= Solution
{parent=d}

Let $N=nn^T$ and $K_{ij}=\varepsilon_{ijk}n_k$. Since $K^2=N-I$, inversion separately on $\mathbb Rn$ and $n^\perp$ gives
$$(A^{-1})_{ij}=\frac{\alpha}{\alpha^2+\gamma^2}\delta_{ij}
+\left(\frac1{\alpha+\beta}-\frac{\alpha}{\alpha^2+\gamma^2}\right)n_in_j
-\frac{\gamma}{\alpha^2+\gamma^2}\varepsilon_{ijk}n_k,$$
provided $(\alpha+\beta)(\alpha^2+\gamma^2)\ne0$.

Solved by gpt-5.6-sol high.

= e
{parent=6c}
{scope}

= Solution
{parent=e}

The line/plane decomposition gives
$$\det A=(\alpha+\beta)(\alpha^2+\gamma^2).$$
Substituting the values in part (a) and simplifying yields
$$\det A=\frac12(a+b+c)\bigl((a-b)^2+(b-c)^2+(c-a)^2\bigr).$$

Solved by gpt-5.6-sol high.

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

= a
{parent=7a}
{scope}

= Solution
{parent=a}

An <eigenvalue> $\lambda$ is a <scalar> for which $\ker(A-\lambda I)$ contains a nonzero <vector>; that kernel is its eigenspace. The characteristic <polynomial> has degree $n$, so the <fundamental theorem of algebra> supplies at least one complex root and hence an <eigenvalue>. Its eigenspace dimension can be any integer from $1$ to $n$.

Solved by gpt-5.6-sol high.

= b
{parent=7a}
{scope}

= Solution
{parent=b}

With $\chi_A(t)=\det(tI-A)$, direct expansion gives
$$\chi_A(t)=(t-3)(t-1)^2(t+1)(t+2).$$
Thus the <eigenvalues> are $3,1,-1,-2$. Kernel calculation gives <eigenspace> dimensions $1,2,1,1$, respectively.

Solved by gpt-5.6-sol high.

= c
{parent=7a}
{scope}

= Solution
{parent=c}

The <eigenvalue> equation is $u_{i+1}=\lambda u_i$, so
$$u=c(1,\lambda,\lambda^2,\ldots)^T.$$
This is a nonzero square-summable <vector> exactly when $c\ne0$ and $|\lambda|<1$. Hence every point of the open unit disc is an <eigenvalue>, with the displayed one-dimensional eigenspace.

Solved by gpt-5.6-sol high.

= d
{parent=7a}
{scope}

= Solution
{parent=d}

For the transpose shift, $(Cu)_1=0$ and $(Cu)_i=u_{i-1}$ for $i\ge2$. If $Cu=\lambda u$, the first equation and the subsequent recurrence force every component to vanish, both for $\lambda=0$ and for $\lambda\ne0$. Thus $C$ has no <eigenvalues>, illustrating that a bounded infinite-dimensional operator need not have one.

Solved by gpt-5.6-sol high.

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

= a
{parent=8a}
{scope}

= Solution
{parent=a}

A complex <matrix> is diagonalisable when it is similar to a diagonal <matrix>, equivalently when the <vector space> has a <basis> of its <eigenvectors>.

Solved by gpt-5.6-sol high.

= b
{parent=8a}
{scope}

= Solution
{parent=b}

If $Av=\lambda v$, then $p(A)v=p(\lambda)v$, proving one inclusion. Conversely, if $\mu$ is an <eigenvalue> of $p(A)$, factor $p(z)-\mu=c\prod_j(z-\lambda_j)$. If none of the $\lambda_j$ were <eigenvalues> of $A$, every $A-\lambda_jI$ would be invertible, making $p(A)-\mu I$ invertible, a contradiction. Thus $\mu=p(\lambda)$ for some <eigenvalue> $\lambda$ of $A$.

Solved by gpt-5.6-sol high.

= c
{parent=8a}
{scope}

= Solution
{parent=c}

Write $A=S\operatorname{diag}(\lambda_1,\ldots,\lambda_n)S^{-1}$. The power <series> gives $e^A=S\operatorname{diag}(e^{\lambda_1},\ldots,e^{\lambda_n})S^{-1}$. Therefore
$$\det(e^A)=\prod_i e^{\lambda_i}=e^{\sum_i\lambda_i}=e^{\operatorname{tr}A}.$$

Solved by gpt-5.6-sol high.

= d
{parent=8a}
{scope}

= Solution
{parent=d}

The <vectors> $(1,0,1)^T,(0,1,0)^T,(1,0,-1)^T$ are <eigenvectors> of $B$ with <eigenvalues> $2,1,0$. Applying the exponential to these eigenspaces gives
$$e^B=\begin{pmatrix}(1+e^2)/2&0&(e^2-1)/2\\0&e&0\\(e^2-1)/2&0&(1+e^2)/2\end{pmatrix}.$$

Solved by gpt-5.6-sol high.

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

= a
{parent=9e}
{scope}

= Solution
{parent=a}

A <sequence> $(a_n)$ is Cauchy when, for every $\varepsilon>0$, some $N$ satisfies $|a_m-a_n|<\varepsilon$ whenever $m,n\ge N$.

Solved by gpt-5.6-sol high.

= b
{parent=9e}
{scope}

= Solution
{parent=b}

The general principle of convergence says that a real <sequence> converges exactly when it is Cauchy. Convergent <sequences> are Cauchy by the triangle inequality. Conversely, a Cauchy <sequence> is bounded, so Bolzano–Weierstrass gives a convergent subsequence $a_{n_k}\to a$; the Cauchy property then forces the entire <sequence> to converge to $a$.

Solved by gpt-5.6-sol high.

= c
{parent=9e}
{scope}

= i
{parent=c}
{scope}

= Solution
{parent=i}

True. For $m>n$, $|a_m-a_n|\le\sum_{j=n}^{m-1}d_j$. Convergence of $\sum d_j$ makes this tail arbitrarily small, so $(a_n)$ is Cauchy and hence convergent.

Solved by gpt-5.6-sol high.

= ii
{parent=c}
{scope}

= Solution
{parent=ii}

False. The <sequence> $a_n=(-1)^n/n$ converges to zero, but $|a_{n+1}-a_n|=1/n+1/(n+1)$, whose <series> diverges.

Solved by gpt-5.6-sol high.

= d
{parent=9e}
{scope}

= Solution
{parent=d}

Write
$$\frac{j}{j^2+j+1}=\frac1j+O(j^{-2})$$
uniformly for $j\ge1$. The accumulated error from $j=n+1$ to $2n$ tends to zero, while the corresponding <harmonic sum> tends to $\log2$. Hence $b_n\to\log2$.

Solved by gpt-5.6-sol high.

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

= a
{parent=10e}
{scope}

= Solution
{parent=a}

The statement $\lim_{t\to x}f(t)=\ell$ means that for every $\varepsilon>0$ there is $\delta>0$ such that $t\in[0,1]$ and $0<|t-x|<\delta$ imply $|f(t)-\ell|<\varepsilon$.

Solved by gpt-5.6-sol high.

= b
{parent=10e}
{scope}

= Solution
{parent=b}

The <function> is continuous at $x$ exactly when $\lim_{t\to x}f(t)=f(x)$.

Solved by gpt-5.6-sol high.

= c
{parent=10e}
{scope}

= Solution
{parent=c}

If $x_n\to x$, continuity of $f$ gives $f(x_n)\to f(x)$, and continuity of $g$ at $f(x)$ then gives $g(f(x_n))\to g(f(x))$. The sequential criterion proves continuity of $g\circ f$ at $x$.

Solved by gpt-5.6-sol high.

= d
{parent=10e}
{scope}

= Solution
{parent=d}

Both <functions> are continuous at every $x>0$. At zero, $f_1(1/(\pi/2+2\pi n))=1$ while $f_1(1/(3\pi/2+2\pi n))=-1$, so $f_1$ is discontinuous. Since $|x\sin(1/x)|\le x\to0$, $f_2$ is continuous at zero and hence everywhere.

Solved by gpt-5.6-sol high.

= e
{parent=10e}
{scope}

= i
{parent=e}
{scope}

= Solution
{parent=i}

Let $S=\{0\}\cup\{1/n:n\ge1\}$ and let $D$ be the Dirichlet <function>, equal to $1$ on rationals and $0$ on irrationals. Then $f(x)=d(x,S)D(x)$ is continuous exactly on $S$: the distance factor squeezes it to zero on $S$, while away from $S$ it is a positive continuous factor times an everywhere-discontinuous <function>.

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= ii
{parent=e}
{scope}

= Solution
{parent=ii}

Using the preceding $f$, define $g(0)=f(0)+1$ and $g(x)=f(x)$ for $x>0$. Changing the value only at zero makes zero discontinuous while preserving continuity at every $1/n$ and discontinuity everywhere else. Thus the continuity set is precisely $\{1/n:n\ge1\}$.

Solved by gpt-5.6-sol high.

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

= a
{parent=11e}
{scope}

= Solution
{parent=a}

Differentiability at $x$ means that the <limit>
$$f\prime(x)=\lim_{h\to0}\frac{f(x+h)-f(x)}h$$
exists as a finite real number.

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= b
{parent=11e}
{scope}

= Solution
{parent=b}

If $f$ is continuous on $[x,y]$ and <differentiable> on $(x,y)$, then some $c\in(x,y)$ satisfies $f(y)-f(x)=f\prime(c)(y-x)$.

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= c
{parent=11e}
{scope}

= Solution
{parent=c}

If $f\prime\ge0$, the <mean value theorem> gives $f(y)-f(x)=f\prime(c)(y-x)\ge0$. Conversely, if $f$ is increasing, every difference quotient with positive or negative increment is nonnegative; taking its <limit> gives $f\prime(x)\ge0$.

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= d
{parent=11e}
{scope}

= Solution
{parent=d}

For $x>0$, apply the <mean value theorem> to $[0,x]$: $f(x)/x=f\prime(c)$ for some $c<x$. Since $f\prime$ is increasing, $f(x)/x\le f\prime(x)$. Therefore
$$\left(\frac{f(x)}x\right)\prime=\frac{xf\prime(x)-f(x)}{x^2}\ge0,$$
and part (c) proves that the quotient is increasing.

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

= a
{parent=12e}
{scope}

= Solution
{parent=a}

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

Solved by gpt-5.6-sol high.

= b
{parent=12e}
{scope}

= Solution
{parent=b}

The fundamental theorem gives
$$g(b)-f(b)=g(a)-f(a)+\int_a^b(g\prime-f\prime)\,dx\ge0,$$
which is the desired inequality.

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= c
{parent=12e}
{scope}

= Solution
{parent=c}

The assumptions imply $f\ge0$. Since $(f^2)\prime=2ff\prime\le2f$ and both sides vanish at zero, integration gives $f(x)^2\le2F(x)$ where $F(x)=\int_0^xf$. Now
$$\frac d{dx}\left(F(x)^2-\int_0^xf(t)^3dt\right)=2Ff-f^3=f(2F-f^2)\ge0.$$
The bracket vanishes at zero, proving $\int_0^xf^3\le(\int_0^xf)^2$.

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= d
{parent=12e}
{scope}

= Solution
{parent=d}

No. Take the <constant function> $f(t)=c>0$, which has $f\prime=0$. The claimed inequality becomes $c^3x\le c^2x^2$, false whenever $0<x<c$.

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