Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2025 / Paper 2

The prescribed orientation orders the two edges from $A$ as $\mathbf{c}-\mathbf{a}$ followed by $\mathbf{b}-\mathbf{a}$. The vector area is half their cross product:The sign agrees with the right-hand rule for the traversal $A\to C\to B\to A$.

For a constant vector field, the flux integral is its dot product with the vector area. Thus

No. Every one of the three vectors has equal first and second Cartesian coordinates, so their linear span is contained in the plane $x=y$. That linear subspace has dimension two and cannot equal ${\mathbb{R}}^3$.

More explicitly, call the three given vectors $\mathbf{u},\mathbf{v},\mathbf{w}$ in their listed order. ThenThis nontrivial linear relation proves linear dependence, whereas a basis must be linearly independent. Being unit vectors alone does not ensure that they form a basis.

Move the right-hand side to the left. The resulting homogeneous linear system has coefficient matrixA nonzero solution exists exactly when its determinant vanishes:Here the upper-left entry of the original matrix is $-1$, as verified in the original PDF; that entry is corrupted in the local OCR.

The determinant condition reduces to $3-2b=0$, givingAs a check, the first linear equation gives $y=2x$, and at this value of $b$ the second linear equation is satisfied for every such pair, including nonzero pairs.

The product-to-sum formula immediately gives the Fourier series:In the convention $f(x)=a_0/2+{\sum }_{n\ge 1}(a_n\cos nx+b_n\sin nx)$, the only nonzero Fourier coefficients are $a_2=a_8=1/2$. This is a finite trigonometric polynomial, and the identity holds at both endpoints as well.

The Kronecker delta restricts the Einstein summation convention to equal indices, giving the dot product

By the definitions of matrix multiplication, transpose and trace,Relabelling the summed indices givesthe Frobenius inner product of these real matrices.

The complementary function solves the associated homogeneous linear differential equation. Substituting $y=e^{mx}$ gives the characteristic equationIts distinct polynomial roots giveThe forcing affects the particular solution, which is not needed for the complementary function.

Use spherical coordinates $x=s\sin \theta \cos \varphi$, $y=s\sin \theta \sin \varphi$, $z=s\cos \theta$. The density becomes ${\rho }_{0}s\sin \theta$, and the Jacobian determinant supplies $dV=s^2\sin \theta dsd\theta d\varphi$. Therefore the mass is the volume integralUsing ${\sin }^2\theta =(1-\cos 2\theta )/2$, the three integrals are $2\pi$, $\pi /2$ and $1/4$. Hence

The divergence sums the corresponding partial derivatives:

By the Cartesian formula for the curl,

An eigenvalue makes the shifted matrix singular, so its characteristic polynomial vanishes:Thus the characteristic equation is

Applying the quadratic formula to the characteristic polynomial gives the two eigenvalues

The three tosses are independent events, each with head probability $1/2$. Consequently

The successful ordered outcomes are $TTH$, $THT$ and $HTT$. Each has probability $1/8$ by independence, and these events are mutually exclusive. Adding their probabilities givesEquivalently, the binomial distribution gives $\left(\genfrac{}{}{0ex}{}{3}{2}\right)(1/2{)}^2(1/2)=3/8$.

Using the complex conjugate, $b\bar{z}+\bar{b}z=2\mathrm{Re}(\bar{b}z)$. The locus is therefore the straight lineIn the complex plane, the quantity $\mathrm{Re}(\bar{b}z)/|b|$ is the signed scalar projection of $z$ in the direction $b$. Thus the straight line is perpendicular to $b$ and passes through $-cb/(2|b{|}^2)$. This description uses no separation of $z$ into real coordinates.

Multiplying the complex numbers gives $b\bar{z}+\bar{b}z=2(b_{1}x+b_{2}y)$. Hence the real linear equation is

Completing the square directly with the complex modulus givesFor $|b{|}^2>c$, this is the circle with centre $-b$ and radius $\sqrt{|b{|}^2-c}$. Equality gives just the point $-b$, and $|b{|}^2<c$ gives the empty locus. In particular, the nondegenerate answer is

Since $z\bar{z}=x^2+y^2$, the real quadratic equation isCompleting the square makes its geometry explicit:

A squared complex modulus is nonnegative. Conversely, every nonnegative right-hand side of $|z+b{|}^2=|b{|}^2-c$ can be attained. The necessary and sufficient condition is therefore

There is a single point exactly when the radius is zero:Then $|z+b{|}^2=0$ forces this unique value of the complex number $z$.

Under the given rotation, $\bar{z}=\bar{\alpha }\bar{w}$ and $z\bar{z}=|\alpha {|}^{2}w\bar{w}=w\bar{w}$. Collecting terms givesSince $\overline{b\bar{\alpha }}=\bar{b}\alpha$, the transformed coefficients are

For the real positive dilation factor, $\bar{z}=\alpha \bar{w}$ and $z\bar{z}={\alpha }^{2}w\bar{w}$. ThusandThe original PDF specifies a dilation; the differently named transformation in the OCR is an error.

For the translation, expand the products of complex numbers and collect the terms in $w\bar{w}$, $\bar{w}$, $w$ and the constant:It follows thatThe last coefficient is real because $b\bar{\alpha }+\bar{b}\alpha =2\mathrm{Re}(\bar{b}\alpha )$.

For the reciprocal Möbius transformation, substitute $z=1/w$, $\bar{z}=1/\bar{w}$ and multiply by $w\bar{w}$, which is nonzero on the stipulated domain. This yieldsConsequentlyFor a nondegenerate original circle, $c=0$ says that it passes through the origin, and its image is a straight line. The restriction $w\ne 0$ must be retained: multiplying through does not authorize adding that excluded point. Degenerate choices can give a constant or an identically zero equation instead of a straight line.

Write $N=g+r$. Successive conditional probabilities account for the changing bag under sampling without replacement:On this event no equality of the two colour counts has occurred, so the game's stopping rule does not interrupt these three draws.

After one draw the two counts are either $(1,0)$ or $(0,1)$ and cannot agree. ThereforeThis is the elementary parity restriction behind the first return time interpretation of this game.

Success after two draws occurs on the two mutually exclusive events $GR$ and $RG$. With $N=g+r$, sampling without replacement gives

The equality must be a first return time: an order starting $GR$ or $RG$ would already have stopped at draw two. The only successful four-draw orders are therefore $GGRR$ and $RRGG$. Their probabilities under sampling without replacement add toHere $N=g+r$. Counting all balanced four-draw orders would include games that have already stopped.

The two possible colour orders are $GGRR$ and $RRGG$. Thus the answer isThey correspond to the positive and negative four-step excursions with first return time four.

Suppose the first draw is green. The excess of green over red must remain positive after draws one through five and return to zero at draw six. After deleting the first green and final red, the remaining four steps must form a Dyck path: they have equal colour counts and never go below zero relative to the new starting level. There are $C_2=2$ such orders, $GGRR$ and $GRGR$, giving $GGGRRR$ and $GGRGRR$ in the original game.

Interchanging green and red gives $RRRGGG$ and $RRGRGG$. Thus there areorders. More generally, the Catalan number $C_{m-1}$ counts each sign of excursion with first return time $2m$.

Let $X_1$ and $X_2$ denote the first and second labels. Sampling with replacement restores the bag before the second draw, so conditioning on $X_1=x$ only specifies the threshold. For $1\le x\le g$, there are $x$ qualifying green balls and $x$ qualifying red balls. For $g<x\le r$, every green ball and the first $x$ red balls qualify. Hence the requested conditional probability isThe formulas agree at $x=g$ and give one at $x=r$. Labels outside the possible first-draw range do not define a conditioning event of positive probability.

Only a green ball carries label four. After it is removed, nine of the remaining nineteen balls are green. The conditional probability is

The specified colour tells us that one green ball has been removed. Thus sampling without replacement again leaves nine green balls among nineteen:The numerical label supplies no further change once this first colour is known.

There are two equally likely balls bearing label eight, one of each colour. The conditional probability that the removed ball was green is therefore $1/2$. By the law of total probability,Both possible first colours must be included because the observed label alone does not determine the colour.

Differentiate the second ordinary differential equation and use the first to eliminate $x^{\prime }$:Thus the reduced second-order linear differential equation isNo solutions are lost: given a solution $y$, define $x=y^{\prime }-\cos t$; then both original ordinary differential equations hold.

The characteristic equation of the reduced homogeneous linear differential equation is $m^2-\beta =0$. Its polynomial roots change type with the sign of $\beta$, givingFor $\beta =0$, the zero polynomial root is repeated, so the second independent solution is $t$. For $\beta <0$, the polynomial roots are the complex conjugate pair $\pm i\sqrt{-\beta }$, and the displayed sine and cosine give a real basis of solutions. The constants are arbitrary real numbers. In every case $x(t)=y^{\prime }(t)-\cos t$ reconstructs the corresponding general solution of the coupled system.

The differential operator on the left is $(D-3{)}^2$, where $D=d/dt$. Its repeated characteristic root suggests writing $y=e^{3t}v$. By the product rule,so the ordinary differential equation reduces to $v^{\prime \prime }=3(t+1)$. Two antiderivatives giveThe boundary conditions imply $v(0)=1$ and $v(1)=0$, hence $B=1$ and $A=-3$. ThereforeThe calculation of $v^{\prime \prime }$ verifies the ordinary differential equation, and the bracket takes the values one and zero at the two endpoints. These boundary conditions determine both constants uniquely.

To distinguish the old and new coordinate descriptions, write $u(x,y)=U(\xi (x,y),\eta (x,y))$. Since ${\xi }_x={\eta }_x=1$, the multivariable chain rule givesEquivalently, the differential operator is ${\partial }_x={\partial }_{\xi }+{\partial }_{\eta }$.

The coordinate partial derivatives are ${\xi }_y=1/\sqrt{y}$ and ${\eta }_y=-1/\sqrt{y}$. The multivariable chain rule therefore givesSince $\xi -\eta =4\sqrt{y}>0$, the expression entirely in the new coordinates isThe inverse coordinate formulas are $x=(\xi +\eta )/2$ and $y=(\xi -\eta {)}^2/16$, with domain $\xi >\eta$.