Year of the natural sciences course of the University of Cambridge

The physics course entries follow the Cavendish Laboratory's 2025–26 teaching list.^[ref]

Past examination papers: Physics Teaching Information System past-paper search (University sign-in required). Specific paper years and their solution availability must be checked inside the system.

A public release includes physics papers for 2017–2019 and 2021–2022, with examiner reports and some answers. ^[ref]

Set $t=2^x>0$. The exponential function identity $4^x=(2^x{)}^2$ reduces the problem to the quadratic equationPositivity excludes $t=-1$, so $2^x=2$. Since the exponential function $2^x$ is strictly increasing, the unique answer is $\boxed{x=1}$.

Join the centre of the unit circle to adjacent vertices of the regular polygon. Each of the $n$ resulting triangles has two sides of length $1$ and included angle $2\pi /n$. Bisecting one gives a right triangle, so each polygon side, a chord of a circle, has length $2\sin (\pi /n)$. Thus the perimeter is $2n\sin (\pi /n)$.

Each of these triangles has area $\frac{1}{2}\sin (2\pi /n)$, so the polygon's area is $\frac{n}{2}\sin (2\pi /n)$. Comparing the two expressions gives

Adding the vectors component by component givesConsequently $\boxed{a=0}$. The requirement that the vectors be nonzero then implies $b\ne 0$.

Since $\mathbf{v}=-\mathbf{u}$ and $\mathbf{u}\ne 0$, the vectors point in opposite directions. More explicitly, their dot product givesHence their angle is $\boxed{\theta =\pi \text{ radians}=180^{\circ }}$.

The derivative isA stationary point at $x=1$ requires $2/a^2-2a^2=0$, or $a^4=1$. Since $a>0$, $\boxed{a=1}$.

For the requested parameter value $a=2$, the function graph has two branches, symmetric about the vertical axis because $f$ is an even function. Its derivatives areThe stationary points are $\boxed{(-2,2)\text{ and }(2,2)}$, both local minima and also the lowest points of their respective branches. The function tends to $+\infty$ as $x\to 0^{\pm }$, so $x=0$ is a vertical asymptote; it also tends to $+\infty$ as $|x|\to \infty$. The negative branch decreases to $(-2,2)$ and then rises towards the vertical asymptote; the positive branch falls from the vertical asymptote to $(2,2)$ and then rises.

The double-angle identity for the cosine gives ${\cos }^{2}x-1=(\cos 2x-1)/2$. Termwise integration therefore givesHere $C$ is an arbitrary constant in the antiderivative.

The derivative of the tangent is ${\sec }^{2}x$, and the order of the endpoints must be retained. Thus the definite integral is

Group consecutive terms in pairs. The $j$th pair, for $j=1,\dots ,25$, isThere are $25$ such pairs in this finite sum, so the answer is $\boxed{25}$.

This geometric series has first term $1/27$ and common ratio $1/3$, whose absolute value is below one. Consequently its sum is

Writing $v=y-1$ turns the first-order linear differential equation into $v^{\prime }=v$. Its general solution is $v=C{e}^x$, so $y=1+C{e}^x$. The initial condition gives $C=-1$, hence

The general solution of the first-order linear differential equation is $y=1+C{e}^x$. The initial condition $y(0)=1$ gives $C=0$, so the solution of this initial value problem is the constant function $y(x)=1$. Thus $\boxed{y(1)=1}$.

For the partial fraction decomposition, multiply by $(x-1)(x-2)$ to obtain the polynomial identityComparing the $x^2$ coefficients gives $a=1$. Substitution of $x=1$ gives $2=-b$, while $x=2$ gives $5=c$. ThereforeThe original identity holds for $x\ne 1,2$; evaluating the multiplied polynomial identity at those values is valid.

Subtracting the two circle equations cancels the common terms and gives $-4x=0$. With $x=0$, either circle equation gives $(y-1{)}^2=0$. Thus the unique common point isThe two circles touch externally there.

The circles have centres $(1,1)$ and $(-1,1)$. Their Euclidean distance is

The two throws are independent, giving $36$ equally likely ordered outcomes in the sample space. A total strictly below $4$ occurs only for $(1,1)$, $(1,2)$ and $(2,1)$. Hence the probability is

The favourable outcomes are $(j,j+1)$ and $(j+1,j)$ for $j=1,\dots ,5$. These are ten distinct ordered outcomes among the $36$ equally likely elements of the sample space, so the probability is

This is a plane through the origin in three-dimensional Euclidean space, with nonzero normal vector $(1,2,-3{)}^T$:

For example, chooseBoth vectors satisfy the plane equation. They are linearly independent vectors: in $\alpha \mathbf{u}+\beta \mathbf{v}=0$, the second coordinate gives $\alpha =0$ and the third gives $\beta =0$.

No. Every vector in the plane can be written asThe two linearly independent vectors in part (ii) therefore form a vector space basis for this vector subspace. Its dimension is $2$, so any three vectors in it have linear dependence.

The dot product with the given unit vector is $\mathbf{r}\cdot \hat{\mathbf{n}}=(x+z)/\sqrt{2}$. Subtracting the corresponding orthogonal projection givesTaking its squared Euclidean norm yields the equivalent Cartesian equation

The removed vector is the orthogonal projection of $\mathbf{r}$ onto the line through the origin in direction $\hat{\mathbf{n}}$. The remaining vector measures the perpendicular displacement from that line. Fixing its Euclidean norm at $2$ therefore describes an infinite circular cylinder of radius $2$, with axis $x=z$, $y=0$.

The plane equation $x+z=0$ has normal vector $(1,0,1{)}^T=\sqrt{2}\hat{\mathbf{n}}$. Thus its normal vector is parallel to the cylinder's axis and the plane is perpendicular to that axis. It passes through the origin, so the intersection is a circle of radius $2$ centred at the origin in that plane.

Indeed, with the orthonormal basis $\mathbf{e}=(1,0,-1{)}^T/\sqrt{2}$ and $\mathbf{f}=(0,1,0{)}^T$ of the plane, write $\mathbf{r}=s\mathbf{e}+t\mathbf{f}$. The cylinder equation becomes $s^2+t^2=4$. Equivalently, the intersection has the parametric equation

Use the distance from a point to a plane formula with the unit normal $\hat{\mathbf{n}}$. Since the plane passes through the origin, the distance isFor a direct check, the closest point on a plane is $\mathbf{p}-(\mathbf{p}\cdot \hat{\mathbf{n}})\hat{\mathbf{n}}=(1,\sqrt{2},-1{)}^T$, whose displacement from $\mathbf{p}$ has Euclidean norm $\sqrt{2}$.

Let $\mathbf{d}={\mathbf{a}}_2-{\mathbf{a}}_1$ and $\mathbf{w}={\mathbf{v}}_1\times {\mathbf{v}}_2$. The assumed linear independence gives $\mathbf{w}\ne 0$. Every displacement joining one line to the other isIts dot product with $\mathbf{w}$ is always $\mathbf{d}\cdot \mathbf{w}$. The component in that direction is therefore fixed, so orthogonal decomposition gives $\Vert \mathbf{s}\Vert \ge |\mathbf{d}\cdot \mathbf{w}|/\Vert \mathbf{w}\Vert$.

The two linearly independent vectors ${\mathbf{v}}_1,{\mathbf{v}}_2$ span the plane perpendicular to $\mathbf{w}$. Suitable ${\lambda }_1,{\lambda }_2$ cancel the component of $\mathbf{d}$ in that plane, attaining the lower bound. Thus the distance between skew lines formula, also valid when the lines intersect, gives

The nonzero direction vectors now have linear dependence, so the lines are parallel or coincident. Subtract from $\mathbf{d}={\mathbf{a}}_2-{\mathbf{a}}_1$ its orthogonal projection along ${\mathbf{v}}_1$; the remaining component is the shortest joining displacement. The distance between parallel lines isThe second expression follows from the relation between the cross product and the perpendicular component. It includes distance zero for coincident lines.

For the stipulated linearly independent vectors ${\mathbf{v}}_1,{\mathbf{v}}_2$, the denominator in part (i) is nonzero. ThusEquivalently, ${\mathbf{a}}_2-{\mathbf{a}}_1$ belongs to the linear span of ${\mathbf{v}}_1,{\mathbf{v}}_2$. Geometrically the lines lie in one plane, and, since their direction vectors are not parallel, they intersect at a unique point. The linear combination expressing ${\mathbf{a}}_2-{\mathbf{a}}_1$ gives the two parameters of that intersection.

Use the exponential series to obtain the Maclaurin series of the hyperbolic sine and hyperbolic cosine:Dividing these power series gives the hyperbolic tangent expansionThe displayed remainder uses Big O notation as $x\to 0$.

Set $\ell =\ln 2$. Applying the exponential series and the Maclaurin series of the sine givesMultiply the two power series, retaining every contribution through degree four:Since $0<\ln 2<1$, the coefficient of $x^4$ is nonzero, so these are indeed the first two nonzero terms. The remainder is in Big O notation.

Write $F(x)=\ln (e^{ax}-b)$. Direct differentiation, using the chain rule, givesThe stated parameter restrictions ensure that the logarithm is defined near $x=1$, and that $F^{\prime }(1)>0$ and $F^{\prime \prime }(1)<0$. With $h=x-1$, the Taylor series beginsIf $e^a-b\ne 1$, the constant term is nonzero, and the first two nonzero terms areThis covers both $a>\ln (1+b)$ and $0<a<\ln (1+b)$; the constant term is respectively positive or negative.

If $a=\ln (1+b)$, which is possible precisely with $0<a<\ln 2$ and $b=e^a-1$, the constant term vanishes. The first two nonzero terms are thenNo further case is needed, since the linear and quadratic coefficients cannot vanish under the given restrictions. All remainders use Big O notation.

Apply integration by parts twice. Taking an antiderivative of $e^{x/2}$ to be $2e^{x/2}$ givesand the second application gives $\int x{e}^{x/2}dx=2x{e}^{x/2}-4e^{x/2}+C$. Therefore the required definite integral is

The factor $x^2$ is an even function and $\sin x$ is an odd function, so their product is an odd function. Its definite integral over the symmetric interval vanishes:Indeed the substitution rule $x=-t$ changes the integral over $[-\pi /2,0]$ into the negative of the integral over $[0,\pi /2]$.

A partial fraction decomposition givesFor the second term the substitution rule $t=x^2+1$, $dt=2xdx$ gives a logarithm. ThusThe constant in the antiderivative may be chosen independently on the two intervals separated by $0$.

Split the integral intoThe first is the standard antiderivative $\mathrm{arsinh}x=\ln (x+\sqrt{1+x^2})$ from the inverse hyperbolic functions. For the second, the substitution rule $t=1+x^2$ gives $\frac{1}{2}\int t^{-1/2}dt=\sqrt{1+x^2}$. HenceThe logarithm's argument is positive for every real $x$.

Use the tangent half-angle substitution $t=\tan (x/2)$, so $\cos x=(1-t^2)/(1+t^2)$ and $dx=2dt/(1+t^2)$. ThenThe inverse tangent antiderivative givesThis expression applies on each interval $((2j-1)\pi ,(2j+1)\pi )$ for an integer $j$. Although $5+3\cos x\ge 2$ everywhere, the chosen inverse tangent expression jumps at the endpoints because the substitution has a pole there. A global continuous antiderivative is obtained by adding $j\pi /2$ on that interval, using one common $C$, and extending by continuity at the endpoints.

A probability density function must have total integral one. Put $h=\pi /6$. The term $ax$ is an odd function, soThus $\boxed{b=1}$.

With $b=1$, nonnegativity of the probability density function at both $t$ and $-t$ requiresThis is equivalent to $|a|\le {\inf }_{0<t\le h}\cos t/t$. The derivativeshows that this infimum occurs at $t=h$. Therefore the necessary and sufficient restriction isSufficiency follows from the same decreasing-ratio argument at every interior point. The endpoint values of $a$ are allowed: the probability density function merely vanishes at one endpoint.

For $h=\pi /6$, the expected value isbecause $x\cos x$ is an odd function. Comparing with the specified expected value gives $\boxed{a=1}$, which lies in the permitted parameter range.

Here $a=0$, so the probability density function is the even function $\cos x$ on $[-h,h]$, where $h=\pi /6$. Hence $\mathbb{E}[X]=0$ by oddness, and the variance is just $\mathbb{E}[X^2]$. Two applications of integration by parts giveThus