Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2014 / Paper 1

Write the rational expression as $(x^2+4{)}^{-1}$ and apply the chain rule:The denominator is positive for every real $x$, so the derivative is defined throughout the real line.

The outer exponential function differentiates to itself, and the inner sine has derivative cosine. The chain rule gives

For positive $a$, write $a^{-x}=\exp (-x\ln a)$. Differentiating by the chain rule givesThe natural logarithm of the constant base supplies the factor; it must not be omitted when $a\ne e$.

Set $u=\ln x$, so $du=dx/x$. The real domain is $x>1$, ensuring $u>0$. The integration by substitution followed by integration by parts givesThus

Since the derivative of $\ln \cos x$ is $-\tan x$ on this interval,

Use the real antiderivative $\ln |x|$, valid on either half-line away from zero:The negative answer is consistent with a negative integrand and increasing integration limits.

The specified range ensures $\cos y\ne 0$, so the separable differential equation can be written ${\sec }^{2}ydy=\sin xdx$. Integrating gives $\tan y=C-\cos x$. The range selects the principal inverse tangent, henceEvery finite $C$ produces a solution within the required range. Constant solutions at the excluded values $y=\pm \pi /2$ are not part of this family or of the question's domain.

The general solution of this first-order linear ordinary differential equation is $y=C{e}^{3x}$. Its initial condition gives $C=3$, soDifferentiation verifies $y^{\prime }=3y$ and substitution at zero gives the required value.

Substituting the line into the hyperbola givesThe two polynomial roots give all intersections:Both satisfy the original line as well as the curve.

Put the equation into standard ellipse form:The centre is $(1,0)$, with horizontal semiaxis $\sqrt{3}$ and vertical semiaxis $\sqrt{3/2}$. Its four axis vertices are $(1\pm \sqrt{3},0)$ and $(1,\pm \sqrt{3/2})$; the ellipse meets the vertical coordinate axis at $(0,\pm 1)$.

The sphere has surface area $4\pi R^2=400\pi {\mathrm{m}}^2$. Dividing the given surface-patch area by that total gives the dimensionless fractionor about $0.796\%$ of the surface. The patch area is already measured on the surface, so no planar projection or spherical-cap conversion is needed.

Interpret the stated side lengths as all eight edges of the square pyramid. The equal lateral edges put its apex above the base centre. The base-centre-to-vertex distance is half the diagonal, $1/\sqrt{2}\mathrm{m}$. The Pythagorean theorem therefore givesUsing the pyramid volume formula with a $1{\mathrm{m}}^2$ base,

Where the original expression is defined, multiplication by $\cos x$ givesThus the graph repeats with period $\pi$ and has vertical asymptotesBetween consecutive asymptotes the extended cotangent branch decreases from $+\infty$ to $-\infty$, since $y^{\prime }=-1/(\cos x+\sin x{)}^2<0$. It passes through $(k\pi ,1)$ and $(\pi /4+k\pi ,1/2)$.

It is essential to preserve the domain when simplifying a function. At $x=\pi /2+k\pi$, the original tangent is undefined. The finite limiting value is zero, so the graph has holes at $(\pi /2+k\pi ,0)$, rather than actual zeros. The simplified expression extends across those points, but the original one does not.

This exponential function is positive everywhere, with $y(0)=1$, $y\to \infty$ as $x\to -\infty$ and $y\to 0$ as $x\to \infty$. Its derivatives areIt is strictly decreasing overall, with a horizontal stationary inflection point at $(0,1)$. A second, nonstationary inflection point occurs at $x=(2/3{)}^{1/3}$, where $y=e^{-2/3}$. The only horizontal asymptote is $y=0$ toward positive infinity; the graph is not symmetric about the vertical axis.

Differentiate the polynomial and factor:The stationary points therefore have horizontal coordinatesFor a check, $y^{\prime \prime }=6x-4$ is negative at the first and positive at the second, giving a local maximum and a local minimum respectively.

The second derivative is $y^{\prime \prime }=6x$, so $y^{\prime \prime }(1)=6>0$. By the second-derivative test, the stationary point is a strict local minimum, at $(1,5)$. It is not a global minimum, since this cubic is unbounded below toward negative infinity.

On $0\le x\le 1/2$, the quadratic polynomial is positive. The required area is therefore its definite integral above the horizontal axis:The units are square coordinate units.

Real points require $x\ge 0$, and the two branches areThis is a semicubical parabola, symmetric about the horizontal axis and with a cusp at the origin. The parametrization $x=t^2$, $y=t^3$ traces both branches. Their limiting slopes at the origin are zero, so the common tangent direction is the positive horizontal axis, even though the curve has a cusp there.

The original line meets the horizontal axis at $(-3/2,0)$. A perpendicular line has slope $-1/2$, the negative reciprocal of $2$, soSubstituting $x=-3/2$ verifies the required common point.

For the partial fraction decomposition, write $A/(x-4)+B/(x+9)$. Clearing denominators gives $13(x+1)=A(x+9)+B(x-4)$. Substitution at $x=4$ and $x=-9$ yields $A=5$ and $B=8$, hence

Expanding the determinant along the first row givesThe matrix trace is the sum of the diagonal entries, $\mathrm{tr}A=1+4+1=6$. Since the determinant is the product of the eigenvalues, at least one eigenvalue is zero. Thus $A$ is singular; the determinant alone does not specify the two other eigenvalues.

The symmetry between the first and third coordinates makes $(1,0,-1{)}^{\mathsf{T}}$ an eigenvector, with eigenvalue $-6$. In the remaining invariant plane, write a vector as $(s,t,s{)}^{\mathsf{T}}$. Its image is $(8s-4t,-8s+4t,8s-4t{)}^{\mathsf{T}}$, so the reduced eigenvalue equation has determinantThe corresponding directions have $t=2s$ for $\lambda =0$ and $t=-s$ for $\lambda =12$. Normalizing givesTheir squared unnormalized norms are $2,6,3$. The pairwise unnormalized dot products are $1-1=0$, $1-1=0$, and $1-2+1=0$, so these unit eigenvectors are mutually orthogonal. The eigenvalues sum to $6$ and have product zero, agreeing with part (a).

Expand an arbitrary vector in the orthonormal eigenbasis:Its image has no component along ${\mathbf{e}}_0$, soAlternatively, symmetry gives $(A\mathbf{r})\cdot {\mathbf{e}}_0=\mathbf{r}\cdot A{\mathbf{e}}_0=0$. This is image-kernel orthogonality for an adjoint, with $A^{\mathsf{T}}=A$.

Resolve the input along the three perpendicular eigendirections. The linear map removes the component along $(1,2,1)$, stretches the component along $(1,-1,1)$ by $12$, and reverses and stretches the component along $(1,0,-1)$ by $6$. Every output lies in the plane perpendicular to $(1,2,1)$. A nonzero input in the kernel of a linear map $A$ is sent to zero; other nonzero inputs remain in that output plane. Equivalently, take the orthogonal projection onto this plane, then scale by the two signed eigenvalues along its perpendicular eigendirections.

The complex number $-1+i$ has complex modulus $\sqrt{2}$ and argument $3\pi /4$. A cube root has modulus $(\sqrt{2}{)}^{1/3}=2^{1/6}$, and its argument is one third of $3\pi /4+2k\pi$. Hence the three distinct roots areTheir displayed arguments are $\pi /4$, $11\pi /12$, and $19\pi /12$; the last is equivalently $-5\pi /12$ in the principal range.

Use the exponential definition of the hyperbolic tangent:The complex exponential has modulus one here, so $\mathrm{Re}z=0$. Its arguments satisfy $2\mathrm{Im}z=-\pi /2+2k\pi$. Thus all solutions areThe denominator is then $1-i\ne 0$, so no pole was admitted by cross-multiplication. Substitution also follows from $\tanh (i\theta )=i\tan \theta$ and the period $i\pi$.

Divide the complex polynomial by its known factor $z-(2+i)$:The quadratic factors further as $(z-1)(z-1-i)$, because its two proposed roots sum to $2+i$ and multiply to $1+i$. Therefore the remaining solutions areAll three roots are distinct. Complex coefficients do not require nonreal roots to occur in conjugate pairs.

By De Moivre's theorem, $(\cos \theta +i\sin \theta {)}^4=\cos 4\theta +i\sin 4\theta$. Take real parts of the binomial expansion:Eliminate ${\sin }^2\theta$ using $1-{\cos }^2\theta$ and collect powers:

An integrating factor is $e^{3x}$. Multiplication gives $(e^{3x}y{)}^{\prime }=8e^{3x}$, so $y=8/3+C{e}^{-3x}$. The initial condition gives $C=4/3$. Therefore

First write the forcing as $\sin x\cos x$. An integrating factor for the linear ordinary differential equation is $e^{-\sin x}$, givingSet $u=\sin x$ inside the antiderivative. ThenThe initial value forces $C=1$, henceThe resulting expression is explicit and valid for all real $x$; the substitution does not impose a global monotonicity condition on $\sin x$ when verifying an antiderivative.

The characteristic polynomial of the homogeneous equation is $r^2+7r+12=(r+3)(r+4)$. Thus the complementary function is $C_1{e}^{-3x}+C_2{e}^{-4x}$. The forcing resonates with the first term, so a resonant exponential particular solution is $y_p=Kx{e}^{-3x}$. Substitution gives $y_p^{\prime \prime }+7y_p^{\prime }+12y_p=K{e}^{-3x}$, hence $K=2$.

The full solution is $y=(C_1+2x)e^{-3x}+C_2{e}^{-4x}$. At zero, the two initial conditions give $C_1+C_2=1$ and $2-3C_1-4C_2=0$. Solving yields $C_1=2$, $C_2=-1$, so

The nonparallel direction unit vectors have nonzero cross product. LetFor any joining vector $\mathbf{s}=\mathbf{c}-\mathbf{a}+\mu \hat{\mathbf{d}}-\lambda \hat{\mathbf{b}}$, its normal component is fixed:Thus its length is at least the absolute value of that component. Since the two line directions span ${\mathbf{n}}^{\perp }$, suitable $\lambda ,\mu$ cancel the entire component in that plane, attaining the bound. Therefore the distance between skew lines, including the zero-distance intersecting case, isThe numerator is a scalar triple product. The nonparallel hypothesis is essential to the denominator and to the spanning argument.

Choose any nonzero auxiliary vector $\mathbf{q}$ not collinear with $\mathbf{p}$. Then $\mathbf{p},\mathbf{q},\mathbf{p}\times \mathbf{q}$ form a basis, so writeTaking the dot product with $\mathbf{p}$ gives $\alpha |\mathbf{p}{|}^2+\beta \mathbf{p}\cdot \mathbf{q}=k$. This separates the two cases requested in the hint.

If $\mathbf{p}\cdot \mathbf{q}\ne 0$, choose $\alpha ,\gamma$ freely and solve for $\beta$:If $\mathbf{p}\cdot \mathbf{q}=0$, the coefficient of $\mathbf{p}$ is fixed and the other two are free:These formulas describe the whole plane with normal $\mathbf{p}$, including $k=0$. Completeness follows from the basis expansion, not only from checking that the displayed vectors satisfy the equation. The parametrization of a plane with an auxiliary vector also gives a single formula valid in both cases.

Use the convention $x=r\sin \theta \cos \varphi$, $y=r\sin \theta \sin \varphi$, $z=r\cos \theta$, with $\theta$ measured from the positive vertical axis. The squared radius isSince the horizontal radial distance is $1/\sqrt{1+t^2}>0$ for finite $t$,The horizontal angle advances with $t$. Thus the spherical coordinates areAn unwrapped continuous azimuth is simply $\varphi =t$; a principal-angle convention introduces harmless jumps by $2\pi$. The polar angle stays in $(0,\pi )$ and approaches the two poles only as $t\to \pm \infty$.

The radial interval is nonempty exactly when $1+\cos \varphi \ge 3/2$, or $\cos \varphi \ge 1/2$. Choose the representative angular range $-\pi /3\le \varphi \le \pi /3$. The region of integration is thereforeIt is the right-hand cap inside the cardioid and outside the circle of radius $3/2$, symmetric about the horizontal axis. Its boundary intersections have coordinates $(3/4,\pm 3\sqrt{3}/4)$, and the cap reaches $(2,0)$. The radial segment on the horizontal axis runs from $3/2$ to $2$.

The polar area element is $rdrd\varphi$. Integrating first in radius and using symmetry givesThis is the specialization $a=1$, $c=3/2$ of the polar cap between a cardioid and a circle area formula.

Substitute $x=r\cos \varphi$, $y=r\sin \varphi$ and include the Jacobian determinant factor $r$. The integrand times the area element becomesBoth terms containing one factor $\sin \varphi$ are odd in $\varphi$, while the radial bounds and angular interval are even. They therefore integrate to zero. The remaining double integral isThe domain stays away from $r=0$, so the denominator of the original integrand causes no singularity in this region.

Translate the variables by $u=x-x_0$, $v=y-y_0$, $w=z-z_0$. The numerator becomes $u+x_0$. The part odd in $u$ integrates to zero, while the constant part factors into three Gaussian integrals. ThusThe absolute integrability of a polynomial times a Gaussian justifies this separation by Fubini's theorem. Neither $y_0$ nor $z_0$ affects the result; when $x_0=0$, odd symmetry makes the integral zero.

The gradient equations factor asFor $y=0$ they give $x=\pm 5$. Otherwise $y=4x/3$, which gives $x=\pm 3$ and the matching $y=\pm 4$. Thus there are exactly four stationary points.

The Hessian matrix isAt $(5,0)$ it is $60I$, and at $(-5,0)$ it is $-60I$. At $(3,4)$ and $(-3,-4)$ its determinant is $-3600<0$. The second-derivative test therefore yieldsThe extrema are local: along $y=0$ the cubic is unbounded both above and below.

The partial derivatives give $x^3=9y$ and $y^3=9x$. If either coordinate is zero, both vanish. Otherwise substitution gives $x^8=3^8$, so the real stationary points are $(0,0)$, $(3,3)$ and $(-3,-3)$.

The Hessian matrix isAt the origin its eigenvalues are $36$ and $-36$, so it is a saddle at level zero. At either of the other points its eigenvalues are $72$ and $144$, so both are strict minima at level $-162$. In fact,with equality only at those two points. HenceFor levels $-162<c<0$, two separate closed contours surround the two minima. At $c=0$ the contours form two lobes, in the first and third quadrants, meeting at the origin. Their limiting tangent directions there are the coordinate axes because the leading term is $-36xy$. On the diagonal the outer zero-level points are $(\pm 3\sqrt{2},\pm 3\sqrt{2})$. Positive levels form one closed outer contour enclosing both wells; there are no contours below $-162$. The graph is symmetric under exchanging $x,y$ and under reversing both signs, but not under a quarter-turn. These features are the $a=3$ case of the quartic double well with bilinear coupling.