Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2018

Take the base-ten logarithm of both positive sides. The exponentiation rule for logarithms gives $x{\log }_{10}2=y$, hence

Substitution in the logarithmic expression gives $x\simeq 23.8/0.301=79.069\dots$. To two significant figures,

Move the constant term to the left and factor the cubic polynomial:Setting each factor to zero gives all three polynomial roots:

The quadratic formula gives the two polynomial roots $(-3\pm \sqrt{5})/2$. Its leading coefficient is positive, so the quadratic polynomial is strictly negative between them. Therefore

Matching the two parametric equations at an intersection gives $\cos \varphi =-t$ and $\sin \varphi =1+t$. The Pythagorean trigonometric identity therefore requiresFor $t=0$, the sine is one and the cosine is zero; for $t=-1$, the cosine is one and the sine is zero. Keeping the stated half-open angular interval givesThe corresponding points are $(0,1)$ and $(2,0)$; both satisfy the original parametric equations.

Use integration by parts, taking $u=x$ and $dv=\sin 3xdx$:

Since $d({\sin }^{2}x)/dx=2\sin x\cos x=\sin 2x$, integration by substitution $u={\sin }^{2}x$ immediately givesHere $C$ is an arbitrary constant of integration.

The two labelled regions in the original diagram are complementary circular sectors. Their areas add to $\pi r^2$ and satisfy $A=2B$, so $B=\pi r^2/3$. The sector with central angle $\theta$ has area $r^2\theta /2$. Hence

The two radii enclose the angle $2\pi /3$, so the triangle has area $T=\frac{1}{2}{r}^2\sin (2\pi /3)=\sqrt{3}{r}^2/4$. From the sector calculation, $r^2=3B/\pi$. Consequently

A space diagonal of the unit cube has displacement vector $(1,1,1)$. Its Euclidean norm, or two successive applications of the Pythagorean theorem, givesThe requested diagonal joins opposite vertices; $\sqrt{2}$ would be the diagonal of a face.

Choose two distinct space-diagonal directions $\mathbf{u}=(1,1,1)$ and $\mathbf{v}=(-1,1,1)$. Their dot product is one and their norms are both $\sqrt{3}$. Thus the acute angle between their unoriented lines isAny two distinct space diagonals have an absolute dot product of one when represented with components $\pm 1$, so they give the same acute angle.

The binomial theorem givesThus, to three decimal places,The first three nonconstant terms already determine this rounding; the remaining positive terms total less than $1.3\times 10^{-6}$.

The initial term at $n=-1$ is three, and the common ratio is $1/3$. The convergent geometric series therefore sums to

The product rule gives $y^{\prime }=e^x(\cos x-\sin x)$. The exponential function never vanishes, so a stationary point requires $\tan x=1$. In the given interval this means $x=\pi /4$. Substitution givesThe derivative changes from positive to negative there, making it a strict local maximum.

Completing the squares gives the standard equation of a circle:Its centre is therefore

In the completed-square equation, the right side is the squared radius of the circle. Taking its positive square root gives

Equating the two polynomials gives $x^3-3x^2+3x-1=(x-1{)}^3=0$. Thus the only intersection has $x=1$ and $y=4$:

The two derivatives are $3x^2+3$ and $6x$, both equal to six at $x=1$. Their common tangent line through $(1,4)$ is $y-4=6(x-1)$, or $y=6x-2$. Setting $x=0$ gives its intersection with the vertical axis:

The factor $1+i$ has modulus $\sqrt{2}$ and principal argument $\pi /4$. Multiplication by $e^{i\pi /6}$ adds $\pi /6$ to its complex argument, while the real factor $1/2$ halves its modulus. HenceTaking the principal logarithm for $w$ givesThis complex number lies in the second quadrant. Its modulus follows from the sum of squared components, and its principal argument follows by the second-quadrant correction:All arguments of this fixed $w$ add integer multiples of $2\pi$ to the displayed one.

If $\ln z$ is interpreted as the multivalued complex logarithm, there is instead a family $w_k=-a+i{b}_k$, where $a=\frac{1}{2}\ln 2>0$ and $b_k=5\pi /12+2\pi k$. Then $|w_k|=\sqrt{a^2+b_k^2}$, and the principal argument is $\pi -\arctan (b_k/a)$ for $b_k>0$, or $-\pi -\arctan (b_k/a)$ for $b_k<0$. This records the logarithm-branch choice that the paper leaves implicit.

Using the exponential definition of hyperbolic tangent, set $q=e^{2z}$:Here $q+1=1-i\ne 0$, so the resulting points are not poles. Taking all values of the complex logarithm gives $2z=i(-\pi /2+2\pi k)$. ThusEquivalently $\tanh (iu)=i\tan u$, so these are exactly the imaginary-axis solutions of $\tan u=-1$.

By De Moivre's theorem, $\cos 5\theta$ is the real part of $(\cos \theta +i\sin \theta {)}^5$. Writing $c=\cos \theta$ and $s=\sin \theta$, the binomial theorem givesUse the Pythagorean trigonometric identity $s^2=1-c^2$ and expand:Therefore

The region is an unbounded wedge in the first quadrant, lying between the rays $y=x/2$ and $y=x$. Both boundaries and the origin are included. For integration in the opposite order, the same region is $y\ge 0$ with $y\le x\le 2y$.

For fixed $x>0$, integrate first over $y$:The integrand is nonnegative, so this order of the double integral is legitimate even before its finiteness is known. Applying the scaled Gaussian integral ${\int }_0^{\infty }{e}^{-a{x}^2}dx=\sqrt{\pi }/(2\sqrt{a})$ yields

Reverse the order in the nonnegative double integral, using $0\le y<\infty$ and $y\le x\le 2y$. For $y>0$, an antiderivative in $x$ is $-(x/y+1/y^2)e^{-xy}$. ThusLet $J$ denote the requested improper integral. Its numerator is $y^2+O(y^4)$ near zero, so its apparent singularity is removable; it also decays at infinity. We may therefore separate the preceding expression into the convergent integralsIt follows thatThe two terms divided separately by $x^2$ would each have a divergent integral at zero; the cancellation in their difference is essential.

Scale the ellipsoid to a unit ball usingwith $0\le r\le 1$, $0\le \theta \le \pi$ and $0\le \varphi <2\pi$. This is spherical coordinates followed by a diagonal linear transformation, so the Jacobian determinant is $abc{r}^2\sin \theta$. The required volume integral is

The ellipsoid and the exponential factor are invariant under separate reflections of any coordinate. Split the integrand into its two polynomial terms. The term $x^4{y}^{2}z$ times that exponential is odd in $z$, so its volume integral vanishes by reflecting $z\mapsto -z$. The other term $x{y}^2{z}^4$ times the exponential is odd in $x$, so its volume integral vanishes by reflecting $x\mapsto -x$. Both integrands are continuous on a bounded region. Therefore odd-integrand cancellation by reflection gives

For $x\ne 0$, divide by $x^2$ to obtain $y^{\prime }+y/x=1/x^2$, a first-order linear ordinary differential equation. An integrating factor is $x$ on either interval not crossing zero, since the product rule givesIntegration and division by $x$ give the explicit general solutionThe original equation would read $0=1$ at zero, so no solution can include that point.

Rewrite the linear ordinary differential equation as $y^{\prime }+y=e^{3x}$. Multiplication by its integrating factor $e^x$ gives $(e^{x}y{)}^{\prime }=e^{4x}$. Integrating yieldsThe general solution consists of this particular solution and the homogeneous solution $C{e}^{-x}$.

The original PDF has $8y^3-4x$ in the denominator and $dy/dx$ on the left. Rearrange it into the differential formThe cross partial derivatives of its coefficients are both four, so it is an exact differential form. Integrating its first coefficient in $x$ gives a potential $F=\frac{5}{2}{x}^2+4xy+h(y)$. Matching $F_y=4x-8y^3$ gives $h(y)=-2y^4$. Therefore an implicit general solution isDifferentiating verifies $y^{\prime }=(5x+4y)/(8y^3-4x)$. A solution graph is understood only on intervals where its denominator is nonzero; the printed quotient does not define a derivative when $8y^3=4x$.

For real solutions with $x>0$, the square root requires $y\ge 0$. First retain the solution $y\equiv 0$, which would be lost by dividing by a square root of $y$.

For $y>0$, use the homogeneous first-order differential equation substitution $v=\sqrt{y/x}>0$, so $y=x{v}^2$, $\sqrt{xy}=xv$ and $y^{\prime }=v^2+2xv{v}^{\prime }$. The equation becomesThis is now a separable differential equation. Integration gives $-2/v+2\ln v=-\ln x+C$. Since $x{v}^2=y$, the positive solutions satisfyThe implicit family is sufficient here because only parts (a) and (b) request an explicit expression. It also shows why the nonnegative choice of $v$ matters: $\sqrt{xy}$ denotes the positive square root.

The exponential function is strictly positive, so the zero level set is determined entirely by its prefactor:This is a rectangular hyperbola with branches in the first and third quadrants.

The sign of $f$ is the sign of $1-4xy$. ThereforeThe positive region includes both coordinate axes, the second and fourth quadrants, and the portions of the first and third quadrants on the origin side of the hyperbola. The two negative regions lie beyond its branches. The left-hand panel shows these regions; its markers locate the stationary points calculated next.

Set both components of the gradient to zero. After dividing out the nonzero exponential, they are equivalent toAdding and subtracting gives $(x+y)(3-4xy)=0$ and $(x-y)(1+4xy)=0$. If $x=y$, the first equation reduces to $x(3-4x^2)=0$, giving the origin and the two diagonal points. If $x=-y$, it reduces to $x(4x^2-1)=0$, giving the origin and the two opposite-diagonal points. The alternative simultaneous values $xy=3/4$ and $xy=-1/4$ are incompatible, so these cases exhaust the possibilities. Thus the five stationary points areThey are marked in both panels of the figure.

Evaluate the Hessian matrix at the five stationary points. At the origin it iswhose eigenvalues are $-6$ and $2$, so the origin is a saddle point. At the two points with $x=y=\pm \sqrt{3}/2$,whose eigenvalues are $12e^{-3/2}$ and $8e^{-3/2}$. Both are positive, so these are strict local minima. At the two points with $x=-y=\pm 1/2$,whose eigenvalues are $-8e^{-1/2}$ and $-4e^{-1/2}$. Both are negative, so these are strict local maxima. In summary,The quadratic changes along the orthogonal directions $(1,1)$ and $(1,-1)$ also provide a direct justification of these classifications.

The completed contour plot is shown in the right-hand panel. Its geometry follows from the symmetries $f(-x,-y)=f(x,y)$ and $f(y,x)=f(x,y)$, together with $f\to 0$ as $x^2+y^2\to \infty$. For orientation, the two local minima have value $-2e^{-3/2}$, the two local maxima have value $2e^{-1/2}$, and the saddle point has value one.

Negative level sets between $-2e^{-3/2}$ and zero form separate closed curves around the two local minima in the first and third quadrants. Positive level sets below one form a single closed curve enclosing the origin and both local maxima. At level one, a critical level set crosses itself at the origin and surrounds the two local maxima in two lobes. Between one and $2e^{-1/2}$, the level sets split into separate closed curves around those local maxima. The zero level set remains the two unbounded branches $xy=1/4$.

Near the origin, the Taylor expansion is $f=1-x^2-4xy-y^2+O((x^2+y^2{)}^2)$. Consequently the crossing directions at the saddle height satisfy $y=(-2\pm \sqrt{3})x$, giving the local shape of the critical level set.

The first $n$ terms of Taylor's theorem, with the Taylor remainder in mean-value form, arewhere $\xi$ lies between $a$ and $x$ for $x\ne a$. Sufficient hypotheses are continuous derivatives through order $n-1$ on the interval between the endpoints and an $n$th derivative in its interior. At $x=a$ the remainder is zero. The finite Taylor polynomial ends at degree $n-1$; this identity does not require convergence of an infinite Taylor series.

Since $a_0=f(0)\ne 0$, write $1/f(x)=b_0+b_{1}x+b_2{x}^2+O(x^3)$ and multiply the two power series. Equating coefficients with the identity $f(x)/f(x)=1$ givesSolving successively yieldsThese are the constant, linear and quadratic terms; particular coefficients may vanish. For a convergent power series defining $f$ near zero, continuity and $a_0\ne 0$ ensure its reciprocal is also defined in a neighborhood of zero.

Multiply the Maclaurin series of hyperbolic cosine by the binomial series for the reciprocal square root:The quadratic terms cancel, while the quartic coefficient is $1/24-1/4+3/8=1/6$. HenceThe absence of odd powers follows from the function's even symmetry.