Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2022 / Paper 1

The difference of two cubes identity gives the real polynomial factorizationExpanding the right side cancels the two mixed terms of each degree. The quadratic factor satisfies $a^2+ab+b^2=(a+b/2{)}^2+3b^2/4$.

For real cube roots, set $u=a^{1/3}$ and $v=b^{1/3}$, so $(ab{)}^{1/3}=uv$. The sum of two cubes identity gives

Completing the square under the square root gives $\sqrt{(x-1{)}^2}=|x-1|$. Thus the equation is $|x-1|=2$, and the definition of absolute value gives $x-1=\pm 2$. Hence

The domain excludes $x=0$. Rearranging the inequality givesA sign check on the intervals separated by $-1$ and $0$ shows that the ratio is nonnegative when $x\le -1$ or $x>0$. The point $-1$ is allowed because it makes the numerator zero. Therefore

The original PDF puts both $x+y$ in the exponent in the second equation. Since the exponential function $3^t$ is injective, $3^{x+y}=1$ gives $x+y=0$. Put $u=3^x>0$, so $3^y=1/u$. The first equation then givesThus $u=1$, so $x=0$ and $y=-x=0$. The unique real solution of the simultaneous equations is

Pair adjacent terms in the finite sum:There are $25$ pairs, each equal to $2$. The result is $\boxed{50}$.

Put $r=\sin (\pi /4)=1/\sqrt{2}$. The geometric series has first term $r$ and common ratio $-r$, with absolute value less than one. Therefore its sum is

The product rule givesFor $x>0$, the derivative is positive before $3$ and negative after $3$. Also $y(0)=0$ and $x^3{e}^{-x}\to 0$ as $x\to \infty$. Thus the global maximum on the requested range is

For $x>0$, the denominator is positive, so the function graph has the same sign as $\sin x$. Its crossings of the horizontal axis areEach crossing is genuine because the derivative there is $(-1{)}^n/(n\pi {)}^2\ne 0$. Near zero, $\sin x\sim x$, so $y\sim 1/x\to +\infty$; the origin is excluded and is not a crossing. The oscillations alternate sign and decay towards zero as $x\to \infty$, with absolute value at most $1/x^2$.

Use the substitution rule with $u=\sqrt{x}$, for which $du=dx/(2\sqrt{x})$ on $x>0$. The integral becomes $\int e^{u}du$, giving the antiderivative

Completing the square givesThus the circle has

The dot product isBoth vectors have squared Euclidean norm $a^2+(a+1{)}^2=2a^2+2a+1$. Under the given condition $a^2+a=1/2$, their dot product is $1$ and each squared Euclidean norm is $2$. Hence

Use the full set of complex logarithm values of $i$:By the definition of complex exponentiation, the inner power can therefore take any value $i^i=\exp [-(\pi /2+2\pi k)]$, a positive real number. Its complex logarithm values are $-(\pi /2+2\pi k)+2\pi im$, where $m\in \mathbb{Z}$. ThusAll possible real parts and imaginary parts are consequentlyChoosing the principal complex logarithm at both stages gives the particular value $-i$. The independent logarithm choices are needed because the question asks for all possible values.

The inner power takes the values $r_k=e^{-(\pi /2+2\pi k)}$. For each such real exponent, all values of the outer complex exponentiation are obtained from all complex logarithm values of its base $i$:Writing ${\Theta }_{k,m}=(\pi /2+2\pi m)e^{-(\pi /2+2\pi k)}$, the answer isAll values lie on the unit circle. The principal choice is $k=m=0$, giving $\cos [(\pi /2)e^{-\pi /2}]+i\sin [(\pi /2)e^{-\pi /2}]$. In particular the two nestings in parts (i) and (ii) are different: complex exponentiation is not associative.

The complex modulus $|z-(-1-i)|$ is the Euclidean distance from $z$ to $-1-i$. Thus the locus isa circle with centre $(-1,-1)$ and radius $8$. Its extreme horizontal coordinates are $-9$ and $7$, and its extreme vertical coordinates are also $-9$ and $7$.

Write $z=x+iy$ and use the complex conjugate $z^{\ast }=x-iy$. ThenThus the real part is stretched by a factor of two while the imaginary part is unchanged. If $u=X+iY$, substitution into the circle equation givesThis is an ellipse with centre $(-2,-1)$, horizontal semiaxis $16$ and vertical semiaxis $8$. It is the second panel in the figure in part (b).

Adding $4+3i$ is a translation by $(4,3)$ in the complex plane. The ellipse of part (i) therefore keeps its semiaxes and axis directions, with new centre $(2,2)$. Writing $v=X+iY$, its equation isThe third panel in the figure in part (b) shows the translated ellipse.

Multiplication by $i$ is a counterclockwise rotation through $\pi /2$: it maps coordinates $(X,Y)$ to $(-Y,X)$. The centre $(2,2)$ therefore becomes $(-2,2)$, and the long axis becomes vertical. For $w=X+iY$, the resulting ellipse isIts horizontal semiaxis is $8$ and its vertical semiaxis is $16$, as shown in the final panel of the figure in part (b).

By De Moivre's theorem, $\sin 5\theta$ is the imaginary part of $(\cos \theta +i\sin \theta {)}^5$. The binomial theorem gives, with $s=\sin \theta$,Use ${\cos }^2\theta =1-s^2$ and simplify:The requested equation cancels the $16s^5$ term, leaving $5s(1-4s^2)=0$. Thus $s=0$ or $s=\pm 1/2$. The complete list in $0\le \theta <2\pi$ is

The stated $x$ bounds depend on $y$, so $y$ must be held fixed while the integral over $x$ is performed. The intended iterated integral isThe original PDF places the two sets of bounds beside the opposite integral signs relative to the printed $dxdy$. That placement cannot be read literally as a conventional nested integral; the dependence of the lower $x$ bound on $y$, and the explicit instruction to integrate over $x$ first, specify the intended region unambiguously.

The inequalities describing the region are $0\le y\le 1$ and $2y\le x\le 2$. It is the closed triangle bounded by $y=0$, $x=2$ and $x=2y$, with verticesA horizontal slice runs from the sloping edge $x=2y$ to the vertical edge $x=2$.

With $y$ held fixed, the inner antiderivative givesThe remaining integral is therefore

For a vertical slice through the triangle, $0\le x\le 2$ and $0\le y\le x/2$. Reversing the order by Fubini's theorem givesThe agreement checks that both orders describe the same region.

The PDF diagram includes a solid base of thickness $d$, occupying $-d\le z\le 0$ and $0\le r\le R$. Above it, $0\le z\le h$, the metal lies between the two specified curved surfaces. In cylindrical coordinates, the Jacobian determinant contributes a factor $r$, so the mass isThe base contributes $\pi \rho R^{2}d$. The wall contributesConsequentlyThe base thickness $d$ comes from the diagram; omitting it would omit part of the cup's volume.

Because the original differential equation divides by $xy$, its domain requires $x\ne 0$ and $y\ne 0$. Set $u=y^2$, so $u^{\prime }=2y{y}^{\prime }$. The equation becomes the first-order linear differential equationAn integrating factor is $x^2$, since $(x^{2}u{)}^{\prime }=x^2{u}^{\prime }+2xu$. ThusReturning to $y$ gives the explicit general real solutionsEach sign is fixed on a connected interval where $x\ne 0$ and the radicand is strictly positive. The strict condition excludes $y=0$, where the original differential equation is undefined. There is no additional zero solution.

Where $\sin x\ne 0$, divide by it to obtain $y^{\prime }+2\cot xy=\csc x$. The integrating factor is ${\sin }^{2}x$, and the product rule givesThe initial condition gives $C=1$. HenceThe simplified expression has a differentiable extension through $x=0$, with $y(0)=1/2$. Direct substitution verifies the original undivided differential equation there as well: its left side is $2y(0)=1$. Thus the maximal interval containing $\pi /2$ on which this solution is finite is $\boxed{-\pi <x<\pi }$; the apparent exclusion of $0$ arose only from division by $\sin x$. This is continuation through a zero of the leading ODE coefficient.

An exact differential form must have $P_y=Q_x$. Direct partial differentiation givesThus $(1-a)xy\cos (ax)=0$ must hold identically. Since $xy\cos (ax)$ is not identically zero for a fixed nonzero $a$, the only possibility isThe explicit potential in part (ii) verifies sufficiency.

For $a=1$, integrate $Q=2xy\sin x$ with respect to $y$ to obtainIts other partial derivative is $f_x=y^2\sin x+x{y}^2\cos x+C^{\prime }(x)$. Comparing with $P$ gives $C^{\prime }(x)=0$. Thus the most general potential on the plane is

The chain rule gives $f_x=f_u{u}_x+f_v{v}_x$ and $f_y=f_u{u}_y+f_v{v}_y$. Since $u_x=2x$, $u_y=2y$, $v_x=2x{y}^2$ and $v_y=2x^{2}y$, the answer isEvery partial derivative of $f$ on the right is evaluated at $u=1+x^2+y^2$, $v=1+x^2{y}^2$.

Apply the product rule and the chain rule once more to each expression in part (i). The second derivative chain rule givesandFor example, the coefficient of $f_{uv}$ in the mixed partial derivative is $u_x{v}_y+v_x{u}_y=4x^{3}y+4x{y}^3$, while the final term is $v_{xy}{f}_v=4xy{f}_v$. The symmetry $f_{uv}=f_{vu}$ is used in combining the mixed terms under the usual twice-continuously-differentiable interpretation.

For $f(u,v)=\ln u+\ln v$, the needed partial derivatives areSubstituting these into part (ii) givesSince $v-x^2{y}^2=1$, the last two terms combine. Thus

First express the function in the original variables:The chain rule givesTaking its partial derivative with respect to $y$ yieldsThe numerator in the second fraction reduces to $4xy$, reproducingBoth logarithms are defined for every real $x,y$ because their arguments are at least one.

For $a\ne 0$, factor out $|a|$, not $a$, and apply the binomial series with exponent $1/2$:For each fixed nonzero $a$, these are the first three nonzero terms of the Taylor series about zero; the full binomial series converges when $|x|<|a|$.

For $a=0$, the function is $|x|$. Its right derivative at zero is $1$ and its left derivative is $-1$, so no Taylor series about zero exists when $a=0$. The one-sided formulas are exactly $x$ for $x\ge 0$ and $-x$ for $x\le 0$; they do not form a single two-sided Taylor expansion, nor can they supply three nonzero terms. The parameter case explicitly included in the question must therefore be treated as this exception.

Write $t_j=t^{(j)}(0)$ and $s=1+t_0>0$. The chain rule givesThe condition $t_3{t}_1<0$ ensures $t_1\ne 0$. Since $t_0\ne 0$, also $s\ne 1$ and $\ln s\ne 0$. Hence the first two nonzero terms are the constant and linear terms:The inequality $s>0$, together with continuity of $t$, makes the logarithm defined on a neighbourhood of zero. The remainder is written in Big O notation.

The second derivative chain rule givesWith $t_0=0$, the constant term vanishes, $f^{\prime }(0)=t_1\ne 0$, and $f^{\prime \prime }(0)=t_2-t_1^2\ne 0$ by the specified case. Thus the first two nonzero terms in the Taylor expansion are

Here $t_0=0$ and $t_2=t_1^2$, so the constant and quadratic terms vanish, while the linear coefficient remains $t_1\ne 0$. A further derivative givesConsequently $f^{(3)}(0)=t_3-3t_1{t}_2+2t_1^3=t_3-t_1^3$. Since $t_3$ and $t_1$ have opposite signs, $t_3$ and $t_1^3$ also have opposite signs, and their difference is nonzero. The first two nonzero terms are thereforeThis is where the full sign condition $t_3{t}_1<0$ rules out a further cancellation in the Taylor expansion.

A probability density function must have total integral one. The numerator is an even function, soThis is positive, ensuring nonnegativity of the probability density function.

Put $N=6-2e^{-2}$. The function graph is symmetric about the vertical axis, zero outside $[-2,2]$, and has its unique maximum at $x=0$ with height $2/N$. On $0<x<2$, it decreases as $(1+e^{-x})/N$; the negative side is its reflection. The one-sided interior endpoint values are $(1+e^{-2})/N$, so the probability density function jumps to zero at both ends of its support. It also has a corner at its central maximum because of the absolute value in the exponent.

Values assigned exactly at $x=\pm 2$ do not affect any probability.

Since the probability density function is an even function, $xf(x)$ is an odd function. Its integral over the symmetric support vanishes, so

The expected value is zero, so the variance is $\mathbb{E}[X^2]$. Symmetry givesThe polynomial term contributes $8/3$. Two applications of integration by parts giveso its definite value on $[0,2]$ is $2-10e^{-2}$. Therefore

Let $F$ be the cumulative distribution function of $X$. Normalisation of the proposed probability density function requires $M={\int }_{-2}^{2}F(y)dy$. The integrated cumulative distribution function on a bounded interval identity givesFor a direct symmetry proof, the continuous symmetric probability distribution satisfies $F(-y)=1-F(y)$. HenceIn particular $g(y)=F(y)/2$ is nonnegative and has total integral one over its stated support.