Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2023

The binomial theorem gives $(1+x{)}^8={\sum }_{k=0}^8(\genfrac{}{}{0ex}{}{8}{k})x^k$. The required binomial coefficients are thereforeHere $[x^k]$ denotes the coefficient of $x^k$ in the polynomial.

The given polynomial root supplies the factor $2x-3$. Polynomial division givesSolving the remaining quadratic equation, the other two polynomial roots are

Use the first of the simultaneous equations to put $y=1-x$. Substitution into the second givesThus $x=0$ or $x=3$. The corresponding values of $y$ give exactly

Completing the square transforms the equation intoComparison with the standard circle equation gives

The addition formula for the sine givesEquating the sine and cosine coefficients gives $A\cos \varphi =\sqrt{3}$ and $A\sin \varphi =1$. Squaring and adding yields $A^2=4$, so $A=2$ because $A>0$. Both coefficients are positive, so $\varphi$ is in the first quadrant. Consequently

This is a finite geometric series with common ratio $r=1/\sqrt{2}$. Since $r^{10}=1/32$,Thus $\boxed{p=31,q=31}$.

The chain rule and the quotient rule giveThe exponential function and denominator are positive, so the stationary points occur exactly at $x=\pm 1$. Their values areThe derivative is negative for $|x|>1$ and positive for $|x|<1$, so $x=-1$ is a local minimum and $x=1$ a local maximum. They are also global extrema, since $-1/2\le x/(1+x^2)\le 1/2$.

Apply implicit differentiation, remembering that $y$ depends on $x$:Since $1+e^y>0$, division is always legitimate. Therefore

The partial fraction decomposition isTaking antiderivatives and using the positive endpoints givesHence $\boxed{n=2}$.

The function graphs intersect when $2x=4x{e}^{-x}$, so $x=0$ or $x=\ln 2$. For $0<x<\ln 2$, the exponential function satisfies $e^{-x}>1/2$, making $4x{e}^{-x}$ the upper graph. The enclosed area is thereforeThe antiderivative of $x{e}^{-x}$ follows by integration by parts.

Direct multiplication, using $i^2=-1$, givesThus the real part is $\boxed{18}$ and the imaginary part is $\boxed{1}$.

Multiply numerator and denominator by the complex conjugate $4-3i$:Thus the real part is $\boxed{6/25}$ and the imaginary part is $\boxed{-17/25}$.

Since $z^2=(x^2-y^2)+2ixy$, equating its real part and imaginary part gives $x^2-y^2=2xy$, orThe quadratic equation in $y$ factors as $[y-(-1+\sqrt{2})x][y-(-1-\sqrt{2})x]=0$. Thus the set is the union of two straight lines through the origin:Their slopes multiply to $-1$, so they are perpendicular. Their angles to the real axis are $\pi /8$ and $-3\pi /8$, modulo $\pi$. The origin is included.

The denominator requires $z\ne 0$. Write $z^2=u+iv$, where $u=x^2-y^2$ and $v=2xy$. Multiplication by the nonzero denominator givesso $u=0$. Conversely, if $u=0$ and $z\ne 0$, then $v\ne 0$ and $v/(iv)=-i$. Thus the required Argand diagram consists exactly ofThese are two perpendicular straight lines, with the origin removed as shown in the figure in part (i).

For real $x$, $\cosh x>0$, so multiplication by the hyperbolic cosine is reversible. The identities $\cosh x-\sinh x=e^{-x}$ and $\mathrm{sech}x=1/\cosh x$ giveThus $e^{-2x}=3$. The exponential function is injective, so the unique real solution is

Set $w=e^z\ne 0$. The hyperbolic cosine identity gives $w+w^{-1}=2i$, hence the quadratic equationLet $L=\ln (1+\sqrt{2})>0$. The first root has modulus $1+\sqrt{2}$ and complex argument $\pi /2$; the second has modulus $\sqrt{2}-1=e^{-L}$ and complex argument $-\pi /2$. Taking all complex logarithm values givesThe two families are distinct because their real parts are respectively $L$ and $-L$. This exhausts the solutions because both roots for $w$ and all possible complex arguments have been included.

At height $z$, the circular sector has radius $r(z)=R{e}^{-z/H}$ and angle $\Theta (z)=2\pi -\pi z/H$. Its area is $\Theta (z)r(z{)}^2/2$, so slicing by horizontal cross-sections gives the volume integralUse the substitution rule $t=z/H$. ThenAn integration by parts gives the antiderivative $(t/4-3/8)e^{-2t}$, whose endpoint difference is $(3-e^{-2})/8$. Therefore

At each height, the removed circular sector has the same angle and $4/5$ of the original radius. Its area and hence its volume are therefore multiplied by $(4/5{)}^2=16/25$. Equating the removed volume to that of the reformed solid sphere givesConsequently

At height $z$, the horizontal triangle has half-width $L\sqrt{1-z/(2H)}$ along $x$ and depth $D(1-z/H)$ along $y$. These shrink with $z$, so its orthogonal projection onto the $(x,y)$ plane is the base triangle with vertices $(-L,0)$, $(L,0)$ and $(0,D)$:The orthogonal projection onto the $(y,z)$ plane is the triangle with vertices $(0,0)$, $(D,0)$ and $(0,H)$:The orthogonal projection onto the $(x,z)$ plane is bounded by $z=0$, $z=H$ and the two curvesEquivalently, these curves lie on the parabola $z=2H(1-x^2/L^2)$, truncated at height $H$. In particular the top width is $\sqrt{2}L$, so this projection has a horizontal upper edge from $-L/\sqrt{2}$ to $L/\sqrt{2}$; it is not triangular. The top horizontal cross-section has zero depth, consistent with this edge.

The horizontal cross-section is a triangle, so its area is half its width times its depth:Thus the volume integral, with $t=z/H$, isFor the substitution rule, put $u=1-t/2$. The endpoint values are $1$ and $1/2$, and $1-t=2u-1$, givingTherefore $\boxed{V=\frac{(4+2\sqrt{2})LDH}{15}}$.

On an interval with $x\ne 0$, division by $x^2$ gives the separable differential equation $y^{\prime }=(4/x^2-1/x)y^2$. For $y\ne 0$, take antiderivatives:Renaming the constant yields the explicit nonzero familyFor $x\ne 0$, the interval of definition must avoid zeros of the denominator. Division by $y^2$ excluded the additional solution $\boxed{y(x)\equiv 0}$, which must also be included.

The original undivided differential equation also makes sense at $x=0$, where it requires $y(0)=0$. For continuation through a zero of the leading ODE coefficient, rewrite the nonzero family asIt extends differentiably with $y(0)=0$ and $y^{\prime }(0)=1/4$, since $y(x)/x\to 1/4$. One may choose the integration constants independently on the two sides of zero and obtain a differentiable solution through zero. The zero solution also extends through zero, with derivative $0$; joining a nonzero branch on just one side to a zero branch on the other would fail differentiability.

For this first-order linear differential equation, an integrating factor on an interval avoiding zeros of $\cos x$ is $\cos x$. The product rule then givesThe initial condition gives $3/\sqrt{2}=1/\sqrt{2}+C$, hence $C=\sqrt{2}$. ThusThe displayed interval is the maximal one containing the specified initial point, since the coefficient $\tan x$ is undefined at its endpoints.

Use the substitution $v=y/x$ for a homogeneous first-order differential equation. Since $y^{\prime }=v+x{v}^{\prime }$, it givesWhere $\sin v\ne 0$, this separable differential equation integrates to $\ln |\sin v|=\ln |x|+C$, hence $\sin v=C_{1}x$. One can obtain the same relation without division by $\sin v$: the equation gives $(\sin v{)}^{\prime }=(\sin v)/x$, so $\sin v=C_{1}x$, including $\sin v\equiv 0$.

All real branches of the inverse sine are captured byEach branch is used on an interval with $x\ne 0$ and $|Cx|<1$, with fixed $k$. The strict inequality excludes zeros of $\cos (y/x)$, where the original tangent is undefined. When $C=0$, the formula includes all straight-line solutions $y=k\pi x$. Keeping both families of inverse sine branches is essential for the two different initial conditions below.

The initial condition gives $C=\sin (\pi /3)=\sqrt{3}/2$. Since $\pi /3$ belongs to the principal inverse sine range, choose $k=0$ in part (i). The particular solution isThe upper endpoint is excluded because $y/x\to \pi /2$ there and the original tangent ceases to be defined.

Again $C=\sin (2\pi /3)=\sqrt{3}/2$, but the principal inverse sine would give $\pi /3$ at $x=1$. The required branch is $v=\pi -\arcsin (Cx)$, corresponding to $k=1$. ThereforeAt $x=1$ the bracket is $2\pi /3$, as required. Both the initial condition and the original differential equation are thus satisfied on the stated branch.

For an exact differential form, equality of the mixed partial derivatives requires $P_y=Q_x$. HereEquality for all $x>0$ and real $y$ forces $1/a=1/2$, so $\boxed{a=2}$. To construct the potential, integrate $P$ with respect to $x$:Differentiation with respect to $y$ gives $v_y=Q+C^{\prime }(y)$, so $C^{\prime }(y)=0$. Thus the most general potential on the connected domain $x>0$ isIts explicit partial derivatives verify sufficiency as well as necessity.

The stated total differential is that of $\pi r^{2}h$, so $V=\pi r^{2}h+C$. The zero-volume condition gives $C=0$. For positive $r,h$, divide the total differential by this volume to obtainFor the specified small changes, $dV/V\approx 2(0.01)+0.02=0.04$, so the volume increases by approximately $4\%$. The fractional expression requires $V>0$; at a degenerate zero-volume state it is undefined. For finite changes, the exact multiplier is $(1.01{)}^2(1.02)$, whose difference from $1$ is $4.0502\%$; the total differential gives its first-order approximation.

Let $t=y/x$. The product rule and the chain rule give the partial derivativesHence the terms containing ${\varphi }^{\prime }$ cancel:This also follows from the Euler theorem for homogeneous functions, because $u(\lambda x,\lambda y)=\lambda u(x,y)$ for $\lambda >0$.

The chain rule expresses the coordinate differential operators asThe coefficients are constant, so a second application givesandAdding cancels the mixed partial derivatives, and $a^2+b^2=1$ givesThe transformation has rotation matrix $\left(\begin{array}{cc}a& -b\\ b& a\end{array}\right)$, so this identity expresses the rotational invariance of the Laplacian.

The Maclaurin series $\sinh t=t+t^3/3!+t^5/5!+O(t^7)$, applied with $t=x^2$ and then multiplied by $x$, givesThese are the first three nonzero terms. The remainder is written in Big O notation.

Put $u=\ln (1+x)=x-x^2/2+x^3/3+O(x^4)$. Since $u=O(x)$, the outer Taylor series requires $u^2=x^2-x^3+O(x^4)$ and $u^3=x^3+O(x^4)$. ThereforeBoth logarithms are defined on a neighbourhood of zero, so this composition of power series is valid there.

Factor the leading term out of the Maclaurin series for the sine:The binomial theorem gives, through degree four inside the bracket,The coefficient of $x^4$ is $1/20+5/12=7/15$. Multiplying back by $x^6$ yields