Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2009 / Paper 2

If $y\ne 0$, the equation would imply $b=-(x/y)a$, contradicting the assumed nonparallel directions. Hence $y=0$. Then $xa=0$ and $a\ne 0$ force $x=0$. This proves that the two vectors are linearly independent.

The function $x^2\sin x$ is an odd function, since $x^2$ is even and the sine is odd. Its integral on the symmetric interval therefore cancels:

The double-angle identity gives ${\cos }^{2}x-{\sin }^{2}x=\cos (2x)$. Thus

The gradients are $\nabla f=(1,1,1)$ and $\nabla g=(2x,2y,2z)$. Equality of their three components therefore gives the unique point

Zero is an eigenvalue exactly when the determinant vanishes. Here $\det A=1-{\alpha }^2$, soFor either value the other eigenvalue is two, as follows from the trace.

Factorization gives $y=(x-2{)}^2(x+1)$, so the zeros are $x=-1$ and $x=2$, with a double root at two. The derivatives are $y^{\prime }=3x(x-2)$ and $y^{\prime \prime }=6(x-1)$. ThusThe second derivative changes sign at one, confirming the inflection point. The curve rises from $-\infty$, crosses at $-1$, reaches its local maximum, falls to touch the axis at two, then rises to $+\infty$. Neither local extremum is a global extremum on the whole real line.

Expanding gives $u=2x^2+2c^2{t}^2$, so $u_{tt}=4c^2$ and $u_{xx}=4$. Hence $u_{tt}=c^2{u}_{xx}$, proving that it solves the one-dimensional wave equation. Equivalently its two summands are oppositely travelling profiles, as in D'Alembert's formula.

At $x=1$, the natural logarithm has value zero and derivative $1/x=1$. Its tangent line is therefore

The logarithm must be expanded about two, not about one. FactorApplying the supplied logarithm expansion to the bracket gives $\log 2+(x/2+x^2/4)-\frac{1}{2}(x/2{)}^2+O(x^3)$. Thus, through second order,In fact the cubic coefficient vanishes, since $\log (1+e^x)=x/2+\log [2\cosh (x/2)]$ and the second term is even; the Maclaurin series remainder can be sharpened to $O(x^4)$.

The inequality is equivalent to $|\cos \theta |>1/\sqrt{2}$. The cosine exceeds $1/\sqrt{2}$ close to zero and $2\pi$, and is less than $-1/\sqrt{2}$ close to $\pi$. Within the specified interval this givesThe four equality angles are excluded because the inequality is strict; zero and $2\pi$ satisfy it and are included.

For $\lambda \ge 0$, the exponential power series gives the normalization of the Poisson distribution:For $\lambda =0$, the limiting law assigns probability one to $k=0$ and zero to every positive integer $k$. Negative $\lambda$ would not define the stated probability distribution.

The direction vector is $b-a=(1,2,-2)$. The vector equation and symmetric Cartesian equation of the line areThe $(x,y)$-plane has $z=0$, which gives $t=2$. Thus its intersection with the line has position

The plane normals are $n_1=(2,-1,-1)$ and $n_2=(3,-1,-3)$. Their cross product is $(2,3,1)$, perpendicular to both normals and hence parallel to the intersection line. Setting $z=0$ in both equations gives the common point $(1,-1,0)$. ThereforeThe normals are not parallel, so their intersection is a unique line.

From the line point $r_0=(1,-1,0)$ to $P=(2,2,1)$, take $w=P-r_0=(1,3,1)$ and line direction $d=(2,3,1)$. Orthogonal projection gives the nearest parameter $t=(w\cdot d)/(d\cdot d)=12/14=6/7$. The perpendicular displacement has squared Euclidean normHence the distance from a point to a line in three dimensions isThe nearest point is $(19/7,11/7,6/7)$; subtracting it from $P$ gives a vector orthogonal to $d$, verifying the minimum.

The characteristic equation is $r^2+\lambda r+k=0$, whose roots are $r_{\pm }=(-\lambda \pm \sqrt{{\lambda }^2-4k})/2$. For distinct real roots, $y=C_1{e}^{r_{+}t}+C_2{e}^{r_{-}t}$; at the repeated-root threshold ${\lambda }^2=4k$, $y=(C_1+C_{2}t)e^{-\lambda t/2}$. In the complex-root case put $\Omega =\sqrt{k-{\lambda }^2/4}$ to obtainThis is precisely the oscillatory regime of the damped harmonic oscillator. The zero solution is of course nonoscillatory; the regime describes its nontrivial motions.

The damped angular frequency is $\Omega =\sqrt{k-{\lambda }^2/4}$. The oscillation envelope is proportional to $e^{-\lambda t/2}$. A decrease by a quarter leaves three quarters of the original amplitude, so $e^{-\lambda t/2}=3/4$. ThereforeThis concerns the envelope; it does not ask when the instantaneous displacement first equals three quarters of a chosen value.

Let $D=1+{\omega }^2$, with $\omega >0$. A particular solution of the forced harmonic oscillator is $y_p=e^{-t}(at+b)$. Substitution gives $Da=2$ and $Db-2a=0$, hence $a=2/D$ and $b=4/D^2$. Adding $A\cos \omega t+B\sin \omega t$ and imposing the two initial conditions gives $A=-4/D^2$ and $B=2(1-{\omega }^2)/(\omega D^2)$. ThusThe first term tends to zero. The amplitude of the surviving sinusoid isAs an independent interpretation, the residual amplitude of an oscillator after an integrable force is ${\omega }^{-1}|{\int }_0^{\infty }F(t)e^{-i\omega t}dt|$. Here that integral is $2/(1+i\omega {)}^2$, giving the same result.

Put $X=x-1$ and $H=e^{-(X^2+y^2)/2}>0$. The gradient components are $z_x=y(1-X^2)H$ and $z_y=X(1-y^2)H$. If either $X$ or $y$ vanishes, both must vanish. Otherwise both squared coordinates must be one. Therefore the five stationary points in the original $(x,y)$ coordinates are

At $(X,y)=(0,0)$ the Hessian matrix is $\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)$, with eigenvalues $1$ and $-1$, so $(x,y)=(1,0)$ is a saddle point. At the other four points the mixed second derivative vanishes and the two diagonal entries both equal $-2Xy/e$. HenceThe four extrema are global: $|z|=(|X|e^{-X^2/2})(|y|e^{-y^2/2})\le e^{-1}$ because each factor has maximum $e^{-1/2}$ at absolute argument one.

The zero level set consists of the lines $x=1$ and $y=0$. The height is positive where $(x-1)y>0$ and negative where that product is negative. In each of the four resulting regions, closed nonzero contours surround its extremum; levels closer to zero extend further from the extremum towards the zero-level axes and towards large distance. The height tends to zero at infinity. The figure marks the four extrema and central saddle point, and includes the downhill path used below.

For the graph surface $Z=z(x,y)$, an upward normal is $(-z_x,-z_y,1)$. At $x=y=1$, let $q=e^{-1/2}$; then $z_x=q$ and $z_y=0$. Normalizing gives the unit normal vectorIts negative is the equally valid downward normal.

A tangent displacement has the form $(dx,dy,z_{x}dx+z_{y}dy)$. At the specified point, $z_y=0$ and $z_x=q>0$, so the fastest decrease of height per unit surface distance has $dy=0$ and $dx<0$. The unit downhill tangent is $(-1,0,-q)/\sqrt{1+q^2}$. Therefore, for speed $u$ measured along the surface,Its Euclidean norm is $u$, and it is orthogonal to the surface normal. The general construction is steepest descent on a graph surface; using a horizontal speed equal to $u$ would overestimate the full three-dimensional speed.

The horizontal direction of steepest descent on a graph surface is parallel to $-\nabla z$. On the line $y=1$, the component $z_y=X(1-y^2)H$ vanishes identically, so that line is invariant under the downhill path. Starting at $X=0$, $z_x=(1-X^2)e^{-(X^2+1)/2}>0$ for $-1<X<0$, so $x$ decreases towards zero. The path reaches the minimumAt the minimum no strictly downhill direction remains. A constant-speed parametrization reaches it after the finite arc length divided by $u$; the unnormalized gradient flow approaches the same endpoint only asymptotically in its own time parameter.

Expanding the determinant along the first row givesIn particular the matrix is invertible.

The cofactor matrix is $\left(\begin{array}{ccc}-3& -1& -1\\ 5& -1& 3\\ -1& 1& 1\end{array}\right)$. Transpose it to form the adjugate matrix, and divide by the determinant $-4$:Direct matrix multiplication gives $A{A}^{-1}=I$, checking all entries.

A homogeneous square linear system has a nonzero solution exactly when its coefficient matrix is singular. HereThusFor this value, the second equation gives $x=-2y$ and the first gives $z=y$; the third then holds automatically. All solutions are $t(-2,1,1)$, with $t\ne 0$ for the nonzero ones.

The coefficient matrix is a Vandermonde matrix, with determinantTherefore a nonzero solution exists exactly whenFor example if $a=b$, the first two columns are equal and $(1,-1,0)$ is a solution. If exactly two parameters agree, the distinct remaining column is independent of the repeated one, so the kernel has dimension one. If all three agree, only $x+y+z=0$ remains, giving a two-dimensional kernel.

Use the real Fourier series convention $g=a_0/2+{\sum }_{n\ge 1}[a_n\cos (n\pi x)+b_n\sin (n\pi x)]$ for period two. Since $g$ is an even function, $b_n=0$, and $a_0={\int }_{-1}^1{x}^{2}dx=2/3$. Twice applying integration by parts, with $k=n\pi$, givesHenceThe periodic extension is continuous at its endpoints, so the series takes the specified endpoint values as well as the interior values.

Apply Parseval's identity in the same coefficient convention:The left-hand side is ${\int }_{-1}^1{x}^{4}dx=2/5$. Thus $2/5=2/9+(16/{\pi }^4){\sum }_{n\ge 1}{n}^{-4}$, giving

Again the function is even, so all sine coefficients vanish. Its constant term is $a_0/2=\sinh 1$. With $k=n\pi$, an antiderivative isThus $a_n=2(-1{)}^n\sinh 1/(1+n^2{\pi }^2)$. In the requested factored notation,This is the period-two specialization of the Fourier series of periodic hyperbolic sine and cosine. The function values match at the endpoints, although its derivative has a jump in the periodic extension.

In Cartesian coordinates, the divergence is the sum of the three component derivatives. Therefore

For $r=\sqrt{x^2+y^2+z^2}>0$, the chain rule gives ${\partial }_x(r^{-1})=-x/r^3$, with analogous formulas for the other components. HenceOn the right the numerator denotes the position vector, while its denominator is the cube of its magnitude.

The vector $m$ is constant, so $\nabla (m\cdot r)=m$. Also $\nabla (r^{-3})=-3r/r^5$, by the same chain rule. The product rule therefore yieldsThe three components follow by differentiating $(\alpha x+\beta y+\gamma z)(x^2+y^2+z^2{)}^{-3/2}$.

For constant $m$, the curl of a cross product identity reduces toCommuting mixed second partial derivatives gives $(m\cdot \nabla )\nabla \varphi =\nabla (m\cdot \nabla \varphi )$. Hence the requested sum equals $m\Delta \varphi$, which is zero because $\varphi$ is harmonic. The argument assumes the indicated second derivatives are continuous on the domain.

From the preceding gradient, ${\partial }_x^2(r^{-1})=-r^{-3}+3x^2{r}^{-5}$. Summing the three Cartesian second partial derivatives givesThus the reciprocal radius is harmonic away from the origin. Its negative gradient isThe curl vanishes because $E$ is a gradient field and mixed second derivatives commute. The exclusion of the origin is essential: the potential and field are singular there.

For a continuous random variable with probability density function $p$, its expected value and variance areThe mean requires an integrable first absolute moment, and a finite variance requires a finite second moment. When these hold, $\mathrm{Var}X=\mathbb{E}[X^2]-(\mathbb{E}X{)}^2$.

Take $\sigma >0$ and substitute $u=(x-\mu )/\sigma$. The density becomes the standard Gaussian distribution density $\phi (u)=e^{-u^2/2}/\sqrt{2\pi }$. Its normalization follows from the supplied Gaussian integral after scaling by $\sqrt{2}$. Since $u\phi (u)$ is an integrable odd function, its integral is zero, and therefore $\mathbb{E}X=\mu$.

For the second moment, ${\phi }^{\prime }=-u\phi$. Integration by parts givesThe endpoint term vanishes by Gaussian decay. Consequently

The conditional probability product rule gives $P(A\mid B)P(B)=P(A\cap B)$. Applying it once more givesThe conditionals in the original expression require their conditioning events to have positive probability.

The events $B$ and $\overline{B}$ form a partition. The law of total probability therefore givesAs written, both conditionals require nonzero conditioning probabilities.

The numerator $P(B\mid A)P(A)$ equals $P(A\cap B)$, while $P(A\mid B)=P(A\cap B)/P(B)$. Thus, wherever the original quotient is defined,In particular a zero numerator and denominator cannot be cancelled; the original expression requires $P(A\mid B)>0$. This is the rearrangement of Bayes' theorem.

The two-event inclusion-exclusion principle subtracts the overlap counted twice:

This is exactly the definition of conditional probability, provided $P(A)>0$:

Under a uniform shuffle, every unordered five-card hand is equally likely, giving $\left(\genfrac{}{}{0ex}{}{52}{5}\right)$ possible hands. Choose the triple's rank in 13 ways and its suits in $\left(\genfrac{}{}{0ex}{}{4}{3}\right)$ ways; then choose a different pair rank in 12 ways and its suits in $\left(\genfrac{}{}{0ex}{}{4}{2}\right)$ ways. Each full house determines these choices uniquely, so there is no further ordering factor. ThereforeThe count uses binomial coefficients because the order of cards within a hand is irrelevant.