The positive-domain graph starts with limiting value $f(0^{+})=1$ and limiting slope $-1$, then decreases towards the horizontal asymptote $f=0$, with $f(q)\sim 1/(2q)$ for large $q$. In fact direct differentiation givesThe first inequality follows from $e^s>1+s$ for $s>0$. Thus the graph is strictly decreasing and convex upward, with $0<f(q)<1$ and no stationary points. The integral representation $f(q)={\int }_0^1(1-e^{-s/q})ds$ also shows its positivity.

These properties belong to the reusable exponential spherical mass factor.

For $q=r_0/R\to \infty$, expand the exponential function in powers of $1/q$:The constant and linear terms cancel in $1-q+q{e}^{-1/q}$, givingFor $q\to 0^{+}$, $e^{-1/q}$ is exponentially small:The small-$q$ correction $q{e}^{-1/q}$ is smaller than every positive power of $q$ beyond the displayed linear term. In dimensional terms, the large-$q$ leading mass is $2\pi A{R}^2/r_0$, while the small-$q$ limit is $4\pi AR$.

In spherical coordinates, with polar angle $0\le \theta \le \pi$ and azimuth $0\le \phi <2\pi$, the volume element isThe mass density is radial, so integrating over both angles givesWriting $q=r_0/R>0$, this becomesAlthough the mass density diverges like $A/(r_{0}r)$ near the origin, multiplication by the spherical volume element gives an integrable expression. Thus the central singularity does not produce infinite mass.

Put $s=t^2-t^{-2}$. Part (b)(ii) gives $u=s+2i$, soThe parameter $s$ increases strictly from $-15/4$ to $15/4$. Therefore the imaginary part $Y=4s$ increases from $-15$ to $15$, while the real part is $X=s^2-4$. The endpoints areThe real axis crossing is at $s=0$, giving $\boxed{(-4,0)}$ at $t=1$. The two imaginary axis crossings satisfy $s^2=4$, givingIn order, the trajectory passes from its lower endpoint through $(0,-8)$, then the leftmost point $(-4,0)$, then $(0,8)$, to its upper endpoint. It is a right-opening parabola segment, traversed continuously upward as shown in the right panel.

Squaring givesThus $\mathrm{Im}u=2$ is constant, while $d(\mathrm{Re}u)/dt=2t+2t^{-3}>0$. The trajectory is the horizontal line segmentIt crosses the imaginary axis at $\boxed{(0,2)}$, when $t=1$, and never crosses the real axis. The middle panel shows the rightward direction.

Write $z=X+iY$. Its coordinates are $X=t$, $Y=1/t$, so $XY=1$. For the given positive range of $t$, this traces a segment of the first-quadrant hyperbola, moving rightward and downward:The complex conjugate is $z^{\ast }=t-i/t$. It is the reflection in the real axis, tracing $XY=-1$ in the fourth quadrant, rightward and upward:Neither trajectory crosses either axis, because $t$ and $1/t$ remain nonzero. The left panel of the figure marks the endpoints and directions.

Square the result of part (a)(i), keeping the cross term:Hence

Expand the square using $i^2=-1$:Thus the real part and imaginary part are

For the $C$ obtained above, the characteristic polynomial isThus the eigenvalues are $5$ and $-5$. For $\mu =5$, $(C-5I)\mathbf{v}=0$ gives $v_2=-3v_1$; for $\mu =-5$, it gives $v_1=3v_2$. Normalizing the eigenvectors yieldsBoth are unit vectors, their dot product is zero, and either overall sign is allowed for each eigenvector.

The original PDF requires $B$ itself to be an orthogonal skew-symmetric matrix; it does not require $C$ to be diagonal. A real two-dimensional skew-symmetric matrix has the form $B=\left(\begin{array}{cc}0& u\\ -u& 0\end{array}\right)$. The orthogonal matrix condition gives $u^2=1$, so choose $u=1$. Then $C=B^{\mathsf{T}}AB$ is forced:Direct multiplication verifies $B^{\mathsf{T}}=-B$, $B^{\mathsf{T}}B=I$ and $BC{B}^{\mathsf{T}}=A$. Choosing $-B$ gives the same $C$. Replacing $B$ by a matrix of eigenvectors to make $C$ diagonal would generally fail the printed antisymmetry requirement.

A three-dimensional rotation matrix is an orthogonal matrix with determinant $+1$. Here $\det A=b(a^2-d^2)$. For the branch $d=0$, the determinant condition gives $b=1$; for the branch $a=0$, it gives $b=-1$. HenceThe first is the identity rotation; the remaining three are half-turns. Their fixed axes are respectively the $y$ axis, the direction $(1,0,-1)$ and the direction $(1,0,1)$. The identity is included under the usual rotation matrix convention.

Put $d=c-1$. The matrix is real and symmetric, so orthogonality means $A^{\mathsf{T}}A=A^2=I$. Multiplication gives the scalar conditionsIf $d=0$, then $c=1$ and $a=\pm 1$; if $a=0$, then $d=\pm 1$ and $c=0$ or $2$. In both cases $b=\pm 1$. The eight parameter triples are thereforeThese conditions are sufficient as well as necessary, since they make every entry of $A^{\mathsf{T}}A$ equal to the corresponding entry of the identity matrix.

Choose the trigonometric substitution $x=\sin u$ with $-\pi /2\le u\le \pi /2$. The cosine is nonnegative there, so $\sqrt{1-x^2}=\cos u$ and $dx=\cos udu$. The double-angle formula then givesSince $u=\arcsin x$ and $\sin 2u=2x\sqrt{1-x^2}$,The formula is an antiderivative on $(-1,1)$ and has continuous endpoint values at $\pm 1$.

For integration by parts, choose $u=\ln x$ and $dv=x^{-4}dx$. Then $du=dx/x$ and $v=-1/(3x^3)$, givingHere $C$ is an arbitrary constant of integration.

Use integration by substitution $u=\cos x$, $du=-\sin xdx$. The denominator factorizes as $(u-2)(u-3)$, and partial fraction decomposition givesIntegrating and restoring $x$ yieldsFor real $x$, both $2-\cos x$ and $3-\cos x$ are positive, so no absolute-value signs are needed in this real logarithm. The denominator never vanishes.

Write $N={\sum }_{j=1}^{10}{X}_j$, where the independent Bernoulli random variables indicate rottenness. Since $X_j^2=X_j$, their variances are $E[X_j^2]-E[X_j{]}^2=p-p^2$. The independence makes all cross covariances zero, soThis standard deviation vanishes at $p=0$ and $p=1$, when the count is certain.

Let $N$ count rotten pieces. Each item supplies an independent Bernoulli trial with success probability $p$, so $N$ has a binomial distribution with parameters $10,p$. There are $\left(\genfrac{}{}{0ex}{}{10}{n}\right)$ choices of the rotten items, and each such choice has probability $p^n(1-p{)}^{10-n}$. HenceFor $p=0$ or $p=1$, interpret this as the corresponding distribution concentrated at zero or ten; no ambiguous $0^0$ evaluation is needed.

Let $I_j$ be the indicator random variable of an apple in lunch $j$. The number remaining is $R=5-I_1-I_2-I_3$. Each expected value $E[I_j]=P(A_j)=1/2$. The linearity of expectation therefore givesAlthough the indicator random variables are dependent under sampling without replacement, the linearity of expectation needs no independence assumption.

By the inclusion-exclusion principle for probabilities,After an initial banana, five of the nine remaining pieces are apples, so the intersection probability is $(1/2)(5/9)=5/18$. ConsequentlyAs a check, the complementary event is an apple followed by a banana, which also has probability $5/18$.

Each position has apple probability $1/2$. Applying Bayes' theorem to the preceding conditional probability givesThis also agrees with exchangeability of the second and third positions.