There are $99$ possible stopping points, so the equal defective-ball probabilities are $d_k=1/99$. With $g_0>0$, the desired conditional probability givesThus, for a fixed good-ball stopping law,Since $g_0\le 1$, the smallest possible production probability over admissible stopping laws isIf strict positivity of every $g_k$ is additionally imposed in this general case, then $g_0<1$ and one only has the infimum $1/2$, approached as $g_0\to 1$. The paper states that strict-positivity assumption explicitly for part (a), not for the general case in part (b). Without choosing or bounding $g_0$, the exact fixed-law answer remains $1/(1+g_0)$.

The event includes stopping positions zero through $k$. Applying Bayes' theorem to their joint and marginal probabilities givesThis conditional probability is defined whenever its conditioning event has positive probability. At $k=n$, it reduces to $p$.

For $0\le k\le n$, sum the marginal stopping probabilities over the selected tail:Equivalently it is $p(1-{\sum }_{j=0}^k{g}_j)+(1-p)(1-{\sum }_{j=0}^k{d}_j)$. At $k=n$, both empty tail sums are zero, as required.

The condition $X>x_0$ selects positions $x_1$ and $x_2$. Their combined good-ball joint probability is $p(g_1+g_2)$, and their total probability is $p(g_1+g_2)+(1-p)(d_1+d_2)$. Thus

Divide the relevant joint probability by the marginal stopping probability. By Bayes' theorem,The denominator is positive under the stated assumptions.

The good and defective cases are disjoint and exhaustive. The law of total probability therefore gives

Multiply the production probability by the conditional stopping probability:

Distinguish the production type as well as the stopping position. The sample space has the six outcomesHere $x_k$ in an event means $X=x_k$. Their joint probabilities are $p{g}_k$ and $(1-p)d_k$, respectively. The conditional stopping laws obey ${\sum }_{k=0}^2{g}_k={\sum }_{k=0}^2{d}_k=1$; with the stated strict positivity all six outcomes are possible.

The arc length factor is the norm of the tangent:Hence $ds=10dt$ and $\Phi =xy=\cos 6t\sin 6t=\frac{1}{2}\sin 12t$. The scalar line integral is thereforeThis uses scalar arc length, rather than the oriented displacement used in the vector line integrals of part (a).

Substituting the endpoint parameter values givesThe curve lies on the unit circular cylinder $x^2+y^2=1$. Its azimuth increases from zero to $\pi /2$ while its height increases linearly, so it is one quarter-turn of a rising helix. The arrow in the sketch indicates increasing parameter.

Parametrize the circle by $\mathbf{r}(t)=(\cos t,\sin t)$, $0\le t\le 2\pi$. Its tangent is ${\mathbf{r}}^{\prime }=(-\sin t,\cos t)=\mathbf{F}(\mathbf{r}(t))$, so the dot product $\mathbf{F}\cdot {\mathbf{r}}^{\prime }$ is one. For the other field,Integrating around the full circle gives

Starting at $(1,0)$, traverse the right, top, left and bottom sides in that order. For $\mathbf{F}$, the differential expression is $-ydx+xdy$. The four line integral contributions are respectively $1,1,0,0$, so the total is two.

For $\mathbf{G}$, the expression is $2x{y}^{2}dx+2y{x}^{2}dy$. The right side contributes ${\int }_0^{1}2ydy=1$, and the top side contributes ${\int }_1^{0}2xdx=-1$; the other sides contribute zero. Therefore

At $(1,2)$, the gradient is $(4,2)$. The requested direction must be normalized to the unit vector $(1,1)/\sqrt{2}$. The directional derivative is thereforeUsing $(1,1)$ without normalization would instead measure change per unit of a differently scaled path parameter.

In the positive quadrant, both components of the gradient are positive. Along $y=x$ the arrows point along $(1,1)$; near the horizontal axis they point mainly upwards, and near the vertical axis mainly rightwards. The level sets are the hyperbolas $xy=f/2$, including the coordinate axes for level zero. The gradient is orthogonal to each regular level set and points towards increasing values of $f$.

The double-angle formula gives $r^2\sin 2\varphi =2(r\cos \varphi )(r\sin \varphi )=2xy$. Therefore

Rationalize the positive difference:Since the denominator is at least $2n^2$,The upper bound is a constant multiple of the convergent p-series with exponent two. The comparison test therefore provesThe asymptotic equivalence of its summand to $a^2/(2n^2)$ gives the same conclusion by limit comparison test.

For the positive terms $a_n=n!/2^n$, the ratio is $a_{n+1}/a_n=(n+1)/2\to \infty$. The ratio test therefore gives divergence. More directly, the terms increase by a factor at least two for all $n\ge 3$, so they fail the necessary term test for divergence condition $a_n\to 0$. Hence

For $m=0$, continuity fails. Take $x_k=(\pi /2+2\pi k{)}^{-1}$ and ${\tilde{x}}_k=(3\pi /2+2\pi k{)}^{-1}$. Both tend to zero, but the corresponding function values are $1$ and $-1$. Thus there is no limit, and the function is not differentiable either, since differentiability would imply continuity.

For $m=1$, the bound $|x\sin (1/x)|\le |x|$ and the squeeze theorem give continuity at zero. Its difference quotient is $\sin (1/h)$, however, which has no limit by the same two sequences. It is therefore not differentiable there.

For $m=2$, the bound $|x^2\sin (1/x)|\le x^2$ proves continuity. Its difference quotient is $h\sin (1/h)$, whose absolute value is bounded by $|h|$, so the derivative exists and equals zero. The complete classification is

For statement (1), continuity at the interior point meansExplicitly, for every $\epsilon >0$ there exists $\delta >0$ such that $|x-x_0|<\delta$ implies $|f(x)-f(x_0)|<\epsilon$ for points in the domain. This is the epsilon-delta definition of continuity.

For statement (2), differentiability means the two-sided difference quotient has a finite limit:The same finite value must be obtained from positive and negative $h$. Differentiability implies continuity, whereas the converse need not hold.

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