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The specified range ensures $\cos y\ne 0$, so the separable differential equation can be written ${\sec }^{2}ydy=\sin xdx$. Integrating gives $\tan y=C-\cos x$. The range selects the principal inverse tangent, henceEvery finite $C$ produces a solution within the required range. Constant solutions at the excluded values $y=\pm \pi /2$ are not part of this family or of the question's domain.

Use the real antiderivative $\ln |x|$, valid on either half-line away from zero:The negative answer is consistent with a negative integrand and increasing integration limits.

Since the derivative of $\ln \cos x$ is $-\tan x$ on this interval,

Assume positive corridor widths $a,b$. Write the line through the two wall-contact endpoints in intercept form. Requiring it to pass through the inner corner givesIts squared length is $F=(a+X{)}^2+(b+Y{)}^2$. Apply Lagrange multipliers to this function subject to $G=XY-ab=0$:Multiplying by $X$ and $Y$ respectively eliminates the multiplier, giving $X(a+X)=Y(b+Y)$. Substitute $Y=ab/X$ and multiply by $X^2$:The branch $X=-a$ is geometrically inadmissible. Thus $X=a^{1/3}{b}^{2/3}$ and $Y=a^{2/3}{b}^{1/3}$. If $A=a^{1/3}$ and $B=b^{1/3}$, the two intercepts become $A(A^2+B^2)$ and $B(A^2+B^2)$, soAlong $XY=ab$ in the positive quadrant, $F\to \infty$ as $X\to 0^{+}$ or $X\to \infty$, and this is its only stationary point. It is therefore the minimum of the longest sticks that fit at each orientation. A stick being turned must pass through every intermediate orientation, so this narrowest clearance is precisely the maximum carryable length, rather than a maximum of $F$. This is the ladder around a right-angled corner problem, with a zero-width stick and wall contact allowed as in the ideal geometry.

Maximizing the squared distance avoids differentiating a square root. The Lagrange multiplier equations for $f=x^2+y^2$ areAdding and subtracting the first two gives $(2-3\lambda )(x+y)=0$ and $(2-\lambda )(x-y)=0$. The feasible stationary points therefore lie on $x=y$ or $x=-y$. The first case gives $(x,y)=(\pm 2/\sqrt{3},\pm 2/\sqrt{3})$ and squared distance $8/3$. The second gives $(2,-2)$ and $(-2,2)$, with squared distance $8$.

The ellipse is compact and its constraint gradient does not vanish on the curve, so an extremum exists among these points. A direct global check is $4=f+xy\ge f/2$, using $xy\ge -f/2$. Equality occurs when $x=-y$. Consequently

Set $u=\ln x$, so $du=dx/x$. The real domain is $x>1$, ensuring $u>0$. The integration by substitution followed by integration by parts givesThus

For positive $a$, write $a^{-x}=\exp (-x\ln a)$. Differentiating by the chain rule givesThe natural logarithm of the constant base supplies the factor; it must not be omitted when $a\ne e$.

Index the terms from $n=0$:Its absolute successive-term ratio tends to zero for every fixed $x$, so it converges everywhere. Divide by $x$ temporarily and use the exponential series divided by a shifted index with $k=3$:ThereforeAt $x=0$ both the original series and this formula are zero, so the temporary division loses no value. The termwise integration is justified by uniform convergence of the exponential series on the finite integration interval.

Multiply the Maclaurin series $e^{-x}=1-x+x^2/2+O(x^3)$ by the binomial series $(1+x{)}^{-1/2}=1-x/2+3x^2/8+O(x^3)$. The constant, linear and quadratic coefficients are therefore

Write the formal Maclaurin series as ${\sum }_{n\ge 0}{a}_n{x}^n$, where $a_n=f^{(n)}(0)/n!$. If the successive nonzero coefficient ratios have a limit, the ratio test givesso the series converges absolutely for $|x|<R$, whereIf that ratio does not exist, the Cauchy-Hadamard theorem gives the general formula $R^{-1}=\mathrm{limsup}|a_n{|}^{1/n}$. The radius of a Maclaurin series from successive derivatives concerns convergence of the series; equality to the original function additionally requires analyticity or a vanishing Taylor remainder. Mere differentiability, or even smoothness, does not ensure that equality.

For the specified logarithm,The absolute successive-term ratio tends to $|x|/2$, so the interval of absolute convergence is $(-2,2)$. At $x=2$ the nonconstant terms are the alternating harmonic series, which converges conditionally to $\ln 2$, giving the full sum $\ln 4$. At $x=-2$ all nonconstant terms are $-1/n$, so the series diverges to $-\infty$; the original logarithm is also undefined there. Thus the full real convergence interval is $(-2,2]$.

When only a tiny fraction of either species is removed, the trout fraction among the remaining fish stays nearly constant at $p=K/N$. Successive catches are then approximately independent trials with that fixed success probability. The count therefore approaches the binomial distributionThe approximation comes from negligible depletion, not from an exact independence property of sampling without replacement.

The three preceding probabilities satisfyTheir expected value and second moment areSubtracting the squared mean givesThe value $\mu =2(2/10)$ also follows from linearity of expectation. The variance includes the finite-population reduction caused by sampling without replacement.

Count unordered sets of distinct labelled fish: there are $\left(\genfrac{}{}{0ex}{}{N}{M}\right)$ possible hauls. To obtain exactly $j$ trout, choose $j$ labels from the $K$ trout and $M-j$ from the $N-K$ carp, giving $\left(\genfrac{}{}{0ex}{}{K}{j}\right)\left(\genfrac{}{}{0ex}{}{N-K}{M-j}\right)$ possibilities. Uniform sampling without replacement therefore gives the hypergeometric distributionOutside this integer range the probability is zero. If ordered catches were counted instead, every haul would have $M!$ orders and that factor would cancel from the probability ratio.

Here equal catchability means equal selection probability for each individual remaining fish. It does not mean that the two species have probability $1/2$ on each draw regardless of their populations. This interpretation is the one that yields the stated hypergeometric law.

Write $p_j=\mathbb{P}(X=j)$, with ${\sum }_{j=0}^N{p}_j=1$. The expected value, variance, and strict lower-tail probability areThe upper summation limit $Y-1$ is essential because the inequality is strict and the variable is integer-valued.

Take the odd extension $H(x)=x(\pi -|x|)$ on $[-\pi ,\pi ]$ and then extend with period $2\pi$. It agrees with the required arch for $0\le x\le \pi$ and is continuous at its joins. Its Fourier sine series hasFor clarity, let $h=x(\pi -x)$, so $h^{\prime }=\pi -2x$ and $h^{\prime \prime }=-2$. Two integrations by parts, with the first endpoint term zero, giveConsequently $b_n=4[1-(-1{)}^n]/(\pi n^3)$: the even coefficients vanish and the odd ones are $8/(\pi n^3)$. This provesThe Fourier sine series of an odd quadratic arch converges to this continuous extension; its coefficients are also absolutely summable. Finally evaluate at $x=\pi /2$. The left side is ${\pi }^2/4$ and the sine factors are $(-1{)}^n$, soThe evaluation is at a continuity point, so it does not invoke a jump midpoint.