There is no singularity at the lower endpoint one. Put $t=\log (1+x)$, so $dx=e^{t}dt$. The nonnegative integral becomesFor $\alpha =1$, it converges as a reciprocal-square integral. For $\alpha >1$, exponential decay makes it converge. For $\alpha <1$, its integrand eventually grows without bound, so its integral diverges. ThusThis is the logarithmic borderline in a power-tail integrability test.

For positive integers $n,m$, let $E_{n,m}=\{x:f(x)\ge 1/n\}\cap [-m,m]$. This measurable set has finite measure, and $f\ge n^{-1}{1}_{E_{n,m}}$. Monotonicity of the Lebesgue integral givesso every $E_{n,m}$ is a Lebesgue null set. The set where $f>0$ is the countable union of these sets over $n,m$. Part (a) makes it null, henceThe bounded truncation ensures that only integrable indicator functions are used with the stated comparison rule.

A Lebesgue null set $E\subset \mathbb{R}$ can, for every $ϵ>0$, be covered by countably many open intervals whose total length is less than $ϵ$. Equivalently, its Lebesgue outer measure is zero.

If $E_j$ is null, choose for it such a cover of total length below $ϵ2^{-j}$, $j\ge 1$. The union of these countably many countable covers covers ${\bigcup }_j{E}_j$ and has total length below $ϵ$. Since $ϵ$ is arbitrary, a countable union of null sets is null.

The two primal constraints are equalities, so let the two dual variables $u,v$ be free. Primal feasibility restricts $x$ to the nonnegative orthant, givingThus the feasible region lies above all four lines:Its lower boundary follows the third line up to $(5/9,7/9)$, the fourth line from there to $(10/3,-1/3)$, and then the first line. The second line never determines the envelope. Along these three pieces $u+v$ respectively has slopes $-3$, $3/5$, and $3/4$, so its unique minimum is at the first corner:

At this dual point the first two inequalities are strict, while the last two are equalities. complementary slackness therefore suggests $x_1=x_2=0$. The remaining primal equations become $4x_3+2x_4=300$, $x_3+5x_4=300$, giving $x_3=x_4=50$. This is nonnegative and satisfies both equations. Its objective function is $3\cdot 50+5\cdot 50=400$, equal to the feasible dual value. weak duality provesNo nonnegativity restriction was imposed on the dual variables; the negative-$v$ second corner is part of the feasible boundary, even though it is not optimal.

With primal constraints $Ax=b$, the dual variable is unrestricted in sign:Membership of $x$ in the nonnegative orthant still yields $c^{T}x\le y^{T}Ax$, but the next step is now the equality $y^{T}Ax=y^{T}b$, so no sign condition on $y$ is needed. Equivalently an equality constraint can be replaced by two opposite inequalities whose nonnegative multipliers combine into one free multiplier.

The dual isweak duality says that any feasible primal $x$ and dual $y$ satisfy $c^{T}x\le b^{T}y$. Indeed,The first inequality uses $x\ge 0$ and $A^{T}y-c\ge 0$, and the second uses $y\ge 0$ and $b-Ax\ge 0$. This proves the theorem. In particular, feasible solutions with equal objective function values are optimal for both problems.

Start the augmenting-path algorithm from the zero flow. With integer capacities and an integer electric current flow, every residual capacity is an integer, so a path bottleneck is a positive integer. Augmenting by it preserves integrality and increases the flow value by at least one. The value is bounded by the finite total capacity leaving the source, so the process terminates. At termination there is no augmenting path, hence the flow is maximum by the max-flow min-cut theorem. Its value is an integer.

Equivalently a minimum cut has capacity equal to the sum of integer arc capacities, so the maximum flow value equals an integer. Thus integer capacities guarantee an integer optimal volume, even though arbitrary feasible real-valued flows need not themselves have integer entries.

flow conservation determines the unspecified initial arcs:Every entry respects capacity, and the initial value is four. In the residual network take the augmenting pathThe step $4\to 3$ is the reverse residual arc of the used arc $3\to 4$. Its use cancels some existing flow rather than sending new physical flow on a nonexistent original arc. The residual capacities along the path are $2,1,2,2,1$, so augment by their minimum, one.

The resulting original-arc flows areTheir value is five. The cut separating the sink from all other nodes has capacity $c_{46}+c_{56}=3+2=5$. Every feasible flow is bounded by this cut capacity, so the value-five flow is optimal.

True. Each connected component of a forest is a tree. If the component sizes are $n_1,\dots ,n_k$, their edge counts are $n_1-1,\dots ,n_k-1$, including isolated vertices as size-one components. Summing gives $m=n-k$, hence

True. Choose a root of the tree and color every vertex according to the parity of its graph distance from that root. The unique-path property implies that the endpoints of each edge have distances differing by one. Thus every edge crosses the two color classes, proving bipartite graph.

False. The four-cycle $C_4=K_{2,2}$ is bipartite, with alternating color classes, but contains a cycle and therefore is not a tree.

False. In $K_{r,s}$ the $r$ vertices in one class have degree $s$, while the $s$ vertices in the other have degree $r$. For positive $r,s$, connectedness is automatic, and Euler's criterion requires both $r$ and $s$ to be even. Evenness of their sum alone allows both to be odd. For example $K_{1,3}$ has $r+s=4$ but all four degrees are odd, so it has no Eulerian cycle.

True. Represent a vertex of $Q_4$ by a four-bit string, with an edge changing exactly one bit. Color according to the parity of the sum of the bits. Every edge joins opposite parities, giving a proper two-coloring. Since the graph has edges, one color is impossible, so $\chi (Q_4)=2$.

False. In a Prüfer code, the degree of a label is one plus its number of occurrences. Here labels $1,\dots ,n-2$ occur once and labels $n-1,n$ do not occur. Thus the former have degree two and the latter degree one. The decoded tree is the pathwith the evident single-edge interpretation when $n=2$. Independently, no vertex of a simple tree on $n$ vertices can have degree $n$ at all. The degree claim in the statement is therefore impossible.

False. The cube has eight vertices and twelve edges. Euler formula for a connected planar graph for its connected plane graph gives $8-12+F=2$, soThe nested-square embedding has four faces between the squares, one inside the smaller square and the unbounded exterior face.

True. Draw two nested squares and join corresponding corners. This is the hypercube graph: each square gives one four-cycle and the four connecting edges give the remaining coordinate direction. The edges have no crossings, furnishing a plane graph.

True. Label the vertices $1,\dots ,n$. The cycle $1,2,\dots ,n,1$ uses existing edges of the complete graph and visits every vertex once before returning. It is a Hamiltonian cycle for every $n\ge 3$.

False. A connected graph has an Eulerian cycle exactly when every vertex has even degree. In $K_n$ every degree is $n-1$, so it has an Eulerian cycle precisely when $n$ is odd. The complete graph $K_4$ is a counterexample to the statement for all $n\ge 3$.

Take the curl of the Euler equations for an inviscid fluid. The gradient on the right has zero curl. Use the stated vector identity with $\mathbf{F}=\mathbf{\omega }$, $\mathbf{G}=\mathbf{u}$:incompressible flow gives $\nabla \cdot \mathbf{u}=0$, while a curl always has zero divergence, so $\nabla \cdot \mathbf{\omega }=0$. The remaining terms giveFor a two-dimensional velocity independent of $z$, $\mathbf{\omega }=(0,0,v_x-u_y)$ and $(\mathbf{\omega }\cdot \nabla )\mathbf{u}={\omega }_z{\partial }_z\mathbf{u}=0$. Thus the scalar vorticity is conserved following each fluid particle. It can still vary from particle to particle; this is advection, not a claim of spatially uniform vorticity.