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An integrating factor is $\mu (x)=\cosh x$, since ${\mu }^{\prime }/\mu =\tanh x$. ThusFor negative $x$ the integral vanishes; for positive $x$ it is $\sinh x$. ConsequentlyBoth branches give $y(0)=1$. This is Heaviside forcing in a first-order integrating-factor equation: the solution is continuous and piecewise differentiable, while its one-sided derivatives at zero are $0$ and $1$. The Heaviside step function's value at zero does not affect the integral solution. The equation holds classically on the two open half-lines and in the weak sense across zero.

The left branch rises from zero at $-\infty$ to one. On the right,so its unique maximum is $y=\sqrt{2}$ at $x=\mathrm{arsinh}1$, after which it decreases to the horizontal asymptote $y=1$ from above.

Fix any starting village and let $A$ be the pickup distance, $B$ the subsequent passenger distance. From each vertex, the four possible pickup distances are $0,5,10,15$, each with probability $1/4$. From each pickup village, the three other possible destinations have distances $5,10,15$, each with probability $1/3$.

Sharing the pickup village does not by itself give independence. Here it does hold because the conditional distribution of $B$ is the same for every pickup village: for all allowed $a,b$,Thus $D=A+B$ has the taxi distance convolution on a rectangular road cycle. Counting the twelve equally likely pairs givesNo other distances are possible. The probability distribution is the same for every starting village, so no assumption about the previous customer's destination is needed.

The separate moments are $\mathbb{E}A=15/2$, $\mathbb{E}B=10$, $\mathrm{Var}A=125/4$ and $\mathrm{Var}B=50/3$. Independence therefore givesAs a check from the displayed probability mass function, $\mathbb{E}{D}^2=2125/6$ and subtracting $(35/2{)}^2$ gives the same variance.

A loaded gun remains loaded after its visit with probability $1/2$. An unloaded gun is loaded and then not fired with probability $(3/4)(1/2)=3/8$. The visited gun's new loaded probability is thereforePreservation requires $p^{\prime }=p$, givingThis is product stationarity under single-coordinate Markov updates: the visited coordinate has the same Bernoulli distribution afterward, and its independent local random decisions do not affect the other coordinates. Thus the originally independent indicator random variables remain jointly independent after each scheduled visit, not merely equal in their marginal probabilities.

Let $X_j$ be the indicator random variable that gun $j$ is loaded. Then $N={\sum }_{j=1}^m{X}_j$ has binomial distribution with parameters $m,3/7$. By linearity of expectation and independence of the indicator random variables,

For a uniform random point in a disk, the event $R<x$ occupies a disk of area $\pi x^2$ when $0<x<1$. ThusThe radial probability density function is $f_R(r)=2r$ on $(0,1)$. Integrating givesChoose the polar angle $\Theta \in [0,2\pi )$. The Cartesian density is $1/\pi$ in the disk, and the polar coordinates Jacobian determinant is $r$. HenceThe product support and product density prove that $R$ and $\Theta$ are independent, with uniform angle.

Reflection symmetry gives $\mathbb{E}X=\mathbb{E}Y=0$ and $\mathbb{E}(XY)=0$, so $\mathrm{Cov}(X,Y)=0$. They are nevertheless not independent. The events $X>3/4$ and $Y>3/4$ each have positive probability, but their intersection is empty inside the disk, because it would require $X^2+Y^2>18/16>1$. Independence would give a positive product probability for that intersection.

Except at the probability-zero center, $X/R=\cos \Theta$ and $Y/R=\sin \Theta$. ThereforeAn arbitrary definition at the center does not change these expected values. This example distinguishes uncorrelated Cartesian coordinates from independent polar coordinates.

The line through the origin is $l_0=\mathrm{span}(u)$, containing zero and closed under addition and scalar multiplication, so it is a vector subspace. If $l_{x_1}=l_{x_2}$, then $x_1\in l_{x_2}$ and $x_1=x_2+\lambda u$ for some real $\lambda$. Conversely that equality gives the same set after shifting the line parameter. ThusChanging representatives to $x^{\prime }=x+su$, $y^{\prime }=y+tu$ changes their sum by $(s+t)u$ and a scalar multiple by $\alpha su$. The line classes therefore do not change, proving both operations well-defined. Associativity follows from $l_{(x+y)+z}=l_{x+(y+z)}$. The zero vector of this quotient vector space is the entire line $l_0$, because $l_x+l_0=l_x$.

For a basis $(u,b_1,b_2)$ of ${\mathbb{R}}^3$, every vector is $x=su+t{b}_1+v{b}_2$, so $l_x=t{l}_{b_1}+v{l}_{b_2}$. If this combination equals $l_0$, then $t{b}_1+v{b}_2=su$; independence of the three original basis vectors forces $t=v=s=0$. Hence $(l_{b_1},l_{b_2})$ is a basis of the space of parallel lines as a quotient vector space.

For the specified direction, add $zu$ to $(x,y,z{)}^T$ to obtain the representative $(x+z,y+3z,0{)}^T$. Thus its coordinates in the two standard line classes are $(x+z,y+3z{)}^T$. The two supplied image classes have coordinates $(1,1{)}^T$ and $(-3,1{)}^T$, while their input coordinates are $(1,-1{)}^T$ and $(1,1{)}^T$. If $a_1,a_2$ are the matrix columns, then $a_1-a_2=(1,1{)}^T$ and $a_1+a_2=(-3,1{)}^T$. Solving gives

Complete the square in the real-valued complex plane equation:Dividing by $\alpha$ gives center and radiusIf the radicand is zero, the locus is the single point $c$, a degenerate circle. If it is negative, the locus is empty. The absolute value of $\alpha$ is essential even when $\alpha <0$.

For the requested sketch, the five centers and radii are respectively $(-1/2,\sqrt{3}/2)$, $(-1,\sqrt{2})$, $(-2,\sqrt{6})$, $(2,\sqrt{2})$ and $(1,0)$, all centers on the real axis.

The original PDF's map is $f(z)=(\beta /\bar{\beta })\bar{z}$; the converted TeX drops the conjugate on its denominator. Write $\beta =|\beta |e^{i\theta }$, so $f(z)=e^{2i\theta }\bar{z}$. This is the reflection symmetry of a complex circle equation. It fixes $c=\beta /\alpha$ and satisfiesThus it maps every point of the circle into the same circle. Conversely, for any point $w$ on that circle, $z=f(w)$ also lies on it and $f(z)=w$. This proves $f(C)=C$, including the surjectivity required by the question.

For any three nonempty circles, fix $u\in C_3$. The triangle inequality gives, for every $z\in C_1,w\in C_2$,Taking the maximum over $z,w$ proves the requested inequality.

For centers $c_1,c_2$ and radii $r_1,r_2$, the maximum separation of two circles is bounded above by $|c_1-c_2|+r_1+r_2$. If the centers differ, choose $z=c_1+r_{1}v$ and $w=c_2-r_{2}v$, where $v=(c_1-c_2)/|c_1-c_2|$, to attain the bound. If the centers coincide, any unit $v$ gives opposite points attaining $r_1+r_2$. ThereforeThe function $m$ is not a metric on positive-radius circles, since $m(C,C)=2r$, even though it satisfies this triangle-type inequality.

If $A{x}_p=b$, then $Ax=b$ is equivalent to $A(x-x_p)=0$. Thus the solution set is the affine solution space of a linear equation $\boxed{x_p+\ker A}$, and every vector of that form solves the equation.

For the symmetric matrix case, let $A{u}_i={\lambda }_i{u}_i$. Symmetry givesDistinct eigenvalues therefore imply $u_i\cdot u_j=0$. Normalizing the three vectors gives an orthonormal basis. Expand $x={\sum }_i{t}_i{u}_i$ and $b={\sum }_i{b}_i{u}_i$, with $b_i=b\cdot u_i$. The equation $(A-{\lambda }_{k}I)x=b$ becomes $({\lambda }_i-{\lambda }_k)t_i=b_i$ for every $i$. Thus solvability is equivalent to $b_k=0$, and all solutions areThis proves the claimed orthogonality condition and the full solution family rather than only a particular solution.

For the numerical symmetric matrix, direct multiplication gives $A{u}_1=0$ for $u_1=(1,1,1{)}^T/\sqrt{3}$, and $A{u}_2=-\sqrt{2}{u}_2$ for $u_2=(1,-1,0{)}^T/\sqrt{2}$. Their orthogonal complement is spanned by $u_3=(1,1,-2{)}^T/\sqrt{6}$, and multiplication gives $A{u}_3=\sqrt{3}{u}_3$. The supplied right side has $b\cdot u_1=0$, so the linear system is solvable. Its other coefficients are $b\cdot u_2=2$ and $b\cdot u_3=3\sqrt{2}$. HenceEquivalently, all solutions are

To find the kernel of a linear map, perform the row operations $R_2\leftarrow R_2-R_1$ and $R_3\leftarrow R_3-(1+\alpha )R_1$, followed by $R_3\leftarrow R_3-R_2$. The equivalent homogeneous equations areThe resulting determinant is ${\alpha }^2(3-\alpha )$. If $\alpha \notin \{0,3\}$ all coordinates vanish. At $\alpha =0$ only $x-y+z=0$ remains; at $\alpha =3$ the equations give $y=2z$, $x=-5z$. Therefore kernel bases areThe empty family is the basis of the zero vector subspace.

For the change of basis, let $B_0,C_0$ be the matrices whose columns are the two ordered bases in standard coordinates. Then $v=B_0[v{]}_B=C_0[v{]}_C$, so $[v{]}_B=S[v{]}_C$ with $S=B_0^{-1}{C}_0$. If $A$ and $A^{\prime }$ represent the same linear map in the two bases, $[\Phi (v){]}_B=A[v{]}_B$ and also $[\Phi (v){]}_B=S{A}^{\prime }[v{]}_C$. Therefore $S{A}^{\prime }=AS$, andFor the concrete bases, their vectors satisfy $c_1=b_1+b_2$, $c_2=b_2+b_3$, $c_3=b_1+b_2+b_3$. These coordinate columns giveAt $\alpha =0$, every row of $A_0$ is $(1,-1,1)$, so every row of $A_{0}S$ is $(0,0,1)$. Multiplying on the left by the displayed inverse givesThe original PDF asks for this full product; its final factor $S$ is missing from the converted TeX.

Suppose $f$ is continuous on $[a,b]$, differentiable on $(a,b)$ and satisfies $f(a)=f(b)$, with $a<b$. Apply the extreme value theorem to both $f$ and $-f$ to obtain a maximum and a minimum. If $f$ is constant, every interior derivative is zero. Otherwise some value differs from the common endpoint value, so either a maximum or a minimum occurs at an interior point $c$. At an interior maximum, difference quotients are nonpositive from the right and nonnegative from the left; since the derivative exists, it must be zero. The minimum case is identical with the inequalities reversed. This proves Rolle's theorem: $\boxed{f^{\prime }(c)=0}$ for some $c\in (a,b)$.

For the mean value theorem, subtract the endpoint chord:Then $g(a)=g(b)=0$, so Rolle's theorem supplies $c\in (a,b)$ with $g^{\prime }(c)=0$. HenceIf $f^{\prime }(t)>0$ everywhere, applying the mean value theorem to any $x<y$ gives $f(y)-f(x)=f^{\prime }(c)(y-x)>0$. Thus $f$ is strictly increasing. The converse is false: $\boxed{f(t)=t^3}$ is strictly increasing but $f^{\prime }(0)=0$.

The definition of a convergent sequence isTo prove the product rule, convergence first gives an eventual bound $|u_n|\le |l|+1=:M$. ThenChoose $n$ sufficiently large that $|v_n-k|<ϵ/(2M)$ and $|u_n-l|<ϵ/[2(|k|+1)]$, as well as the bound on $u_n$. The right side is less than $ϵ$, proving $u_n{v}_n\to lk$ directly from the limit of a sequence definition.

If $l\ne 0$, eventually $|u_n-l|<|l|/2$, so the reverse triangle inequality gives $|u_n|>|l|/2$. ConsequentlyThus $1/u_n\to 1/l$. The condition $u_n\ne 0$ ensures the reciprocal sequence is defined at every index; the nonzero limit supplies the eventual uniform bound away from zero. Finally $\boxed{u_n=1/(n+1)\to 0}$ has every term nonzero, showing that this termwise property alone does not prevent a zero limit.

A family $x_1,\dots ,x_n$ is linearly independent if ${\sum }_{i=1}^n{a}_i{x}_i=0$ forces $a_1=\cdots =a_n=0$. For the three specified vectors, direct substitution givesThe coefficients are not all zero, so the vectors are linearly dependent.

For the implication question, using the usual interpretation that a basis contains the listed family as independent entries, the answers in order (i)–(x) areThe key distinctions are that the unrestricted zero combination always exists, whereas linear dependence requires a nontrivial combination; and dependence does not necessarily let the specified third vector be solved in terms of the first two. For example $x=y=e_1$, $z=e_2$ makes (iii) and (iv) false, while $x=e_1,y=e_2,z=e_3$ disproves (vi).

There is a set-versus-family qualification in (v). If “contains” is interpreted as literal set containment, repeated nonzero vectors need only occur once in the basis. Then $x=y=e_1,z=e_2$ is dependent as a family but all three belong to the basis $\{e_1,e_2\}$, so (v) becomes false. With three distinct vectors, or with family-containment as above, basis extension gives (v) true. Implication (x) remains true on either reading.

Using the Levi-Civita symbol and the contraction of two Levi-Civita symbols, the $i$th component of the vector triple product isThus the triple-product identity is $a\times (b\times c)=(a\cdot c)b-(a\cdot b)c$.

Apply this with the unit vector $n$ to obtain $\Pi (x)=x-(n\cdot x)n$. Dot products and scalar multiplication are linear in $x$, so $\Pi (sx+ty)=s\Pi (x)+t\Pi (y)$. Its matrix representation isThe fixed-vector equation is equivalent to $(n\cdot x)n=0$. Since $n\ne 0$, all its solutions are $x\in n^{\perp }$. Also $\boxed{\Pi (n)=0}$. Every vector decomposes as $x=\Pi (x)+(n\cdot x)n$, with $\Pi (x)$ orthogonal to $n$. Hence $\Pi$ is the orthogonal projection onto the plane through the origin normal to $n$: it fixes that plane and removes the perpendicular component.