Parameterize the straight line by $\mathbf{r}(t)=(t,t,t{)}^T$, $0\le t\le 1$. Then ${\mathbf{r}}^{\prime }=(1,1,1{)}^T$ and the line integral becomes

The vector $\mathbf{c}=\mathbf{p}\times \hat{\mathbf{a}}$ is a normal vector to the span of $\mathbf{p}$ and $\hat{\mathbf{a}}$. Since $\mathbf{c}\ne 0$, these vectors determine a plane. Thus the missing words are

the line and the origin.

Indeed this plane consists of all linear combinations of $\mathbf{p}$ and the line's direction vector.

Decompose $\mathbf{p}$ into its orthogonal projections parallel and perpendicular to $\hat{\mathbf{a}}$. Since $\mathbf{p}\times \hat{\mathbf{a}}$ removes the parallel component,The last vector is the displacement to the point of the straight line nearest the origin. HenceThe original PDF has $c=|\mathbf{c}|$, not the squared magnitude inserted by the OCR.

The unit vector $\hat{\mathbf{a}}$ specifies the direction of the straight line. Reversing its sign, together with the sign of $\mathbf{c}$, leaves the same straight line.

Take the cross product with $\hat{\mathbf{q}}$:Thus one choice isConversely, this cross product equation implies $(\mathbf{r}-\mathbf{p})\times \hat{\mathbf{q}}=0$, so $\mathbf{r}-\mathbf{p}$ is parallel to $\hat{\mathbf{q}}$. It therefore describes precisely the original straight line.

For $x>-1$, interpret the endpoint $-1$ by a one-sided limit. Part (iii) shows that $g$ extends continuously there with value zero, so both nested integrals converge. Let $F(x)$ denote the nested expression inside the derivative. The fundamental theorem of calculus, applied twice with moving lower endpoints, givesBy part (iii), $g^{\prime }(x)=1$, henceA direct check is ${\int }_s^{-1}(t+1)dt=-(s+1{)}^2/2$, so $F(x)=(x+1{)}^3/6$, with the same third derivative.

At $x=0$, the integrand vanishes for $0<t<1$, so $I(0)=0$ andSolving the first-order linear differential equation from part (i) on $(-1,\infty )$ gives $g=C(x+1)$. The initial condition fixes $C=1$, soEquivalently, the defining exponent is the logarithmic integral representation of the natural logarithm, $I(x)=\ln (x+1)$.

The required range isNear $t=1$, the ratio $(t^x-1)/\ln t$ tends to $x$, so it causes no convergence problem. Near $t=0$, one useful identity isFor $x>-1$, this bounds its absolute value by $|x|t^{\min (x,0)}$, which is integrable on $(0,1)$. It also supplies the domination used in part (i).

For $x=-1$, the integrand is negative and behaves as $-1/[t|\ln t|]$ near zero; its improper integral diverges. For $x<-1$ its negative singularity is even stronger, and comparison with that same divergent integral again rules out convergence. Thus the original finite real-valued definition of $g$ has precisely the stated domain.

Write $I(x)=\ln g(x)$. For $x>-1$, differentiation under the integral sign givesIndeed, on any compact parameter interval $c\le x\le d$ with $c>-1$, the differentiated integrand is bounded by the integrable function $t^c$. The dominated convergence theorem therefore justifies the operation despite the improper lower endpoint. The apparent singularity at $t=1$ is removable.

Since $g=e^I$, the chain rule gives the first-order linear differential equation

Holding $x$ fixed when taking the antiderivative with respect to $t$ givesUse the product rule and the chain rule to differentiate this explicit expression:This agrees term by term with part (i).

For the Leibniz integral rule, the upper endpoint has derivative $2\pi x$ and the lower endpoint has derivative $1$. The integrand has partial derivative $4xt+\sin t$. ThereforeThe last integral is $[2x{t}^2-\cos t{]}_x^{\pi x^2}=2{\pi }^2{x}^5-2x^3-\cos (\pi x^2)+\cos x$. Combining terms gives

Compare the positive terms with those of the harmonic series:By the limit comparison test, the given series has the same convergence behaviour as $\sum 1/n$, which diverges. Thus the series diverges to $+\infty$.

For the positive terms $a_n=n^4/3^n$, the consecutive ratio isThe ratio test therefore proves that the series converges absolutely. More explicitly, for all sufficiently large $n$ the ratio is at most a fixed $q<1$, so the tail is bounded by a convergent geometric series.

The difference quotient at zero isIf $p>1$, its absolute value is at most $|h{|}^{p-1}\to 0$, so $f$ is differentiable at zero with derivative $0$.

To exclude $p\le 1$, use $h_n=1/(2\pi n)$ and $k_n=1/(2\pi n+\pi /2)$. The difference quotient equals $(2\pi n{)}^{1-p}$ along the first sequence, and $0$ along the second. For $p=1$ the first value is $1$, while for $p<1$ it is unbounded. Hence a common finite limit is impossible. Therefore

The bound $|\cos (1/x)|\le 1$ gives $|f(x)-f(0)|\le |x{|}^p$ when $x\ne 0$. If $p>0$, given any $\epsilon >0$, choose $\delta ={\epsilon }^{1/p}$. Then $0<|x|<\delta$ impliesthe inequality also holds at $x=0$. This proves continuity using the epsilon-delta definition of continuity.

For necessity, take the positive sequence $x_n=1/(2\pi n)\to 0$. Then $\cos (1/x_n)=1$, so $f(x_n)=(2\pi n{)}^{-p}$. For $p=0$ this equals $1$; for $p<0$ it tends to $+\infty$. In neither case does it tend to $f(0)=0$, so continuity fails. ConsequentlyThis is an instance of the general continuity and differentiability thresholds for a damped oscillation.

The epsilon-delta definition of continuity says that $f$ is continuous at $a$ ifHere $a$ belongs to the domain of $f$, and the implication is required for every $x$ in that domain. The number $\delta$ may depend on $\epsilon$ and $a$, but must work for all such $x$ simultaneously.

For each possibility in part (i), squaring the diagonal entries gives $M^2=I$. Equivalently, the linear transformation fixes a vector space basis after being applied twice. ThereforeGeometrically, reflecting twice in the same plane returns every vector to itself.

The three coordinate basis vectors are eigenvectors, so $M$ is diagonal in those coordinates. Exactly one coordinate is multiplied by $-1$, while the other two are fixed. The three possibilities areThey are respectively reflections in the coordinate planes $x=0$, $y=0$ and $z=0$. Each is a Householder reflection with unit normal equal to the coordinate eigenvector carrying the eigenvalue $-1$.

Subtracting the second row from the third gives a third row $(0,2,0)$, so expansion along it shows $\det B=-2\ne 0$. Thus the matrix inverse exists and every right-hand side has a unique solution.

To calculate it, temporarily solve $B\mathbf{x}=\mathbf{b}$ with general $\mathbf{b}=(b_1,b_2,b_3{)}^T$. Subtracting the second equation from the third and from the first givesBack-substitution into the second equation givesThereforeFor the given right-hand side, matrix multiplication yields the unique answer

The first coordinate direction is already an eigenvector of $A$ with eigenvalue $4$. On the last two coordinates, the matrix is $\left(\begin{array}{cc}2& 4\\ 4& 2\end{array}\right)$, with characteristic polynomialThus the three eigenvalues are $4,6,-2$. For $\lambda =6$, the last two components are equal; for $\lambda =-2$, they are opposite. Normalising by the Euclidean norm gives the complete listEach sign-reversed vector is an equally valid choice. The three displayed eigenvectors are mutually orthogonal, as expected for a real symmetric matrix with distinct eigenvalues.