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At $A=(1/2,1/4)$, the gradient is $(-3/64,-9/32{)}^T$. The position vector has length $\sqrt{5}/4$, so its normalized direction is $(2,1{)}^T/\sqrt{5}$. Hence the directional derivative is

The gradient is

Three vectors form a basis of three-dimensional space exactly when their scalar triple product is nonzero:It is the determinant of the matrix with these columns, so this condition is equivalent to linear independence and spanning the whole space.

Take the dot product of the expansion of $\mathbf{y}$ with $\mathbf{v}\times \mathbf{w}$. The terms containing $\mathbf{v}$ and $\mathbf{w}$ vanish, leaving $\mathbf{y}\cdot (\mathbf{v}\times \mathbf{w})=\alpha \Delta$. Cycling the same argument yieldsThe three cross-product numerators use the reciprocal basis.

First $(1,-1,0{)}^T\times (0,0,1{)}^T=(-1,-1,0{)}^T$. Taking the remaining cross product givesThe order of the two cross products matters because the operation is not associative.

Expanding the cross product componentwise givesIts dot product with each original vector is zero, as required.

Put $r=\sin (\pi /4)=1/\sqrt{2}$. The geometric series has first term $r$ and common ratio $-r$, with absolute value less than one. Therefore its sum is

Pair adjacent terms in the finite sum:There are $25$ pairs, each equal to $2$. The result is $\boxed{50}$.

First suppose $\alpha >0$. Differentiation under the integral sign, justified on compact positive parameter intervals by an integrable bound, givesThe first integral is $f(\alpha )/\alpha$. For $\alpha \ne 1$, use the partial fraction decompositionThe standard inverse tangent antiderivatives then giveThe same value holds at $\alpha =1$ by continuity, or by direct integration. Hence the proposed first-order linear differential equation is valid on the positive parameter range:Divide by $\alpha$ to recognise a quotient rule derivative:To fix $C$, the value $f(0)=0$ alone is insufficient, since every displayed solution tends to zero there. Instead considerThis limit follows from the dominated convergence theorem: for $0<\alpha \le 1$ and $0<x<1$, the logarithm's absolute value is bounded by $2|\ln x|+\ln 2$; for $x\ge 1$, it is at most $\ln (1+x^2)$. After division by $1+x^2$, these are integrable. The limiting improper integral is absolutely convergent, and the substitution rule $x=1/t$ changes it into its own negative. Its value is therefore zero, giving $C=0$. ThusThe question calls $\alpha$ a real parameter, but this last formula and the proposed differential equation do not hold on the entire real axis. The defining integral is odd in $\alpha$, because its logarithm depends on ${\alpha }^2$. Applying the positive-parameter result to $|\alpha |$ therefore gives the correct extensionwith $f(0)=0$. In particular, for $\alpha <0$, the right-hand side of the differential equation must be $-\pi \alpha /(1-\alpha )$; for every $\alpha \ne 0$ the unified expression is $\pi |\alpha |/(1+|\alpha |)$. The logarithmic quadratic parameter integral therefore provides the requested identity on its valid range and the corrected answer elsewhere.

Under the usual regularity assumptions, the Leibniz integral rule isThere is a genuine domain problem in the following integral as printed in the PDF. PutThe point $t=2$ is a simple pole, withFor every $x>4$, the lower endpoint $\sin (1/x)$ is less than $2$ and the upper endpoint $\sqrt{x}$ is greater than $2$. The interval therefore crosses the pole. Since $q(t)=33/[7(t-2)]+O(1)$ near $2$, the two one-sided improper integrals diverge logarithmically, with opposite signs. Their divergences do not define an ordinary integral. Consequently the ordinary limit in the question is undefined as written: its underlying integral has no finite value for any $x\ge 4$.

A separately specified Cauchy principal value does exist for $x>4$. Under that interpretation, differentiating a principal-value integral across a fixed simple pole givesThe first term isand the second tends to zero because $q$ is bounded near $t=0$. Thus the principal-value version has limit $\boxed{1}$. The two conclusions are distinct: this regularised answer requires the additional principal-value convention, which the printed question does not state.

Let $S_N={\sum }_{n=1}^N(-1{)}^{n+1}/n$. Integrate the finite geometric series identity ${\sum }_{k=0}^{N-1}(-t{)}^k=[1-(-t{)}^N]/(1+t)$ over $0\le t\le 1$:The remainder has absolute value at most ${\int }_0^1{t}^{N}dt=1/(N+1)\to 0$. Hence the partial sums converge, withThis proves both convergence and the value of the alternating harmonic series without substituting an endpoint into an unproved infinite power series. Its convergence is conditional because the corresponding harmonic series of absolute values diverges.

Expand the right-hand side in the orthonormal basis from part (b). Its coefficients areThus $\mathbf{b}=2\sqrt{2}{\mathbf{e}}_4+\sqrt{3}{\mathbf{e}}_6$. If $\mathbf{x}=c_0{\mathbf{e}}_0+c_4{\mathbf{e}}_4+c_6{\mathbf{e}}_6$, then matrix multiplication gives $A\mathbf{x}=4c_4{\mathbf{e}}_4+6c_6{\mathbf{e}}_6$. Matching coefficients fixes $c_4=\sqrt{2}/2$ and $c_6=\sqrt{3}/6$, while $c_0$ is arbitrary. Consequently all solutions areGeometrically, the solution set is an affine line in a vector space parallel to the kernel of $A$. The zero eigenvalue makes the component in that direction invisible to $A$; the other two components are uniquely determined. The displayed particular solution is perpendicular to the kernel, so it is also the point of this line closest to the origin.

The vector $(1,-1,0{)}^T$ is an eigenvector with eigenvalue $4$. The complementary subspace of vectors with first two components equal is preserved by $A$: a vector $(s,s,t{)}^T$ maps to $(4s+2t,4s+2t,4s+2t{)}^T$. There, the eigenvalue equations reduce to the $2\times 2$ matrix $\left(\begin{array}{cc}4& 2\\ 4& 2\end{array}\right)$, whose characteristic polynomial is $\lambda (\lambda -6)$. For $\lambda =6$, $t=s$; for $\lambda =0$, $t=-2s$.

Normalising by the Euclidean norm, a complete orthonormal basis of eigenvectors isSign changes of any of these eigenvectors are equally valid. The eigenvalues also check part (a): their sum is $10$ and the sum of their squares is $52$.

Adding the diagonal entries gives $\mathrm{tr}A=4+4+2=10$. Expansion along the first row gives $\det A=4(8-4)+2(0-8)=0$. For the remaining trace, direct matrix multiplication givesThus $\mathrm{tr}(A^2)=20+20+12=52$. By multiplicativity of the determinant, $\det (A^2)=(\det A{)}^2=0$. Altogether,