Use $g=x^2+y^2+z^2$. The Lagrange multiplier equations areTaking the dot product with $(x,y,z{)}^T$ gives $2T=2\lambda$, so $\lambda =T$.

If $\lambda \ne 0$, the first and third equations give $x=z$. The middle equation then gives $x=\lambda y$, so $2{\lambda }^2=1$. The norm constraint gives $x^2=z^2=1/4$ and $y^2=1/2$. Thus the hottest points and their temperature areThe coldest points areFor $\lambda =0$, the equations give $y=0$, $z=-x$, and hence the other two constrained stationary points areThey are saddle points of the restriction to the sphere: with $u=(x+z)/\sqrt{2}$ and $v=y$ as local tangent coordinates there, $T=\sqrt{2}uv$ changes sign in every neighbourhood. The surface is compact and smooth, so the listed extreme temperatures are global. Here “zero gradient on the surface” means zero surface gradient; the ambient gradient at the hottest and coldest points is normal to the sphere, rather than zero.

At a regular constrained extremum, every tangent vector to the level set $g=A$ gives zero directional derivative of $f$. Thus $\nabla f$ is perpendicular to the tangent plane, and must be a scalar multiple of its nonzero normal vector $\nabla g$. The Lagrange multiplier equations areOne solves these four equations for $x,y,z,\lambda$, evaluates $f$ at the candidates, and classifies or compares the values as required. Points where $\nabla g=0$, and any boundary or limiting cases of the feasible domain, require separate checks because the regularity argument does not cover them.

Evaluate at $x=\pi /2$, where the original expression gives $f(\pi /2)=1-{\pi }^2/4$. The Fourier series givesRearranging provesThe convergence at this point is justified by the continuity properties noted in part (b).

Separate the already finite sine terms from the even part $g(x)=x^2-\pi |x|$. Since $g$ is an even function, it has no sine coefficients. Its constant coefficient isFor $n\ge 1$, integration by parts givesTherefore the cosine Fourier coefficients areOnly even frequencies survive: $a_{2m}=1/m^2$. The sine Fourier coefficients are $b_1=1$, $b_2=-2$, with all others zero. HenceThe periodic function is continuous at zero and at its endpoint joins, and its coefficients are absolutely summable. Thus this Fourier series represents it everywhere, including those joins.

The displayed matrix is not orthogonal: its first column has squared Euclidean norm three, whereas every column of an orthogonal matrix must have squared length one.

For the general assertion, put $C=AB$. In the Einstein summation convention, orthogonality gives $A_{ki}{A}_{kj}={\delta }_{ij}$ and $B_{ki}{B}_{kj}={\delta }_{ij}$. The entries of $C$ satisfyThis is exactly $C^{T}C=I$, proving that the matrix product $AB$ is orthogonal. The Kronecker delta contracts the two middle indices without changing the multiplication order of the matrices.

Use cylindrical radius $\rho =\sqrt{x^2+y^2}$. The upper cone is $z=1-\rho$ and the lower elliptic paraboloid is $z={\rho }^2-1$, both with $0\le \rho \le 1$. They meet along the unit circle in $z=0$. The enclosed solid isThus the coordinate ranges on the complete surface areThese separate ranges do not mean every point of their Cartesian product lies on the surface. A vertical cross-section has a straight upper cone profile and a curved lower paraboloid profile.

The characteristic equation of the associated homogeneous linear differential equation is $m^2+7m+12=(m+3)(m+4)=0$. A constant particular integral is $1/2$, soThe first boundary condition gives $A+B=-1/2$. The second gives $A{e}^{-1}+B{e}^{-4/3}=-e^{-1}/2$. Eliminating $A$ yields $B(e^{-4/3}-e^{-1})=0$, hence $B=0$ and $A=-1/2$. Therefore

On a nonconstant branch with $\sin y\ne 0$, regard $x$ as a function of $y$. The equation becomes the linear ordinary differential equationAn integrating factor is $\sin y$, givingTaking an antiderivative yields the implicit familyOn a regular branch this can also be written $x=\sin y+C/\sin y$. The latter form loses the constant solutions $y=n\pi$, which must be restored separately.

The implicit relation is a first integral of an ordinary differential equation: along any differentiable solution of the original equation,Conversely, a differentiable graph satisfying this relation solves the original equation wherever $\cos y\ne 0$. The original equation is undefined when $\cos y=0$; points at which $2\sin y-x=0$ also need separate attention before solving explicitly for $y^{\prime }$. This is why dividing immediately by $\tan y$ can lose valid solutions and obscure nonuniqueness.

For $n$ independent seeds the probability of at least one success is $1-(1-x{)}^n$. Requiring it to be at least $0.95$ gives $(1-x{)}^n\le 0.05$. For $0<x<1$, taking logarithms and dividing by the negative number $\log (1-x)$ reverses the inequality:The smallest permitted integer is thereforeThe ceiling function is necessary because seeds are counted discretely. If $x=0$, no finite number suffices; if $x=1$, one seed suffices. The target is interpreted as at least a ninety-five percent chance, since exact equality may not be attainable with an integer count.