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The first curve is a circle of radius $1$ centred at $(2,1)$. A line through the origin with equation $kx-y=0$ intersects it precisely when the perpendicular distance from that centre is at most the radius. The distance from a point to a line givesHence the full parameter range isThe two endpoint lines are tangent, at $(2,0)$ and $(6/5,8/5)$ respectively; interior values give two real intersections. Both endpoints are therefore included.

The original PDF contains the product $\sin x(1+4\cos x)$; the transcription drops the $x$ after the first sine. Using $2\sin x\cos x=\sin 2x$ gives the finite Fourier sine seriesThus $b_1=1$, $b_2=2$, and all other sine coefficients vanish. No cosine or constant terms are present.

The divergence is $y^2+x^2+z^2=r^2$. In spherical coordinates, the angular integral is $4\pi$ and the volume element contributes $r^{2}dr$. Thus

On $x\ge 1$, an integrating factor is $1/x$. Multiplication gives $(y/x{)}^{\prime }=1$, so the general solution is $y=x^2+Cx$. The initial value $y(1)=1$ forces $C=0$. Therefore

Taking each partial derivative of the scalar potential givesThe factor $5y^4$ in the middle component comes from the chain rule. The potential is smooth, so the curl of a gradient identity gives $\boxed{\nabla \times \mathbf{F}=0}$.

The characteristic polynomial is $(1-\lambda {)}^2-4=(\lambda -3)(\lambda +1)$. For $\lambda =3$ the components are equal; for $\lambda =-1$ they are opposite. After normalization, the eigenvalues and associated eigenvectors areAn overall sign change of either normalized eigenvector is equivalent.

The Taylor series of the logarithm givesThe first nonzero term is therefore $\boxed{x^2}$. It is positive on either side of the origin, so the stationary point is a strict local minimum. In fact $\log (1+x^2)\ge 0$, with equality only at zero, making it the global minimum as well.