Since the probability density function is an even function, $xf(x)$ is an odd function. Its integral over the symmetric support vanishes, so

Put $N=6-2e^{-2}$. The function graph is symmetric about the vertical axis, zero outside $[-2,2]$, and has its unique maximum at $x=0$ with height $2/N$. On $0<x<2$, it decreases as $(1+e^{-x})/N$; the negative side is its reflection. The one-sided interior endpoint values are $(1+e^{-2})/N$, so the probability density function jumps to zero at both ends of its support. It also has a corner at its central maximum because of the absolute value in the exponent.

Values assigned exactly at $x=\pm 2$ do not affect any probability.

A probability density function must have total integral one. The numerator is an even function, soThis is positive, ensuring nonnegativity of the probability density function.

For $a\ne 0$, factor out $|a|$, not $a$, and apply the binomial series with exponent $1/2$:For each fixed nonzero $a$, these are the first three nonzero terms of the Taylor series about zero; the full binomial series converges when $|x|<|a|$.

For $a=0$, the function is $|x|$. Its right derivative at zero is $1$ and its left derivative is $-1$, so no Taylor series about zero exists when $a=0$. The one-sided formulas are exactly $x$ for $x\ge 0$ and $-x$ for $x\le 0$; they do not form a single two-sided Taylor expansion, nor can they supply three nonzero terms. The parameter case explicitly included in the question must therefore be treated as this exception.

Where $\sin x\ne 0$, divide by it to obtain $y^{\prime }+2\cot xy=\csc x$. The integrating factor is ${\sin }^{2}x$, and the product rule givesThe initial condition gives $C=1$. HenceThe simplified expression has a differentiable extension through $x=0$, with $y(0)=1/2$. Direct substitution verifies the original undivided differential equation there as well: its left side is $2y(0)=1$. Thus the maximal interval containing $\pi /2$ on which this solution is finite is $\boxed{-\pi <x<\pi }$; the apparent exclusion of $0$ arose only from division by $\sin x$. This is continuation through a zero of the leading ODE coefficient.

Because the original differential equation divides by $xy$, its domain requires $x\ne 0$ and $y\ne 0$. Set $u=y^2$, so $u^{\prime }=2y{y}^{\prime }$. The equation becomes the first-order linear differential equationAn integrating factor is $x^2$, since $(x^{2}u{)}^{\prime }=x^2{u}^{\prime }+2xu$. ThusReturning to $y$ gives the explicit general real solutionsEach sign is fixed on a connected interval where $x\ne 0$ and the radicand is strictly positive. The strict condition excludes $y=0$, where the original differential equation is undefined. There is no additional zero solution.

The PDF diagram includes a solid base of thickness $d$, occupying $-d\le z\le 0$ and $0\le r\le R$. Above it, $0\le z\le h$, the metal lies between the two specified curved surfaces. In cylindrical coordinates, the Jacobian determinant contributes a factor $r$, so the mass isThe base contributes $\pi \rho R^{2}d$. The wall contributesConsequentlyThe base thickness $d$ comes from the diagram; omitting it would omit part of the cup's volume.