Definitions — Question 8

Question 8 – Solution

Strategy. Compute derivatives at the joining point using limits, then test each equation in its original form.

Step 1: First derivative. The pieces meet at $z_a(0)=0$. The difference quotient at $0$ is $h$ for $h<0$ and $ah$ for $h>0$, so $z_a'(0)=0$ for every real $a$. Therefore $$z_a'(x)=\begin{cases}2x,&x<0,\\0,&x=0,\\2ax,&x>0,\end{cases}$$ which is continuous at $0$. Hence

Step 2: Second derivative at the join. The left-hand limit of $[z_a'(h)-z_a'(0)]/h$ is $2$, and the right-hand limit is $2a$. Thus $$\boxed{z_a''(0)\text{ exists exactly when }a=1.}$$

Step 3: Check equation (A). For $x<0$, $xz_a'=x(2x)=2z_a$. For $x>0$, $xz_a'=x(2ax)=2z_a$. At $x=0$, both sides are $0$. Consequently

Step 4: Check equation (B). On $x>0$, $z_a''=2a$, so (B) requires $a=1$. For that value $z_a=x^2$ globally and $z_a''=2$, including at $0$. Thus

Step 5: Domain of the equation. The quotient $2y/x$ is undefined at $x=0$, even when $y(0)=0$. The normalized equation therefore cannot have a solution interval containing $0$. It agrees with (A) on intervals contained in $(-\infty,0)$ or $(0,\infty)$, but division removes the original equation’s valid point $x=0$.