For a positive real number a, let S(a) denote the set of points (x, y) satistying |x|^a + |y|^a = 1 A positive number a is said to be good if the points in S(a) that are closest to the origin lie only on the coordinate axes. Then
(A) all $a$ in $(0, 1)$ are good and others are not good.
(B) all $a$ in $(1,2)$ are good and others are not good.
(C) all $a> 2$ are good and others are not good.
(D) all $a >1$ are good and others are not good.
Is there anyway to determine the nature of the curve for values like a=1/2 or 1/3 or for a>2 and specifically to determine whether a wud be good or bad? I can't see anyway to determine so even for only the 1st quadrant
By symmetry, one need only consider the $x,y \geqslant 0$. The points in the region considered that also lie on the coordinate axes are $(1,0)$ and $(0,1)$, and their distance to the origin is $1$. Notice that in general, the distance to the distance of $(x,y)$ to the origin is $\sqrt{x^2 + y^2}$.
We may hence reframe the definition of good as: $a>0$ is good when all $x,y>0$ with $x^a + y^a = 1$ are such that $\sqrt{x^2 + y^2} > 1$. Because $\sqrt{x^2 + y^2} > 1\iff x^2 + y^2 > 1$, we may then rephrase the problem as
It follows immediately that $2$ is not good. This will be useful later.
Now, write $y^2 = {\left(y^a\right)}^{2/a} = {\left(1-x^a\right)}^{2/a}$, so we can look for the minimum of
$$f(x) = x^2 + {\left(1-x^a\right)}^{2/a},$$
where $0\leqslant x\leqslant 1$. The minimum is hence either when $x=0$, when $x=1$ or when $f'(x) = 0$. On the one hand, we have $f(0) = f(1) = 1$. On the other,
$$\begin{align} f'(x) &= 2x + \frac2a {\left(1-x^a\right)}^{2/a - 1} (-ax^{a-1}) \\&= 2\left(x - {\left(1-x^a\right)}^{2/a - 1} x^{a-1}\right) \end{align}$$
and hence
$$\begin{align} f'(x) = 0 &\iff x = {\left(1-x^a\right)}^{2/a - 1} x^{a-1} \\&\iff x^{2-a} = {\left(1-x^a\right)}^{(2-a)/a} \\&\iff x^{a(2-a)} = {\left(1-x^a\right)}^{(2-a)}. \end{align}$$
If $a \neq 2$, we can proceed as follows:
$$\begin{align} f'(x) = 0 &\iff x^a = 1-x^a \\&\iff 2x^a = 1 \iff x = {(1/2)}^{1/a} \end{align}$$
Finally, with $x_a = {(1/2)}^{1/a}$, we have $f\left(x_a\right) = 2^{1-2/a}$. It follows that
$$f(x_a) > 1 \iff 2^{1-2/a} > 1 \iff 1-\frac2a > 0 \iff a > 2.$$
Hence, when $a>2$, $f(x_a) > 1$ and $x_a$ is a maximum of $f$, so that the minimum is $f(0) = f(1) = 1$. In other words, when $a>2$, $a$ is good.
Conversely, when $a<2$ we have $f(x_a) < 1 = f(0) = f(1)$, so $x_a$ is a minimum of $f$. In this case, it is clear that $a$ is not good.
It follows that $a$ is good if and only if $a>2$.