Given,$A={{(x,y) \in \mathbb{R}^2:x^4+y^2\le1}},B={(x,y) \in \mathbb{R}^2:x^6+y^4\le1}$

Question

Then

A. $B \subseteq A$

B. $A \subseteq B$

C.Each of the sets $A-B,B-A \space and \space A\cap B \space $ is not empty

D.None of the above

I am absolutely clueless on this as I tried to draw two graphs but cannot understand the concept. Will anybody please care to explain?

Graph of $x^6+y^4 \le 1$

Graph of $x^4+y^2\le 1$

From this I can conclude that $A \subseteq B$

Answer 1

Hint: $\;x^4 \le 1 - y^2 \le 1\,$, so $\,x^4 \le 1 \iff |x| \le 1\,$, and the same goes for $\,|y| \le 1\,$. Then use that for all $|a| \le 1 \implies a^6 \le a^4\le a^2\,$.