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ISI2015-MMA-31

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Consider the sets defined by the real solutions of the inequalities

$$A = \{(x,y):x^2+y^4 \leq 1 \} \:\:\:\:\:\:\:\: B = \{ (x,y):x^4+y^6 \leq 1\}$$

Then

  1. $B \subseteq A$
  2. $A \subseteq B$
  3. Each of the sets $A – B, \: B – A$ and $A \cap B$ is non-empty
  4. none of the above
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Answer: B

 

Remark: for $w,z\in(0,1)$, $w^m< w^n; w^m>z^m$ for $m>n; w>z;  m,n\in\mathbb{N}$.

 

We note that A, B are symmetric sets. Hence, we can consider the values of $x$ and $y$ only in the first quadrant.

Further, we consider the extreme values of the set A and set B i.e. 

$$A_1=\{(x,y):x^2+y^4=1\}$$ and $$B_1=\{(x,y):x^4+y^6=1\}$$

Moreover,  for fixed $x\in (0,1)$ \begin{align*}x^2>& x^4\\ \Rightarrow-x^2<&-x^4\\\Rightarrow 1-x^2<&1-x^4\\\Rightarrow y_{A_1}^4<&y_{B_1}^6\\\Rightarrow y_{A_1}^4<&y_{B_1}^4 \enspace \text{[see Remark]}\\\Rightarrow y_{A_1}<&y_{B_1}\end{align*}

 

Thus, $A\subseteq B$.

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