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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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1 Answer

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Here first both $|x|,|y|\leq 1$ as otherwise both A and B will have values greater than 1 hence would not satisfy the inequality.

Now, Let $(x,y)\in A$ then $x^2+y^4 \leq 1$

$\Rightarrow$ $x^4+y^6 = x^2 *x^2 +y^2*y^4 \leq 1^2 *x^2 +1^2*y^4 = x^2 + y^4 \leq 1$

$\Rightarrow$ $x^4+y^6 \leq 1$

Thus $(x,y)\in B$

Hence every point in A is also in B or $A \subseteq B$.

We can also prove the above by,

$x^4 \leq x^2$ since $|x|\leq 1$ , similarly $y^6\leq y^4$ as $|y|\leq 1$ ,

$\Rightarrow$ $x^4+y^6 \leq 1$.

Now since $x^4+y^6<x^2+y^4 \leq 1$ there exists a point $(x_1,y_1)$ such that $|x_1|,|y_1| \leq 1$

for which the value of  $x_1^4+y_1^6<1$

but $x_1^2+y_1^4>1$ ,

hence $B\not\subset A$ $\Rightarrow$ $B \neq A$.

This point $(x_1,y_1)$ must be along the boundaries as can be seen from the below image that as the value of the equations in A and B both increases value with increase in the value of $(x,y)$ due to even powers. 

 

 

Thus the answer is Option (B)

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