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1 vote
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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
in Set Theory & Algebra
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280 views
0
How??
0

@ankitgupta.1729 

Can you please try this question.

0

 

$A\subseteq B$ Thus answer is (B) . This can be proved by finding the maximum Radius of curvature of both the equations in , since in all the 4 quadrants the equations behave similarly because of having even powers.

Edit : Added Strike.

0
But how would you theoretically justified.

I mean without taking the help of such plotter.
0

I have mentioned that already- "This can be proved by finding the maximum Radius of curvature of both the equations in  , since in all the 4 quadrants the equations behave similarly because of having even powers.".

 

Edit : Added Strike

0
Ok
0
Kindly check the answer, I have striked off the above comments.

1 Answer

2 votes

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)


edited by
1
Great!

Thanks for the answer.

I wish I could make it as the best answer.
0
What (x,y) contain in A??

(0,1),(1,0),(0,0),(-1,0),(0,-1)

right??

Now, what will be B) contain that is not in A??
1

All those points that you mentioned are present in both A as well as B.

A point that is in B but not in A is (0.8,0.8) which can be found using the  Plotter.

0
yes, and how do ugot it?

Without diagram, can I get it?
0
This is an inequality so many points will satisfy it ...hence it is difficult to find out exactly such points without trial and error or using plotter...or may be by using some method to solve system of non linear equations(I am not exactly sure if we can apply those methods for inequalities as well).
0

 

Can you prove (Mathematically) this statement you wrote?

0

See it is like, if I say 3<5 then there must always be a value 4 which lies between 3 and 5.Likewise there exists a point $(x_1,y_1)$ such that the value of $x_1^4+y_1^6 < 1$ and  $x_1^2+y_1^4> 1$ just because $x^4+y^6 < x^2+y^4 $ for $|x|,|y|\leq 1$ 

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