JEE MAIN - Mathematics (2018 (Offline) - No. 6)
Two sets A and B are as under :
A = {($$a$$, b) $$ \in $$ R $$ \times $$ R : |$$a$$ - 5| < 1 and |b - 5| < 1};
B = {($$a$$, b) $$ \in $$ R $$ \times $$ R : 4($$a$$ - 6)2 + 9(b - 5)2 $$ \le $$ 36 };
Then
A = {($$a$$, b) $$ \in $$ R $$ \times $$ R : |$$a$$ - 5| < 1 and |b - 5| < 1};
B = {($$a$$, b) $$ \in $$ R $$ \times $$ R : 4($$a$$ - 6)2 + 9(b - 5)2 $$ \le $$ 36 };
Then
neither A $$ \subset $$ B nor B $$ \subset $$ A
B $$ \subset $$ A
A $$ \subset $$ B
A $$ \cap $$ B = $$\phi $$ ( an empty set )
Explanation
Given,
$$4{\left( {a - 6} \right)^2} + 9{\left( {b - 5} \right)^2} \le 36$$
Let $$a - 6 = x$$ and $$b - 5 = y$$
$$\therefore\,\,\,\,$$ $$4{x^2} + 9{y^2} \le 36$$
$$ \Rightarrow \,\,\,\,{{{x^2}} \over 9} + {{{y^2}} \over 4} \le 1$$
This is a equation of ellipse.
This ellipse will look like this,
_en_6_1.png)
According to set A,
$$\left| {a - 5} \right| < 1$$
as $$a - 6 = x$$ then $$a - 5 = x + 1$$
$$\therefore\,\,\,$$ $$\left| {x + 1} \right| < 1$$
$$ \Rightarrow \,\,\, - 1 < x + 1 < 1$$
$$ \Rightarrow \,\,\, - 2 < x < 0$$
$$\left| {b - 5} \right| < 1$$
as $$b - 5 = y$$
$$\therefore\,\,\,$$ $$\left| y \right| < 1$$
$$ \Rightarrow \,\,\,\, - 1 < y < 1$$
This will look like this,
_en_6_2.png)
By combining both those graphs it will look like this,
_en_6_3.png)
To check entire set A is inside of B or not put any paint of set A on the equation $${{{x^2}} \over 9} + {{{y^2}} \over 4} \le 1$$ and if this inequality satisfies, then it is inside B.
By looking at the graph you can surely say (0, 1) or (0, $$-$$1) inside the graph. But to check point ($$-2$$, 1) or ($$-$$2, $$-$$1) is inside of the ellipse or not, put on the $${{{x^2}} \over 9} + {{{y^2}} \over 4} \le 1.$$
Putting ($$-$$2, 1) on the inequality,
LHS $$ = {{\left( { - 2} \right){}^2} \over 9} + {{{1^2}} \over 4}$$
$$ = {4 \over 9} + {1 \over 4}$$
$$ = {{25} \over {36}} < 1$$
$$\therefore\,\,\,$$ Inequality holds.
So, $$\left( { - 2,1} \right)$$ is inside the ellipse.
Similarly by checking we can see $$\left( { - 2, - 1} \right)$$ is also inside the ellipse.
Hence, we can say entire set A is inside of the set B.
$$\therefore\,\,\,\,$$ A $$ \subset $$ B
$$4{\left( {a - 6} \right)^2} + 9{\left( {b - 5} \right)^2} \le 36$$
Let $$a - 6 = x$$ and $$b - 5 = y$$
$$\therefore\,\,\,\,$$ $$4{x^2} + 9{y^2} \le 36$$
$$ \Rightarrow \,\,\,\,{{{x^2}} \over 9} + {{{y^2}} \over 4} \le 1$$
This is a equation of ellipse.
This ellipse will look like this,
According to set A,
$$\left| {a - 5} \right| < 1$$
as $$a - 6 = x$$ then $$a - 5 = x + 1$$
$$\therefore\,\,\,$$ $$\left| {x + 1} \right| < 1$$
$$ \Rightarrow \,\,\, - 1 < x + 1 < 1$$
$$ \Rightarrow \,\,\, - 2 < x < 0$$
$$\left| {b - 5} \right| < 1$$
as $$b - 5 = y$$
$$\therefore\,\,\,$$ $$\left| y \right| < 1$$
$$ \Rightarrow \,\,\,\, - 1 < y < 1$$
This will look like this,
By combining both those graphs it will look like this,
To check entire set A is inside of B or not put any paint of set A on the equation $${{{x^2}} \over 9} + {{{y^2}} \over 4} \le 1$$ and if this inequality satisfies, then it is inside B.
By looking at the graph you can surely say (0, 1) or (0, $$-$$1) inside the graph. But to check point ($$-2$$, 1) or ($$-$$2, $$-$$1) is inside of the ellipse or not, put on the $${{{x^2}} \over 9} + {{{y^2}} \over 4} \le 1.$$
Putting ($$-$$2, 1) on the inequality,
LHS $$ = {{\left( { - 2} \right){}^2} \over 9} + {{{1^2}} \over 4}$$
$$ = {4 \over 9} + {1 \over 4}$$
$$ = {{25} \over {36}} < 1$$
$$\therefore\,\,\,$$ Inequality holds.
So, $$\left( { - 2,1} \right)$$ is inside the ellipse.
Similarly by checking we can see $$\left( { - 2, - 1} \right)$$ is also inside the ellipse.
Hence, we can say entire set A is inside of the set B.
$$\therefore\,\,\,\,$$ A $$ \subset $$ B
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