QJEEM18Ma774a: JEE Main 2018 Mathematics - Class 11 JEEM18Ma774a Solution: Two sets A and B are as under : A = {( , b) R R :

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,



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