JEE Main 2018 — Sets And Relations Question with Solution

Given,

$$4{\left(a-6\right)}^2+9{\left(b-5\right)}^2\le 36$$

Let $$a-6=x$$ and $$b-5=y$$

$$\therefore$$ $$4x^2+9y^2\le 36$$

$$\Rightarrow \frac{x^2}{9}+\frac{y^2}{4}\le 1$$

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 $$\frac{x^2}{9}+\frac{y^2}{4}\le 1$$ 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 $$\frac{x^2}{9}+\frac{y^2}{4}\le 1.$$

Putting ($$-$$2, 1) on the inequality,

LHS $$=\frac{\left(-2\right){}^2}{9}+\frac{1^2}{4}$$

$$=\frac{4}{9}+\frac{1}{4}$$

$$=\frac{25}{36}<1$$

$$\therefore$$ 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