Question

Two sets $A,B$ are as under:
$A = \{ (a,b) \in R \times R:\left| {a - 5} \right| < 1{\text{ and }}\left| {b - 5} \right| < 1\} :$
$B = \{ (a,b) \in R \times R:4{(a - 6)^2} + 9{(b - 5)^2} \leqslant 36\} .$
Then
A. $A \cap B = \phi $(empty set)
B. neither $A \subset B$nor$B \subset A$
C. $B \subset A$
D. $A \subset B$

Answer 1

Hint: At first, take the set$A$. Solve the given condition and find the range of$a{\text{ and b}}$. Then check whether those values satisfy the conditions given in the set$B$. Also, check for the same values not included in the set $A$ that may satisfy the set $B.$

Complete step-by-step solution:
Here in this question, we are given definitions of two sets A and B. With this information of two sets, we need to find the correct option among the given four choices.
So for finding the correct option, we must first find the elements in both sets A and B.
Let us consider the set $A$ which is:
$A = \{ (a,b) \in R \times R:\left| {a - 5} \right| < 1{\text{ and }}\left| {b - 5} \right| < 1\} :$
Now let us check for the given conditions,
$ \Rightarrow \left| {a - 5} \right| < 1{\text{ and }}\left| {b - 5} \right| < 1$
The symbol $'\left| {} \right|'$ represents the absolute value function. We will solve these to find the range of $a,b$
$ \Rightarrow \left| {a - 5} \right| < 1{\text{ , }}\left| {b - 5} \right| < 1$
Now we will solve this to find the range of $a$
$ \Rightarrow \left| {a - 5} \right| < 1$
We can resolve this modulus function to get the following expression:
\[ \Rightarrow a - 5 < 1,5 - a < 1\]
This inequality can be further solved as:
$ \Rightarrow a < 6$and $4 > a$
Hence $a \in (4,6)$
Similarly, we can solve for the range of ‘b’
$ \Rightarrow \left| {b - 5} \right| < 1$
The modulus function can be resolved as:
$ \Rightarrow b - 5 < 1$ and $5 - b < 1$
This can be further solved
$ \Rightarrow b < 6$ and $b < 4$
Hence $b \in (4,6)$
Therefore the range of$a,b$, we get from the set $A$ is $a \in (4,6)$ and $b \in (4,6)$
Now we will consider set$B = \{ (a,b) \in R \times R:4{(a - 6)^2} + 9{(b - 5)^2} \leqslant 36\} .$
The condition given in the set$B$ is:
$4{(a - 6)^2} + 9{(b - 5)^2} \leqslant 36$$ - - - - - (1)$
Now we will use the value from the range received from a set$A$ such as$(5,5)$.
But what is the reason for that?
It is the reason to check whether the set$A,B$ does have anything in common.
If $(5,5)$ satisfies, it means either$A \subset B,B \subset A$.
Substituting $a = 5,b = 5$ in (1)
$ \Rightarrow 4{(a - 6)^2} + 9{(b - 5)^2} \leqslant 36 \Rightarrow 4{(5 - 6)^2} + 9{(5 - 5)^2} \leqslant 36$
On solving the parenthesis, we get:
$ \Rightarrow 4 \leqslant 36$
It satisfies that $(5,5)$ belongs to set$B$.
As $A,B$ have a point in common therefore $A \cap B \ne \phi $ which means that option A is eliminated.
Hence we will check the point outside the range of $A$ like $(6,6)$
Substituting it in (1)
$ \Rightarrow 4{(6 - 6)^2} + 9{(6 - 5)^2} \leqslant 36 \Rightarrow 9 \leqslant 36$
which satisfies the equation
Therefore, we can conclude that $A \subset B$

Hence option D is correct.

Note: In mathematics, the absolute value or modulus of a real number ‘x’, denoted $\left| x \right|$ , is the non-negative value of x without regard to its sign. Namely,$\left| x \right| = x$ if x is positive, and $\left| x \right| = - x$ if x is negative (in which case $ - x$ is positive), and $\left| 0 \right| = 0$.