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Question: Two sets \(A,B\) are as under: \(A = \\{ (a,b) \in R \times R:\left| {a - 5} \right| < 1{\text{ an...

Two sets A,BA,B are as under:
A=(a,b)∈R×R:∣a−5∣<1 and ∣b−5∣<1:A = \\{ (a,b) \in R \times R:\left| {a - 5} \right| < 1{\text{ and }}\left| {b - 5} \right| < 1\\} :
B=(a,b)∈R×R:4(a−6)2+9(b−5)2⩽36.B = \\{ (a,b) \in R \times R:4{(a - 6)^2} + 9{(b - 5)^2} \leqslant 36\\} .
Then
A. A∩B=ϕA \cap B = \phi (empty set)
B. neither A⊂BA \subset BnorB⊂AB \subset A
C. B⊂AB \subset A
D. A⊂BA \subset B

Explanation

Solution

At first, take the setAA. Solve the given condition and find the range ofa and ba{\text{ and b}}. Then check whether those values satisfy the conditions given in the setBB. Also, check for the same values not included in the set AA that may satisfy the set B.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 AA which is:
A=(a,b)∈R×R:∣a−5∣<1 and ∣b−5∣<1: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,
⇒∣a−5∣<1 and ∣b−5∣<1\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,ba,b
⇒∣a−5∣<1 , ∣b−5∣<1\Rightarrow \left| {a - 5} \right| < 1{\text{ , }}\left| {b - 5} \right| < 1
Now we will solve this to find the range of aa
⇒∣a−5∣<1\Rightarrow \left| {a - 5} \right| < 1
We can resolve this modulus function to get the following expression:
⇒a−5<1,5−a<1\Rightarrow a - 5 < 1,5 - a < 1
This inequality can be further solved as:
⇒a<6\Rightarrow a < 6and 4>a4 > a
Hence a∈(4,6)a \in (4,6)
Similarly, we can solve for the range of ‘b’
⇒∣b−5∣<1\Rightarrow \left| {b - 5} \right| < 1
The modulus function can be resolved as:
⇒b−5<1\Rightarrow b - 5 < 1 and 5−b<15 - b < 1
This can be further solved
⇒b<6\Rightarrow b < 6 and b<4b < 4
Hence b∈(4,6)b \in (4,6)
Therefore the range ofa,ba,b, we get from the set AA is a∈(4,6)a \in (4,6) and b∈(4,6)b \in (4,6)
Now we will consider setB=(a,b)∈R×R:4(a−6)2+9(b−5)2⩽36.B = \\{ (a,b) \in R \times R:4{(a - 6)^2} + 9{(b - 5)^2} \leqslant 36\\} .
The condition given in the setBB is:
4(a−6)2+9(b−5)2⩽364{(a - 6)^2} + 9{(b - 5)^2} \leqslant 36 −−−−−(1) - - - - - (1)
Now we will use the value from the range received from a setAA such as(5,5)(5,5).
But what is the reason for that?
It is the reason to check whether the setA,BA,B does have anything in common.
If (5,5)(5,5) satisfies, it means eitherA⊂B,B⊂AA \subset B,B \subset A.
Substituting a=5,b=5a = 5,b = 5 in (1)
⇒4(a−6)2+9(b−5)2⩽36⇒4(5−6)2+9(5−5)2⩽36\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:
⇒4⩽36\Rightarrow 4 \leqslant 36
It satisfies that (5,5)(5,5) belongs to setBB.
As A,BA,B have a point in common therefore A∩B≠ϕA \cap B \ne \phi which means that option A is eliminated.
Hence we will check the point outside the range of AA like (6,6)(6,6)
Substituting it in (1)
⇒4(6−6)2+9(6−5)2⩽36⇒9⩽36\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⊂BA \subset B

Hence option D is correct.

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