Two sets A and B are as under: A = \ (a, b) ∈ R × R: |a - 5| B = \...

Two sets

A

and

B

are as under:

A = \left\{\left(a, b\right) \in R \times R: \left|a - 5\right| < 1\text{and}\left|b - 5\right| < 1\right\};

B = \left\{\left(a, b\right) \in R \times R: 4{\left(a - 6\right)}^{2} + 9{\left(b - 5\right)}^{2} \le 36\right\} .

Then : $Two sets A and B are as under: A = \ (a, b) ∈ R × R: |a - 5| B = \ (a, b) ∈ R × R: 4 (a - 6) ^ 2 + 9 (b - 5) ^ 2 ≤ 36\$ $Two sets A and B are as under: A = \ (a, b) ∈ R × R: |a - 5| B = \ (a, b) ∈ R × R: 4 (a - 6) ^ 2 + 9 (b - 5) ^ 2 ≤ 36\$

Options

Aneither $A ⊂ B$ nor $B ⊂ A$
B $B ⊂ A$
C $A ⊂ B$
D $A \cap B = ϕ$ (an empty set)

Correct answer

C. $A ⊂ B$

Step-by-step solution

Given sets, $A = \left\{\left(a, b\right) \in R \times R: \left|a - 5\right| < 1\text{and}\left|b - 5\right| < 1\right\}$

$A = \left\{\left(a, b\right) \in R \times R: - 1 < a - 5 < 1\text{and} - 1 < b - 5 < 1\right\}$

$A = \left\{\left(a, b\right) \in R \times R: 4 < a < 6\text{and}4 < b < 6\right\}$

$B = \dfrac{{\left(a - 6\right)}^{2}}{9} + \dfrac{{\left(b - 5\right)}^{2}}{4} \le 1$

For set $A,$

$Two sets A and B are as under: A = \ (a, b) ∈ R × R: |a - 5| B = \ (a, b) ∈ R × R: 4 (a - 6) ^ 2 + 9 (b - 5) ^ 2 ≤ 36\$

For corner coordinates of the set $A$ is \left(4, 4\right), \left(4, 6\right), \left(6, 4\right), \left(6, 3\right) . $For set$ B, $<img alt="" src="https://cdn-question-pool.Quantrex.app/pyq/jee_main/4509ca30-3013-418e-bee8-90e2b47732bf-e0dbeab1-9c71-4364-97d7-4bdd2a7cf983-image.png" style="width: 343px; height: 265px;" />Now, Put$ \left(4, 4\right) $in the equation of the set$ B, $$ \dfrac{4}{9} + \dfrac{1}{4} \le 1,$ $\left(4, 4\right)$ lies in the set $B .$

Similarly, $\left(4, 6\right),$ \left(6, 4\right), \left(6, 3\right) $also lies in the set$ B . $So,$ A ⊂ B$.

Practice Ellipse on Quantrex Academy →

More from Ellipse