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

Question

Two sets A and B are as under: A = (a, b) ∈ R × R: |a - 5| < 1 and |b - 5| < 1 ; B = (a, b) ∈ R × R: 4(a - 6)2 + 9(b - 5)2 ≤ 36 . Then (A) B ⊂ A  (B) A ⊂ B  (C) A ∩ B = φ (an empty set) (D) neither A ⊂ B nor B ⊂ A

Tardigrade Admin · Accepted Answer

As, |a-5|<1 and |b-5|<1 ⇒ 4< a , b <6 and ((a-6)2/9)+((b-5)2/4) ≤ 1 Taking axes as a-axis and b-axis <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/mathematics/7feb06db276b9548945273bc1fd0b3c0-.png" /> The set A represents square PQRS inside set B representing ellipse and hence A ⊂ B.

Q. Two sets $A$ and $B$ are as under:
$A = {(a, b) \in R \times R : ∣ a - 5∣ < 1$ and $∣ b - 5∣ < 1}$ ;
$B = {(a, b) \in R \times R : 4 (a - 6) 2 + 9 (b - 5)^{2} \leq 36}$ . Then

$B \subset A$

$A \subset B$

$A \cap B = ϕ$ (an empty set)

neither $A \subset B$ nor $B \subset A$

As, $∣ a - 5∣ < 1$ and $∣ b - 5∣ < 1$
$\Rightarrow 4 < a, b < 6$ and $\frac{( a - 6 ) ^{2}}{9} + \frac{( b - 5 ) ^{2}}{4} \leq 1$
Taking axes as $a$ -axis and $b$ -axis

The set $A$ represents square $PQRS$ inside set $B$ representing ellipse and hence $A \subset B$ .

Q. Two sets A and B are as under: A={(a,b)∈R×R:∣a−5∣<1 and ∣b−5∣<1}; B={(a,b)∈R×R:4(a−6)2+9(b−5)2≤36}. Then

B⊂A

A⊂B

A∩B=ϕ (an empty set)

neither A⊂B nor B⊂A