Two sets A and B are as under:A={(a,b)∈R×R:|a−5|<1,|b−5|<1} ; B={(a,b)∈R×R:4(a−6)2+9(b−5)2≤36} , then number of points of the form (a, b) where a & b are integers ∈ B – A is

A={(a,b)∈R×R:|a−5|<1,|b−5|<1} Let a−5=x,b−5=ySet A contains all points inside |x|<1,|y|<1B={(a,b)∈R×R:4(a−6)2+9(b−5)2≤36} set B contains all points inside or on (x−1)29+y24=1∴(±1,±1) lies inside the ellipseHence A⊂B.

Two sets A and B are as under:A={(a,b)∈R×R:|a−5|<1,|b−5|<1} ; B={(a,b)∈R×R:4(a−6)2+9(b−5)2≤36} , then number of points of the form (a, b) where a & b are integers ∈ B – A is

A={(a,b)∈R×R:|a−5|<1,|b−5|<1} Let a−5=x,b−5=ySet A contains all points inside |x|<1,|y|<1B={(a,b)∈R×R:4(a−6)2+9(b−5)2≤36} set B contains all points inside or on (x−1)29+y24=1∴(±1,±1) lies inside the ellipseHence A⊂B.

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Two sets A and B are as under:
$A = {(a, b) \in R \times R : | a - 5 | < 1, | b - 5 | < 1}$ ; $B = {(a, b) \in R \times R : 4 {(a - 6)}^{2} + 9 {(b - 5)}^{2} \leq 36}$ , then number of points of the form (a, b) where a & b are integers $\in$ B – A is

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Detailed Solution

$A = {(a, b) \in R \times R : | a - 5 | < 1, | b - 5 | < 1}$
Let $a - 5 = x, b - 5 = y$
Set A contains all points inside $| x | < 1, | y | < 1$
$B = {(a, b) \in R \times R : 4 {(a - 6)}^{2} + 9 {(b - 5)}^{2} \leq 36}$
set B contains all points inside or on $\frac{{(x - 1)}^{2}}{9} + \frac{y^{2}}{4} = 1$
$∴ (\pm 1, \pm 1)$ lies inside the ellipse
Hence $A \subset B$ .