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

(C) $\mathrm{A} \subset \mathrm{B}$

(A) neither $\mathrm{A} \subset \mathrm{B}$ nor $\mathrm{B} \subset \mathrm{A}$

(B) $\mathrm{B} \subset \mathrm{A}$

(D) $\mathrm{A} \cap \mathrm{B}=\phi$ (an empty set)

JEE Main · 2018 · 08 Apr 2018 · Offline · Sets Relations and Functions

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 :

Answer: (C)

Set

A

is the open square

4 < a < 6, 4 < b < 6

, and set

B

is the filled ellipse

\frac{( a - 6 ) ^{2}}{9} + \frac{( b - 5 ) ^{2}}{4} \leq 1

centred at

(6, 5)

with semi-axes

3

and

2

. Testing the extreme corner

(4, 4)

of the square in the ellipse,

\frac{( 4 - 6 ) ^{2}}{9} + \frac{( 4 - 5 ) ^{2}}{4} = \frac{4}{9} + \frac{1}{4} = \frac{25}{36} < 1

, and every point of the square satisfies the inequality at least as comfortably, so

A \subset B

JEE Main · 2018 · 08 Apr 2018 · Offline · Single correct option