MCQ
Let
$A=\{(\alpha, \beta) \in \mathbf{R} \times \mathbf{R}:|\alpha-1| \leq 4$ and $|\beta-5| \leq 6\}$
and
$B=\left\{(\alpha, \beta) \in \mathbf{R} \times \mathbf{R}: 16(\alpha-2)^{2}+9(\beta-6)^{2} \leq 144\right\}$.
Then
  • A
    $\mathrm{B} \subset \mathrm{A}$
  • B
    $\mathrm{A} \cup \mathrm{B}=\{(\mathrm{x}, \mathrm{y}):-4 \leq \mathrm{x} \leq 4,-1 \leq \mathrm{y} \leq 11\}$
  • C
    neither $\mathrm{A} \subset \mathrm{B}$ nor $\mathrm{B} \subset \mathrm{A}$
  • D
    $\mathrm{A} \subset \mathrm{B}$

Answer

A. $\mathrm{B} \subset \mathrm{A}$
A : $|x-1| \leq 4$ and $|y-5| \leq 6$
$\Rightarrow-4 \leq \mathrm{x}-1 \leq 4 \Rightarrow-6 \leq y-5 \leq 6$
$\Rightarrow-3 \leq \mathrm{x} \leq 5 \quad \Rightarrow-1 \leq \mathrm{y} \leq 11$
B : $16(x-2)^{2}+9(y-6)^{2} \leq 144$
B : $\frac{(x-2)^{2}}{9}+\frac{(y-6)^{2}}{16} \leq 1$
Image
From Diagram $\mathrm{B} \subset \mathrm{A}$

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