Question

Let $A = \{(x,y) : y = e^{2x}, x \in R\}$ and
$B = \{(x, y ) : y = e^{-2x} \,\forall \,x \in R\}$ then $A \cap B$ is

• A. Not a set
• B. Singleton set
• C. Empty set
• D. None of these

Answer 1

Answer: (B)

Given $A = \left\{\left(x, y\right) : y = e^{2x}, \forall\,x \in R\right\}$
and $B = \left\{\left(x, y \right) : y = e^{-2x}, \forall\, x \in R\right\}$
$\Rightarrow e^{2x} = \frac{1}{e^{2x}}$
$\Rightarrow e^{4x} = 1$
$\Rightarrow e^{4x} = e^{0}$
$\Rightarrow x = 0$ and $y = 1$
As $e^{2x}$ and $e^{-2x}$ both intersect each other at $(0, 1)$, so $A \cap B$ is a singleton set.