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.