Let A = {(x, y) : y = e^x, x in R} and B = {( | vedclass

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(B) To find the intersection $A \cap B$,we set the $y$-values equal: $e^x = e^{-x}$.Multiplying both sides by $e^x$,we get $e^{2x} = 1$.Taking the nat

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$A \cap B 
eq \phi$

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(B) To find the intersection $A \cap B$,we set the $y$-values equal: $e^x = e^{-x}$.Multiplying both sides by $e^x$,we get $e^{2x} = 1$.Taking the natural logarithm on both sides,$2x = 0$,which implies $x = 0$.Substituting $x = 0$ into either equation,we get $y = e^0 = 1$.Thus,the point $(0, 1)$ belongs to both sets $A$ and $B$.Since there exists at least one common point,$A \cap B 
eq \phi$.

Let $A = \{(x, y) : y = e^x, x \in R\}$ and $B = \{(x, y) : y = e^{-x}, x \in R\}$. Then:

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