If the sets A and B are defined as A = {(x, y | vedclass

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(C) Given the sets $A = \{(x, y) : y = e^x, x \in R\}$ and $B = \{(x, y) : y = x, x \in R\}$.To find the intersection $A \cap B$,we need to check if t

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$A \cap B = \phi$

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(C) Given the sets $A = \{(x, y) : y = e^x, x \in R\}$ and $B = \{(x, y) : y = x, x \in R\}$.To find the intersection $A \cap B$,we need to check if there exists any point $(x, y)$ that satisfies both $y = e^x$ and $y = x$.This implies $e^x = x$,or $e^x - x = 0$.Let $f(x) = e^x - x$. The derivative $f'(x) = e^x - 1$.Setting $f'(x) = 0$ gives $x = 0$.The minimum value of $f(x)$ is $f(0) = e^0 - 0 = 1$.Since the minimum value of $e^x - x$ is $1$,which is greater than $0$,the equation $e^x = x$ has no real solutions.Therefore,there is no common element between sets $A$ and $B$.Thus,$A \cap B = \phi$.

If the sets $A$ and $B$ are defined as $A = \{(x, y) : y = e^x, x \in R\}$ and $B = \{(x, y) : y = x, x \in R\}$,then:

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