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

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(B) To find $A \cap B$,we need to find the points $(x, y)$ that satisfy both equations: $y = e^x$ and $y = e^{-x}$. Equating the two expressions for $

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$A \cap B 
eq \phi $

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(B) To find $A \cap B$,we need to find the points $(x, y)$ that satisfy both equations: $y = e^x$ and $y = e^{-x}$. Equating the two expressions for $y$,we get: $e^x = e^{-x}$ Taking the natural logarithm on both sides: $x = -x$ $2x = 0 \implies x = 0$. Substituting $x = 0$ into either equation: $y = e^0 = 1$. Thus,the intersection point is $(0, 1)$. Since $(0, 1) \in A$ and $(0, 1) \in B$,the intersection $A \cap B = \{ (0, 1) \} $. Therefore,$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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