If `f(x)={{:((e^((1+(1)/(x)))-a)/(e^((1)/(x))+1),":",xne0),(b,":",x=0):}` (where a and b are arbitrary constants) is continuous at x = 0, then the value of `a^(2)` is equal to
(use e = 2.7)

Answer

To solve the problem, we need to ensure that the function \( f(x) \) is continuous at \( x = 0 \). This means that the left-hand limit (LHL) and right-hand limit (RHL) as \( x \) approaches 0 must equal the value of the function at \( x = 0 \), which is \( b \).

### Step 1: Calculate the Left-Hand Limit (LHL)
The left-hand limit as \( x \) approaches 0 from the left (denoted as \( x \to 0^- \)) is given by:
\[ \text{LHL} = \lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} \frac{e^{(1 + \frac{1}{x})} - a}{e^{\frac{1}{x}} + 1} \]
Substituting \( x = -h \) (where \( h \to 0^+ \)):
\[ \text{LHL} = \lim_{h \to 0^+} \frac{e^{(1 - \frac{1}{h})} - a}{e^{-\frac{1}{h}} + 1} \]
As \( h \to 0^+ \), \( e^{-\frac{1}{h}} \to 0 \) and \( e^{(1 - \frac{1}{h})} \to \infty \):
\[ \text{LHL} = \lim_{h \to 0^+} \frac{\infty - a}{0 + 1} = \infty \]

### Step 2: Calculate the Right-Hand Limit (RHL)
The right-hand limit as \( x \) approaches 0 from the right (denoted as \( x \to 0^+ \)) is given by:
\[ \text{RHL} = \lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} \frac{e^{(1 + \frac{1}{x})} - a}{e^{\frac{1}{x}} + 1} \]
Substituting \( x = h \) (where \( h \to 0^+ \)):
\[ \text{RHL} = \lim_{h \to 0^+} \frac{e^{(1 + \frac{1}{h})} - a}{e^{\frac{1}{h}} + 1} \]
As \( h \to 0^+ \), \( e^{\frac{1}{h}} \to \infty \):
\[ \text{RHL} = \lim_{h \to 0^+} \frac{\infty - a}{\infty + 1} = 1 \]

### Step 3: Set LHL and RHL Equal to \( b \)
For continuity at \( x = 0 \):
\[ \text{LHL} = \text{RHL} = b \]
From our calculations:
\[ \infty = 1 = b \]
This indicates that \( a \) must be chosen such that the limits balance out.

### Step 4: Solve for \( a \)
From the continuity condition:
\[ -a = b \]
Since we have \( b = 1 \):
\[ -a = 1 \implies a = -1 \]

### Step 5: Calculate \( a^2 \)
Now, we find \( a^2 \):
\[ a^2 = (-1)^2 = 1 \]

### Conclusion
Thus, the value of \( a^2 \) is:
\[ \boxed{1} \]