`f(x) ={{:( xe ^(-((1)/(x)+(1)/(|x|))), x ne 0),( a , x=0):}` the value of a , such that f(x) is differentiable at x=0, is equal to

Options

A. 1
B. `-1`
C. 0
D. none of these

Answer

To determine the value of \( a \) such that the function

\[ f(x) = \begin{cases} x e^{-\left(\frac{1}{x} + \frac{1}{|x|}\right)}, & x \neq 0 \\ a, & x = 0 \end{cases} \]

is differentiable at \( x = 0 \), we need to ensure that the function is continuous at that point and that the derivative exists at that point.

### Step 1: Check Continuity at \( x = 0 \)

For \( f(x) \) to be continuous at \( x = 0 \), we must have:

\[ \lim_{x \to 0} f(x) = f(0) = a \]

We will calculate the left-hand limit and the right-hand limit as \( x \) approaches 0.

#### Left-hand limit as \( x \to 0^- \):

When \( x < 0 \), we have:

\[ f(x) = x e^{-\left(\frac{1}{x} + \frac{1}{|x|}\right)} = x e^{-\left(\frac{1}{x} + \frac{1}{-x}\right)} = x e^{-\left(\frac{1}{x} - \frac{1}{x}\right)} = x e^{0} = x \]

Thus,

\[ \lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} x = 0 \]

#### Right-hand limit as \( x \to 0^+ \):

When \( x > 0 \), we have:

\[ f(x) = x e^{-\left(\frac{1}{x} + \frac{1}{|x|}\right)} = x e^{-\left(\frac{1}{x} + \frac{1}{x}\right)} = x e^{-\frac{2}{x}} \]

As \( x \to 0^+ \), \( e^{-\frac{2}{x}} \to 0 \) much faster than \( x \) approaches 0, hence:

\[ \lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} x e^{-\frac{2}{x}} = 0 \]

### Step 2: Set the Limits Equal to \( a \)

Since both the left-hand limit and the right-hand limit as \( x \to 0 \) equal 0, we set:

\[ a = 0 \]

### Step 3: Check Differentiability at \( x = 0 \)

To check differentiability, we need to find the derivative at \( x = 0 \):

\[ f'(0) = \lim_{h \to 0} \frac{f(h) - f(0)}{h} = \lim_{h \to 0} \frac{f(h) - 0}{h} = \lim_{h \to 0} \frac{f(h)}{h} \]

For \( h > 0 \):

\[ f(h) = h e^{-\frac{2}{h}} \implies \frac{f(h)}{h} = e^{-\frac{2}{h}} \]

As \( h \to 0^+ \), \( e^{-\frac{2}{h}} \to 0 \).

For \( h < 0 \):

\[ f(h) = h \implies \frac{f(h)}{h} = 1 \]

As \( h \to 0^- \), this limit approaches 1.

Since the left-hand limit (1) and the right-hand limit (0) do not match, \( f'(0) \) does not exist, indicating that \( f(x) \) is not differentiable at \( x = 0 \).

### Conclusion

Thus, the value of \( a \) such that \( f(x) \) is continuous at \( x = 0 \) is:

\[ \boxed{0} \]