Let `f _(n) x+ f _(n) (y ) = (x ^(n)+y ^(n))/(x ^(n) y ^(n))AA x, y in R-{0}.` where `n in N` and `g (x) = max { f_(2) (x), f _(3) (x),(1)/(2)} AA x in R - {0}`
The number of values of x for which g(x) is non-differentiable `(x in R - {0}):`

Options

A. 2
B. 4
C. 5
D. 1

Answer

To solve the problem, we need to analyze the function \( f_n(x) \) and then determine the function \( g(x) \) based on the maximum of certain functions. Finally, we will find the points where \( g(x) \) is non-differentiable.

### Step 1: Determine \( f_n(x) \)

Given the equation:

\[ f_n(x) + f_n(y) = \frac{x^n + y^n}{x^n y^n} \]

for \( x, y \in \mathbb{R} \setminus \{0\} \).

By setting \( y = 1 \), we can simplify the equation:

\[ f_n(x) + f_n(1) = \frac{x^n + 1}{x^n} \]

Let \( f_n(1) = c \). Then we have:

\[ f_n(x) + c = \frac{x^n + 1}{x^n} \]

This leads to:

\[ f_n(x) = \frac{x^n + 1}{x^n} - c = \frac{x^n + 1 - cx^n}{x^n} = \frac{(1 - c)x^n + 1}{x^n} \]

To find \( c \), we can set \( x = 1 \):

\[ f_n(1) = \frac{1^n + 1}{1^n} - c = 2 - c \]

Setting \( c = 2 - c \) gives \( c = 1 \). Therefore:

\[ f_n(x) = \frac{1}{x^n} \]

### Step 2: Define \( g(x) \)

Now we define \( g(x) \):

\[ g(x) = \max\{f_2(x), f_3(x), \frac{1}{2}\} \]

Substituting the expressions for \( f_2(x) \) and \( f_3(x) \):

\[ g(x) = \max\left\{\frac{1}{x^2}, \frac{1}{x^3}, \frac{1}{2}\right\} \]

### Step 3: Analyze the behavior of \( g(x) \)

We need to analyze the intervals where each function is dominant:

1. **For \( x < -1 \)**:
- \( \frac{1}{x^2} < \frac{1}{2} \) and \( \frac{1}{x^3} < \frac{1}{2} \)
- Thus, \( g(x) = \frac{1}{2} \)

2. **For \( -1 < x < 0 \)**:
- \( \frac{1}{x^2} > \frac{1}{2} \) and \( \frac{1}{x^3} < \frac{1}{2} \)
- Thus, \( g(x) = \frac{1}{x^2} \)

3. **For \( 0 < x < 1 \)**:
- \( \frac{1}{x^2} > \frac{1}{2} \) and \( \frac{1}{x^3} > \frac{1}{2} \)
- Thus, \( g(x) = \frac{1}{x^2} \)

4. **For \( x > 1 \)**:
- \( \frac{1}{x^2} < \frac{1}{2} \) and \( \frac{1}{x^3} < \frac{1}{2} \)
- Thus, \( g(x) = \frac{1}{2} \)

### Step 4: Identify points of non-differentiability

The function \( g(x) \) can change its form at the points where the maximum switches between the functions. The critical points are:

- At \( x = -1 \): Transition from \( \frac{1}{2} \) to \( \frac{1}{x^2} \)
- At \( x = 1 \): Transition from \( \frac{1}{x^2} \) to \( \frac{1}{2} \)

### Conclusion

Thus, \( g(x) \) is non-differentiable at two points, \( x = -1 \) and \( x = 1 \).

The number of values of \( x \) for which \( g(x) \) is non-differentiable is:

\[ \boxed{2} \]