If `f (x)= max. (x ^(4), x ^(2) ,(1)/(81))AAx in [0, oo),` then the sum of square of reciprocal of all the values of x where f (x) is non-differentiable, is equal to:

Options

A. 1
B. 81
C. 82
D. `82/81`

Answer

To solve the problem, we need to analyze the function \( f(x) = \max(x^4, x^2, \frac{1}{81}) \) for \( x \in [0, \infty) \) and find the points where it is non-differentiable.

### Step 1: Identify the functions involved
The function consists of three parts:
1. \( x^4 \)
2. \( x^2 \)
3. \( \frac{1}{81} \)

### Step 2: Find the points of intersection
We need to find where these functions intersect, as these points will help us identify where the maximum function changes.

**Intersection of \( x^2 \) and \( x^4 \):**
Set \( x^2 = x^4 \):
\[ x^4 - x^2 = 0 \implies x^2(x^2 - 1) = 0 \]
This gives us \( x = 0 \) and \( x = 1 \).

**Intersection of \( x^2 \) and \( \frac{1}{81} \):**
Set \( x^2 = \frac{1}{81} \):
\[ x = \sqrt{\frac{1}{81}} = \frac{1}{9} \]

### Step 3: Determine the intervals for \( f(x) \)
Now we can analyze the function in the intervals determined by the intersection points:
1. \( [0, \frac{1}{9}) \)
2. \( [\frac{1}{9}, 1) \)
3. \( [1, \infty) \)

- **For \( x \in [0, \frac{1}{9}) \)**:
- \( x^2 \) is greater than \( \frac{1}{81} \) and \( x^4 \) is less than \( \frac{1}{81} \).
- Thus, \( f(x) = x^2 \).

- **For \( x \in [\frac{1}{9}, 1) \)**:
- \( x^2 \) is greater than \( \frac{1}{81} \) and less than \( x^4 \).
- Thus, \( f(x) = x^2 \).

- **For \( x \in [1, \infty) \)**:
- \( x^4 \) is greater than both \( x^2 \) and \( \frac{1}{81} \).
- Thus, \( f(x) = x^4 \).

### Step 4: Identify non-differentiable points
The function \( f(x) \) is non-differentiable at the points where the maximum function changes:
1. At \( x = \frac{1}{9} \) (change from \( \frac{1}{81} \) to \( x^2 \))
2. At \( x = 1 \) (change from \( x^2 \) to \( x^4 \))

### Step 5: Calculate the sum of squares of the reciprocals
Now we need to find the sum of the squares of the reciprocals of these points:
1. For \( x = 1 \):
\[ \text{Reciprocal} = 1 \quad \Rightarrow \quad \text{Square} = 1^2 = 1 \]
2. For \( x = \frac{1}{9} \):
\[ \text{Reciprocal} = 9 \quad \Rightarrow \quad \text{Square} = 9^2 = 81 \]

### Step 6: Final calculation
Now sum the squares:
\[ 1 + 81 = 82 \]

### Conclusion
The sum of the square of the reciprocal of all the values of \( x \) where \( f(x) \) is non-differentiable is \( \boxed{82} \).