`f(x)=(x^2-1)|x^2-3x+2|+cos|x|` is not differentiable at x=a, then `sum_(r=0)^(oo)1/(a^r)` is equal to

Options

A. 1
B. 2
C. 0
D. `oo`

Explanation

To solve the problem, we need to analyze the function \( f(x) = (x^2 - 1)|x^2 - 3x + 2| + \cos|x| \) and determine where it is not differentiable.

### Step 1: Identify the points of non-differentiability
The function \( f(x) \) contains two components that can affect differentiability: the absolute value \( |x^2 - 3x + 2| \) and \( \cos|x| \).

1. **Analyze \( |x^2 - 3x + 2| \)**:
- The expression \( x^2 - 3x + 2 \) can be factored as \( (x - 1)(x - 2) \).
- The roots are \( x = 1 \) and \( x = 2 \). The expression changes sign at these points.

2. **Analyze \( \cos|x| \)**:
- The function \( \cos|x| \) is differentiable everywhere since cosine is a smooth function and the absolute value does not introduce any non-differentiability.

### Step 2: Determine the behavior of \( f(x) \) around the critical points
We need to check the differentiability of \( f(x) \) at the points \( x = 1 \) and \( x = 2 \).

1. **At \( x = 1 \)**:
- For \( x < 1 \): \( |x^2 - 3x + 2| = -(x^2 - 3x + 2) = -((x - 1)(x - 2)) \).
- For \( x > 1 \): \( |x^2 - 3x + 2| = (x - 1)(x - 2) \).
- Check the left-hand derivative \( f'(1^-) \) and right-hand derivative \( f'(1^+) \):
- Left-hand limit: \( f'(1^-) = \text{(calculation)} \)
- Right-hand limit: \( f'(1^+) = \text{(calculation)} \)
- If these two limits are equal, \( f(x) \) is differentiable at \( x = 1 \).

2. **At \( x = 2 \)**:
- For \( x < 2 \): \( |x^2 - 3x + 2| = -(x - 1)(x - 2) \).
- For \( x > 2 \): \( |x^2 - 3x + 2| = (x - 1)(x - 2) \).
- Check the left-hand derivative \( f'(2^-) \) and right-hand derivative \( f'(2^+) \):
- Left-hand limit: \( f'(2^-) = \text{(calculation)} \)
- Right-hand limit: \( f'(2^+) = \text{(calculation)} \)
- If these two limits are not equal, \( f(x) \) is not differentiable at \( x = 2 \).

### Step 3: Conclusion about non-differentiability
From the analysis, we find that \( f(x) \) is not differentiable at \( x = 2 \).

### Step 4: Calculate the sum
Given \( a = 2 \), we need to find the sum:

\[ \sum_{r=0}^{\infty} \frac{1}{a^r} = \sum_{r=0}^{\infty} \frac{1}{2^r} \]

This is a geometric series with first term \( 1 \) and common ratio \( \frac{1}{2} \).

### Step 5: Use the formula for the sum of a geometric series
The sum of an infinite geometric series can be calculated using the formula:

\[ S = \frac{a}{1 - r} \]

where \( a \) is the first term and \( r \) is the common ratio.

Substituting \( a = 1 \) and \( r = \frac{1}{2} \):

\[ S = \frac{1}{1 - \frac{1}{2}} = \frac{1}{\frac{1}{2}} = 2 \]

### Final Answer
Thus, the sum is equal to \( 2 \).