Let `f(x)=(sinx)/x` and g(x)=|x.f(x)|+||x-2|-1|. Then, in the interval `(0,3pi)` g(x) is :

Options

A. not differentiable at 2 points
B. not differentiable at 4 points
C. not differentiable at 5 points
D. None of these

Explanation

To solve the problem, we need to analyze the function \( g(x) = |x \cdot f(x)| + ||x - 2| - 1| \) where \( f(x) = \frac{\sin x}{x} \) in the interval \( (0, 3\pi) \).

### Step 1: Rewrite the function \( g(x) \)

Given:
\[ f(x) = \frac{\sin x}{x} \]
We can substitute \( f(x) \) into \( g(x) \):
\[ g(x) = |x \cdot f(x)| + ||x - 2| - 1| \]
Substituting \( f(x) \):
\[ g(x) = |x \cdot \frac{\sin x}{x}| + ||x - 2| - 1| \]
This simplifies to:
\[ g(x) = |\sin x| + ||x - 2| - 1| \]

### Step 2: Analyze the components of \( g(x) \)

1. **Analyzing \( |\sin x| \)**:
- The function \( |\sin x| \) is continuous and differentiable everywhere except at points where \( \sin x = 0 \), which occurs at \( x = n\pi \) for \( n \in \mathbb{Z} \).
- In the interval \( (0, 3\pi) \), the points where \( \sin x = 0 \) are \( x = \pi \) and \( x = 2\pi \).

2. **Analyzing \( ||x - 2| - 1| \)**:
- The expression \( |x - 2| \) is continuous and differentiable everywhere except at \( x = 2 \).
- The expression \( ||x - 2| - 1| \) will have points of non-differentiability where \( |x - 2| = 1 \), which occurs at:
- \( x - 2 = 1 \) ⇒ \( x = 3 \)
- \( x - 2 = -1 \) ⇒ \( x = 1 \)

### Step 3: Identify points of non-differentiability

From the analysis:
- \( |\sin x| \) is non-differentiable at \( x = \pi \) and \( x = 2\pi \).
- \( ||x - 2| - 1| \) is non-differentiable at \( x = 1 \), \( x = 2 \), and \( x = 3 \).

### Step 4: Combine the points

The points where \( g(x) \) is non-differentiable in the interval \( (0, 3\pi) \) are:
- From \( |\sin x| \): \( x = \pi, 2\pi \)
- From \( ||x - 2| - 1| \): \( x = 1, 2, 3 \)

### Step 5: List all unique points

The unique points of non-differentiability are:
- \( x = 1 \)
- \( x = 2 \)
- \( x = 3 \)
- \( x = \pi \)
- \( x = 2\pi \)

Thus, there are a total of 5 points where \( g(x) \) is not differentiable in the interval \( (0, 3\pi) \).

### Final Answer

The function \( g(x) \) is not differentiable at 5 points in the interval \( (0, 3\pi) \).

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