If `f(x) = {sinx , x lt 0 and cosx-|x-1| , x leq 0` then `g(x) = f(|x|)` is non-differentiable for

Options

A. no value of x
B. exctly one value of x
C. exactly three values of x
D. none of these

Answer

To solve the problem, we need to analyze the function \( f(x) \) and then determine the points where \( g(x) = f(|x|) \) is non-differentiable.

### Step 1: Define the function \( f(x) \)

The function \( f(x) \) is defined as follows:
\[ f(x) = \begin{cases} \sin x & \text{if } x < 0 \\ \cos x - |x - 1| & \text{if } x \geq 0 \end{cases} \]

### Step 2: Analyze the function \( f(x) \)

1. For \( x < 0 \), \( f(x) = \sin x \).
2. For \( x \geq 0 \), we need to simplify \( f(x) = \cos x - |x - 1| \):
- When \( 0 \leq x < 1 \), \( |x - 1| = 1 - x \), so:
\[ f(x) = \cos x - (1 - x) = \cos x + x - 1 \]
- When \( x = 1 \), \( |x - 1| = 0 \), so:
\[ f(1) = \cos(1) \]
- When \( x > 1 \), \( |x - 1| = x - 1 \), so:
\[ f(x) = \cos x - (x - 1) = \cos x + 1 - x \]

### Step 3: Define \( g(x) = f(|x|) \)

Now, we define \( g(x) \):
\[ g(x) = f(|x|) = \begin{cases} f(-x) & \text{if } x < 0 \\ f(x) & \text{if } x \geq 0 \end{cases} \]

### Step 4: Analyze \( g(x) \)

1. For \( x < 0 \) (which means \( -x > 0 \)):
\[ g(x) = f(-x) = \cos(-x) - | -x - 1 | = \cos x - | -x - 1 | \]
- For \( -x < 1 \) (i.e., \( x > -1 \)), \( |-x - 1| = -(-x - 1) = x + 1 \):
\[ g(x) = \cos x - (x + 1) = \cos x - x - 1 \]
- For \( -x \geq 1 \) (i.e., \( x \leq -1 \)), \( |-x - 1| = -x - 1 \):
\[ g(x) = \cos x - (-x - 1) = \cos x + x + 1 \]

2. For \( x \geq 0 \):
- If \( 0 \leq x < 1 \):
\[ g(x) = \cos x + x - 1 \]
- If \( x = 1 \):
\[ g(1) = \cos(1) \]
- If \( x > 1 \):
\[ g(x) = \cos x + 1 - x \]

### Step 5: Identify non-differentiable points

The points where \( g(x) \) can be non-differentiable are at the points where the definition of \( g(x) \) changes, which are:
1. \( x = 0 \) (transition from \( x < 0 \) to \( x \geq 0 \))
2. \( x = 1 \) (transition in the definition of \( f(x) \))
3. \( x = -1 \) (transition in the definition of \( g(x) \) for negative values)

### Conclusion

Thus, \( g(x) \) is non-differentiable at the points:
\[ x = -1, \quad x = 0, \quad x = 1 \]