Let `f(x)={{:(-2",",-3lexle0),(x-2",",0ltxle3):}andg(x)=f(|x|)+|f(x)|`
Which of the following statement is/are correct?
1. g(x) is differentiable x=0.
g(x) is differentiable at x=2.
Select the correct aswer using the code given below:

Options

A. 1 only
B. 2 only
C. Both 1 and 2
D. Neither 1 nor 2

Explanation

To solve the problem, we need to analyze the function \( g(x) = f(|x|) + |f(x)| \) and determine the differentiability of \( g(x) \) at \( x = 0 \) and \( x = 2 \).

### Step 1: Define the function \( f(x) \)
The function \( f(x) \) is defined as follows:
- \( f(x) = -2 \) for \( -3 \leq x \leq 0 \)
- \( f(x) = x - 2 \) for \( 0 < x \leq 3 \)

### Step 2: Evaluate \( g(x) \) at \( x = 0 \)
To find \( g(0) \):
\[ g(0) = f(|0|) + |f(0)| = f(0) + |f(0)| \]
Since \( f(0) = -2 \):
\[ g(0) = -2 + |-2| = -2 + 2 = 0 \]

### Step 3: Calculate the left-hand derivative of \( g(x) \) at \( x = 0 \)
For \( x < 0 \) (let \( x = -h \) where \( h \to 0^+ \)):
\[ g(-h) = f(-h) + |f(-h)| \]
Since \( -h \) falls in the range \( -3 \leq -h \leq 0 \):
\[ f(-h) = -2 \quad \text{and} \quad |f(-h)| = |-2| = 2 \]
Thus,
\[ g(-h) = -2 + 2 = 0 \]
Now, calculate the left-hand derivative:
\[ \lim_{h \to 0^+} \frac{g(-h) - g(0)}{-h} = \lim_{h \to 0^+} \frac{0 - 0}{-h} = 0 \]

### Step 4: Calculate the right-hand derivative of \( g(x) \) at \( x = 0 \)
For \( x > 0 \):
\[ g(h) = f(h) + |f(h)| \]
Since \( 0 < h \leq 3 \):
\[ f(h) = h - 2 \quad \text{and} \quad |f(h)| = |h - 2| \]
For \( 0 < h < 2 \), \( |h - 2| = 2 - h \):
\[ g(h) = (h - 2) + (2 - h) = 0 \]
For \( 2 < h \leq 3 \), \( |h - 2| = h - 2 \):
\[ g(h) = (h - 2) + (h - 2) = 2(h - 2) \]
Now, calculate the right-hand derivative:
- For \( 0 < h < 2 \):
\[ \lim_{h \to 0^+} \frac{g(h) - g(0)}{h} = \lim_{h \to 0^+} \frac{0 - 0}{h} = 0 \]
- For \( h \to 2^+ \):
\[ g(2) = 2(2 - 2) = 0 \]
Thus, the right-hand derivative at \( x = 0 \) is also \( 0 \).

### Step 5: Conclusion for \( x = 0 \)
Since both left-hand and right-hand derivatives at \( x = 0 \) are equal (both are \( 0 \)), \( g(x) \) is differentiable at \( x = 0 \).

### Step 6: Evaluate \( g(x) \) at \( x = 2 \)
To find \( g(2) \):
\[ g(2) = f(2) + |f(2)| \]
Since \( 2 \) falls in the range \( 0 < x \leq 3 \):
\[ f(2) = 2 - 2 = 0 \quad \text{and} \quad |f(2)| = |0| = 0 \]
Thus,
\[ g(2) = 0 + 0 = 0 \]

### Step 7: Calculate the left-hand derivative of \( g(x) \) at \( x = 2 \)
For \( x < 2 \) (let \( x = 2 - h \)):
\[ g(2 - h) = f(2 - h) + |f(2 - h)| \]
For \( 1 < 2 - h < 2 \):
\[ f(2 - h) = (2 - h) - 2 = -h \quad \text{and} \quad |f(2 - h)| = |-h| = h \]
Thus,
\[ g(2 - h) = -h + h = 0 \]
Now, calculate the left-hand derivative:
\[ \lim_{h \to 0^+} \frac{g(2 - h) - g(2)}{-(h)} = \lim_{h \to 0^+} \frac{0 - 0}{-h} = 0 \]

### Step 8: Calculate the right-hand derivative of \( g(x) \) at \( x = 2 \)
For \( x > 2 \):
\[ g(2 + h) = f(2 + h) + |f(2 + h)| \]
For \( 2 < 2 + h \leq 3 \):
\[ f(2 + h) = (2 + h) - 2 = h \quad \text{and} \quad |f(2 + h)| = |h| = h \]
Thus,
\[ g(2 + h) = h + h = 2h \]
Now, calculate the right-hand derivative:
\[ \lim_{h \to 0^+} \frac{g(2 + h) - g(2)}{h} = \lim_{h \to 0^+} \frac{2h - 0}{h} = 2 \]

### Step 9: Conclusion for \( x = 2 \)
Since the left-hand derivative at \( x = 2 \) is \( 0 \) and the right-hand derivative is \( 2 \), they are not equal. Therefore, \( g(x) \) is not differentiable at \( x = 2 \).

### Final Answer
1. \( g(x) \) is differentiable at \( x = 0 \) - **True**
2. \( g(x) \) is differentiable at \( x = 2 \) - **False**

Thus, the correct answer is that only the first statement is true.