Let `f(x)={(-,1, -2lexlt0),(x^2,-1,0lexlt2):}` if `g(x)=|f(x)|+f(|x|)` then `g(x)` in `(-2,2)` is
(A) not continuous is (B) not differential at one point (C) differential at all points (D) not differential at two points

Options

A. Differentiable at all points
B. not differentiable at two points
C. Not continuous
D. not differentiable at one point

Answer

To solve the problem, we need to analyze the function \( f(x) \) and then compute \( g(x) \) based on the given definitions.

### Step-by-Step Solution:

1. **Define the function \( f(x) \)**:
\[ f(x) = \begin{cases} -1 & \text{if } -2 \leq x < 0 \\ x^2 - 1 & \text{if } 0 \leq x < 2 \end{cases} \]

2. **Calculate \( |f(x)| \)**:
- For \( -2 \leq x < 0 \): \( f(x) = -1 \) implies \( |f(x)| = 1 \).
- For \( 0 \leq x < 2 \): \( f(x) = x^2 - 1 \) implies \( |f(x)| = x^2 - 1 \) when \( x^2 - 1 \geq 0 \) (which is for \( x \geq 1 \)) and \( |f(x)| = 1 - x^2 \) when \( x < 1 \).

Thus, we can summarize:
\[ |f(x)| = \begin{cases} 1 & \text{if } -2 \leq x < 0 \\ 1 - x^2 & \text{if } 0 \leq x < 1 \\ x^2 - 1 & \text{if } 1 \leq x < 2 \end{cases} \]

3. **Calculate \( f(|x|) \)**:
- For \( -2 \leq x < 0 \): \( |x| = -x \) which is in the range \( [0, 2) \), so \( f(|x|) = f(-x) = -1 \).
- For \( 0 \leq x < 2 \): \( f(|x|) = f(x) = x^2 - 1 \).

Thus, we can summarize:
\[ f(|x|) = \begin{cases} -1 & \text{if } -2 \leq x < 0 \\ x^2 - 1 & \text{if } 0 \leq x < 2 \end{cases} \]

4. **Define \( g(x) = |f(x)| + f(|x|) \)**:
- For \( -2 \leq x < 0 \):
\[ g(x) = 1 + (-1) = 0 \]
- For \( 0 \leq x < 1 \):
\[ g(x) = (1 - x^2) + (x^2 - 1) = 0 \]
- For \( 1 \leq x < 2 \):
\[ g(x) = (x^2 - 1) + (x^2 - 1) = 2x^2 - 2 \]

Thus, we can summarize:
\[ g(x) = \begin{cases} 0 & \text{if } -2 \leq x < 1 \\ 2x^2 - 2 & \text{if } 1 \leq x < 2 \end{cases} \]

5. **Check continuity of \( g(x) \)**:
- At \( x = 0 \): \( g(0^-) = 0 \) and \( g(0^+) = 0 \) ⇒ Continuous.
- At \( x = 1 \): \( g(1^-) = 0 \) and \( g(1^+) = 0 \) ⇒ Continuous.

6. **Check differentiability of \( g(x) \)**:
- For \( -2 < x < 0 \): \( g'(x) = 0 \).
- For \( 0 < x < 1 \): \( g'(x) = 0 \).
- For \( 1 < x < 2 \): \( g'(x) = 4x \).

- At \( x = 0 \):
- Left-hand derivative: \( 0 \)
- Right-hand derivative: \( 0 \)
- Thus, \( g(x) \) is differentiable at \( x = 0 \).

- At \( x = 1 \):
- Left-hand derivative: \( 0 \)
- Right-hand derivative: \( 4 \)
- Thus, \( g(x) \) is not differentiable at \( x = 1 \).

### Conclusion:
The function \( g(x) \) is not differentiable at one point (specifically at \( x = 1 \)). Therefore, the answer is:
**(B) not differentiable at one point.**