Let `g(x)=|x-2| and h(x)=g(g(x))` be two functions, then the value of `h'(-1)+h'(1)+h'(3)+h'(5)` is equal to (where, h' denotes the derivative of h)

Options

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

Answer

To solve the problem, we need to find the value of \( h'(-1) + h'(1) + h'(3) + h'(5) \), where \( h(x) = g(g(x)) \) and \( g(x) = |x - 2| \).

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

The function \( g(x) \) is defined as:
\[ g(x) = |x - 2| \]
This function outputs the distance of \( x \) from 2.

### Step 2: Find \( g(g(x)) \)

Next, we need to compute \( h(x) = g(g(x)) \):
\[ h(x) = g(g(x)) = g(|x - 2|) \]
Now, we need to evaluate \( g(|x - 2|) \):
- If \( |x - 2| \geq 2 \), then \( g(|x - 2|) = | |x - 2| - 2 | \)
- If \( |x - 2| < 2 \), then \( g(|x - 2|) = 2 - |x - 2| \)

### Step 3: Determine the intervals for \( h(x) \)

1. **Case 1**: \( x < 0 \)
- \( |x - 2| = 2 - x \)
- \( g(g(x)) = g(2 - x) = |(2 - x) - 2| = | - x | = x \)

2. **Case 2**: \( 0 \leq x < 2 \)
- \( |x - 2| = 2 - x \)
- \( g(g(x)) = g(2 - x) = |(2 - x) - 2| = | - x | = x \)

3. **Case 3**: \( 2 \leq x < 4 \)
- \( |x - 2| = x - 2 \)
- \( g(g(x)) = g(x - 2) = |(x - 2) - 2| = |x - 4| \)

4. **Case 4**: \( x \geq 4 \)
- \( |x - 2| = x - 2 \)
- \( g(g(x)) = g(x - 2) = |(x - 2) - 2| = |x - 4| \)

### Step 4: Derivative of \( h(x) \)

Now we can find the derivative \( h'(x) \) in each case:

1. **For \( x < 2 \)**: \( h(x) = x \) → \( h'(x) = 1 \)
2. **For \( 2 \leq x < 4 \)**: \( h(x) = |x - 4| \)
- If \( x < 4 \), \( h(x) = 4 - x \) → \( h'(x) = -1 \)
- If \( x \geq 4 \), \( h(x) = x - 4 \) → \( h'(x) = 1 \)

### Step 5: Evaluate \( h'(-1), h'(1), h'(3), h'(5) \)

- \( h'(-1) = 1 \) (since \(-1 < 2\))
- \( h'(1) = 1 \) (since \(1 < 2\))
- \( h'(3) = -1 \) (since \(2 \leq 3 < 4\))
- \( h'(5) = 1 \) (since \(5 \geq 4\))

### Step 6: Calculate the final value

Now we can sum these derivatives:
\[ h'(-1) + h'(1) + h'(3) + h'(5) = 1 + 1 - 1 + 1 = 2 \]

### Final Answer

The value of \( h'(-1) + h'(1) + h'(3) + h'(5) \) is \( \boxed{2} \).