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 \( g(x) = |x - 2| \) and \( h(x) = g(g(x)) \).

### Step 1: Define the function \( g(x) \)
The function \( g(x) = |x - 2| \) can be expressed in piecewise form:
- For \( x \geq 2 \), \( g(x) = x - 2 \)
- For \( x < 2 \), \( g(x) = 2 - x \)

### Step 2: Find \( h(x) = g(g(x)) \)
Now, we need to find \( h(x) \):
1. **Case 1**: When \( x \geq 2 \):
- \( g(x) = x - 2 \)
- Since \( x - 2 \geq 0 \) (because \( x \geq 2 \)), we have:
\[ h(x) = g(g(x)) = g(x - 2) = (x - 2) - 2 = x - 4 \]

2. **Case 2**: When \( x < 2 \):
- \( g(x) = 2 - x \)
- Since \( 2 - x \geq 0 \) (because \( x < 2 \)), we have:
\[ h(x) = g(g(x)) = g(2 - x) = 2 - (2 - x) = x \]

Thus, we can summarize \( h(x) \) as:
\[ h(x) = \begin{cases} x - 4 & \text{if } x \geq 2 \\ x & \text{if } x < 2 \end{cases} \]

### Step 3: Find the derivative \( h'(x) \)
Now, we compute the derivative \( h'(x) \):
1. **For \( x \geq 2 \)**:
- \( h(x) = x - 4 \)
- Therefore, \( h'(x) = 1 \)

2. **For \( x < 2 \)**:
- \( h(x) = x \)
- Therefore, \( h'(x) = 1 \)

Thus, we have:
\[ h'(x) = 1 \quad \text{for all } x \]

### Step 4: Calculate \( h'(-1) + h'(1) + h'(3) + h'(5) \)
Now we can evaluate:
- \( h'(-1) = 1 \) (since \(-1 < 2\))
- \( h'(1) = 1 \) (since \(1 < 2\))
- \( h'(3) = 1 \) (since \(3 \geq 2\))
- \( h'(5) = 1 \) (since \(5 \geq 2\))

Adding these values together:
\[ h'(-1) + h'(1) + h'(3) + h'(5) = 1 + 1 + 1 + 1 = 4 \]

### Final Answer
The value of \( h'(-1) + h'(1) + h'(3) + h'(5) \) is \( 4 \).

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