If `f(x)={(x^(2),xle1),(1-x,xgt1):}` & composite function `h(x)=|f(x)|+f(x+2)`, then

Options

A. `h(x)=2x^(2)+4x+4 AAxle-1`
B. `h(x)=x^(2)+x+1 AA-1ltxle1`
C. `h(x)=x^(2)-x-1 AA-1ltxle1`
D. `h(x)=-2AAxgt1`

Explanation

To solve the problem, we need to evaluate the function \( h(x) = |f(x)| + f(x+2) \) where the function \( f(x) \) is defined piecewise as follows:

\[ f(x) = \begin{cases} x^2 & \text{if } x \leq 1 \\ 1 - x & \text{if } x > 1 \end{cases} \]

### Step 1: Determine \( |f(x)| \)

1. For \( x \leq 1 \):
- \( f(x) = x^2 \)
- Therefore, \( |f(x)| = x^2 \) (since \( x^2 \) is always non-negative).

2. For \( x > 1 \):
- \( f(x) = 1 - x \)
- Since \( 1 - x \) is negative for \( x > 1 \), we have \( |f(x)| = -(1 - x) = x - 1 \).

Thus, we can summarize \( |f(x)| \) as:

\[ |f(x)| = \begin{cases} x^2 & \text{if } x \leq 1 \\ x - 1 & \text{if } x > 1 \end{cases} \]

### Step 2: Determine \( f(x+2) \)

Next, we need to evaluate \( f(x+2) \):

1. For \( x + 2 \leq 1 \) (which implies \( x \leq -1 \)):
- \( f(x+2) = (x+2)^2 \)

2. For \( x + 2 > 1 \) (which implies \( x > -1 \)):
- \( f(x+2) = 1 - (x + 2) = -x - 1 \)

Thus, we can summarize \( f(x+2) \) as:

\[ f(x+2) = \begin{cases} (x + 2)^2 & \text{if } x \leq -1 \\ -x - 1 & \text{if } x > -1 \end{cases} \]

### Step 3: Combine \( |f(x)| \) and \( f(x+2) \) to find \( h(x) \)

Now, we can express \( h(x) = |f(x)| + f(x+2) \):

1. **For \( x \leq -1 \)**:
- \( h(x) = |f(x)| + f(x+2) = x^2 + (x + 2)^2 \)
- Simplifying gives:
\[ h(x) = x^2 + (x^2 + 4x + 4) = 2x^2 + 4x + 4 \]

2. **For \( -1 < x \leq 1 \)**:
- \( h(x) = |f(x)| + f(x+2) = x^2 + (-x - 1) \)
- Simplifying gives:
\[ h(x) = x^2 - x - 1 \]

3. **For \( x > 1 \)**:
- \( h(x) = |f(x)| + f(x+2) = (x - 1) + (-x - 1) \)
- Simplifying gives:
\[ h(x) = -2 \]

### Final Result

Thus, the complete piecewise function for \( h(x) \) is:

\[ h(x) = \begin{cases} 2x^2 + 4x + 4 & \text{if } x \leq -1 \\ x^2 - x - 1 & \text{if } -1 < x \leq 1 \\ -2 & \text{if } x > 1 \end{cases} \]