Let `f(x)=x+1 and phi(x)=x-2.` Then the value of x satisfying `|f(x)+phi(x)|=|f(x)|+|phi(x)|` are :

Options

A. `(-oo,1]`
B. `[2,oo)`
C. `(-oo,-2]`
D. `[1,oo)`

Answer

To solve the problem, we need to analyze the equation given:

\[ |f(x) + \phi(x)| = |f(x)| + |\phi(x)| \]

where \( f(x) = x + 1 \) and \( \phi(x) = x - 2 \).

### Step 1: Substitute the functions into the equation

First, we substitute \( f(x) \) and \( \phi(x) \) into the equation:

\[ | (x + 1) + (x - 2) | = |x + 1| + |x - 2| \]

This simplifies to:

\[ |2x - 1| = |x + 1| + |x - 2| \]

### Step 2: Analyze the absolute values

The equality \( |a| = |b| + |c| \) holds true when \( a, b, c \) are of the same sign or when both sides are zero. We need to consider the signs of \( 2x - 1 \), \( x + 1 \), and \( x - 2 \).

### Step 3: Find critical points

The critical points are where each expression inside the absolute values is zero:

1. \( 2x - 1 = 0 \) gives \( x = \frac{1}{2} \)
2. \( x + 1 = 0 \) gives \( x = -1 \)
3. \( x - 2 = 0 \) gives \( x = 2 \)

These points divide the number line into intervals:

- \( (-\infty, -1) \)
- \( [-1, \frac{1}{2}) \)
- \( [\frac{1}{2}, 2) \)
- \( [2, \infty) \)

### Step 4: Test each interval

1. **Interval \( (-\infty, -1) \)**:
- Here, \( 2x - 1 < 0 \), \( x + 1 < 0 \), \( x - 2 < 0 \)
- Thus, \( |2x - 1| = -(2x - 1) = -2x + 1 \)
- \( |x + 1| = -(x + 1) = -x - 1 \)
- \( |x - 2| = -(x - 2) = -x + 2 \)
- Plugging these into the equation gives:
\[ -2x + 1 = (-x - 1) + (-x + 2) \implies -2x + 1 = -2x + 1 \]
- This is true for all \( x < -1 \).

2. **Interval \( [-1, \frac{1}{2}) \)**:
- Here, \( 2x - 1 < 0 \), \( x + 1 \geq 0 \), \( x - 2 < 0 \)
- Thus, \( |2x - 1| = -2x + 1 \), \( |x + 1| = x + 1 \), \( |x - 2| = -x + 2 \)
- Plugging these into the equation gives:
\[ -2x + 1 = (x + 1) + (-x + 2) \implies -2x + 1 = 3 \]
- This is not true for any \( x \) in this interval.

3. **Interval \( [\frac{1}{2}, 2) \)**:
- Here, \( 2x - 1 \geq 0 \), \( x + 1 \geq 0 \), \( x - 2 < 0 \)
- Thus, \( |2x - 1| = 2x - 1 \), \( |x + 1| = x + 1 \), \( |x - 2| = -x + 2 \)
- Plugging these into the equation gives:
\[ 2x - 1 = (x + 1) + (-x + 2) \implies 2x - 1 = 2 \]
- This simplifies to \( 2x = 3 \) or \( x = \frac{3}{2} \), which is valid in this interval.

4. **Interval \( [2, \infty) \)**:
- Here, \( 2x - 1 \geq 0 \), \( x + 1 \geq 0 \), \( x - 2 \geq 0 \)
- Thus, \( |2x - 1| = 2x - 1 \), \( |x + 1| = x + 1 \), \( |x - 2| = x - 2 \)
- Plugging these into the equation gives:
\[ 2x - 1 = (x + 1) + (x - 2) \implies 2x - 1 = 2x - 1 \]
- This is true for all \( x \geq 2 \).

### Step 5: Combine results

From our analysis, the values of \( x \) satisfying the original equation are:

- All \( x < -1 \)
- \( x = \frac{3}{2} \)
- All \( x \geq 2 \)

Thus, the final solution is:

\[ (-\infty, -1] \cup \left\{ \frac{3}{2} \right\} \cup [2, \infty) \]