The absolute valued function f is defined as `f(x) = {{:(x,, x ge 0),(-x ,, x lt 0):}}` and fractional part function g(x) as g(x) = x-[x], graphically the number of real solution(s) of the equation f(x) = g(x) is obtained by finding the point(s) of interaction of the graph of y = f(x) and y = g(x).
The number of solution (s) `|x-1| - |x+2| = k`, when `-3 lt k lt 3`

Options

A. 0
B. 1
C. 2
D. 3

Answer

To solve the equation \( |x - 1| - |x + 2| = k \) for the range \( -3 < k < 3 \), we will analyze the function step by step.

### Step 1: Identify Critical Points
The absolute value functions change at the points where the expressions inside the absolute values are zero. Thus, we find the critical points:
- \( x - 1 = 0 \) gives \( x = 1 \)
- \( x + 2 = 0 \) gives \( x = -2 \)

The critical points divide the real line into three intervals:
1. \( (-\infty, -2) \)
2. \( [-2, 1] \)
3. \( (1, \infty) \)

### Step 2: Analyze Each Interval
We will analyze the function \( f(x) = |x - 1| - |x + 2| \) in each of these intervals.

#### Interval 1: \( x < -2 \)
In this interval, both expressions inside the absolute values are negative:
- \( |x - 1| = -(x - 1) = -x + 1 \)
- \( |x + 2| = -(x + 2) = -x - 2 \)

Thus, we have:
\[ f(x) = (-x + 1) - (-x - 2) = -x + 1 + x + 2 = 3 \]

#### Interval 2: \( -2 \leq x < 1 \)
In this interval, \( x - 1 \) is negative and \( x + 2 \) is non-negative:
- \( |x - 1| = -(x - 1) = -x + 1 \)
- \( |x + 2| = x + 2 \)

Thus, we have:
\[ f(x) = (-x + 1) - (x + 2) = -x + 1 - x - 2 = -2x - 1 \]

#### Interval 3: \( x \geq 1 \)
In this interval, both expressions are non-negative:
- \( |x - 1| = x - 1 \)
- \( |x + 2| = x + 2 \)

Thus, we have:
\[ f(x) = (x - 1) - (x + 2) = x - 1 - x - 2 = -3 \]

### Step 3: Summary of Function Behavior
We can summarize the function \( f(x) \) as follows:
- For \( x < -2 \), \( f(x) = 3 \)
- For \( -2 \leq x < 1 \), \( f(x) = -2x - 1 \)
- For \( x \geq 1 \), \( f(x) = -3 \)

### Step 4: Finding Intersections with \( k \)
Now we need to find the number of solutions to \( f(x) = k \) for \( -3 < k < 3 \).

1. **For \( k = 3 \)**:
- In the interval \( (-\infty, -2) \), \( f(x) = 3 \) gives one solution.

2. **For \( k = -3 \)**:
- In the interval \( [1, \infty) \), \( f(x) = -3 \) gives one solution.

3. **For \( -3 < k < 3 \)**:
- In the interval \( [-2, 1) \), we solve \( -2x - 1 = k \):
\[ -2x = k + 1 \implies x = -\frac{k + 1}{2} \]
This will yield one solution as long as \( -2 \leq -\frac{k + 1}{2} < 1 \).

### Step 5: Conclusion
Thus, for \( -3 < k < 3 \), we have:
- One solution from \( f(x) = 3 \) for \( x < -2 \)
- One solution from \( f(x) = -3 \) for \( x \geq 1 \)
- One solution from \( f(x) = k \) in the interval \( [-2, 1) \)

Overall, there are **three solutions** for the equation \( |x - 1| - |x + 2| = k \) when \( -3 < k < 3 \).