Let `f : R to R` be defined by
`f(x) = |2-x| - |x+1|`
The number of integral values of a for which `f(x) =a` has exactly one solution is:

Options

A. 3
B. 5
C. 6
D. 8

Answer

To solve the problem, we need to analyze the function \( f(x) = |2 - x| - |x + 1| \) and determine the number of integral values of \( a \) for which the equation \( f(x) = a \) has exactly one solution.

### Step 1: Identify critical points
The function involves absolute values, so we need to identify the points where the expressions inside the absolute values change sign. The critical points are:
- \( 2 - x = 0 \) → \( x = 2 \)
- \( x + 1 = 0 \) → \( x = -1 \)

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

### Step 2: Define \( f(x) \) in each interval
Now, we will express \( f(x) \) in each of these intervals.

**Interval 1: \( x < -1 \)**
- Here, both \( 2 - x \) and \( x + 1 \) are positive.
- Thus, \( f(x) = (2 - x) - (-(x + 1)) = 2 - x + x + 1 = 3 \).

**Interval 2: \( -1 \leq x \leq 2 \)**
- Here, \( 2 - x \) is positive and \( x + 1 \) is non-negative.
- Thus, \( f(x) = (2 - x) - (x + 1) = 2 - x - x - 1 = 1 - 2x \).

**Interval 3: \( x > 2 \)**
- Here, both \( 2 - x \) and \( x + 1 \) are positive.
- Thus, \( f(x) = (2 - x) - (x + 1) = 2 - x - x - 1 = 1 - 2x \).

### Step 3: Analyze the function
Now we have:
- \( f(x) = 3 \) for \( x < -1 \)
- \( f(x) = 1 - 2x \) for \( -1 \leq x \leq 2 \)
- \( f(x) = 1 - 2x \) for \( x > 2 \)

### Step 4: Find the range of \( f(x) \)
1. For \( x < -1 \), \( f(x) = 3 \).
2. For \( -1 \leq x \leq 2 \), the function \( 1 - 2x \) decreases from \( 3 \) (at \( x = -1 \)) to \( -3 \) (at \( x = 2 \)).
3. For \( x > 2 \), \( f(x) \) continues to decrease from \( -3 \) as \( x \) increases.

### Step 5: Determine the values of \( a \)
The function \( f(x) \) achieves the following:
- \( f(x) = 3 \) at \( x < -1 \) (exactly one solution).
- \( f(x) \) decreases continuously from \( 3 \) to \( -3 \) in the interval \( [-1, 2] \) (exactly one solution for each \( a \) in \( (3, -3) \)).
- \( f(x) = -3 \) at \( x = 2 \) (exactly one solution).

### Step 6: Count the integral values of \( a \)
The range of \( a \) for which \( f(x) = a \) has exactly one solution is \( (-3, 3) \). The integral values in this range are:
- \( -2, -1, 0, 1, 2 \)

Thus, there are **5 integral values of \( a \)** for which \( f(x) = a \) has exactly one solution.

### Final Answer
The number of integral values of \( a \) for which \( f(x) = a \) has exactly one solution is **5**.

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