If `(x - p)/(x^(2) - 3x + 2)` takes all real values for ` x in R` then the range of P is

Answer

To find the range of \( p \) such that the expression \( \frac{x - p}{x^2 - 3x + 2} \) takes all real values for \( x \in \mathbb{R} \), we will follow these steps:

### Step 1: Define the expression
Let:
\[ y = \frac{x - p}{x^2 - 3x + 2} \]

### Step 2: Rearrange the equation
Rearranging gives us:
\[ y(x^2 - 3x + 2) = x - p \]
This can be rewritten as:
\[ yx^2 - 3yx + 2y - x + p = 0 \]
or:
\[ yx^2 + (-3y - 1)x + (2y + p) = 0 \]

### Step 3: Identify coefficients
In this quadratic equation in \( x \), the coefficients are:
- \( A = y \)
- \( B = -3y - 1 \)
- \( C = 2y + p \)

### Step 4: Apply the condition for real roots
For \( x \) to take all real values, the discriminant \( D \) of this quadratic must be non-negative:
\[ D = B^2 - 4AC \geq 0 \]
Substituting the coefficients:
\[ (-3y - 1)^2 - 4y(2y + p) \geq 0 \]

### Step 5: Expand and simplify the discriminant
Expanding the discriminant:
\[ (9y^2 + 6y + 1) - (8y^2 + 4py) \geq 0 \]
This simplifies to:
\[ y^2 + (6 - 4p)y + 1 \geq 0 \]

### Step 6: Analyze the quadratic inequality
For the quadratic \( y^2 + (6 - 4p)y + 1 \geq 0 \) to hold for all \( y \), its discriminant must be less than or equal to zero:
\[ (6 - 4p)^2 - 4 \cdot 1 \cdot 1 \leq 0 \]

### Step 7: Solve the discriminant inequality
Calculating the discriminant:
\[ (6 - 4p)^2 - 4 \leq 0 \]
This simplifies to:
\[ (6 - 4p)^2 \leq 4 \]
Taking square roots gives:
\[ |6 - 4p| \leq 2 \]
This leads to two inequalities:
1. \( 6 - 4p \leq 2 \)
2. \( 6 - 4p \geq -2 \)

### Step 8: Solve the inequalities
1. From \( 6 - 4p \leq 2 \):
\[ -4p \leq -4 \implies p \geq 1 \]
2. From \( 6 - 4p \geq -2 \):
\[ -4p \geq -8 \implies p \leq 2 \]

### Conclusion: Determine the range of \( p \)
Combining both results, we find:
\[ 1 \leq p \leq 2 \]
Thus, the range of \( p \) is:
\[ \boxed{[1, 2]} \]