For `p gt 2 and x in R`, if the number of natural numbers in the range of `f(x)=(x^(2)+2x+p)/(x^(2)+2x+2)` is 3, then the value of p is equal to

Options

A. 3
B. 4
C. 5
D. 6

Answer

To solve the problem, we need to analyze the function given and determine the value of \( p \) such that the number of natural numbers in the range of the function \( f(x) = \frac{x^2 + 2x + p}{x^2 + 2x + 2} \) is exactly 3.

### Step 1: Simplify the function
The function can be rewritten as:
\[ f(x) = \frac{x^2 + 2x + p}{x^2 + 2x + 2} \]
Let \( y = f(x) \), then we have:
\[ y = \frac{x^2 + 2x + p}{x^2 + 2x + 2} \]
Cross-multiplying gives:
\[ y(x^2 + 2x + 2) = x^2 + 2x + p \]
Rearranging leads to:
\[ yx^2 + 2yx + 2y - x^2 - 2x - p = 0 \]
This can be rearranged to:
\[ (y - 1)x^2 + (2y - 2)x + (2y - p) = 0 \]

### Step 2: Analyze the quadratic equation
For \( f(x) \) to have real values for all \( x \), the discriminant of this quadratic must be non-negative:
\[ D = (2y - 2)^2 - 4(y - 1)(2y - p) \geq 0 \]
Expanding the discriminant:
\[ D = 4(y - 1)^2 - 4(y - 1)(2y - p) \]
Factoring out \( 4(y - 1) \):
\[ D = 4(y - 1)((y - 1) - (2y - p)) = 4(y - 1)(p - y) \]
For the discriminant to be non-negative, we need:
1. \( y - 1 \geq 0 \) (i.e., \( y \geq 1 \))
2. \( p - y \geq 0 \) (i.e., \( y \leq p \))

### Step 3: Determine the range of \( y \)
From the conditions above, we find that:
\[ 1 \leq y \leq p \]
This means the range of \( f(x) \) is from 1 to \( p \).

### Step 4: Count the natural numbers in the range
We want the number of natural numbers in the interval \( [1, p] \) to be exactly 3. The natural numbers in this range are \( 2, 3, 4 \) if \( p \) is at least 4. This implies:
\[ p \geq 4 \]
To have exactly 3 natural numbers, \( p \) must be less than or equal to 5. Therefore:
\[ 4 \leq p < 5 \]

### Step 5: Determine the exact value of \( p \)
Since \( p \) must be greater than 2 and we want exactly 3 natural numbers, the only integer value that satisfies \( p < 5 \) is:
\[ p = 5 \]

### Conclusion
Thus, the value of \( p \) is:
\[ \boxed{5} \]