If `f(x)=Pi_(k=1)^(999)(x^(2)-47x+k)`. then product of all real roots of `f(x)=0` is

Options

A. `550!`
B. `551!`
C. `552!`
D. `999!`

Answer

To solve the problem, we need to find the product of all real roots of the polynomial function defined as:

\[ f(x) = \prod_{k=1}^{999} (x^2 - 47x + k) \]

### Step 1: Understanding the Roots
Each factor in the product \( (x^2 - 47x + k) \) is a quadratic polynomial. The roots of this polynomial can be found using the quadratic formula:

\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

For our polynomial \( x^2 - 47x + k \), we have:
- \( a = 1 \)
- \( b = -47 \)
- \( c = k \)

### Step 2: Finding the Discriminant
The discriminant \( D \) of the quadratic is given by:

\[ D = b^2 - 4ac = (-47)^2 - 4(1)(k) = 2209 - 4k \]

For the roots to be real, the discriminant must be non-negative:

\[ 2209 - 4k \geq 0 \]

### Step 3: Solving for \( k \)
Rearranging the inequality gives:

\[ 4k \leq 2209 \implies k \leq \frac{2209}{4} = 552.25 \]

Since \( k \) must be an integer and ranges from 1 to 999, the maximum integer value for \( k \) that satisfies this condition is \( k = 552 \).

### Step 4: Identifying Valid \( k \) Values
Thus, \( k \) can take values from 1 to 552. For each \( k \) in this range, the quadratic \( x^2 - 47x + k \) has real roots.

### Step 5: Product of Roots for Each Quadratic
The product of the roots of a quadratic \( ax^2 + bx + c = 0 \) is given by:

\[ \text{Product of roots} = \frac{c}{a} = k \]

For each \( k \) from 1 to 552, the product of the roots of \( x^2 - 47x + k \) is \( k \).

### Step 6: Total Product of All Real Roots
The total product of all real roots of \( f(x) = 0 \) is the product of \( k \) for \( k = 1 \) to \( 552 \):

\[ \text{Total Product} = 1 \times 2 \times 3 \times \ldots \times 552 = 552! \]

### Conclusion
Thus, the product of all real roots of \( f(x) = 0 \) is:

\[ \boxed{552!} \]