If `f(x)=(x-p)(x-q)(x-r)` where `plt qlt ltr,` are real numbers, then the application, of Rolle's theorem on f leasds to

Options

A. `(p+q+r)(pq+qr+rp)=3`
B. `(p+q+r)^(2)=3(pq+qr+rp)`
C. `(p+q+r)^(2)gt3(pq+qr+rp)`
D. `(p+q+r)^(2)lt3(pq+qr+rp)`

Explanation

To solve the problem, we will apply Rolle's theorem to the function \( f(x) = (x - p)(x - q)(x - r) \), where \( p < q < r \) are distinct real numbers.

### Step-by-Step Solution:

1. **Identify the Function and Its Roots:**
The given function is:
\[ f(x) = (x - p)(x - q)(x - r) \]
The roots of this polynomial are \( p, q, \) and \( r \). Therefore, we have:
\[ f(p) = f(q) = f(r) = 0 \]

2. **Apply Rolle's Theorem:**
According to Rolle's theorem, if a function is continuous on a closed interval and differentiable on the open interval, and if the function takes the same value at the endpoints, then there exists at least one \( c \) in the open interval such that:
\[ f'(c) = 0 \]
Here, we can apply this theorem on the intervals \( [p, q] \) and \( [q, r] \).

3. **Finding the Derivative:**
We need to find the derivative \( f'(x) \). Using the product rule or expanding the polynomial, we can differentiate:
\[ f(x) = x^3 - (p + q + r)x^2 + (pq + qr + rp)x - pqr \]
Thus, the derivative is:
\[ f'(x) = 3x^2 - 2(p + q + r)x + (pq + qr + rp) \]

4. **Finding Roots of the Derivative:**
By applying Rolle's theorem, we know that there is at least one root \( c_1 \) in \( (p, q) \) and at least one root \( c_2 \) in \( (q, r) \). Therefore, \( f'(x) \) has at least two distinct real roots.

5. **Analyzing the Roots:**
The quadratic equation \( f'(x) = 0 \) can be analyzed using the discriminant:
\[ D = [2(p + q + r)]^2 - 4 \cdot 3 \cdot (pq + qr + rp) \]
For \( f'(x) \) to have distinct real roots, the discriminant must be greater than zero:
\[ 4(p + q + r)^2 - 12(pq + qr + rp) > 0 \]
This simplifies to:
\[ (p + q + r)^2 < 3(pq + qr + rp) \]

6. **Conclusion:**
Thus, the application of Rolle's theorem leads us to conclude that:
\[ (p + q + r)^2 < 3(pq + qr + rp) \]