If `q^2 - 4pr =0 , p gt 0` then the domain of the function `f(x) = log(p x^3 +(p+q)x^2 +(q+r) x + r)` is (a) `R-{-q/2p}` (b) `R-[(-oo,-1]uu{-q/(2p)}]` (c) `R-[(-oo,-1)nn{-1/(2p)}]` (d) `R`

Options

A. `R-{-q/(2p)}`
B. `R-[(-oo,-1]uu{-1/(2p)}]`
C. `R-[(-oo,-1)nn{-1/(2p)}]`
D. `R`

Explanation

To find the domain of the function \( f(x) = \log(px^3 + (p+q)x^2 + (q+r)x + r) \) given that \( q^2 - 4pr = 0 \) and \( p > 0 \), we need to ensure that the argument of the logarithm is positive.

### Step-by-Step Solution:

1. **Identify the Expression Inside the Logarithm**:
The function is given as:
\[ f(x) = \log(px^3 + (p+q)x^2 + (q+r)x + r) \]
We need to find when the expression \( px^3 + (p+q)x^2 + (q+r)x + r > 0 \).

2. **Finding a Root**:
We will check if there are any roots of the polynomial. By substituting \( x = -1 \):
\[ p(-1)^3 + (p+q)(-1)^2 + (q+r)(-1) + r = -p + (p+q) - (q+r) + r \]
Simplifying this gives:
\[ -p + p + q - q - r + r = 0 \]
Thus, \( x = -1 \) is a root.

3. **Factoring the Polynomial**:
Since \( x = -1 \) is a root, we can factor the polynomial as:
\[ px^3 + (p+q)x^2 + (q+r)x + r = (x + 1)(px^2 + (p + q - p)x + (q + r - r)) \]
Simplifying gives:
\[ = (x + 1)(px^2 + qx + r) \]

4. **Finding the Roots of the Quadratic**:
The quadratic \( px^2 + qx + r \) can be solved using the quadratic formula:
\[ x = \frac{-q \pm \sqrt{q^2 - 4pr}}{2p} \]
Given \( q^2 - 4pr = 0 \), the roots simplify to:
\[ x = \frac{-q}{2p} \]
This means \( px^2 + qx + r \) has a double root at \( x = \frac{-q}{2p} \).

5. **Analyzing the Sign of the Quadratic**:
Since \( p > 0 \), the quadratic \( px^2 + qx + r \) opens upwards. Therefore, it is positive for all \( x \) except at its double root \( x = \frac{-q}{2p} \).

6. **Combining Conditions**:
The function \( f(x) \) is defined when:
- \( x + 1 > 0 \) which gives \( x > -1 \)
- \( px^2 + qx + r > 0 \) which is true for all \( x \) except \( x = \frac{-q}{2p} \).

7. **Final Domain**:
Thus, the domain of \( f(x) \) is:
\[ x \in \mathbb{R} \setminus \left\{ -1, \frac{-q}{2p} \right\} \]
This can be expressed as:
\[ (-\infty, -1) \cup (-1, \frac{-q}{2p}) \cup (\frac{-q}{2p}, \infty) \]

### Conclusion:
The correct answer is:
(b) \( \mathbb{R} - [(-\infty, -1] \cup \{-\frac{q}{2p}\}] \)