Let p and q be real numbers such that the function `f(x)={{:(p^(2)+qx","xge1),(-5px^(2)-6","xlt1):}` is derivable for all `x in R`. Then sum of all possible values of p is

Options

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

Explanation

To solve the problem, we need to ensure that the function \( f(x) \) is both continuous and differentiable at \( x = 1 \). The function is defined as follows:

\[ f(x) = \begin{cases} p^2 + qx & \text{if } x \geq 1 \\ -5px^2 - 6 & \text{if } x < 1 \end{cases} \]

### Step 1: Check for Differentiability at \( x = 1 \)

For \( f(x) \) to be differentiable at \( x = 1 \), the left-hand derivative must equal the right-hand derivative at that point.

1. **Right-hand derivative at \( x = 1 \)**:
\[ f'(x) = q \quad \text{for } x \geq 1 \]
Thus,
\[ f'(1^+) = q \]

2. **Left-hand derivative at \( x = 1 \)**:
\[ f'(x) = -10p \quad \text{for } x < 1 \]
Thus,
\[ f'(1^-) = -10p \]

Setting the derivatives equal gives us:
\[ q = -10p \tag{1} \]

### Step 2: Check for Continuity at \( x = 1 \)

For \( f(x) \) to be continuous at \( x = 1 \), the left-hand limit must equal the right-hand limit.

1. **Right-hand limit as \( x \) approaches 1**:
\[ f(1) = p^2 + q \]

2. **Left-hand limit as \( x \) approaches 1**:
\[ f(1) = -5p(1)^2 - 6 = -5p - 6 \]

Setting these equal gives us:
\[ p^2 + q = -5p - 6 \tag{2} \]

### Step 3: Substitute Equation (1) into Equation (2)

Substituting \( q = -10p \) into Equation (2):
\[ p^2 - 10p = -5p - 6 \]

Rearranging this gives:
\[ p^2 - 10p + 5p + 6 = 0 \]
\[ p^2 - 5p + 6 = 0 \]

### Step 4: Factor the Quadratic Equation

We can factor the quadratic:
\[ (p - 2)(p - 3) = 0 \]

Thus, the possible values of \( p \) are:
\[ p = 2 \quad \text{or} \quad p = 3 \]

### Step 5: Find the Sum of All Possible Values of \( p \)

The sum of all possible values of \( p \) is:
\[ 2 + 3 = 5 \]

### Final Answer

The sum of all possible values of \( p \) is \( \boxed{5} \).