let the function f be defined by `f (x)= {{:(p+ qx+ x^(2)"," , x lt 2),( 2 px+3qx ^(2)"," , x ge 2):},` Then:

Options

A. f (x) is continous in R if `3p+10q=4`
B. `f (x)` is differentiable in R is `p=q =(4)/(13)`
C. If `p=-2, q=1,` then f(x) is continous in R
D. f (x) is differentiable in R is `2p+ 11q=4`

Answer

To solve the problem, we need to analyze the function \( f(x) \) defined as:

\[ f(x) = \begin{cases} p + qx + x^2 & \text{if } x < 2 \\ 2px + 3qx^2 & \text{if } x \geq 2 \end{cases} \]

We need to check for continuity and differentiability at \( x = 2 \).

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

For \( f(x) \) to be continuous at \( x = 2 \), the left-hand limit as \( x \) approaches 2 must equal the right-hand limit, and both must equal \( f(2) \).

1. **Left-hand limit** as \( x \) approaches 2:
\[ f(2^-) = p + 2q + 2^2 = p + 2q + 4 \]

2. **Right-hand limit** at \( x = 2 \):
\[ f(2^+) = 2p(2) + 3q(2^2) = 4p + 12q \]

3. **Setting the limits equal for continuity**:
\[ p + 2q + 4 = 4p + 12q \]

Rearranging gives:
\[ 4 - 4p + 2q - 12q = 0 \implies 10q + 3p = 4 \quad \text{(Equation 1)} \]

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

For \( f(x) \) to be differentiable at \( x = 2 \), the left-hand derivative must equal the right-hand derivative.

1. **Right-hand derivative**:
\[ f'(2^+) = \lim_{h \to 0} \frac{f(2+h) - f(2)}{h} = \lim_{h \to 0} \frac{(2p(2+h) + 3q(2+h)^2) - (4p + 12q)}{h} \]

Expanding \( f(2+h) \):
\[ = \lim_{h \to 0} \frac{(4p + 2ph + 3q(4 + 4h + h^2)) - (4p + 12q)}{h} \]
\[ = \lim_{h \to 0} \frac{2ph + 12qh + 3qh^2}{h} = 2p + 12q \]

2. **Left-hand derivative**:
\[ f'(2^-) = \lim_{h \to 0} \frac{f(2-h) - f(2)}{h} = \lim_{h \to 0} \frac{(p + q(2-h) + (2-h)^2) - (p + 2q + 4)}{-h} \]

Expanding \( f(2-h) \):
\[ = \lim_{h \to 0} \frac{(p + 2q - qh + 4 - 4h + h^2) - (p + 2q + 4)}{-h} \]
\[ = \lim_{h \to 0} \frac{-qh - 4h + h^2}{-h} = q + 4 \]

3. **Setting the derivatives equal for differentiability**:
\[ 2p + 12q = q + 4 \]
Rearranging gives:
\[ 2p + 11q = 4 \quad \text{(Equation 2)} \]

### Step 3: Solve the System of Equations

We have two equations:
1. \( 10q + 3p = 4 \) (Equation 1)
2. \( 2p + 11q = 4 \) (Equation 2)

From Equation 1, we can express \( p \) in terms of \( q \):
\[ p = \frac{4 - 10q}{3} \]

Substituting \( p \) into Equation 2:
\[ 2\left(\frac{4 - 10q}{3}\right) + 11q = 4 \]
\[ \frac{8 - 20q}{3} + 11q = 4 \]
Multiplying through by 3 to eliminate the fraction:
\[ 8 - 20q + 33q = 12 \]
\[ 13q = 4 \implies q = \frac{4}{13} \]

Substituting \( q \) back to find \( p \):
\[ p = \frac{4 - 10\left(\frac{4}{13}\right)}{3} = \frac{4 - \frac{40}{13}}{3} = \frac{\frac{52}{13} - \frac{40}{13}}{3} = \frac{12/13}{3} = \frac{4}{13} \]

### Conclusion

Thus, we find:
\[ p = \frac{4}{13}, \quad q = \frac{4}{13} \]