Let `f:RtoR` is given by
`f(x)={(p+qx+x^(2),xlt2),(2px+3qx^(2),xge2):}` 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 piecewise function 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 will check for continuity and differentiability at the point \( x = 2 \).

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

For \( f(x) \) to be continuous at \( x = 2 \), we need:

\[ \lim_{x \to 2^-} f(x) = \lim_{x \to 2^+} f(x) = f(2) \]

Calculating \( f(2) \):

\[ f(2) = 2p + 3q(2^2) = 2p + 12q \]

Calculating \( \lim_{x \to 2^-} f(x) \):

\[ \lim_{x \to 2^-} f(x) = p + q(2) + (2^2) = p + 2q + 4 \]

Calculating \( \lim_{x \to 2^+} f(x) \):

\[ \lim_{x \to 2^+} f(x) = 2p + 12q \]

Setting the limits equal for continuity:

\[ p + 2q + 4 = 2p + 12q \]

Rearranging gives:

\[ p + 2q + 4 - 2p - 12q = 0 \implies -p - 10q + 4 = 0 \implies p + 10q = 4 \quad \text{(Condition 1)} \]

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

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

Calculating LHD at \( x = 2 \):

\[ \text{LHD} = \frac{d}{dx}(p + qx + x^2) \bigg|_{x=2} = q + 2(2) = q + 4 \]

Calculating RHD at \( x = 2 \):

\[ \text{RHD} = \frac{d}{dx}(2px + 3qx^2) \bigg|_{x=2} = 2p + 6q(2) = 2p + 12q \]

Setting LHD equal to RHD for differentiability:

\[ q + 4 = 2p + 12q \]

Rearranging gives:

\[ q + 4 - 2p - 12q = 0 \implies -11q - 2p + 4 = 0 \implies 2p + 11q = 4 \quad \text{(Condition 2)} \]

### Step 3: Solve the System of Equations

Now we have two equations:

1. \( p + 10q = 4 \)
2. \( 2p + 11q = 4 \)

We can solve this system of equations. From the first equation, we can express \( p \):

\[ p = 4 - 10q \]

Substituting into the second equation:

\[ 2(4 - 10q) + 11q = 4 \]

Expanding and simplifying:

\[ 8 - 20q + 11q = 4 \implies -9q = -4 \implies q = \frac{4}{9} \]

Substituting \( q \) back into the first equation to find \( p \):

\[ p + 10\left(\frac{4}{9}\right) = 4 \implies p + \frac{40}{9} = 4 \implies p = 4 - \frac{40}{9} = \frac{36}{9} - \frac{40}{9} = -\frac{4}{9} \]

### Final Result

Thus, the values of \( p \) and \( q \) that make \( f(x) \) continuous and differentiable at \( x = 2 \) are:

\[ p = -\frac{4}{9}, \quad q = \frac{4}{9} \]