Find the values of p and q , so that `f(x)={{:(x^(2)+3x+p, ifxle1),(qx+2,ifx gt1):}` is differentiable at `x = 1`

Explanation

To find the values of \( p \) and \( q \) such that the function

\[ f(x) = \begin{cases} x^2 + 3x + p & \text{if } x \leq 1 \\ qx + 2 & \text{if } x > 1 \end{cases} \]

is differentiable at \( x = 1 \), we need to ensure that the function is continuous and that the derivatives from both sides at \( x = 1 \) are equal.

### Step 1: Ensure Continuity at \( x = 1 \)

For \( f(x) \) to be continuous at \( x = 1 \), we need:

\[ f(1^-) = f(1^+ \]

Calculating \( f(1^-) \):

\[ f(1^-) = 1^2 + 3(1) + p = 1 + 3 + p = p + 4 \]

Calculating \( f(1^+) \):

\[ f(1^+) = q(1) + 2 = q + 2 \]

Setting these equal for continuity:

\[ p + 4 = q + 2 \]

Rearranging gives us our first equation:

\[ p - q = -2 \quad \text{(Equation 1)} \]

### Step 2: Ensure Differentiability at \( x = 1 \)

Next, we need to check the derivatives from both sides at \( x = 1 \).

Calculating the derivative for \( x \leq 1 \):

\[ f'(x) = 2x + 3 \]

Thus,

\[ f'(1^-) = 2(1) + 3 = 5 \]

Calculating the derivative for \( x > 1 \):

\[ f'(x) = q \]

Thus,

\[ f'(1^+) = q \]

Setting these equal for differentiability:

\[ f'(1^-) = f'(1^+) \]

This gives us:

\[ 5 = q \quad \text{(Equation 2)} \]

### Step 3: Solve the Equations

Now we have two equations:

1. \( p - q = -2 \)
2. \( q = 5 \)

Substituting \( q = 5 \) into Equation 1:

\[ p - 5 = -2 \]

Adding 5 to both sides:

\[ p = 3 \]

### Final Values

Thus, the values of \( p \) and \( q \) are:

\[ p = 3, \quad q = 5 \]