If the function `f(x)={(ax+b",",-oo lt x le2),(x^(2)-5x+6",", 2lt x lt 3),(px^(2)+qx+1",",3 le x lt oo):}` is differentiable in `(-oo, oo)`, then:

Options

A. `a=-1, p= -(4)/(9)`
B. `b=2, q= (5)/(3)`
C. `a=1, b=2`
D. `a=-1, q= -(5)/(3)`

Answer

To determine the values of \( a \), \( b \), \( p \), and \( q \) such that the function \( f(x) \) is differentiable over the entire real line, we need to ensure that the function is continuous and that its derivatives match at the points where the piecewise definitions change, specifically at \( x = 2 \) and \( x = 3 \).

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

The function is defined as:
- \( f(x) = ax + b \) for \( -\infty < x \leq 2 \)
- \( f(x) = x^2 - 5x + 6 \) for \( 2 < x < 3 \)

To ensure continuity at \( x = 2 \), we set the left-hand limit equal to the right-hand limit:
\[ \lim_{x \to 2^-} f(x) = \lim_{x \to 2^+} f(x) \]

Calculating the left-hand limit:
\[ \lim_{x \to 2^-} f(x) = a(2) + b = 2a + b \]

Calculating the right-hand limit:
\[ \lim_{x \to 2^+} f(x) = 2^2 - 5(2) + 6 = 4 - 10 + 6 = 0 \]

Setting these equal gives us:
\[ 2a + b = 0 \quad \text{(1)} \]

### Step 2: Ensure continuity at \( x = 3 \)

Next, we check continuity at \( x = 3 \):
- \( f(x) = x^2 - 5x + 6 \) for \( 2 < x < 3 \)
- \( f(x) = px^2 + qx + 1 \) for \( 3 \leq x < \infty \)

Setting the left-hand limit equal to the right-hand limit:
\[ \lim_{x \to 3^-} f(x) = \lim_{x \to 3^+} f(x) \]

Calculating the left-hand limit:
\[ \lim_{x \to 3^-} f(x) = 3^2 - 5(3) + 6 = 9 - 15 + 6 = 0 \]

Calculating the right-hand limit:
\[ \lim_{x \to 3^+} f(x) = p(3^2) + q(3) + 1 = 9p + 3q + 1 \]

Setting these equal gives us:
\[ 9p + 3q + 1 = 0 \quad \text{(2)} \]

### Step 3: Ensure differentiability at \( x = 2 \)

Next, we check the derivatives at \( x = 2 \):
\[ f'(x) = a \quad \text{for } -\infty < x < 2 \]
\[ f'(x) = 2x - 5 \quad \text{for } 2 < x < 3 \]

Setting the derivatives equal at \( x = 2 \):
\[ a = 2(2) - 5 = 4 - 5 = -1 \quad \text{(3)} \]

### Step 4: Ensure differentiability at \( x = 3 \)

Now we check the derivatives at \( x = 3 \):
\[ f'(x) = 2x - 5 \quad \text{for } 2 < x < 3 \]
\[ f'(x) = 2px + q \quad \text{for } 3 \leq x < \infty \]

Setting the derivatives equal at \( x = 3 \):
\[ 2(3) - 5 = 2p(3) + q \]
\[ 6 - 5 = 6p + q \quad \Rightarrow \quad 1 = 6p + q \quad \text{(4)} \]

### Step 5: Solve the equations

Now we have the following equations to solve:
1. \( 2a + b = 0 \)
2. \( 9p + 3q + 1 = 0 \)
3. \( a = -1 \)
4. \( 6p + q = 1 \)

Substituting \( a = -1 \) into equation (1):
\[ 2(-1) + b = 0 \quad \Rightarrow \quad -2 + b = 0 \quad \Rightarrow \quad b = 2 \]

Now substituting \( p \) and \( q \) into equations (2) and (4):
From equation (4):
\[ q = 1 - 6p \]

Substituting into equation (2):
\[ 9p + 3(1 - 6p) + 1 = 0 \]
\[ 9p + 3 - 18p + 1 = 0 \]
\[ -9p + 4 = 0 \quad \Rightarrow \quad 9p = 4 \quad \Rightarrow \quad p = \frac{4}{9} \]

Now substituting \( p \) back to find \( q \):
\[ q = 1 - 6\left(\frac{4}{9}\right) = 1 - \frac{24}{9} = 1 - \frac{8}{3} = \frac{3}{3} - \frac{8}{3} = -\frac{5}{3} \]

### Final Values
Thus, we have:
- \( a = -1 \)
- \( b = 2 \)
- \( p = \frac{4}{9} \)
- \( q = -\frac{5}{3} \)