If the function `f(x)={{:(ax+b, -ooltxle2),(x^(2)-5x+6,2ltxlt3),(px^(2)+qx+1,3lexleoo):}` 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)`

Explanation

To solve the problem, we need to ensure that the function \( f(x) \) is continuous and differentiable at the points where the piecewise function changes, specifically at \( x = 2 \) and \( x = 3 \).

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

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

To ensure continuity at \( x = 2 \), we need:
\[ \lim_{x \to 2^-} f(x) = \lim_{x \to 2^+} f(x) = f(2) \]

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

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

Setting the limits equal for continuity:
\[ 2a + b = 0 \quad \text{(1)} \]

### 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 \( x \geq 3 \)

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

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

Setting the limits equal for continuity:
\[ 9p + 3q + 1 = 0 \quad \text{(2)} \]

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

For differentiability at \( x = 2 \), we need:
\[ f'(2^-) = f'(2^+) \]

Calculating the left-hand derivative:
\[ f'(x) = a \quad \text{for } x < 2 \]

Calculating the right-hand derivative:
\[ f'(x) = 2x - 5 \quad \text{for } 2 < x < 3 \]
\[ f'(2^+) = 2(2) - 5 = 4 - 5 = -1 \]

Setting the derivatives equal:
\[ a = -1 \quad \text{(3)} \]

### Step 4: Substitute \( a \) into Equation (1)

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

### Step 5: Ensure Differentiability at \( x = 3 \)

For differentiability at \( x = 3 \):
\[ f'(3^-) = f'(3^+) \]

Calculating the left-hand derivative:
\[ f'(3^-) = 2(3) - 5 = 6 - 5 = 1 \]

Calculating the right-hand derivative:
\[ f'(x) = 2px + q \quad \text{for } x \geq 3 \]
\[ f'(3^+) = 2p(3) + q = 6p + q \]

Setting the derivatives equal:
\[ 1 = 6p + q \quad \text{(4)} \]

### Step 6: Substitute \( q \) from Equation (2)

From equation (2), we have:
\[ 9p + 3q + 1 = 0 \implies 3q = -9p - 1 \implies q = -3p - \frac{1}{3} \]

Substituting \( q \) into equation (4):
\[ 1 = 6p + (-3p - \frac{1}{3}) \implies 1 = 3p - \frac{1}{3} \]
\[ 3p = 1 + \frac{1}{3} = \frac{4}{3} \implies p = \frac{4}{9} \]

### Step 7: Substitute \( p \) into \( q \)

Now substituting \( p = \frac{4}{9} \) into \( q = -3p - \frac{1}{3} \):
\[ q = -3 \left(\frac{4}{9}\right) - \frac{1}{3} = -\frac{12}{9} - \frac{3}{9} = -\frac{15}{9} = -\frac{5}{3} \]

### Final Values

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

### Conclusion

The values of \( a, b, p, \) and \( q \) are:
- \( a = -1 \)
- \( b = 2 \)
- \( p = \frac{4}{9} \)
- \( q = -\frac{5}{3} \)