If `f(x) = {(px + q, :x le 2),(x^2 - 5x + 6, : 2 < x < 3),(ax^2 + bx + 1, : x ge 3):}`
is differentiable everywhere, then `|p| + |q| + |1/a| + |1/b|` is equal to

Answer

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

\[ f(x) = \begin{cases} px + q & \text{if } x \leq 2 \\ x^2 - 5x + 6 & \text{if } 2 < x < 3 \\ ax^2 + bx + 1 & \text{if } x \geq 3 \end{cases} \]

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

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

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

Calculating the left-hand limit:

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

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:

\[ 2p + q = 0 \quad \text{(1)} \]

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

For continuity at \( x = 3 \):

\[ \lim_{x \to 3^-} f(x) = \lim_{x \to 3^+} f(x) \]

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:

\[ \lim_{x \to 3^+} f(x) = a(3^2) + b(3) + 1 = 9a + 3b + 1 \]

Setting these equal gives us:

\[ 9a + 3b + 1 = 0 \quad \text{(2)} \]

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

For \( f(x) \) to be differentiable at \( x = 2 \):

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

Calculating the left-hand derivative:

\[ f'(x) = p \quad \text{for } x < 2 \]

Calculating the right-hand derivative:

\[ f'(x) = 2x - 5 \quad \text{for } 2 < x < 3 \]

Evaluating at \( x = 2 \):

\[ \lim_{x \to 2^+} f'(x) = 2(2) - 5 = 4 - 5 = -1 \]

Setting these equal gives us:

\[ p = -1 \quad \text{(3)} \]

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

Substituting \( p = -1 \) into equation (1):

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

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

For differentiability at \( x = 3 \):

\[ \lim_{x \to 3^-} f'(x) = \lim_{x \to 3^+} f'(x) \]

Calculating the left-hand derivative:

\[ f'(x) = 2x - 5 \quad \text{for } 2 < x < 3 \]

Evaluating at \( x = 3 \):

\[ \lim_{x \to 3^-} f'(x) = 2(3) - 5 = 6 - 5 = 1 \]

Calculating the right-hand derivative:

\[ f'(x) = 2ax + b \quad \text{for } x \geq 3 \]

Evaluating at \( x = 3 \):

\[ \lim_{x \to 3^+} f'(x) = 2a(3) + b = 6a + b \]

Setting these equal gives us:

\[ 6a + b = 1 \quad \text{(5)} \]

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

From equation (2):

\[ 9a + 3b + 1 = 0 \implies 3b = -9a - 1 \implies b = -3a - \frac{1}{3} \]

Substituting \( b \) into equation (5):

\[ 6a + (-3a - \frac{1}{3}) = 1 \]

This simplifies to:

\[ 3a - \frac{1}{3} = 1 \implies 3a = 1 + \frac{1}{3} = \frac{4}{3} \implies a = \frac{4}{9} \quad \text{(6)} \]

### Step 7: Substitute \( a \) back to find \( b \)

Substituting \( a = \frac{4}{9} \) into \( b = -3a - \frac{1}{3} \):

\[ b = -3\left(\frac{4}{9}\right) - \frac{1}{3} = -\frac{12}{9} - \frac{3}{9} = -\frac{15}{9} = -\frac{5}{3} \quad \text{(7)} \]

### Step 8: Calculate \( |p| + |q| + \frac{1}{|a|} + \frac{1}{|b|} \)

Now we have:

- \( p = -1 \) so \( |p| = 1 \)
- \( q = 2 \) so \( |q| = 2 \)
- \( a = \frac{4}{9} \) so \( \frac{1}{|a|} = \frac{9}{4} \)
- \( b = -\frac{5}{3} \) so \( \frac{1}{|b|} = \frac{3}{5} \)

Now we can compute:

\[ |p| + |q| + \frac{1}{|a|} + \frac{1}{|b|} = 1 + 2 + \frac{9}{4} + \frac{3}{5} \]

Finding a common denominator (20):

\[ = 1 + 2 + \frac{45}{20} + \frac{12}{20} = 3 + \frac{57}{20} = \frac{60}{20} + \frac{57}{20} = \frac{117}{20} \]

Thus, the final answer is:

\[ \frac{117}{20} \]

### Final Answer

\[ |p| + |q| + \frac{1}{|a|} + \frac{1}{|b|} = \frac{117}{20} \]