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

Options

A. `(71)/(10)`
B. `(51)/(10)`
C. `(33)/(5)`
D. `(31)/(5)`

Answer

To solve the problem, we need to ensure that the function \( f(x) \) is continuous and differentiable at the points \( x = 2 \) and \( x = 3 \). The function is defined piecewise 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: Continuity at \( x = 2 \)

To ensure continuity at \( x = 2 \), we need:

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

Calculating \( f(2) \):

\[ f(2) = p(2) + q = 2p + q \]

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

\[ \lim_{x \to 2^+} f(x) = 2^2 - 5(2) + 6 = 4 - 10 + 6 = 0 \]

Setting the limits equal for continuity:

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

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

For continuity at \( x = 3 \):

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

Calculating \( f(3) \):

\[ f(3) = a(3^2) + b(3) + 1 = 9a + 3b + 1 \]

Calculating \( \lim_{x \to 3^-} f(x) \):

\[ \lim_{x \to 3^-} f(x) = 3^2 - 5(3) + 6 = 9 - 15 + 6 = 0 \]

Setting the limits equal for continuity:

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

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

For differentiability at \( x = 2 \):

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

Calculating \( f'(2^-) \):

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

Calculating \( f'(2^+) \):

\[ f'(x) = 2x - 5 \quad \text{(for } 2 < x < 3\text{)} \]
\[ f'(2^+) = 2(2) - 5 = 4 - 5 = -1 \]

Setting the derivatives equal for differentiability:

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

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

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

\[ 2(-1) + q = 0 \implies -2 + q = 0 \implies q = 2 \]

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

For differentiability at \( x = 3 \):

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

Calculating \( f'(3^-) \):

\[ f'(3^-) = 2(3) - 5 = 6 - 5 = 1 \]

Calculating \( f'(3^+) \):

\[ f'(x) = 2ax + b \quad \text{(for } x \geq 3\text{)} \]
\[ f'(3^+) = 6a + b \]

Setting the derivatives equal for differentiability:

\[ 1 = 6a + b \quad \text{(4)} \]

### Step 6: Substitute \( b \) from Equation (4) into Equation (2)

From equation (2):

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

Substituting into equation (4):

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

### Step 7: Find \( b \)

Substituting \( a \) back into the equation for \( b \):

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

### Step 8: Calculate \( |p| + |q| + |1/a| + |1/b| \)

Now we have:

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

Calculating the total:

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

Finding a common denominator (which is 20):

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

### Final Answer

Thus, the final answer is:

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