Let `f (x)= {{:(ax (x-1)+b,,, x lt 1),( x+2,,, 1 le x le 3),(px ^(2) +qx +2,,, x gt 3):}` is continous `AA x in R` except `x =1` but `|f (x)|` is differentiable everywhere and `f '(x)` is continous at `x=3 and |a+p+q|=k,` then k=

Answer

To solve the problem, we need to analyze the piecewise function given and ensure continuity and differentiability conditions are satisfied. Let's break down the solution step by step.

### Step 1: Analyze the function at \( x = 3 \)

The function is given as:
\[ f(x) = \begin{cases} ax(x-1) + b & \text{if } x < 1 \\ x + 2 & \text{if } 1 \leq x \leq 3 \\ px^2 + qx + 2 & \text{if } x > 3 \end{cases} \]

Since \( f(x) \) is continuous at \( x = 3 \), we need to check the left-hand limit and right-hand limit at this point.

**Left-hand limit as \( x \) approaches 3:**
\[ \lim_{x \to 3^-} f(x) = 3 + 2 = 5 \]

**Right-hand limit as \( x \) approaches 3:**
\[ \lim_{x \to 3^+} f(x) = p(3^2) + q(3) + 2 = 9p + 3q + 2 \]

Setting these equal for continuity:
\[ 9p + 3q + 2 = 5 \]
This simplifies to:
\[ 9p + 3q = 3 \quad \text{(Equation 1)} \]

### Step 2: Differentiate the function and check at \( x = 3 \)

Next, we need to find the derivatives from both sides at \( x = 3 \).

**For \( x < 3 \):**
\[ f'(x) = 1 \quad \text{(since the derivative of \( x + 2 \) is 1)} \]

**For \( x > 3 \):**
\[ f'(x) = 2px + q \]
At \( x = 3 \):
\[ f'(3) = 6p + q \]

Setting the left-hand derivative equal to the right-hand derivative:
\[ 6p + q = 1 \quad \text{(Equation 2)} \]

### Step 3: Solve the equations

Now we have two equations:
1. \( 9p + 3q = 3 \)
2. \( 6p + q = 1 \)

From Equation 2, we can express \( q \) in terms of \( p \):
\[ q = 1 - 6p \]

Substituting \( q \) into Equation 1:
\[ 9p + 3(1 - 6p) = 3 \]
\[ 9p + 3 - 18p = 3 \]
\[ -9p = 0 \implies p = 0 \]

Now substituting \( p = 0 \) back into the expression for \( q \):
\[ q = 1 - 6(0) = 1 \]

### Step 4: Find \( a \)

Now we need to ensure that \( |f(x)| \) is differentiable everywhere, particularly at \( x = 1 \).

**Left-hand limit at \( x = 1 \):**
\[ \lim_{x \to 1^-} f(x) = a(1)(1-1) + b = b \]

**Right-hand limit at \( x = 1 \):**
\[ \lim_{x \to 1^+} f(x) = 1 + 2 = 3 \]

Setting these equal for continuity:
\[ b = 3 \]

Now we check the differentiability at \( x = 1 \):
\[ f'(x) = 2ax \quad \text{for } x < 1 \]
At \( x = 1 \):
\[ f'(1) = 2a \]

For \( x \geq 1 \):
\[ f'(x) = 1 \]

Setting these equal:
\[ 2a = 1 \implies a = \frac{1}{2} \]

### Step 5: Calculate \( k \)

Now, we have:
- \( a = \frac{1}{2} \)
- \( p = 0 \)
- \( q = 1 \)

Thus,
\[ |a + p + q| = \left|\frac{1}{2} + 0 + 1\right| = \left|\frac{3}{2}\right| = \frac{3}{2} \]

### Final Answer

Therefore, the value of \( k \) is:
\[ \boxed{\frac{3}{2}} \]