Let the given function is differentiable at x = 1.
`f(x)={{:(lim_(nrarroo) (ax(x-1)(cot.(pix)/(4))^(n)+(px^(2)+2))/((cot,(pix)/(4))^(n)+1)",",x in(0,1)uu(1,2)),(0",",x=1):}`
Then the value of `|a+p|` is

Options

A. 4
B. 6
C. 8
D. 10

Explanation

To solve the problem, we need to analyze the given function and find the values of \( a \) and \( p \) such that the function is differentiable at \( x = 1 \).

### Step 1: Analyze the function
The function is given as:
\[ f(x) = \begin{cases} \lim_{n \to \infty} \frac{a x (x-1) \left( \cot\left(\frac{\pi}{4}\right) \right)^n + p x^2 + 2}{\left( \cot\left(\frac{\pi}{4}\right) \right)^n + 1}, & x \in (0, 1) \cup (1, 2) \\ 0, & x = 1 \end{cases} \]

### Step 2: Evaluate the limit as \( n \to \infty \)
1. For \( x > 1 \):
- The term \( x^n \) approaches infinity, hence the limit becomes:
\[ f(x) \to \frac{a x (x-1) \cdot \infty + p x^2 + 2}{\infty + 1} = \infty \]
Thus, \( f(x) \) approaches infinity for \( x > 1 \).

2. For \( x < 1 \):
- The term \( x^n \) approaches 0, hence the limit becomes:
\[ f(x) \to \frac{0 + p x^2 + 2}{1} = p x^2 + 2 \]

### Step 3: Continuity at \( x = 1 \)
For \( f(x) \) to be continuous at \( x = 1 \):
\[ \lim_{x \to 1^-} f(x) = f(1) = 0 \]
Thus,
\[ p(1^2) + 2 = 0 \implies p + 2 = 0 \implies p = -2 \]

### Step 4: Evaluate the left-hand derivative (LHD) at \( x = 1 \)
We need to find:
\[ LHD = \lim_{h \to 0} \frac{f(1-h) - f(1)}{-h} \]
For \( x = 1 - h \) (where \( h \to 0 \)):
\[ f(1-h) = p(1-h)^2 + 2 \]
Substituting \( p = -2 \):
\[ f(1-h) = -2(1 - h)^2 + 2 = -2(1 - 2h + h^2) + 2 = -2 + 4h - 2h^2 + 2 = 4h - 2h^2 \]
Thus,
\[ LHD = \lim_{h \to 0} \frac{(4h - 2h^2) - 0}{-h} = \lim_{h \to 0} \frac{4h - 2h^2}{-h} = \lim_{h \to 0} (-4 + 2h) = -4 \]

### Step 5: Evaluate the right-hand derivative (RHD) at \( x = 1 \)
We need to find:
\[ RHD = \lim_{h \to 0} \frac{f(1+h) - f(1)}{h} \]
For \( x = 1 + h \):
\[ f(1+h) = 0 \text{ (since it approaches infinity)} \]
Thus,
\[ RHD = \lim_{h \to 0} \frac{0 - 0}{h} = 0 \]

### Step 6: Set LHD equal to RHD
Since \( f(x) \) is differentiable at \( x = 1 \), we set:
\[ LHD = RHD \implies -4 = 0 \text{ (which is not possible)} \]
This indicates that we need to check the values of \( a \).

### Step 7: Find \( a \)
From the left-hand side, we have:
\[ LHD = a \]
Thus, we have:
\[ a = -4 \]

### Step 8: Find \( |a + p| \)
Now, we have:
\[ a = -4, \quad p = -2 \]
Thus,
\[ |a + p| = |-4 - 2| = |-6| = 6 \]

### Final Answer
The value of \( |a + p| \) is \( \boxed{6} \).