If the function `f(x) = {{:(a|pi - x|+1"," x le 5),(" is continuous at "),(b|x-pi|+3","xgt5):}`
x = 5, then the value of a - b is

Options

A. `(2)/(pi - 5)`
B. `(2)/(pi + 5)`
C. `(2)/(5 - pi)`
D. `(-2)/(pi + 5)`

Explanation

To solve the problem, we need to ensure that the function \( f(x) \) is continuous at \( x = 5 \). This means that the left-hand limit (LHL), the right-hand limit (RHL), and the function value at that point must all be equal.

### Step 1: Find the function value at \( x = 5 \)
The function is defined as:
\[ f(x) = \begin{cases} a | \pi - x | + 1 & \text{if } x \leq 5 \\ b | x - \pi | + 3 & \text{if } x > 5 \end{cases} \]
At \( x = 5 \):
\[ f(5) = a | \pi - 5 | + 1 \]
Since \( \pi - 5 \) is negative, we have:
\[ | \pi - 5 | = 5 - \pi \]
Thus,
\[ f(5) = a(5 - \pi) + 1 \]

### Step 2: Find the Left-Hand Limit (LHL) as \( x \) approaches 5
For \( x < 5 \):
\[ \text{LHL} = a | \pi - x | + 1 \]
As \( x \) approaches 5 from the left:
\[ \text{LHL} = a | \pi - 5 | + 1 = a(5 - \pi) + 1 \]

### Step 3: Find the Right-Hand Limit (RHL) as \( x \) approaches 5
For \( x > 5 \):
\[ \text{RHL} = b | x - \pi | + 3 \]
As \( x \) approaches 5 from the right:
\[ \text{RHL} = b | 5 - \pi | + 3 \]
Since \( 5 - \pi \) is negative, we have:
\[ | 5 - \pi | = \pi - 5 \]
Thus,
\[ \text{RHL} = b(\pi - 5) + 3 \]

### Step 4: Set LHL equal to RHL
Since the function is continuous at \( x = 5 \), we set LHL equal to RHL:
\[ a(5 - \pi) + 1 = b(\pi - 5) + 3 \]

### Step 5: Rearranging the equation
Rearranging gives us:
\[ a(5 - \pi) - b(5 - \pi) = 3 - 1 \]
\[ (a - b)(5 - \pi) = 2 \]

### Step 6: Solve for \( a - b \)
Dividing both sides by \( 5 - \pi \):
\[ a - b = \frac{2}{5 - \pi} \]

### Final Step: Find the value of \( a - b \)
Thus, the value of \( a - b \) is:
\[ \boxed{\frac{2}{5 - \pi}} \]