If `f(x)= {:{((x^(2) + 3x+p)/(2(x^(2)-1)) , xne 1),(5/4, x = 1):}` is continuous at x =1 then

Options

A. p =2
B. p=0
C. p=-4
D. None of these

Answer

To determine the value of \( p \) such that the function

\[ f(x) = \begin{cases} \frac{x^2 + 3x + p}{2(x^2 - 1)} & \text{if } x \neq 1 \\ \frac{5}{4} & \text{if } x = 1 \end{cases} \]

is continuous at \( x = 1 \), we need to ensure that the limit of \( f(x) \) as \( x \) approaches 1 equals \( f(1) \).

### Step 1: Set up the condition for continuity
For \( f(x) \) to be continuous at \( x = 1 \), we need:

\[ \lim_{x \to 1} f(x) = f(1) \]

Given that \( f(1) = \frac{5}{4} \), we can write:

\[ \lim_{x \to 1} f(x) = \frac{5}{4} \]

### Step 2: Calculate the limit
We will calculate the limit of \( f(x) \) as \( x \) approaches 1:

\[ \lim_{x \to 1} f(x) = \lim_{x \to 1} \frac{x^2 + 3x + p}{2(x^2 - 1)} \]

### Step 3: Substitute \( x = 1 \)
Substituting \( x = 1 \) directly into the limit gives:

\[ \frac{1^2 + 3(1) + p}{2(1^2 - 1)} = \frac{1 + 3 + p}{2(0)} = \frac{4 + p}{0} \]

Since the denominator approaches 0, the limit will be undefined unless the numerator also approaches 0. Thus, we need:

\[ 4 + p = 0 \]

### Step 4: Solve for \( p \)
From the equation \( 4 + p = 0 \):

\[ p = -4 \]

### Step 5: Conclusion
Thus, the value of \( p \) that makes \( f(x) \) continuous at \( x = 1 \) is:

\[ \boxed{-4} \]