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

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

### Step 1: Find the limit as \( x \) approaches 1

We first find 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 2: Simplify the denominator

The denominator can be factored:

\[ 2(x^2 - 1) = 2(x - 1)(x + 1). \]

### Step 3: Substitute \( x = 1 \) into the numerator

Now we substitute \( x = 1 \) into the numerator:

\[ x^2 + 3x + p = 1^2 + 3(1) + p = 1 + 3 + p = 4 + p. \]

### Step 4: Set up the limit expression

Thus, we have:

\[ \lim_{x \to 1} f(x) = \frac{4 + p}{2(1 - 1)(1 + 1)} = \frac{4 + p}{0}. \]

Since the limit must exist, the numerator must also equal zero when \( x = 1 \):

\[ 4 + p = 0. \]

### Step 5: Solve for \( p \)

Solving for \( p \):

\[ p = -4. \]

### Step 6: Verify the limit

Now we can verify that with \( p = -4 \), the limit is:

\[ \lim_{x \to 1} f(x) = \frac{4 - 4}{2(0)(2)} = \frac{0}{0}, \]

which is an indeterminate form. We can apply L'Hôpital's Rule or factor the numerator.

### Step 7: Factor the numerator

The numerator \( x^2 + 3x - 4 \) can be factored as:

\[ (x - 1)(x + 4). \]

Thus,

\[ f(x) = \frac{(x - 1)(x + 4)}{2(x - 1)(x + 1)} \quad \text{for } x \neq 1. \]

### Step 8: Cancel the common factor

Canceling \( (x - 1) \):

\[ f(x) = \frac{x + 4}{2(x + 1)} \quad \text{for } x \neq 1. \]

### Step 9: Evaluate the limit again

Now, we can find the limit as \( x \) approaches 1:

\[ \lim_{x \to 1} f(x) = \frac{1 + 4}{2(1 + 1)} = \frac{5}{4}. \]

### Conclusion

Since

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

the function is continuous at \( x = 1 \) when \( p = -4 \).

Thus, the value of \( p \) is

\[ \boxed{-4}. \]