Let `f(x)=x^(3)-x^(2)-3x-1, g(x)=(x+1)a and h(x)=(f(x))/(g(x))` where h is a rational function such that
(i) It is continuous everywhere except when `x=-1`,
(ii) `lim_(xrarr -1)h(x)=(1)/(2)`.
The value of h(1) is

Options

A. `1//2`
B. `1//4`
C. `-1//2`
D. 1

Explanation

To solve the problem step by step, we will analyze the functions given and find the value of \( h(1) \).

### Step 1: Define the functions
We have:
- \( f(x) = x^3 - x^2 - 3x - 1 \)
- \( g(x) = (x + 1)a \)
- \( h(x) = \frac{f(x)}{g(x)} \)

### Step 2: Factor \( f(x) \)
To find the value of \( h(x) \), we first need to factor \( f(x) \). We can check if \( x + 1 \) is a factor by substituting \( x = -1 \):
\[ f(-1) = (-1)^3 - (-1)^2 - 3(-1) - 1 = -1 - 1 + 3 - 1 = 0 \]
Since \( f(-1) = 0 \), \( x + 1 \) is indeed a factor of \( f(x) \).

### Step 3: Perform polynomial long division
We divide \( f(x) \) by \( x + 1 \):
1. Divide \( x^3 \) by \( x \) to get \( x^2 \).
2. Multiply \( x^2 \) by \( x + 1 \) to get \( x^3 + x^2 \).
3. Subtract: \( f(x) - (x^3 + x^2) = -2x^2 - 3x - 1 \).
4. Divide \( -2x^2 \) by \( x \) to get \( -2x \).
5. Multiply \( -2x \) by \( x + 1 \) to get \( -2x^2 - 2x \).
6. Subtract: \( -2x^2 - 3x - 1 - (-2x^2 - 2x) = -x - 1 \).
7. Divide \( -x \) by \( x \) to get \( -1 \).
8. Multiply \( -1 \) by \( x + 1 \) to get \( -x - 1 \).
9. Subtract: \( -x - 1 - (-x - 1) = 0 \).

Thus, we have:
\[ f(x) = (x + 1)(x^2 - 2x - 1) \]

### Step 4: Rewrite \( h(x) \)
Now, substituting \( f(x) \) into \( h(x) \):
\[ h(x) = \frac{(x + 1)(x^2 - 2x - 1)}{(x + 1)a} \]
For \( x \neq -1 \), we can simplify:
\[ h(x) = \frac{x^2 - 2x - 1}{a} \]

### Step 5: Find the limit as \( x \to -1 \)
We are given that:
\[ \lim_{x \to -1} h(x) = \frac{1}{2} \]
Calculating the limit:
\[ \lim_{x \to -1} h(x) = \frac{(-1)^2 - 2(-1) - 1}{a} = \frac{1 + 2 - 1}{a} = \frac{2}{a} \]
Setting this equal to \(\frac{1}{2}\):
\[ \frac{2}{a} = \frac{1}{2} \]
Cross-multiplying gives:
\[ 4 = a \]

### Step 6: Calculate \( h(1) \)
Now that we have \( a = 4 \), we can find \( h(1) \):
\[ h(1) = \frac{1^2 - 2(1) - 1}{4} = \frac{1 - 2 - 1}{4} = \frac{-2}{4} = -\frac{1}{2} \]

### Final Answer
The value of \( h(1) \) is:
\[ \boxed{-\frac{1}{2}} \]