Let `f:A to B f (x) = (x +a)/(bx ^(2) + cx +2),` where A represent domain set and B represent range set of function `f (x)` a,b,c `inR, f (-1)=0 and y=1` is an asymptote of `y =f (x) and y=g (x)` is the inverse of `f (x).`
g (0) is equal to :

Options

A. `-1`
B. `-3`
C. `-5/2`
D. `-3/2`

Answer

To solve the problem step by step, we need to analyze the function \( f(x) = \frac{x + a}{bx^2 + cx + 2} \) and use the given conditions to find the value of \( g(0) \), where \( g(x) \) is the inverse of \( f(x) \).

### Step 1: Use the condition \( f(-1) = 0 \)

We start by substituting \( x = -1 \) into the function:

\[ f(-1) = \frac{-1 + a}{b(-1)^2 + c(-1) + 2} = 0 \]

This implies:

\[ \frac{-1 + a}{b - c + 2} = 0 \]

For the fraction to equal zero, the numerator must be zero:

\[ -1 + a = 0 \implies a = 1 \]

### Step 2: Substitute \( a \) back into \( f(x) \)

Now we substitute \( a = 1 \) into the function:

\[ f(x) = \frac{x + 1}{bx^2 + cx + 2} \]

### Step 3: Analyze the asymptote condition \( y = 1 \)

We know that \( y = 1 \) is a horizontal asymptote of \( f(x) \). For a rational function, the horizontal asymptote can be determined by comparing the degrees of the numerator and denominator.

Since the degree of the numerator (1) is less than the degree of the denominator (2), we need to adjust the degrees. Thus, we set \( b = 0 \) to make the degrees equal:

\[ f(x) = \frac{x + 1}{cx + 2} \]

### Step 4: Find the limit as \( x \to \infty \)

To satisfy the asymptote condition \( y = 1 \):

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

Calculating this limit:

\[ \lim_{x \to \infty} \frac{x + 1}{cx + 2} = \lim_{x \to \infty} \frac{1 + \frac{1}{x}}{c + \frac{2}{x}} = \frac{1}{c} \]

Setting this equal to 1 gives:

\[ \frac{1}{c} = 1 \implies c = 1 \]

### Step 5: Final form of \( f(x) \)

Now we have:

\[ a = 1, \quad b = 0, \quad c = 1 \]

Thus, the function simplifies to:

\[ f(x) = \frac{x + 1}{x + 2} \]

### Step 6: Find the inverse \( g(x) \)

To find \( g(x) \), we set \( f(x) = y \):

\[ y = \frac{x + 1}{x + 2} \]

Cross-multiplying gives:

\[ y(x + 2) = x + 1 \implies yx + 2y = x + 1 \]

Rearranging terms:

\[ yx - x = 1 - 2y \implies x(y - 1) = 1 - 2y \]

Thus, we find \( x \):

\[ x = \frac{1 - 2y}{y - 1} \]

Therefore, the inverse function is:

\[ g(y) = \frac{1 - 2y}{y - 1} \]

### Step 7: Calculate \( g(0) \)

Now we need to find \( g(0) \):

\[ g(0) = \frac{1 - 2(0)}{0 - 1} = \frac{1}{-1} = -1 \]

### Final Answer

Thus, the value of \( g(0) \) is:

\[ \boxed{-1} \]