If `f (x) = (x ^(2) - 3x +4)/(x ^(2)+ 3x +4),` then complete solution of `0lt f (x) lt 1,` is :

Options

A. `(-oo, oo)`
B. `(0,oo)`
C. ` (-oo,0)`
D. ` (0,1) uu (2, oo)`

Answer

To solve the inequality \( 0 < f(x) < 1 \) for the function \( f(x) = \frac{x^2 - 3x + 4}{x^2 + 3x + 4} \), we will break it down into two parts: solving \( f(x) > 0 \) and \( f(x) < 1 \).

### Step 1: Solve \( f(x) > 0 \)

The function \( f(x) \) is positive when the numerator is positive since the denominator is always positive (as we will see later).

1. **Numerator**: \( x^2 - 3x + 4 > 0 \)

To find the roots of the quadratic equation \( x^2 - 3x + 4 = 0 \):

\[ D = b^2 - 4ac = (-3)^2 - 4 \cdot 1 \cdot 4 = 9 - 16 = -7 \]

Since the discriminant \( D < 0 \), the quadratic has no real roots and opens upwards (as the coefficient of \( x^2 \) is positive). Thus, \( x^2 - 3x + 4 > 0 \) for all \( x \).

### Step 2: Solve \( f(x) < 1 \)

Now we need to solve the inequality \( f(x) < 1 \):

\[ \frac{x^2 - 3x + 4}{x^2 + 3x + 4} < 1 \]

2. **Rearranging the inequality**:

Subtract 1 from both sides:

\[ \frac{x^2 - 3x + 4}{x^2 + 3x + 4} - 1 < 0 \]

This simplifies to:

\[ \frac{x^2 - 3x + 4 - (x^2 + 3x + 4)}{x^2 + 3x + 4} < 0 \]

Simplifying the numerator:

\[ x^2 - 3x + 4 - x^2 - 3x - 4 = -6x \]

Thus, the inequality becomes:

\[ \frac{-6x}{x^2 + 3x + 4} < 0 \]

3. **Analyzing the denominator**:

The denominator \( x^2 + 3x + 4 \) can be checked for positivity:

\[ D = 3^2 - 4 \cdot 1 \cdot 4 = 9 - 16 = -7 \]

Since the discriminant is negative, \( x^2 + 3x + 4 > 0 \) for all \( x \).

4. **Finding the sign of the fraction**:

The inequality \( \frac{-6x}{x^2 + 3x + 4} < 0 \) implies that \( -6x < 0 \), which leads to:

\[ x > 0 \]

### Step 3: Combine the results

Since \( f(x) > 0 \) for all \( x \) and \( f(x) < 1 \) when \( x > 0 \), we conclude:

\[ 0 < f(x) < 1 \text{ for } x > 0 \]

### Final Answer

The complete solution of \( 0 < f(x) < 1 \) is:

\[ x \in (0, \infty) \]