The function `f:R rarr R` defined as `f(x)=(3x^2+3x-4)/(3+3x-4x^2)` is :

Options

A. One ot one buty not onto
B. Onto but not one to one
C. Both one to one and onto
D. Neither one to one nor onto

Answer

To analyze the function \( f(x) = \frac{3x^2 + 3x - 4}{3 + 3x - 4x^2} \), we will determine whether it is one-one (injective) and onto (surjective).

### Step 1: Check if the function is one-one

To check if the function is one-one, we need to see if \( f(x_1) = f(x_2) \) implies that \( x_1 = x_2 \).

Assume \( f(x) = 0 \):
\[ f(x) = 0 \implies \frac{3x^2 + 3x - 4}{3 + 3x - 4x^2} = 0 \]
This implies that the numerator must be zero:
\[ 3x^2 + 3x - 4 = 0 \]

### Step 2: Solve the quadratic equation

We can use the quadratic formula \( x = \frac{-b \pm \sqrt{D}}{2a} \) where \( D = b^2 - 4ac \).

Here, \( a = 3 \), \( b = 3 \), and \( c = -4 \):
\[ D = 3^2 - 4 \cdot 3 \cdot (-4) = 9 + 48 = 57 \]
Since \( D > 0 \), the quadratic equation has two distinct roots.

### Step 3: Conclusion about one-one property

Since there are two distinct values \( x_1 \) and \( x_2 \) such that \( f(x_1) = f(x_2) = 0 \), the function is not one-one (many-to-one).

### Step 4: Check if the function is onto

To check if the function is onto, we need to see if for every \( y \in \mathbb{R} \), there exists an \( x \in \mathbb{R} \) such that \( f(x) = y \).

Assume:
\[ f(x) = y \implies \frac{3x^2 + 3x - 4}{3 + 3x - 4x^2} = y \]
Cross-multiplying gives:
\[ 3x^2 + 3x - 4 = y(3 + 3x - 4x^2) \]
Rearranging this leads to:
\[ (4y - 3)x^2 + (3y - 3)x + (3 - 4) = 0 \]
This is a quadratic equation in \( x \).

### Step 5: Find the discriminant

To ensure that there are real solutions for \( x \), we need the discriminant of this quadratic to be non-negative:
\[ D = (3y - 3)^2 - 4(4y - 3)(-1) \]
Calculating this:
\[ D = (3y - 3)^2 + 4(4y - 3) \]
Expanding:
\[ D = 9y^2 - 18y + 9 + 16y - 12 = 9y^2 - 2y - 3 \]
We need \( D \geq 0 \).

### Step 6: Analyze the quadratic in \( y \)

The quadratic \( 9y^2 - 2y - 3 \) has a discriminant:
\[ D' = (-2)^2 - 4 \cdot 9 \cdot (-3) = 4 + 108 = 112 \]
Since \( D' > 0 \), the quadratic can take all real values, which means there are real solutions for every \( y \).

### Conclusion about onto property

Since the discriminant is non-negative for all \( y \), the function is onto.

### Final Conclusion

The function \( f(x) \) is many-to-one and onto.