Let `f : R rightarrow R` is defined by `f(x) = (x^2 + 2x – 15)/(x^2 – 4x + 19)` then is

Options

A. one-one and onto
B. one-one into
C. many-one and onto
D. many-one and into

Answer

To determine the nature of the function \( f(x) = \frac{x^2 + 2x - 15}{x^2 - 4x + 19} \), we will analyze its properties such as whether it is one-one, onto, many-one, or into.

### Step-by-step Solution:

1. **Identify the function**:
We have \( f(x) = \frac{x^2 + 2x - 15}{x^2 - 4x + 19} \).

2. **Factor the numerator**:
The numerator can be factored as follows:
\[ x^2 + 2x - 15 = (x + 5)(x - 3) \]
So, we rewrite \( f(x) \):
\[ f(x) = \frac{(x + 5)(x - 3)}{x^2 - 4x + 19} \]

3. **Analyze the denominator**:
The denominator \( x^2 - 4x + 19 \) can be analyzed for its nature. The discriminant \( D \) of the quadratic is:
\[ D = (-4)^2 - 4 \cdot 1 \cdot 19 = 16 - 76 = -60 \]
Since \( D < 0 \), the denominator does not have real roots and is always positive.

4. **Find the first derivative \( f'(x) \)**:
We will use the quotient rule to find the derivative:
\[ f'(x) = \frac{(2x + 2)(x^2 - 4x + 19) - (x^2 + 2x - 15)(2x - 4)}{(x^2 - 4x + 19)^2} \]
Simplifying the numerator:
- Let \( u = x^2 + 2x - 15 \) and \( v = x^2 - 4x + 19 \).
- The derivative becomes:
\[ f'(x) = \frac{(2x + 2)v - u(2x - 4)}{v^2} \]

5. **Determine the sign of \( f'(x) \)**:
To check if \( f'(x) \) is always positive or negative, we can evaluate it at specific points. For instance, substituting \( x = 0 \):
\[ f'(0) = \frac{(2)(19) - (-15)(-4)}{(19)^2} = \frac{38 - 60}{361} = \frac{-22}{361} < 0 \]
This indicates that \( f'(x) \) can be negative.

6. **Check for critical points**:
We can also check for other values of \( x \) to see if \( f'(x) \) changes sign. If it does, it indicates that the function is not one-one.

7. **Conclusion on the function's nature**:
Since \( f'(x) \) is not consistently positive or negative, \( f(x) \) is not a one-one function. Additionally, since the function does not cover all real numbers (it has horizontal asymptotes), it is not onto.

Therefore, we conclude that \( f(x) \) is a many-one function and not onto.

### Final Answer:
The function \( f(x) \) is a many-one function and not onto.