Question

The function \( f(x) = \frac{x^2 + 2x - 15}{x^2 - 4x + 9} \), \( x \in \mathbb{R} \) is:

• A. both one-one and onto.
• B. onto but not one-one
• C. neither one-one nor onto
• D. one-one but not onto

Answer 1

The Correct Option is C

Let \( g(x) = x^2 - 4x + 9 \).

The discriminant of \( g(x) \) is calculated as \( D = (-4)^2 - 4(1)(9) = 16 - 36 = -20 \).

Since \( D<0 \), \( g(x)>0 \) for all real numbers \( x \).

Consider the function \( f(x) \):

\[ f(x) = \frac{(x + 5)(x - 3)}{x^2 - 4x + 9}. \]

Evaluating \( f(x) \) at specific points yields:

\[ f(-5) = 0, \quad f(3) = 0. \]

As \( f(x) \) yields the same value at two distinct points, \( -5 \) and \( 3 \), \( f(x) \) is a many-to-one function.

To determine the range of \( f(x) \), we set \( y = f(x) \):

\[ y \cdot (x^2 - 4x + 9) = x^2 + 2x - 15. \]

Rearranging this equation gives a quadratic in \( x \):

\[ x^2(y - 1) - 2x(2y + 1) + (9y + 15) = 0. \]

For \( f(x) \) to produce real values, the discriminant of this quadratic in \( x \) must be non-negative:

\[ D = 4(2y + 1)^2 - 4(y - 1)(9y + 15) \geq 0. \]

Simplifying the discriminant:

\[ D = 4 \left[(2y + 1)^2 - (y - 1)(9y + 15)\right] = 4 \left[4y^2 + 4y + 1 - (9y^2 + 6y - 15)\right] = 4 \left[-5y^2 - 2y + 16\right]. \]

Factoring the expression for \( D \):

\[ D = 4(-5y + 8)(y + 2). \]

The condition \( D \geq 0 \) implies:

\[ -5y + 8 \geq 0 \quad \text{and} \quad y + 2 \geq 0. \]

Solving these inequalities yields the range for \( y \):

\[ y \in \left[-2, \frac{8}{5}\right]. \]

Therefore, the range of \( f(x) \) is \( y \in \left[-2, \frac{8}{5}\right] \).

If the function is defined as \( f : \mathbb{R} \to \mathbb{R} \), option (3) is the correct choice.

\( f(x) \) is not onto. Consequently, \( f(x) \) is neither one-to-one nor onto.