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:

\[ D = (-4)^2 - 4(1)(9) = 16 - 36 = -20. \]

Since \( D < 0 \), \( g(x) > 0 \ \forall x \in \mathbb{R} \).

For \( f(x) \), consider:

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

Evaluate \( f(x) \) at specific points:

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

Since \( f(x) \) takes the same value at two different points \((-5 \text{ and } 3)\), \( f(x) \) is many-one.

Next, find the range of \( f(x) \):

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

Rearrange:

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

For \( f(x) \) to be real, the discriminant of the quadratic in \( x \) must satisfy:

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

Simplify:

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

Expanding and simplifying:

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

Factorize:

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

For \( D \geq 0 \), solve:

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

This gives:

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

Thus, the range of \( f(x) \) is:

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

If the function is defined from \( f : \mathbb{R} \to \mathbb{R} \), then the only correct answer is option (3).

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