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

Answer: (C)

neither one-one nor onto

Answer 2

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.