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.

Answer 2

To analyze the function \( f(x) = \frac{x^2 + 2x - 15}{x^2 - 4x + 9} \), we need to determine whether it is one-one and/or onto.

Step 1: Check if \( f(x) \) is One-One

A function \( f(x) \) is one-one (injective) if for any two distinct elements \( x_1 \) and \( x_2 \) in the domain, the images \( f(x_1) \) and \( f(x_2) \) are distinct. Mathematically, \( f(x_1) = f(x_2) \implies x_1 = x_2 \).

Consider \( f(x_1) = f(x_2) \):

\(\frac{x_1^2 + 2x_1 - 15}{x_1^2 - 4x_1 + 9} = \frac{x_2^2 + 2x_2 - 15}{x_2^2 - 4x_2 + 9}\)

Cross-multiplying yields:

\((x_1^2 + 2x_1 - 15)(x_2^2 - 4x_2 + 9) = (x_2^2 + 2x_2 - 15)(x_1^2 - 4x_1 + 9)\)

This results in a complex polynomial without clear reduction to \( x_1 = x_2 \), suggesting non-injectivity. To confirm, test specific values:

• For \( x_1 = 0 \) and \( x_2 = 1 \):
• \( f(0) = \frac{0^2 + 2(0) - 15}{0^2 - 4(0) + 9} = \frac{-15}{9} = -\frac{5}{3} \)
• \( f(1) = \frac{1^2 + 2(1) - 15}{1^2 - 4(1) + 9} = \frac{1 + 2 - 15}{1 - 4 + 9} = \frac{-12}{6} = -2 \)
• If for \( x_1 \neq x_2 \), \( f(x_1) = f(x_2) \), the function is not one-one. The test confirms different values.

Step 2: Check if \( f(x) \) is Onto

A function \( f(x) \) is onto (surjective) if for every real number \( y \), there is some real \( x \) such that \( f(x) = y \).

Simplifying to check range:

Rearrange: \( y(x^2 - 4x + 9) = x^2 + 2x - 15 \)

Rewriting: \( x^2(y - 1) + x(2 + 4y) + (9y + 15) = 0 \)

This is a quadratic in \( x \). For solutions to exist for each \( y \), \( (2 + 4y)^2 - 4(y - 1)(9y + 15) \) must be non-negative, implying limited \( y \).

As this is not always possible, the function is not onto.

Conclusion

The function \( f(x) = \frac{x^2 + 2x - 15}{x^2 - 4x + 9} \) is neither one-one nor onto.