Question

Let \( f : \mathbb{R}_+ \to [-5, \infty) \) be defined as \( f(x) = 9x^2 + 6x - 5 \), where \( \mathbb{R}_+ \) is the set of all non-negative real numbers. Then, \( f \) is:

• A. one-one
• B. onto
• C. bijective
• D. neither one-one nor onto

Hint:

To confirm if a quadratic function is bijective, examine the function's monotonicity over the given domain and ensure the range covers all possible outputs.

Answer 1

The Correct Option is C

Step 1: Examine the function \( f(x) \).
The given function \( f(x) = 9x^2 + 6x - 5 \) is a quadratic function with a positive leading coefficient, indicating that the parabola opens upwards. Step 2: Determine the domain and range.
The domain of the function is \( \mathbb{R}_+ \), meaning that \( x \geq 0 \). To find the minimum value, we calculate the vertex of the parabola, which occurs at: \[ x = -\frac{b}{2a} = -\frac{6}{2 \cdot 9} = -\frac{1}{3}. \] However, since \( x \geq 0 \), we check \( f(x) \) at \( x = 0 \): \[ f(0) = -5. \] Thus, the range of \( f(x) \) is \( [-5, \infty) \), which shows that the function is onto. Step 3: Verify the one-one property.
Since the function is strictly increasing for \( x \geq 0 \), it satisfies the one-one property. Step 4: Conclusion.
Given that \( f(x) \) is both one-one and onto, the function is bijective.