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:

For quadratic functions, check monotonicity and evaluate the range over the given domain to verify one-one and onto properties.

Answer 1

The Correct Option is C

Step 1: Analyze \( f(x) \).
The function \( f(x) = 9x^2 + 6x - 5 \) is a quadratic equation, which opens upwards as the coefficient of \( x^2 \) is positive. Step 2: Domain and range.
The domain is \( \mathbb{R}_+ \), meaning \( x \geq 0 \). The minimum value of \( f(x) \) occurs at: \[ x = -\frac{b}{2a} = -\frac{6}{2 \cdot 9} = -\frac{1}{3}. \] However, since \( x \geq 0 \), we evaluate \( f(x) \) at \( x = 0 \): \[ f(0) = -5. \] Thus, the range of \( f(x) \) is \( [-5, \infty) \), making \( f(x) \) onto. Step 3: Check one-one property.
Since the function is strictly increasing on \( \mathbb{R}_+ \), it is one-one. Step 4: Conclusion.
As \( f(x) \) is both one-one and onto, it is bijective.