2. Let $$ f: R, ightarrow[-5, x) $$ be defined as $$ f(x)=9 x^{2}+6 x-5 $$, where $$ R_{+} $$, is the set of all non-negative $$ r $$ n numbers. Then, $$ f $$ is: $$ \begin{array}{ll} ext { (d) one-one } & ext { (b) onto } \ ext { (d) bijective } & ext { (d) neither one-one nor onto }\end{array} $$

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2. Let $$ f: R, ightarrow[-5, x) $$ be defined as $$ f(x)=9 x^{2}+6 x-5 $$, where $$ R_{+} $$, is the set of all non-negative $$ r $$ n numbers. Then, $$ f $$ is: $$ \begin{array}{ll}	ext { (d) one-one } & 	ext { (b) onto } \ 	ext { (d) bijective } & 	ext { (d) neither one-one nor onto }\end{array} $$

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Let's analyze the function $$ f: \mathbb{R}_+ \rightarrow [-5, \infty) $$ defined by $$ f(x) = 9x^2 + 6x - 5 $$, where $$ \mathbb{R}_+ $$ is the set of all non-negative real numbers.

### 1. **One-One (Injective)**
A function is **one-one** if each element of the domain maps to a unique element in the codomain.

- **Derivative Analysis:**  
  $$ f'(x) = 18x + 6 $$  
  For $$ x \geq 0 $$, $$ f'(x) \geq 6 $$, which means the function is **strictly increasing** on $$ \mathbb{R}_+ $$.

- **Conclusion:**  
  Since the function is strictly increasing, it is **one-one**.

### 2. **Onto (Surjective)**
A function is **onto** if every element in the codomain has a pre-image in the domain.

- **Range of $$ f(x) $$:**  
  $$ f(0) = 9(0)^2 + 6(0) - 5 = -5 $$  
  As $$ x $$ increases, $$ f(x) $$ increases without bound, approaching $$ \infty $$.

- **Conclusion:**  
  The range of $$ f(x) $$ is $$ [-5, \infty) $$, which exactly matches the codomain. Therefore, $$ f $$ is **onto**.

### 3. **Bijective**
A function is **bijective** if it is both one-one and onto.

- **Conclusion:**  
  Since $$ f $$ is both one-one and onto, it is **bijective**.

**Final Answer:**  
$$ f $$ is **bijective**.

**Answer:**  
$$ f $$ is **bijective**.
$$ f $$ is bijective.

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