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Check if the function is one-one:<\/strong><\/p>
Given \(f(x) = 9x^2 + 6x - 5\). To check if it's one-one, we can analyze its derivative. \(f'(x) = 18x + 6\). Since \(x \in R_{+}\), \(x \geq 0\), so \(f'(x) \geq 6 > 0\). This means the function is strictly increasing for \(x \geq 0\). Therefore, it is one-one.
Check if the function is onto:<\/strong><\/p>
The range of the function is given as \([-5, \infty)\). We need to check if the function actually attains all values in this range. We can rewrite \(f(x)\) as \(f(x) = (3x + 1)^2 - 6\). Since \(x \in R_{+}\), the minimum value of \(3x + 1\) is 1 (when x=0). Therefore, the minimum value of \((3x + 1)^2\) is 1, and the minimum value of \(f(x)\) is \(1 - 6 = -5\). As \(x\) increases from 0, \((3x + 1)^2\) increases, and so does \(f(x)\). Since \(x\) can take any non-negative real value, \(f(x)\) can take any value greater than or equal to -5. Thus, the range of \(f(x)\) is indeed \([-5, \infty)\), which matches the given codomain. Therefore, the function is onto.
Conclusion:<\/strong><\/p>
Since the function is both one-one and onto, it is bijective.
Correct Answer: bijective<\/strong>
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