Correct option is (C) bijective
one-one
Let \(x_1, x_2 \in R +\) and \(f(x_1) = f(x_2)\)
\({9x_1}^2 + 6x_1 -5 = {9x_2}^2 + 6x_2 - 5\)
\({9x_1}^2 + 6x_1 = {9x_2}^2 + 6x_2\) (removing -5 from both side)
\({3x_1}^2 + 2x_1 = {3x_2}^2 + 2x_2\) (dividing 3 on both side)
\({3x_1}^2 - {3x_2}^2 + 2x_1 - 2x_2 = 0\)
\(\Rightarrow 3({x_1}^2 -{x_2}^2) +2(x_1 -x_2) = 0\)
\(3(x_1 - x_2)(x_1 + x_2) + 2(x_1 - x_2) = 0\)
\(x_1 - x_2 = 0\) (OR) \(x_1 + x_2 + 2 = 0\)
Consider \(x_1 + x_2 + 2 = 0 \Rightarrow x_1 = -x_2 - 2\)
As \(x_1, x_2 \in R +\) (non-negative real numbers), it \((x_1 + x_2 + 2 = 0 )\) can not be possible. So \(x_1 - x_2 = 0\Rightarrow x_1 = x_2\)
Hence it is one-one.
onto
Let \(y = f(x)\) and \(y \in[-5, \infty]\)
So \(y = 9x^2 + 6x - 5\)
\(9x^2 + 6x - 5 -y = 0\)
\(x = \frac{-6\pm \sqrt{6^2-4\times 9 \times(-y - 5)}}{2 \times 9}\) (by using formula \(x = \frac{-b\pm \sqrt{b^2-4ac}}{2a}\) for quadratic equation \(ax^2 + bx + c = 0\))
\(x = \frac{-6\pm \sqrt{36+36(y+5)}}{18}\)
\(\Rightarrow x = \frac{-6\pm 6\sqrt{(y+6)}}{18}\)
x has two solution \(x = \frac{-1-\sqrt{y + 6}}3\) and \(x = \frac{-1+\sqrt{y + 6}}3\). We need to prove either one solution (x) exists for every \(y \in [-5, \infty)\). In the solution \(x = \frac{-1-\sqrt{y + 6}}3\), x is the negative number for any y in the given range. Now we left with \(x = \frac{-1+\sqrt{y + 6}}3\). For y = -5, x = 0 and any value greater then -5 x is always non negative real numbers. It proves that every value in the range \(y \in [-5, \infty)\) there exists a value in domain \(\to x \in R+\). Hence it is onto.
So the function is bijective (one-one and onto).
