Function Analysis: $f(x) = \frac{2x^2 - 3x + 2}{3x^2 + x + 3}$
Checking the One-One Property
A function $f$ is one-one if $f(x_1) = f(x_2)$ implies $x_1 = x_2$. Let's test this condition.
Assume $f(x) = f(y)$: $ \frac{2x^2 - 3x + 2}{3x^2 + x + 3} = \frac{2y^2 - 3y + 2}{3y^2 + y + 3} $ Cross-multiplying and simplifying leads to: $ 11(x - y)(xy - 1) = 0 $ This equation implies $x = y$ or $xy = 1$. If we choose $x \neq y$ such that $xy = 1$ (for example, $x=2$ and $y=1/2$), then $f(x) = f(y)$.
- Let $x=2$: $f(2) = \frac{2(2^2) - 3(2) + 2}{3(2^2) + 2 + 3} = \frac{8 - 6 + 2}{12 + 2 + 3} = \frac{4}{17}$.
- Let $y=1/2$: $f(1/2) = \frac{2(1/2)^2 - 3(1/2) + 2}{3(1/2)^2 + 1/2 + 3} = \frac{2(1/4) - 3/2 + 2}{3(1/4) + 1/2 + 3} = \frac{1/2 - 3/2 + 2}{3/4 + 2/4 + 12/4} = \frac{1}{17/4} = \frac{4}{17}$.
Since $f(2) = f(1/2)$ but $2 \neq 1/2$, the function is not one-one.
Checking the Onto Property
A function $f: \mathbf{R} \rightarrow \mathbf{R}$ is onto if its range is equal to its codomain ($\mathbf{R}$). Let's find the range of $f(x)$.
Set $y = f(x)$: $ y = \frac{2x^2 - 3x + 2}{3x^2 + x + 3} $ Rearrange into a quadratic equation in terms of $x$: $ y(3x^2 + x + 3) = 2x^2 - 3x + 2 $ $ 3yx^2 + yx + 3y = 2x^2 - 3x + 2 $ $ (3y - 2)x^2 + (y + 3)x + (3y - 2) = 0 $ For $x$ to be real, the discriminant ($\Delta$) of this quadratic equation must be non-negative ($\Delta \ge 0$). The discriminant is $\Delta = B^2 - 4AC$, where $A = (3y - 2)$, $B = (y + 3)$, and $C = (3y - 2)$. $ \Delta = (y + 3)^2 - 4(3y - 2)(3y - 2) $ $ \Delta = (y^2 + 6y + 9) - 4(9y^2 - 12y + 4) $ $ \Delta = y^2 + 6y + 9 - 36y^2 + 48y - 16 $ $ \Delta = -35y^2 + 54y - 7 $ We require $\Delta \ge 0$, so: $ -35y^2 + 54y - 7 \ge 0 $ $ 35y^2 - 54y + 7 \le 0 $ To find the values of $y$ that satisfy this inequality, we find the roots of $35y^2 - 54y + 7 = 0$. Using the quadratic formula: $ y = \frac{-(-54) \pm \sqrt{(-54)^2 - 4(35)(7)}}{2(35)} = \frac{54 \pm \sqrt{2916 - 980}}{70} = \frac{54 \pm \sqrt{1936}}{70} = \frac{54 \pm 44}{70} $ The roots are $y_1 = \frac{54 - 44}{70} = \frac{10}{70} = \frac{1}{7}$ and $y_2 = \frac{54 + 44}{70} = \frac{98}{70} = \frac{7}{5}$. Since the quadratic $35y^2 - 54y + 7$ opens upwards, the inequality $35y^2 - 54y + 7 \le 0$ holds for $y$ between the roots.
The range of the function is $[\frac{1}{7}, \frac{7}{5}]$.
The codomain is $\mathbf{R}$. Since the range $[\frac{1}{7}, \frac{7}{5}]$ is a strict subset of $\mathbf{R}$, the function is not onto.
Conclusion
The function $f(x)$ is neither one-one nor onto.
Therefore, the correct option is D: neither one-one nor onto.
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