Domain and Range Analysis of Rational Function

To determine the nature of the function $f: \mathbf{R} \rightarrow \mathbf{R}$ defined by $f(x) = \frac{2x^2 - 3x + 2}{3x^2 + x + 3}$

we need to evaluate whether $f$ is one-one (injective) and onto (surjective).

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1. Test for Injectivity (One-One)

A function $f$ is one-one if $f(x_1) = f(x_2) \implies x_1 = x_2$.

Let us check if there exist distinct values $x_1, x_2 \in \mathbf{R}$ such that $f(x_1) = f(x_2) = 1$: $\frac{2x^2 - 3x + 2}{3x^2 + x + 3} = 1$

Cross-multiplying yields: $2x^2 - 3x + 2 = 3x^2 + x + 3$ $x^2 + 4x + 1 = 0$

The discriminant of this quadratic equation is: $D = 4^2 - 4(1)(1) = 16 - 4 = 12 > 0$

Since $D > 0$, there exist two distinct real roots: $x_1 = -2 + \sqrt{3} \quad \text{and} \quad x_2 = -2 - \sqrt{3}$

Because $f(-2 + \sqrt{3}) = f(-2 - \sqrt{3}) = 1$ with $x_1 \neq x_2$, the function is not one-one (it is many-one).

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2. Test for Surjectivity (Onto)

A function $f: \mathbf{R} \rightarrow \mathbf{R}$ is onto if its range is equal to its co-domain ($\mathbf{R}$).

Let $y = f(x)$: $y = \frac{2x^2 - 3x + 2}{3x^2 + x + 3}$

Rearranging the terms to form a quadratic in $x$: $y(3x^2 + x + 3) = 2x^2 - 3x + 2$ $(3y - 2)x^2 + (y + 3)x + (3y - 2) = 0$

Case 1: If $3y - 2 = 0 \implies y = \frac{2}{3}$, the equation becomes: $0 + \left(\frac{2}{3} + 3\right)x + 0 = 0 \implies \frac{11}{3}x = 0 \implies x = 0$ Since $x = 0 \in \mathbf{R}$, $y = \frac{2}{3}$ is in the range.

Case 2: If $3y - 2 \neq 0$, then for $x \in \mathbf{R}$, the discriminant of the quadratic equation must be non-negative ($D \ge 0$): $D = (y + 3)^2 - 4(3y - 2)(3y - 2) \ge 0$ $(y + 3)^2 - 4(3y - 2)^2 \ge 0$

Using the difference of squares identity $A^2 - B^2 = (A - B)(A + B)$ where $A = y + 3$ and $B = 2(3y - 2) = 6y - 4$: $[(y + 3) - (6y - 4)][(y + 3) + (6y - 4)] \ge 0$ $(-5y + 7)(7y - 1) \ge 0$ $(5y - 7)(7y - 1) \le 0$

Solving this inequality gives: $\frac{1}{7} \le y \le \frac{7}{5}$

Thus, the range of $f(x)$ is: $\text{Range} = \left[\frac{1}{7}, \frac{7}{5}\right]$

Since $\text{Range} = \left[\frac{1}{7}, \frac{7}{5}\right] \neq \mathbf{R}$ (the co-domain), the function is not onto.

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Conclusion

The function $f$ is neither one-one nor onto.

Correct Option: D