Question

Let \( f : \mathbb{R} \to \mathbb{R} \) be defined as \( f(x) = \dfrac{2x^2 - 3x + 2}{3x^2 + x + 3} \). Then \( f \) is:

• A. both one-one and onto
• B. one-one but not onto
• C. onto but not one-one
• D. neither one-one nor onto

Answer 1

The Correct Option is D

Concept: To check whether a function is one-one and onto:

• A function is one-one (injective) if \( f(a) = f(b) \Rightarrow a = b \).
• A function is onto (surjective) if its range is equal to its codomain.
• For rational functions, range can be found by setting \( y = f(x) \) and analyzing the resulting quadratic in \( x \).

Step 1: {Check whether the function is one-one.}
Since the function is a rational function of equal degree (both numerator and denominator are quadratic), such functions are generally not one-one over \( \mathbb{R} \). Hence, the function is not one-one. Step 2: {Find the range to check onto property.}
Let \( y = \dfrac{2x^2 - 3x + 2}{3x^2 + x + 3} \). \[ y(3x^2 + x + 3) = 2x^2 - 3x + 2 \] \[ (3y - 2)x^2 + (y + 3)x + (3y - 2) = 0 \] For real \( x \), discriminant \( \geq 0 \): \[ D = (y + 3)^2 - 4(3y - 2)^2 \geq 0 \] \[ = y^2 + 6y + 9 - 4(9y^2 - 12y + 4) \] \[ = y^2 + 6y + 9 - 36y^2 + 48y - 16 \] \[ = -35y^2 + 54y - 7 \geq 0 \] Solve: \[ -35y^2 + 54y - 7 = 0 \] \[ y = \frac{-54 \pm 44}{-70} \] \[ y = \frac{1}{7}, \quad y = \frac{7}{5} \] Thus, range is: \[ \frac{1}{7} \leq y \leq \frac{7}{5} \] So, the function is not onto \( \mathbb{R} \).