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

Step 1: Understanding the Question
We need to determine if the given function \( f(x) \) is injective (one-one) and surjective (onto) over the domain and codomain of real numbers \( \mathbb{R} \).
Step 2: Key Formula or Approach
A function is one-one (injective) if \( f(x_1) = f(x_2) \) implies \( x_1 = x_2 \).
A function is onto (surjective) if its range is equal to its codomain. For \( f : \mathbb{R} \to \mathbb{R} \), the range must be \( \mathbb{R} \).
To find the range of a rational function like this, we set \( y = f(x) \), form a quadratic equation in \( x \), and use the condition that the discriminant must be non-negative (\( D \ge 0 \)) for \( x \) to be real.
Step 3: Detailed Explanation
Checking for one-one property:
Let's check if different values of \(x\) can give the same value of \(y\).
Let \( y = f(x) \). For \( y=1 \), we have:
\[ 1 = \dfrac{2x^2 - 3x + 2}{3x^2 + x + 3} \] \[ 3x^2 + x + 3 = 2x^2 - 3x + 2 \] \[ x^2 + 4x + 1 = 0 \] This quadratic equation has two distinct real roots, \( x = -2 \pm \sqrt{3} \).
Since \( f(-2 + \sqrt{3}) = 1 \) and \( f(-2 - \sqrt{3}) = 1 \), two different inputs give the same output.
Therefore, the function is not one-one.
Checking for onto property:
Let \( y = \dfrac{2x^2 - 3x + 2}{3x^2 + x + 3} \).
To find the range, we rearrange the equation into a quadratic in \(x\):
\[ y(3x^2 + x + 3) = 2x^2 - 3x + 2 \] \[ 3yx^2 + yx + 3y = 2x^2 - 3x + 2 \] \[ (3y - 2)x^2 + (y + 3)x + (3y - 2) = 0 \] For \(x\) to be a real number, the discriminant \( D \) of this quadratic equation must be greater than or equal to zero.
\[ D = (y + 3)^2 - 4(3y - 2)(3y - 2) \ge 0 \] \[ (y + 3)^2 - 4(3y - 2)^2 \ge 0 \] \[ (y^2 + 6y + 9) - 4(9y^2 - 12y + 4) \ge 0 \] \[ y^2 + 6y + 9 - 36y^2 + 48y - 16 \ge 0 \] \[ -35y^2 + 54y - 7 \ge 0 \] Multiplying by -1 and reversing the inequality sign:
\[ 35y^2 - 54y + 7 \le 0 \] The roots of \( 35y^2 - 54y + 7 = 0 \) are \( y = \frac{1}{7} \) and \( y = \frac{7}{5} \).
Since the parabola \( 35y^2 - 54y + 7 \) opens upwards, the expression is less than or equal to zero between the roots.
Thus, the range of \( f(x) \) is \( \left[\frac{1}{7}, \frac{7}{5}\right] \).
The codomain is \( \mathbb{R} \), but the range is a small interval. So, the range is not equal to the codomain.
Therefore, the function is not onto.
Step 4: Final Answer
The function is neither one-one nor onto.