Question

Let \( A \subseteq \mathbb{R}, B \subseteq \mathbb{R} \) and \( f : A \to B \) be defined by \( f(x) = x^2 - 3x + 2 \). If \( f \) is a bijection, then

• A. \( A = (-\infty, 0], B = \left(-\infty, \frac{-1}{4}\right] \)
• B. \( A = \left(-\infty, \frac{3}{2}\right], B = \left[\frac{-1}{4}, \infty \right) \)
• C. \( A = \left[\frac{3}{2}, \infty \right), B = \left(-\infty, \frac{-1}{4} \right] \)
• D. \( A = (-\infty, \infty), B = \left[ \frac{-1}{4}, \infty \right) \)

Hint:

Quadratic functions can be made bijective by restricting the domain to one side of the vertex.

Answer 1

The Correct Option is B

Step 1: Analyze the function.
Given \( f(x) = x^2 - 3x + 2 \), which is a quadratic equation.
This opens upwards and has vertex at \[ x = \frac{-(-3)}{2(1)} = \frac{3}{2}, \quad f\left(\frac{3}{2}\right) = \left(\frac{3}{2}\right)^2 - 3\left(\frac{3}{2}\right) + 2 = \frac{9}{4} - \frac{9}{2} + 2 = \frac{-1}{4}. \] Step 2: Make \( f \) bijective.
A quadratic function is not one-one on the entire domain, but it is one-one if we restrict it to a monotonic interval.
Since the vertex is at \( x = \frac{3}{2} \), restrict \( A = (-\infty, \frac{3}{2}] \) to make it decreasing and thus one-one.
Then the range will be \( \left[\frac{-1}{4}, \infty \right) \) (as values go from minimum upward). Step 3: Confirm bijection.
The function is: - One-one on \( (-\infty, \frac{3}{2}] \) - Onto \( \left[\frac{-1}{4}, \infty \right) \) Hence, bijection is satisfied. \[ \boxed{A = \left(-\infty, \frac{3}{2} \right], B = \left[ \frac{-1}{4}, \infty \right)} \]