Question

If a real valued function \( f: [a, \infty) \to [b, \infty) \) is defined by \( f(x) = 2x^2 - 3x + 5 \) and is a bijection, then find the value of \( 3a + 2b \):

• A. 20
• B. 10
• C. 12
• D. 6

Hint:

For quadratic functions to be bijective, ensure the function is strictly increasing or decreasing, and find the minimum or maximum value using the vertex formula.

Answer 1

The Correct Option is B

We are given the function \( f(x) = 2x^2 - 3x + 5 \) and the domain of the function is \( x \in [a, \infty) \) and its range is \( f(x) \in [b, \infty) \). Since the function is a bijection, it must be both injective and surjective. Let's first find the values of \( a \) and \( b \). 

Step 1: Analyze the function \( f(x) = 2x^2 - 3x + 5 \). The function is a quadratic function, and to ensure it is injective (one-to-one), the function must be strictly monotonic (either strictly increasing or strictly decreasing) on the domain \( [a, \infty) \). Since the coefficient of \( x^2 \) is positive, the function is a parabola opening upwards. Therefore, the function will be strictly increasing after the vertex. The vertex of the parabola occurs at \( x = -\frac{b}{2a} \), where \( a = 2 \) and \( b = -3 \) for the quadratic \( f(x) = 2x^2 - 3x + 5 \). \( x_{\text{vertex}} = \frac{-(-3)}{2(2)} = \frac{3}{4} \) Thus, the function is strictly increasing for \( x \geq \frac{3}{4} \), and we set \( a = \frac{3}{4} \) to ensure the function is injective. 

Step 2: Determine the value of \( b \). To find \( b \), we evaluate the function at \( x = \frac{3}{4} \): \( f\left( \frac{3}{4} \right) = 2\left( \frac{3}{4} \right)^2 - 3\left( \frac{3}{4} \right) + 5 \) \( f\left( \frac{3}{4} \right) = 2\left( \frac{9}{16} \right) - \frac{9}{4} + 5 = \frac{18}{16} - \frac{36}{16} + \frac{80}{16} = \frac{62}{16} = \frac{31}{8} \) Thus, \( b = \frac{31}{8} \). 

Step 3: Calculate \( 3a + 2b \). We know \( a = \frac{3}{4} \) and \( b = \frac{31}{8} \), so: \( 3a + 2b = 3 \times \frac{3}{4} + 2 \times \frac{31}{8} = \frac{9}{4} + \frac{62}{8} = \frac{18}{8} + \frac{62}{8} = \frac{80}{8} = 10 \) Thus, the value of \( 3a + 2b \) is \( 10 \).