Question

Let \( f: \mathbb{R} \rightarrow \mathbb{R} \) be a function defined by \( f(x) = \frac{2x+1}{3} \). If \( \alpha \) is an element in the domain of \( f \) whose image is \( \frac{1}{\alpha} \), then the sum of all possible values of such \( \alpha \) is

• A. \( -\frac{1}{2} \)
• B. \( \frac{1}{2} \)
• C. \( \frac{5}{2} \)
• D. \( 0 \)

Hint:

When dealing with functions and their images, remember to set up the equation based on the given information. For a function \( f(x) \), if the image of \( a \) is \( b \), then \( f(a) = b \). Solving the resulting equation will give the required values. In case of a quadratic equation \( ax^2 + bx + c = 0 \), the sum of the roots is given by \( -\frac{b}{a} \). In this problem, after forming the quadratic equation \( 2\alpha^2 + \alpha - 3 = 0 \), the sum of the roots (possible values of \( \alpha \)) is \( -\frac{1}{2} \).

Answer 1

The Correct Option is A

Step 1 The function is given by \( f(x) = \frac{2x+1}{3} \).

Step 2 The image of \( \alpha \) under the function \( f \) is \( f(\alpha) \).

Step 3 We are given that the image of \( \alpha \) is \( \frac{1}{\alpha} \).

Step 4 Therefore, we have the equation \( f(\alpha) = \frac{1}{\alpha} \).

Step 5 Substituting \( x = \alpha \) into the function definition, we get: $$\frac{2\alpha + 1}{3} = \frac{1}{\alpha}$$
Step 6 To solve for \( \alpha \), we can cross-multiply: $$\alpha(2\alpha + 1) = 3(1)$$
Step 7 $$2\alpha^2 + \alpha = 3$$
Step 8 Rearranging the terms to form a quadratic equation: $$2\alpha^2 + \alpha - 3 = 0$$
Step 9 We need to find the roots of this quadratic equation.
We can use the quadratic formula \( \alpha = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 2 \), \( b = 1 \), and \( c = -3 \).
$$\alpha = \frac{-1 \pm \sqrt{1^2 - 4(2)(-3)}}{2(2)}$$
Step 10 $$\alpha = \frac{-1 \pm \sqrt{1 + 24}}{4}$$
Step 11 $$\alpha = \frac{-1 \pm \sqrt{25}}{4}$$
Step 12 $$\alpha = \frac{-1 \pm 5}{4}$$
Step 13 The two possible values for \( \alpha \) are: $$\alpha_1 = \frac{-1 + 5}{4} = \frac{4}{4} = 1$$ $$\alpha_2 = \frac{-1 - 5}{4} = \frac{-6}{4} = -\frac{3}{2}$$
Step 14 The sum of all possible values of \( \alpha \) is: $$\alpha_1 + \alpha_2 = 1 + \left(-\frac{3}{2}\right) = 1 - \frac{3}{2} = \frac{2 - 3}{2} = -\frac{1}{2}$$