Question

Let $f : \mathbb{R} - \{2\} \rightarrow \mathbb{R} - \{1\}$ defined by $f(x) = \frac{x-3}{x-2}$ and $g : \mathbb{R} \rightarrow \mathbb{R}$ defined by $g(x) = 3x - 2$, then sum of all values of $x$ for which $f^{-1}(x) + g^{-1}(x) = 19/6$ is ______.

• A. 5/2
• B. 7/2
• C. 9/2
• D. 11/2

Hint:

If a question only asks for the "sum of values" or "product of values" resulting from an equation, try to manipulate it into a standard polynomial. You can then use Vieta's formulas to find the answer instantly without having to actually calculate the individual roots!

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
We are given two functions, $f(x)$ and $g(x)$. We must first compute their respective inverse functions, set their sum equal to $19/6$, and then find the sum of the roots of the resulting equation.

Step 2: Detailed Explanation:
First, find the inverse function $f^{-1}(x)$. Set $f(y) = x$:
$$x = \frac{y-3}{y-2}$$
Solve for $y$:
$$x(y-2) = y-3 \implies xy - 2x = y - 3 \implies xy - y = 2x - 3 \implies y(x-1) = 2x-3$$
$$y = f^{-1}(x) = \frac{2x-3}{x-1}$$
Next, find the inverse function $g^{-1}(x)$. Set $g(y) = x$:
$$x = 3y - 2 \implies y = g^{-1}(x) = \frac{x+2}{3}$$
Now, substitute the inverses into the given equation:
$$\frac{2x-3}{x-1} + \frac{x+2}{3} = \frac{19}{6}$$
Find a common denominator and multiply the entire equation by $6(x-1)$ to eliminate fractions:
$$6(2x-3) + 2(x+2)(x-1) = 19(x-1)$$
Expand all terms:
$$12x - 18 + 2(x^2 + x - 2) = 19x - 19$$
$$12x - 18 + 2x^2 + 2x - 4 = 19x - 19$$
Group like terms:
$$2x^2 + 14x - 22 = 19x - 19$$
Move all terms to one side to form a standard quadratic equation:
$$2x^2 + (14x - 19x) + (-22 + 19) = 0$$
$$2x^2 - 5x - 3 = 0$$
The question asks for the sum of all values of $x$ (the roots). According to Vieta's formulas for a quadratic $ax^2 + bx + c = 0$, the sum of roots is $-b/a$:
$$\text{Sum of roots} = -\frac{(-5)}{2} = \frac{5}{2}$$

Step 4: Final Answer:
The sum of all values is 5/2, which corresponds to option (a).