Let `f:R -{3} to R{1} : f(x) = (x-2)/(x-3)and g:R to R , g(x)=2x-3 and f^-1(x)+g^-1(x)=13/2` then sum of all value of x is

Options

A. 2
B. 3
C. 5
D. 7

Answer

To solve the problem, we need to find the values of \( x \) such that \( f^{-1}(x) + g^{-1}(x) = \frac{13}{2} \), where \( f(x) = \frac{x-2}{x-3} \) and \( g(x) = 2x - 3 \).

### Step 1: Find the inverse of \( f(x) \)

Let \( y = f(x) = \frac{x-2}{x-3} \).

Cross-multiplying gives:
\[ y(x - 3) = x - 2 \]
This simplifies to:
\[ yx - 3y = x - 2 \]
Rearranging terms, we have:
\[ yx - x = 3y - 2 \]
Factoring out \( x \) from the left side:
\[ x(y - 1) = 3y - 2 \]
Thus, we can express \( x \) in terms of \( y \):
\[ x = \frac{3y - 2}{y - 1} \]
Therefore, the inverse function is:
\[ f^{-1}(y) = \frac{3y - 2}{y - 1} \]
Replacing \( y \) with \( x \):
\[ f^{-1}(x) = \frac{3x - 2}{x - 1} \]

### Step 2: Find the inverse of \( g(x) \)

Let \( a = g(x) = 2x - 3 \).

Rearranging gives:
\[ a + 3 = 2x \implies x = \frac{a + 3}{2} \]
Thus, the inverse function is:
\[ g^{-1}(a) = \frac{a + 3}{2} \]
Replacing \( a \) with \( x \):
\[ g^{-1}(x) = \frac{x + 3}{2} \]

### Step 3: Set up the equation

We have:
\[ f^{-1}(x) + g^{-1}(x) = \frac{13}{2} \]
Substituting the inverse functions:
\[ \frac{3x - 2}{x - 1} + \frac{x + 3}{2} = \frac{13}{2} \]

### Step 4: Clear the fractions

To eliminate the fractions, multiply through by \( 2(x - 1) \):
\[ 2(3x - 2) + (x + 3)(x - 1) = 13(x - 1) \]
Expanding each term:
\[ 6x - 4 + (x^2 + 3x - x - 3) = 13x - 13 \]
This simplifies to:
\[ 6x - 4 + x^2 + 2x - 3 = 13x - 13 \]
Combining like terms:
\[ x^2 + 8x - 7 = 13x - 13 \]
Rearranging gives:
\[ x^2 - 5x + 6 = 0 \]

### Step 5: Factor the quadratic equation

Factoring:
\[ (x - 2)(x - 3) = 0 \]
Thus, the solutions are:
\[ x = 2 \quad \text{and} \quad x = 3 \]

### Step 6: Find the sum of all values of \( x \)

The sum of the values of \( x \) is:
\[ 2 + 3 = 5 \]

### Final Answer
The sum of all values of \( x \) is \( \boxed{5} \).