Question

If a function \( f: \mathbb{R} \setminus \{1\} \rightarrow \mathbb{R} \setminus \{m\} \) is defined by \( f(x) = \frac{x+3}{x-2} \), then \( \frac{3}{l} + 2m = \)

• A. \( 10 \)
• B. \( 12 \)
• C. \( 8 \)
• D. \( 14 \)

Hint:

For bijections involving rational functions, ensure the function is not undefined at any point within its intended domain or that it avoids specific values in its codomain.

Answer 1

The Correct Option is C

Step 1: Identify the value \( x \) cannot take
The function \( f(x) = \frac{x+3}{x-2} \) is undefined when the denominator is equal to zero. \[ x - 2 = 0 \] \[ x = 2 \] Therefore, \( x \) cannot take the value 2. So, \( l = 2 \). 
Step 2: Determine the values \( f(x) \) cannot take
Let \( y = f(x) \). We want to find the values that \( y \) cannot take. \[ y = \frac{x+3}{x-2} \] \[ y(x-2) = x+3 \] \[ xy - 2y = x+3 \] \[ xy - x = 2y + 3 \] \[ x(y-1) = 2y + 3 \] \[ x = \frac{2y+3}{y-1} \] From this expression for \( x \) in terms of \( y \), we can see that \( y \) cannot take the value 1, as the denominator would be zero. Therefore, \( m = 1 \). 
Step 3: Calculate \( 3l + 2m \)
We have \( l = 2 \) and \( m = 1 \). \[ 3l + 2m = 3(2) + 2(1) = 6 + 2 = 8 \] Therefore, \( 3l + 2m = 8 \).