Question

Check whether $f:\mathbb{R}-\{3\}\rightarrow\mathbb{R}$ defined by $f(x)=\dfrac{x-2}{x-3}$ is onto or not.

Hint:

To test whether a rational function is onto, assume \(y=f(x)\), solve for \(x\), and check if a real solution exists for every \(y\) in the codomain.

Answer 1

To check if the function \( f(x) = \frac{x-2}{x-3} \) is onto, we need to determine if for every \( y \in \mathbb{R} \), there exists an \( x \in \mathbb{R} - \{3\} \) such that \( f(x) = y \). In other words, we need to solve for \( x \) in terms of \( y \) and check whether every real value of \( y \) has a corresponding \( x \). 
Step 1: Solve \( f(x) = y \) for \( x \)
We start with the equation:
\[ \frac{x-2}{x-3} = y \] Multiply both sides by \( x-3 \) to eliminate the denominator: \[ x - 2 = y(x - 3) \] Expand the right-hand side: \[ x - 2 = yx - 3y \] Bring all terms involving \( x \) to one side: \[ x - yx = -3y + 2 \] Factor out \( x \) on the left-hand side: \[ x(1 - y) = -3y + 2 \] Solve for \( x \): \[ x = \frac{-3y + 2}{1 - y} \] Step 2: Analyze the solution
The function is defined for all \( x \in \mathbb{R} - \{3\} \), which means we need to ensure that \( x \neq 3 \). For \( x = 3 \), the denominator \( 1 - y \) must be zero, which happens when \( y = 1 \). Thus, when \( y = 1 \), there is no value of \( x \) such that \( f(x) = 1 \), because the denominator in the expression for \( x \) would be zero, making \( x \) undefined. 
Step 3: Conclusion
Since for \( y = 1 \), there is no corresponding \( x \) in \( \mathbb{R} - \{3\} \), the function \( f(x) = \frac{x-2}{x-3} \) is not onto. The function does not cover all real values because the value \( y = 1 \) is missing from the range of \( f(x) \). 
Therefore, the function is not onto.