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

Step 1: Write the definition of onto function.
A function \(f:A\rightarrow B\) is said to be onto (surjective) if for every \(y\in B\), there exists at least one \(x\in A\) such that \[ f(x)=y \] Thus, to check whether the function is onto, we assume \[ y=f(x) \] and solve for \(x\).
Step 2: Substitute $f(x)$ in terms of $y$.
\[ y=\frac{x-2}{x-3} \] Multiply both sides by \(x-3\): \[ y(x-3)=x-2 \] \[ yx-3y=x-2 \] Step 3: Rearrange the equation.
\[ yx-x=3y-2 \] \[ x(y-1)=3y-2 \] \[ x=\frac{3y-2}{y-1} \] Step 4: Check whether $x$ exists for every real $y$.
The value of \(x\) exists for all real \(y\) except when \[ y-1=0 \] \[ y=1 \] Thus, the function cannot produce the value \(y=1\).
Therefore, the range of the function is \[ \mathbb{R}-\{1\} \] while the codomain is \(\mathbb{R}\).
Step 5: Conclusion.
Since one real value \(y=1\) is not obtained, the function is not onto.
Final Answer:
\[ \boxed{\text{The function is not onto}} \]