The teacher hasn't uploaded a solution for this question yet.
To prove that $f(x)$ is one-one, we need to show that if $f(x_1) = f(x_2)$, then $x_1 = x_2$. Let $x_1, x_2 \in A$ such that $f(x_1) = f(x_2)$. Then, $\frac{x_1 - 3}{x_1 - 5} = \frac{x_2 - 3}{x_2 - 5}$.
Cross-multiplying, we get: $(x_1 - 3)(x_2 - 5) = (x_2 - 3)(x_1 - 5)$ $x_1x_2 - 5x_1 - 3x_2 + 15 = x_1x_2 - 5x_2 - 3x_1 + 15$ $-5x_1 - 3x_2 = -5x_2 - 3x_1$ $2x_2 = 2x_1$ $x_1 = x_2$ Since $f(x_1) = f(x_2)$ implies $x_1 = x_2$, the function $f(x)$ is one-one.
To prove that $f(x)$ is onto, we need to show that for every $y \in B$, there exists an $x \in A$ such that $f(x) = y$. Let $y \in B$, so $y \in R - \{1\}$. We want to find $x \in A$ such that $f(x) = y$, i.e., $\frac{x - 3}{x - 5} = y$.
Solving for $x$: $x - 3 = y(x - 5)$ $x - 3 = yx - 5y$ $x - yx = 3 - 5y$ $x(1 - y) = 3 - 5y$ $x = \frac{3 - 5y}{1 - y}$
We need to check if $x \in A$, i.e., $x \in R - \{5\}$. If $x = 5$, then $\frac{3 - 5y}{1 - y} = 5$ $3 - 5y = 5 - 5y$ $3 = 5$, which is a contradiction. Thus, $x \neq 5$, and $x \in A$.
Since for every $y \in B$, there exists an $x = \frac{3 - 5y}{1 - y} \in A$ such that $f(x) = y$, the function $f(x)$ is onto.
Since $f(x)$ is both one-one and onto, it is a bijective function.
Final Answer: f is one-one and onto
AI generated content. Review strictly for academic accuracy.