Let `A=R-{2}` and `B=R-{1}` if `f:A rarr B` is a function defined by `f(x)=(x-1)/(x-2)` show that f is one-one and onto. Hence find `f^(-1)`.

Answer

To solve the problem, we need to show that the function \( f: A \to B \) defined by \( f(x) = \frac{x-1}{x-2} \) is one-one (injective) and onto (surjective), and then find its inverse \( f^{-1} \).

### Step 1: Show that \( f \) is one-one (injective)

To prove that \( f \) is one-one, we need to show that if \( f(a) = f(b) \), then \( a = b \).

Assume \( f(a) = f(b) \):
\[ \frac{a-1}{a-2} = \frac{b-1}{b-2} \]

Cross-multiplying gives:
\[ (a-1)(b-2) = (b-1)(a-2) \]

Expanding both sides:
\[ ab - 2a - b + 2 = ab - 2b - a + 2 \]

Now, simplifying both sides:
\[ -2a - b + 2 = -2b - a + 2 \]

Subtracting 2 from both sides:
\[ -2a - b = -2b - a \]

Rearranging gives:
\[ -2a + a = -2b + b \]
\[ -a = -b \]

Thus, we have:
\[ a = b \]

Since \( f(a) = f(b) \) implies \( a = b \), we conclude that \( f \) is one-one.

### Step 2: Show that \( f \) is onto (surjective)

To prove that \( f \) is onto, we need to show that for every \( y \in B \), there exists an \( x \in A \) such that \( f(x) = y \).

Let \( y = f(x) = \frac{x-1}{x-2} \). We need to solve for \( x \):
\[ y = \frac{x-1}{x-2} \]

Cross-multiplying gives:
\[ y(x-2) = x-1 \]

Expanding:
\[ yx - 2y = x - 1 \]

Rearranging terms:
\[ yx - x = 2y - 1 \]

Factoring out \( x \):
\[ x(y - 1) = 2y - 1 \]

Thus, we can solve for \( x \):
\[ x = \frac{2y - 1}{y - 1} \]

Now, we need to check if \( x \) is defined for all \( y \in B \). The function is undefined when \( y - 1 = 0 \) (i.e., \( y = 1 \)), which is not in the codomain \( B \). Therefore, for every \( y \in B \), there exists a corresponding \( x \in A \).

Thus, \( f \) is onto.

### Step 3: Find the inverse \( f^{-1} \)

From the previous step, we derived:
\[ x = \frac{2y - 1}{y - 1} \]

Now, we can express the inverse function:
\[ f^{-1}(y) = \frac{2y - 1}{y - 1} \]

To express it in terms of \( x \), we replace \( y \) with \( x \):
\[ f^{-1}(x) = \frac{2x - 1}{x - 1} \]

### Final Result

Thus, we conclude that:
- The function \( f \) is one-one and onto.
- The inverse function is given by:
\[ f^{-1}(x) = \frac{2x - 1}{x - 1} \]