1.

Prove that \( f: R-\{2\} \rightarrow R-\{2\}, f(x)=\frac{2 x-1}{x-2} \) is one-one and onto.

Answer»

\(f(x) = \frac{2 x -1}{x -2}\)

Let \(f(x_1) \ne f(x_2)\) & \(x_1 \ne 2, x_2 \ne 2.\)

\(\therefore \frac{2x_1-1}{x_1-2}\ne \frac{2x_2-1}{x_2-2}\)

⇒ \((2x_1 -1) (x_2-2) \ne (2x_2-1)(x_1-2)\)

⇒ \(2x_1x_2 - 4x_1 - x_2 + 2\ne 2x_1x_2 - 4x_2 - x_1 + 2\)

⇒ \(-3x_1 \ne -3x_2\)

⇒ \(x_1 \ne x_2\)

\(\therefore\) Function f(x) is one-one.

Let \(y = \frac{2x - 1}{x - 2}\) 

⇒ \(xy - 2y = 2x - 1\)

⇒ \(xy - 2x = 2y-1\)

⇒ \(x(y - 2) = 2y - 1\)

⇒ \(x = \frac{2y -1}{y - 2}\)

⇒ \(f^{-1}(y) = \frac{2y - 1}{y - 2}\)

⇒ \(f^{-1}(x) = \frac{2y - 1}{y - 2}\)

\(\therefore \) Range of function f(x) is R-{2}

which is codomain  of f(x).

\(\therefore \) f(x) is onto.

Hence, given function f(x) is both one-one & onto.


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