Let $A=R-\{3\}$ and $B=R-\{1\}$. Consider the function $f:A \rightarrow B$ defined by $f(x)=\dfrac{x-2}{x-3}$. Show that $f$ is one-one and onto; hence find $f^{-1}$.

$f$ is one-one and onto; $f^{-1}(x)=\dfrac{3x-2}{x-1}$

CBSE Class 12 · 2012 · Relations And Functions XIILong Answer

Let

A = R - {3}

and

B = R - {1}

. Consider the function

f : A \to B

defined by

f (x) = \frac{x - 2}{x - 3}

. Show that

f

is one-one and onto; hence find

f^{- 1}

Answer: $f$ is one-one and onto; $f^{- 1} (x) = \frac{3 x - 2}{x - 1}$

One-one: let

f (x_{1}) = f (x_{2})

\frac{x _{1} - 2}{x _{1} - 3} = \frac{x _{2} - 2}{x _{2} - 3} \Rightarrow (x_{1} - 2) (x_{2} - 3) = (x_{2} - 2) (x_{1} - 3)

Expanding and cancelling the common

x_{1} x_{2} + 6

terms:

- 3 x_{1} - 2 x_{2} = - 3 x_{2} - 2 x_{1} \Rightarrow x_{1} = x_{2}

f

is one-one.
Onto: let

y \in B

and solve

y = \frac{x - 2}{x - 3}

for

x

y (x - 3) = x - 2 \Rightarrow x (y - 1) = 3 y - 2 \Rightarrow x = \frac{3 y - 2}{y - 1}

For every

y \neq = 1

this gives a valid

x \in A

, so

f

is onto. Hence

f

is a bijection and invertible.

f^{- 1} (x) = \frac{3 x - 2}{x - 1}

CBSE Class 12 Board · 2012 · Subjective