Show that the function $f$ in $A=R-\left\{\dfrac{2}{3}\right\}$ defined as $f(x)=\dfrac{4x+3}{6x-4}$ is one-one and onto. Hence find $f^{-1}$.

$f$ is one-one and onto; $f^{-1}(x)=\dfrac{4x+3}{6x-4}$ (self-inverse).

CBSE Class 12 · 2013 · Relations And Functions XIILong Answer

Show that the function

f

A = R - {\frac{2}{3}}

defined as

f (x) = \frac{4 x + 3}{6 x - 4}

is one-one and onto. Hence find

f^{- 1}

Answer: $f$ is one-one and onto; $f^{- 1} (x) = \frac{4 x + 3}{6 x - 4}$ (self-inverse).

One-one: let

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

\frac{4 x _{1} + 3}{6 x _{1} - 4} = \frac{4 x _{2} + 3}{6 x _{2} - 4} \Rightarrow (4 x_{1} + 3) (6 x_{2} - 4) = (4 x_{2} + 3) (6 x_{1} - 4)

Expanding, the

24 x_{1} x_{2}

terms cancel and we get

- 16 x_{1} + 18 x_{2} = - 16 x_{2} + 18 x_{1} \Rightarrow 34 x_{1} = 34 x_{2} \Rightarrow x_{1} = x_{2}

f

is one-one.
Onto: solve

y = \frac{4 x + 3}{6 x - 4}

for

x

y (6 x - 4) = 4 x + 3 \Rightarrow x (6 y - 4) = 4 y + 3 \Rightarrow x = \frac{4 y + 3}{6 y - 4}

A valid pre-image exists for each admissible

y

, so

f

is onto and hence a bijection.

f^{- 1} (x) = \frac{4 x + 3}{6 x - 4}

CBSE Class 12 Board · 2013 · Subjective