Question

If $f\left(x\right) = \frac{2x -3}{3x+4} $ then $f^{-1} \left(\frac{-4}{3}\right) = $

• A. $Zero$
• B. $\frac{3}{4}$
• C. $\frac{-2}{3}$
• D. $None\, of \,these$

Answer 1

Answer: (D)

$f\left(x\right) = \frac{2x-3}{3x-4}$
Let $ f\left(x\right) = y = \frac{2x-3}{3x+4}$
On Cross multiplication, we get
$ \Rightarrow 3xy +4y = 2x-3$
$ \Rightarrow x\left(3y -2\right) = -3 - 4y$
$ \Rightarrow x= \frac{-3 - 4 y}{3y -2}$
$ \Rightarrow x - f^{^{-1} } \left(y\right)= \frac{-3 -4y}{3y - 2}$
Put $y =- \frac{4}{3}$ we get
$ f^{-1} \left(- \frac{4}{3}\right) = \frac{-3 -4 \times\left(-\frac{4}{3}\right)}{3\left(- \frac{4}{3}\right) - 2}$
$ = \frac{-3 + \frac{16}{3}}{-4 -2} = \frac{7}{ 3\times \left(-6\right)} = - \frac{7}{18} $