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. $Noneofthese$

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