Q65 / 180 · COMEDK 2014 · Mathematics

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

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

Solution

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

Hence, the correct answer is option: 'D'.

More Functions Questions