Practicing Success

Target Exam

CUET

Subject

-- Mathematics - Section B1

Chapter

Continuity and Differentiability

Question:

If $f(x)=\frac{3x+2}{5x-3}, x\in R -\begin{Bmatrix} \frac{3}{5}\end{Bmatrix},$ then

Options:

$f^{-1}(x)=f(x)$

$f^{-1}(x)=-f(x)$

$f(f(x))=-x$

$f^{-1}(x)=\frac{1}{7}f(x)$

Correct Answer:

$f^{-1}(x)=f(x)$

Explanation:

$f(x)=\frac{3x+2}{5x-3}$, $y=\frac{3x+2}{5x-3}$

so $5xy-3y=3x+2$ so $5xy-3x=3y+2$

so $x(5y-3)=3y+2$

so $x=\frac{3y+2}{5y-3}=f^{-1}(y)⇒f^{-1}(x)=\frac{3x+2}{5x-3}$

so $f(x)=f^{-1}(x)$