## Filters

Q&A - Ask Doubts and Get Answers
Q

# Let f defined from R minus minus 4 over 3 to R be a function defined as f x is equal to 4x over 3x + 4 The inverse of f is the map g defined from R to R minus minus 4 over 3 given by A g y is equal to 3y over 3 minus 4y

Q. 14 Let $f: R - \left\{-\frac{4}{3}\right\} \rightarrow R$ be a function defined as $f(x) = \frac{4x}{3x + 4}$. The inverse of $f$ is the map $g : Range\;f \rightarrow R - \left \{-\frac{4}{3} \right \}$given by

(A) $g(y) = \frac{3y}{3 -4y}$

(B) $g(y) = \frac{4y}{4 -3y}$

(C) $g(y) = \frac{4y}{3 -4y}$

(D) $g(y) = \frac{3y}{3 -4y}$

Answers (1)
Views

$f: R - \left\{-\frac{4}{3}\right\} \rightarrow R$

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

Let f inverse  $g : Range\;f \rightarrow R - \left \{-\frac{4}{3} \right \}$

Let y be element of range f.

Then there is $x \in R - \left\{-\frac{4}{3}\right\}$    such that

$y=f(x)$

$y=\frac{4x}{3x+4}$

$y(3x+4)=4x$

$3xy+4y=4x$

$3xy-4x+4y=0$

$x(3y-4)+4y=0$

$x= \frac{-4y}{3y-4}$

$x= \frac{4y}{4-3y}$

Now , define$g : Range\;f \rightarrow R - \left \{-\frac{4}{3} \right \}$     as $g(y)= \frac{4y}{4-3y}$

$gof(x)= g(f(x))= g(\frac{4x}{3x+4})$

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

$=\frac{16x}{12x+16-12x}$

$=\frac{16x}{16}$

$=x$

$fog(y)=f(g(y))=f(\frac{4y}{4-3y})$

$= \frac{4(\frac{4y}{4-3y})}{3(\frac{4y}{4-3y}) + 4}$

$=\frac{16y}{12y+16-12y}=\frac{16y}{16}$

$=y$

Hence, g is inverse of f and $f^{-1}=g$

Inverse of f is given by   $g(y)= \frac{4y}{4-3y}$.

The correct option is B.

Exams
Articles
Questions