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)

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 , defineg : 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.

 

 

 

 

 

 

 

 

 

 

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