Question

Consider $f:R-\{-\frac{4}{3}\}\to R-\left\{\frac{4}{3}\right\}$ given by $f\left(x\right)=\frac{4x+3}{3x+4}$. Show that $f$ is bijective. Find the inverse of $f$ and hence if the value of $f^{-1}\left(0\right)$ is $A$ and $x$ is $B$ such that $f^{-1}\left(x\right)=2$, then find $(A+B)\times 100$.

Answer 1

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

For one-one function, we have

$f\left(x_1\right)=f\left(x_2\right)⟹x_1=x_2$

Thus, consider $f\left(x_1\right)=f\left(x_2\right)$

$⟹\frac{4x_1+3}{3x_1+4}=$ $\frac{4x_2+3}{3x_2+4}$

$⟹12x_1{x}_2+16x_1+9x_2+12=12x_1{x}_2+9x_1+16x_2+12$

$⟹7x_1=7x_2$

$⟹x_1=x_2$

$\therefore f\left(x\right)$ is one-one function.

For onto function:

Let $f\left(x\right)=y$

$⟹\frac{4x+3}{3x+4}=y$

$⟹x=$ $\frac{3-4y}{3y-4}$

Domain of the above function is $R-\left\{\frac{4}{3}\right\}$, which is range of function $f\left(x\right)$.

So, given function $f\left(x\right)$ is onto.

Hence, given function $f\left(x\right)$ is bijective.

Now, let us find the inverse of $f\left(x\right)$

Let the given function $f\left(x\right)$ be $y$

$\Rightarrow f\left(x\right)=\frac{4x+3}{3x+4}=y$

So we get $f^{-1}\left(y\right)=x=\frac{3-4y}{3y-4}$

So the inverse of function $f$ is $f^{-1}\left(x\right)=\frac{3-4x}{3x-4}$

Also, $A=f^{-1}\left(0\right)$

$\Rightarrow A=f^{-1}\left(0\right)=-\frac{3}{4}$

Given that $f^{-1}\left(x\right)=\frac{3-4x}{3x-4}=2$

$\Rightarrow 3-4x=6x-8$

$\Rightarrow B=x=\frac{11}{10}$

$\therefore 100\times (A+B)=35$