Question

The values of the parameter $\alpha$, for which the function $f\left( x \right)=1+\alpha x$, $\alpha \ne 0$ is the inverse of itself, is
$\text{A}\text{. 1}$
$\text{B}\text{. 2}$
$\text{C}\text{. }-1$
$\text{D}\text{. 0}$

Answer 1

In this question we have been given with the function $f\left( x \right)=1+\alpha x$, $\alpha \ne 0$. We will find the inverse of the function denoted by $f'\left( x \right)$. We will convert the function to its inverse by substituting $f\left( x \right)=y$ and then solving for the value of $x$ and then substituting $x$ wherever $y$is present. we then have to find the value of $\alpha$ when the functions and its inverse are equal to each other therefore, we will equate both of them and solve for the value of $\alpha$.

Complete step by step solution:
We have the function given to us as:
$\Rightarrow f\left( x \right)=1+\alpha x$
Now we will find the inverse of the function. let the inverse of the function be $f'\left( x \right)$
Now let’s consider $f\left( x \right)=y$
On substituting the value of $f\left( x \right)$, we get:
$\Rightarrow 1+\alpha x=y$
Now we will solve for the value of $x$. On transferring $1$ from the left-hand side to the right-hand side, we get:
$\Rightarrow \alpha x=y-1$
On transferring $\alpha$ from the left-hand side to the right-hand side, we get:
$\Rightarrow x=\dfrac{y-1}{\alpha }$
Now we will substitute $y=x$ to get the function in terms of $x$, which is the inverse function therefore, we get:
$\Rightarrow f'\left( x \right)=\dfrac{x-1}{\alpha }$
Now we have to find the value of $\alpha$ such that $f\left( x \right)=f'\left( x \right)$therefore, we get:
$\Rightarrow 1+\alpha x=\dfrac{x-1}{\alpha }$
On transferring $\alpha$ from the right-hand side to the left-hand side, we get:
$\Rightarrow \alpha +{{\alpha }^{2}}x=x-1$
Now on comparing the terms, we get:
$\alpha =-1$
${{\alpha }^{2}}x=x$
Now since $\alpha =-1$ satisfies the equation ${{\alpha }^{2}}x=x$, the value of $\alpha =-1$ is the required solution.

Note: It is to be noted that not all functions have an inverse. The domain and the range of the function should be well defined when finding the inverse. It is to be remembered that the inverse of the inverse function gives the original function itself.