Question

If $f\left( x \right) = \dfrac{{\alpha x}}{{x + 1}},x \ne - 1$ . Then for what value of $\alpha $ is $f\left( {f\left( x \right)} \right) = x$
A. $\sqrt 2 $
B. $ - \sqrt 2 $
C. $1$
D. $ - 1$

Answer 1

Hint: Here we have been given a function in which we have an unknown variable by using an equal two expression we have to find its value. Firstly by using the function in the expression given we will form an equation. Then we will simplify the equation so that we get a quadratic equation and finally we will solve them to get the desired answer.

Complete step-by-step answer:
The function is given as follows:
$f\left( x \right) = \dfrac{{\alpha x}}{{x + 1}},x \ne - 1$…..$\left( 1 \right)$
We have been given the expression as follows:
 $f\left( {f\left( x \right)} \right) = x$…..$\left( 2 \right)$
Substitute value from equation (1) to equation (2) as follows:
$f\left( {\dfrac{{\alpha x}}{{x + 1}}} \right) = x$
Now from equation (1) we can rewrite the above function by replacing $x = \dfrac{{\alpha x}}{{x + 1}}$ as follows:
$\dfrac{{\left( {\alpha \times \dfrac{{\alpha x}}{{x + 1}}} \right)}}{{\left( {\dfrac{{\alpha x}}{{x + 1}} + 1} \right)}} = x$
$ \Rightarrow \dfrac{{\left( {\dfrac{{{\alpha ^2}x}}{{x + 1}}} \right)}}{{\left( {\dfrac{{\alpha x + x + 1}}{{x + 1}}} \right)}} = x$
Simplifying the above equation further we get,
$ \Rightarrow \dfrac{{{\alpha ^2}x}}{{\alpha x + x + 1}} = x$
$ \Rightarrow {\alpha ^2}x = x\left( {\alpha x + x + 1} \right)$
Take all the terms on one side as follows:
$ \Rightarrow {\alpha ^2}x - x\left( {\alpha x + x + 1} \right) = 0$
$ \Rightarrow {\alpha ^2}x - \alpha {x^2} - {x^2} - x = 0$
Take $ - {x^2}$ common from the terms having it and $x$ from the other two terms as follows:
$ \Rightarrow - {x^2}\left( {\alpha + 1} \right) + x\left( {{\alpha ^2} - 1} \right) = 0$
$ \Rightarrow - {x^2}\left( {\alpha + 1} \right) + x\left( {\alpha - 1} \right)\left( {\alpha + 1} \right) = 0$
Take $\alpha + 1$ common from both sides as follows:
$ \Rightarrow \left( {\alpha + 1} \right)\left( { - {x^2} + x\left( {\alpha - 1} \right)} \right) = 0$
So we get the values as,
$\alpha + 1 = 0$ and $ - {x^2} + x\alpha - x = 0$
So we get the value of $\alpha = - 1$ from the first equation.
Hence the correct option is (D).
So, the correct answer is “Option D”.

Note: In this type of question it is necessary to substitute the value from the function given to the expression given for doing the right calculation if that one step is done incorrectly the whole answer will come different. As we only want the value of $\alpha $ so there is no need to solve the second value obtained at the end but if value of $x$ is also asked we can solve it as follows:
Replace $\alpha = - 1$ we get,
$ - {x^2} - x - x = 0$
$ \Rightarrow - {x^2} - 2x = 0$
Taking $ - x$ common we get,
$ \Rightarrow - x\left( {x + 2} \right) = 0$
$ - x = 0$ and $x + 2 = 0$
So we get the value as $x = 0,x = - 2$