Question

If \[f\left( x \right)=\dfrac{2018x-2019}{x+\lambda }\] and \[f\left( f\left( x \right) \right)=x\] , then \[\lambda \] is equal to
(A) 2018
(B) 2019
(C) -2019
(D) -2018

Answer 1

Hint: First of all, find \[f\left( f\left( x \right) \right)\] by using the basic procedure that when \[x\] in \[f\left( x \right)\] is replaced by \[f\left( x \right)\] then, \[f\left( f\left( x \right) \right)\] is obtained. Since \[f\left( f\left( x \right) \right)\] is equal to \[x\] so, make the obtained \[f\left( f\left( x \right) \right)\] equal to \[x\] . Now, use the formula, \[\left( {{a}^{2}}-{{b}^{2}} \right)=\left( a+b \right)\left( a-b \right)\] and simplify it further. Finally, solve for \[\lambda \] .

Complete step by step answer:
According to the question, we are given a function \[f\left( x \right)\] , and the value of \[f\left( f\left( x \right) \right)\] is equal to \[x\] . Using all information, we are asked to find the value of the unknown term \[\lambda \] .
The given function is \[f\left( x \right)=\dfrac{2018x-2019}{x+\lambda }\] ……………………………………..(1)
Also, \[f\left( f\left( x \right) \right)=x\] …………………………………….(2)
In the above equation, we can observe that we require \[f\left( f\left( x \right) \right)\] .
We know the procedure to find \[f\left( f\left( x \right) \right)\] when \[f\left( x \right)\] is given i.e when \[x\] in \[f\left( x \right)\] is replaced by \[f\left( x \right)\] then, \[f\left( f\left( x \right) \right)\] is obtained.
Now, on replacing x by \[f\left( x \right)\] in equation (1), we get
\[f\left( f\left( x \right) \right)=\dfrac{2018f\left( x \right)-2019}{f\left( x \right)+\lambda }\] …………………………………(3)
Using equation (1) and on replacing \[f\left( x \right)\] by \[f\left( x \right)=\dfrac{2018x-2019}{x+\lambda }\] in equation (3), we get
\[\begin{align}
  & \Rightarrow f\left( f\left( x \right) \right)=\dfrac{2018f\left( x \right)-2019}{f\left( x \right)+\lambda } \\
 & \Rightarrow f\left( f\left( x \right) \right)=\dfrac{2018\left( \dfrac{2018x-2019}{x+\lambda } \right)-2019}{\left( \dfrac{2018x-2019}{x+\lambda } \right)+\lambda } \\
 & \Rightarrow f\left( f\left( x \right) \right)=\dfrac{\dfrac{{{\left( 2018 \right)}^{2}}x-2018\times 2019-2019\left( x+\lambda \right)}{\left( x+\lambda \right)}}{\dfrac{\left( 2018x-2019 \right)+\lambda x+{{\lambda }^{2}}}{\left( x+\lambda \right)}} \\
\end{align}\]
\[\Rightarrow f\left( f\left( x \right) \right)=\dfrac{{{\left( 2018 \right)}^{2}}x-2018\times 2019-2019\left( x+\lambda \right)}{\left( 2018x-2019 \right)+\lambda x+{{\lambda }^{2}}}\] ………………………………………(4)
But, from equation (2), we have \[f\left( f\left( x \right) \right)=x\] .
Now, on replacing \[f\left( f\left( x \right) \right)\] by \[x\] in equation (4), we get
\[\begin{align}
  & \Rightarrow x=\dfrac{{{\left( 2018 \right)}^{2}}x-2018\times 2019-2019\left( x+\lambda \right)}{\left( 2018x-2019 \right)+\lambda x+{{\lambda }^{2}}} \\
 & \Rightarrow x\left\{ \left( 2018x-2019 \right)+\lambda x+{{\lambda }^{2}} \right\}={{\left( 2018 \right)}^{2}}x-2018\times 2019-2019\left( x+\lambda \right) \\
 & \Rightarrow 2018{{x}^{2}}-2019x+\lambda {{x}^{2}}+{{\lambda }^{2}}x={{\left( 2018 \right)}^{2}}x-2018\times 2019-2019x-2019\lambda \\
 & \Rightarrow \left( 2018+\lambda \right){{x}^{2}}+{{\lambda }^{2}}x={{\left( 2018 \right)}^{2}}x-2018\times 2019-2019\lambda \\
 & \Rightarrow \left( 2018+\lambda \right){{x}^{2}}+x\left( {{\lambda }^{2}}-{{2018}^{2}} \right)+2018\times 2019+2019\lambda =0 \\
\end{align}\]
\[\Rightarrow \left( 2018+\lambda \right){{x}^{2}}+x\left( {{\lambda }^{2}}-{{2018}^{2}} \right)+2019\left( 2018+\lambda \right)=0\] ………………………………..(5)
We know the formula that \[\left( {{a}^{2}}-{{b}^{2}} \right)=\left( a+b \right)\left( a-b \right)\] …………………………………………..(6)
Now, from equation (5) and equation (6), we get
\[\Rightarrow \left( 2018+\lambda \right){{x}^{2}}+x\left( \lambda -2018 \right)\left( \lambda +2018 \right)+2019\left( 2018+\lambda \right)=0\] ………………………………………….(7)
In the above equation, we can observe that the term \[\left( \lambda +2018 \right)\] can be taken as common from the whole.
Now, on taking the term \[\left( \lambda +2018 \right)\] out as common from the whole, we get
\[\Rightarrow \left( 2018+\lambda \right)\left\{ {{x}^{2}}+x\left( \lambda -2018 \right)+2019 \right\}=0\]
So, either \[\left( 2018+\lambda \right)=0\] or \[\left\{ {{x}^{2}}+x\left( \lambda -2018 \right)+2019 \right\}=0\]
Here, we can ignore the term \[\left\{ {{x}^{2}}+x\left( \lambda -2018 \right)+2019 \right\}=0\] . Since we don’t know any restrictions on the value of x and we are asked to find the value of \[\lambda \] so, it is just too complex to find the value of \[\lambda \] using this equation.
Now, let us proceed with \[\left( 2018+\lambda \right)=0\] .
\[\begin{align}
  & \Rightarrow \left( 2018+\lambda \right)=0 \\
 & \Rightarrow \lambda =-2018 \\
\end{align}\]
Here, we have the value of \[\lambda \] .
Therefore, the value of \[\lambda \] is -2018.

So, the correct answer is “Option D”.

Note: Whenever this type of question appears where \[f\left( x \right)\] is given and there is a requirement of \[f\left( f\left( x \right) \right)\] . Use the basic procedure to find \[f\left( f\left( x \right) \right)\] i.e, to calculate \[f\left( f\left( x \right) \right)\] , just replace x by \[f\left( x \right)\] in the function of \[f\left( x \right)\] and calculate \[f\left( f\left( x \right) \right)\] .