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

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

& \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{aligned}$$ $$\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{aligned} & \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{aligned}$$ $$\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{aligned} & \Rightarrow \left( 2018+\lambda \right)=0 \\\ & \Rightarrow \lambda =-2018 \\\ \end{aligned}$$ Here, we have the value of $$\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)$$ .