The Domain of f(x) =log_{1 /2}.log₄.log₃[(x-lambda)²] (where [ ] denotes the greatest integer function) is (-infinity,2] U [6,infinity) , then thevalue of lambda is:-

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$$\begin{gathered} \mathrm{We} \mathrm{have} \\ \mathrm{f}\left(\mathrm{x}\right)={\log }_{\frac{1}{2}}{\log }_4{\log }_3\left[{\left(\mathrm{x}-\mathrm{\lambda }\right)}^2\right] \\ \mathrm{Quantity} \mathrm{inside} \log \mathrm{must} \mathrm{be} \mathrm{positive}, \mathrm{hence} \\ {\left(\mathrm{x}-\mathrm{\lambda }\right)}^2>0 \\ \Rightarrow \mathrm{x}\in \left(-\infty ,\infty \right)-\left\{\mathrm{\lambda }\right\}....\left(\mathrm{i}\right) \\ {\log }_3\left[{\left(\mathrm{x}-\mathrm{\lambda }\right)}^2\right]>0 \\ \Rightarrow {\left(\mathrm{x}-\mathrm{\lambda }\right)}^2>1 \\ \Rightarrow {\left(\mathrm{x}-\mathrm{\lambda }\right)}^2-1>0 \\ \Rightarrow \left\{\mathrm{x}-\mathrm{\lambda }-1\right\}\left\{\mathrm{x}-\mathrm{\lambda }+1\right\}>0 \\ \Rightarrow \left\{\mathrm{x}-\left(\mathrm{\lambda }+1\right)\right\}\left\{\mathrm{x}-\left(\mathrm{\lambda }-1\right)\right\}>0 \\ \Rightarrow \mathrm{x}\in \left(-\infty ,\mathrm{\lambda }-1\right)\cup \left(\mathrm{\lambda }+1,\infty \right)....\left(\mathrm{ii}\right) \\ {\log }_4{\log }_3\left[{\left(\mathrm{x}-\mathrm{\lambda }\right)}^2\right]>0 \\ {\log }_4\left[{\left(\mathrm{x}-\mathrm{\lambda }\right)}^2\right]>1 \\ \Rightarrow {\left(\mathrm{x}-\mathrm{\lambda }\right)}^2>4 \\ \Rightarrow {\left(\mathrm{x}-\mathrm{\lambda }\right)}^2-2^2>0 \\ \Rightarrow \left\{\mathrm{x}-\mathrm{\lambda }-2\right\}\left\{\mathrm{x}-\mathrm{\lambda }+2\right\}>0 \\ \Rightarrow \left\{\mathrm{x}-\left(\mathrm{\lambda }+2\right)\right\}\left\{\mathrm{x}-\left(\mathrm{\lambda }-2\right)\right\}>0 \\ \Rightarrow \mathrm{x}\in \left(-\infty ,\mathrm{\lambda }-2\right)\cup \left(\mathrm{\lambda }+2,\infty \right)....\left(\mathrm{iii}\right) \\ \mathrm{Taking} \mathrm{intersection} \mathrm{of} \left(\mathrm{i}\right),\left(\mathrm{ii}\right),\left(\mathrm{iii}\right) \mathrm{get} \\ \Rightarrow \mathrm{x}\in \left(-\infty ,\mathrm{\lambda }-2\right)\cup \left(\mathrm{\lambda }+2,\infty \right) \mathrm{when} \mathrm{\lambda }=4 \\ \Rightarrow \mathrm{x}\in \left(-\infty ,4-2\right)\cup \left(4+2,\infty \right) \\ \Rightarrow \mathrm{x}\in \left(-\infty ,2\right)\cup \left(6,\infty \right) \\ \mathrm{Hence} \mathrm{option}\left(4\right) \mathrm{is} \mathrm{correct}. \end{gathered}$$

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