question_answer

Direction: (Q. Nos. 87) For \[x\in R,\,f(x)\] is defined as \[f(x)\,=\,\left\{ \begin{matrix}    x+1, & 0\le x\le 2  \\    x-4, & x\ge 2  \\ \end{matrix} \right.\] . For  \[x\in R,\,\,|x|\,=\,\left\{ \begin{matrix}    x, & x\ge 0  \\    -x, & x<0  \\ \end{matrix} \right.\]

For \[x\ge 3,\] the solution set of \[\{f(x)+|x-2|\}\] \[f(x)\le 0\] lies in

A)  \[(4,\,\,\infty )\]                             

B)  \[(-\,\infty ,\,\,3)\]

C)  \[[3,\,\,4]\]                       

D)  None of these

Correct Answer: C

Solution :

 \[\left\{ f(x)+|x-2| \right\}\,f(x)\,\le 0\] \[\Rightarrow \]               \[(x-4+x-2)\,(x-4)\le 0\] \[\Rightarrow \]               \[2(x-3)\,(x-4)\le 0\] \[\Rightarrow \]               \[3\le x\le 4\]