question_answer

The roots of\[|x-2{{|}^{2}}+|x-2|-6=0\]are:

A) \[4,\,\,2\]                                           

B) \[0,\,\,4\]

C) \[-1,\,\,3\]                         

D) \[5,\,\,1\]

Correct Answer: B

Solution :

Case I: When\[x<2\] \[\therefore \]  \[{{(x-2)}^{2}}-(x-2)-6=0\] \[\Rightarrow \]               \[{{x}^{2}}+4-4x-x-4=0\] \[\Rightarrow \]               \[{{x}^{2}}-5x=0\] \[\Rightarrow \]               \[x=0,\,\,x=5\] But we have\[x<2\] \[\therefore \]We take\[x=0\] Case II: When\[x\ge 2\] \[\therefore \]  \[{{(x-2)}^{2}}+(x-2)-6=0\] \[\Rightarrow \]               \[{{x}^{2}}+4-4x+x-8=0\] \[\Rightarrow \]               \[{{x}^{2}}-3x-4=0\] \[\Rightarrow \]               \[(x+1)(x-4)=0\] \[\Rightarrow \]               \[x=-1,\,\,4\] \[\Rightarrow \]               \[x=4\]                                 \[(\because \,\,x\ge 2)\] \[\therefore \]Required roots are\[0,\,\,4\]. Alternative Solution: Let          \[y=|x-2|\]                 \[{{y}^{2}}+y-6=0\] \[\Rightarrow \]               \[{{y}^{2}}+3y-2y-6=0\] \[\Rightarrow \]               \[(y+3)(y-2)=0\] \[\Rightarrow \]               \[y=-3,\,\,2\]. It is clear from the figure that line \[y=2\] intersect the curve \[y=\,\,|x-2|\] at two distinct points whose \[x-\]coordinates are\[0,\,\,\,4\].