A) \[\{-1,1\}\]
B) \[[2,\infty )\]
C) \[[0,2)\]
D) \[\{0,2\}\]
Correct Answer: D
Solution :
[d] Case 1: Let \[0\le x<2,\]then \[\left\langle x \right\rangle =(x+1)\] and the equation becomes \[{{(x+1)}^{2}}+x=(x+1)+{{x}^{2}}\] \[\Rightarrow 2x=0\Rightarrow x=0\] Case 2: Let \[x\ge 2\] then \[\left\langle x \right\rangle =\left| x-4 \right|\] and the equation becomes \[{{\left| x-4 \right|}^{2}}+x=\left| x-4 \right|+{{x}^{2}}\] \[\Rightarrow {{x}^{2}}-8x+16+x=\left| x-4 \right|+{{x}^{2}}\] \[\Rightarrow \left| x-4 \right|=16-7x\] \[\therefore x-4=\pm (16-7x),\] provided \[16-7x\ge 0\] \[\therefore \,\,\,\,x=\frac{5}{2}\,\,or\,\,2,\] But for \[x=\frac{5}{2},\,\,\,\,16-7x<0,\] hence rejected \[\therefore x=2.\]The solution set is \[\{0,2\}\]
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