A) \[\{-1,1\}\]
B) \[\left[ 2,\infty \right)\]
C) \[\left[ 0,2 \right)\]
D) \[\{0,2\}\]
Correct Answer: D
Solution :
| Case1: Let \[0\le x<2\], then \[\left\| x \right\|=x+1\]and the equation becomes |
| \[{{(x+1)}^{2}}+x=(x+1)+{{x}^{2}}\Rightarrow 2x=0\] |
| \[\Rightarrow 2x=0\Rightarrow x=0\] |
| Case 2: Let \[x\ge 2\], then \[\left\| x \right\|=\left| x-4 \right|\]and the equation becomes \[{{\left| x-4 \right|}^{2}}+x=\left| x-4 \right|+{{x}^{2}}\] |
| \[\Rightarrow \]\[{{x}^{2}}-8x+16+x=\left| x-4 \right|+{{x}^{2}}\]\[\Rightarrow \]\[\left| x-4 \right|=16-17x\] |
| \[\therefore \]\[x-4=\pm (16-7x)\], provided \[16-7x\ge 0\] |
| \[\therefore \]\[x=\frac{5}{2}\]or 2, but for\[x=\frac{5}{2},16-7x\ge 0,\] |
| Hence rejected |
| \[\therefore \]\[x=2.\] |
| The solution set is\[\,\{0,2\}\]. |
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