A) 1
B) 2
C) 3
D) 4
Correct Answer: D
Solution :
Given \[|x{{|}^{2}}-3|x|+2=0\] Here we consider two cases \[viz.\,\,x<0\]and \[x>0\] Case I: \[x<0\] This gives \[{{x}^{2}}+3x+2=0\] Þ \[(x+2)(x+1)=0\,\,\Rightarrow x=-2,-1\] Also \[x=-1,-2\]satisfy \[x<0,\]so\[x=-1\], - 2 is solution in this case. Case II: \[x>0\]. This gives \[{{x}^{2}}-3x+2=0\] Þ \[(x-2)(x-1)=0\,\,\Rightarrow x=2,1\], so\[x=2\], 1 is solution in this case. Hence the number of solutions are four i.e. \[x=-1,\,1,\,2,\,-2\] Aliter: \[|x{{|}^{2}}-3|x|+2=0\] Þ \[(|x|-1)(|x|-2)=0\] Þ \[|x|=1\]and\[|x|=2\]Þ \[x=\pm 1,x=\pm 2\].
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