A) \[(2,3)\]
B) \[(0.27,1)\cup (2,2.73)\]
C) \[(0.5,1.5)\cup (2,2.5)\]
D) \[\left( 2-\sqrt{3},1 \right)\cup \left( 3,2+\sqrt{3} \right)\]
Correct Answer: D
Solution :
(d): \[{{x}^{2}}-4x+2>-1\] or \[{{x}^{2}}-4x+3>0\] Since the coefficient of x2 is positive and the roots of \[{{x}^{2}}-4x+3=0\] are 1 and 3, therefore, \[{{x}^{2}}-4x+3=0\] for \[x<1\] and \[x>3\], i.e., \[x\in \left( -\infty ,1 \right)\cup \left( 3,\infty \right)\]?????.(i) Again, \[{{x}^{2}}-4x+2<1\] or \[{{x}^{2}}-4x+1<0\] Since the coefficient of \[{{x}^{2}}\] is positive and the roots of \[{{x}^{2}}-4x+1=0\] are \[2-\sqrt{3}\] and \[2+\sqrt{3}\], therefore, \[{{x}^{2}}-4x+1<0\] for \[2-\sqrt{3}<x<2+\sqrt{3}\], i.e. \[x\in \left( 2-\sqrt{3},2+\sqrt{3} \right)\]??????.(ii) Combining (i) and (ii), we get \[2-\sqrt{3}<x<1\]or \[3<x<2+\sqrt{3}\], which are the required ranges of the values of x.
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