A) 2
B) 3
C) 5
D) 9
Correct Answer: B
Solution :
(b): We have, \[x=\frac{1}{2-\sqrt{3}}=\frac{1}{2-\sqrt{3}}\times \frac{2+\sqrt{3}}{2+\sqrt{3}}=\frac{2-\sqrt{3}}{{{2}^{2}}-{{(\sqrt{3})}^{2}}}\] \[=\frac{2+\sqrt{3}}{4-3}=2+\sqrt{3}\] \[\Rightarrow x-2=\sqrt{3}\] \[\Rightarrow {{\left( x-2 \right)}^{2}}={{\left( \sqrt{3} \right)}^{2}}\] \[{{x}^{2}}-4x+4=3\Rightarrow {{x}^{2}}-4x+1=0\] \[\therefore {{x}^{3}}-2{{x}^{2}}-7x+5\] \[=x\left( {{x}^{2}}-4x+1 \right)+2({{x}^{2}}-4x+1)+3\] \[=x\times 0+2\times 0+3=3\]
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