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