A) \[14\]
B) \[8\]
C) \[4\]
D) \[16\]
Correct Answer: B
Solution :
We have \[x=\frac{1}{2-\sqrt{3}}\] \[\Rightarrow \]\[x=\frac{1}{2-\sqrt{3}}\times \frac{2+\sqrt{3}}{2+\sqrt{3}}\] \[x-2=\sqrt{3}\] On squaring both sides \[{{x}^{2}}-4x+4-3\](or) \[{{x}^{2}}-4x+1=0\] Now on dividing \[{{x}^{3}}-2{{x}^{2}}-7x+10\,by\,{{x}^{2}}-4x+1,\] we get remainder 8. Hence, by using division algorithm we find \[{{x}^{3}}-2{{x}^{2}}-7x+10=({{x}^{2}}-4x+1)\text{(quotient)}+8\] On putting \[{{x}^{2}}-4x+1=0\]we get the value of \[{{x}^{3}}-2{{x}^{2}}-7x+10\,as\,8.\]
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