A) \[0.2\]
B) \[4.8\]
C) \[1.02\]
D) \[5.2\]
Correct Answer: B
Solution :
Given that \[x=\frac{\left[ \begin{align} & 729+6(2)(243)+15(4)(81)+20 \\ & (8)(27)+15(16)(9)+6(32)(3)+64 \\ \end{align} \right]}{1+4(4)+6(16)+4(64)+256}\] \[=\frac{\begin{align} & ^{6}{{C}_{0}}{{3}^{6}}{{+}^{6}}{{C}_{1}}\cdot 2{{+}^{6}}{{C}_{2}}{{3}^{4}}\cdot {{2}^{2}}{{+}^{6}}{{C}_{3}}{{3}^{3}}\cdot {{2}^{3}} \\ & {{+}^{6}}{{C}_{4}}{{3}^{2}}\cdot {{2}^{4}}{{+}^{6}}{{C}_{5}}3\cdot {{2}^{5}}{{+}^{6}}{{C}_{6}}{{2}^{6}} \\ \end{align}}{^{4}{{C}_{0}}{{1}^{4}}{{+}^{4}}{{C}_{1}}4{{+}^{4}}{{C}_{2}}{{4}^{2}}{{+}^{4}}{{C}_{3}}{{4}^{3}}{{+}^{4}}{{C}_{4}}{{4}^{4}}}\] \[\Rightarrow \] \[x=\frac{{{(3+2)}^{6}}}{{{(1+4)}^{4}}}={{5}^{2}}\] \[\Rightarrow \] \[\sqrt{x}=5\] \[\therefore \] \[\sqrt{x}-\frac{1}{\sqrt{x}}=5-\frac{1}{5}=\frac{24}{5}\] \[=4.8\]
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