A) 0.2
B) 4.8
C) 1.02
D) 5.2
E) 25
Correct Answer: B
Solution :
\[x=\frac{\left[ \begin{align} & 729+6(2)(243)+15(4)(81)+20 \\ & \times (8)(27)+15(16)(9)+6(32)(3)+64 \\ \end{align} \right]}{1+4(4)+6(16)+4(64)+256}\] \[=\frac{\left[ \begin{align} & ^{6}{{C}_{0}}{{(3)}^{6}}{{+}^{6}}{{C}_{1}}{{3}^{5}}.2{{+}^{6}}{{C}_{2}}{{3}^{4}}{{.2}^{2}}{{+}^{6}}{{C}_{3}} \\ & \times {{3}^{3}}{{2}^{2}}{{+}^{6}}{{C}_{4}}{{.3}^{2}}{{.2}^{4}}{{+}^{6}}{{C}_{5}}{{3.2}^{5}}{{+}^{6}}{{C}_{6}}{{2}^{6}} \\ \end{align} \right]}{\left[ \begin{align} & ^{4}{{C}_{0}}{{1}^{4}}{{+}^{4}}{{C}_{1}}4{{+}^{4}}{{C}_{2}}{{4}^{2}}{{+}^{4}}{{C}_{3}}{{4}^{3}} \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{{+}^{4}}{{C}_{4}}{{4}^{4}} \\ \end{align} \right]}\] \[\Rightarrow \] \[x=\frac{{{(3+2)}^{6}}}{{{(1+4)}^{4}}}={{5}^{2}}\] \[\therefore \] \[\sqrt{x}=5\] \[\therefore \] \[\sqrt{x}-\frac{1}{\sqrt{x}}=5-\frac{1}{5}\] \[=\frac{24}{5}=4.8\]
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