| \[x+\frac{1}{2x}=2,\]then find the value of \[8{{x}^{2}}+\frac{1}{{{x}^{3}}}.\] [SSC (10+2) 2013] |
A) 48
B) 88
C) 40
D) 44
Correct Answer: C
Solution :
| Given, \[x+\frac{1}{2x}=2\] |
| On multiplying by 2 both sides, we get |
| \[2x+\frac{2}{2x}=4\]\[\Rightarrow \]\[2x+\frac{1}{x}=4\] |
| On cubing both sides, we get |
| \[{{\left( 2x+\frac{1}{x} \right)}^{3}}={{4}^{3}}\] |
| \[\Rightarrow \]\[{{(2x)}^{3}}+{{\left( \frac{1}{x} \right)}^{3}}+3\times 2x\times \frac{1}{x}\left( 2x+\frac{1}{x} \right)=64\] |
| \[[\because {{(a+b)}^{3}}={{a}^{3}}+{{b}^{3}}+3ab\,\,(a+b)]\] |
| \[\Rightarrow \]\[8{{x}^{2}}+\frac{1}{{{x}^{3}}}+3\times 2x\times \frac{1}{x}\left( 2x+\frac{1}{x} \right)=64\] |
| \[\Rightarrow \]\[8{{x}^{3}}+\frac{1}{{{x}^{3}}}+6\times 4=64\] |
| \[\Rightarrow \]\[8{{x}^{3}}+\frac{1}{{{x}^{3}}}=64-24=40\] |
You need to login to perform this action.
You will be redirected in
3 sec
