| If \[\frac{x}{2{{x}^{2}}+5x+2}=\frac{1}{6},\]then the value of \[\left( x+\frac{1}{x} \right)\] is |
A) \[2\]
B) \[\frac{1}{2}\]
C) \[-\frac{1}{2}\]
D) \[-\,\,2\]
Correct Answer: B
Solution :
| \[\frac{x}{2{{x}^{2}}+5x+2}=\frac{1}{6}\] |
| \[\Rightarrow \] \[2{{x}^{2}}+5x+2=6x\] |
| \[\Rightarrow \]\[2{{x}^{2}}+2=6x-5x=x\] |
| \[2\,\,({{x}^{2}}+1)={{x}^{2}}\] |
| On dividing by 2, we get |
| \[{{x}^{2}}+1=\frac{x}{2}\] |
| On dividing by x, we get |
| \[x+\frac{1}{x}=\frac{1}{2}\] |
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