A) \[-\,2\]
B) \[-\frac{a}{2}\]
C) \[\frac{2}{a}\]
D) \[-\frac{2}{a}\]
Correct Answer: D
Solution :
| [d] Given,\[x+\frac{a}{x}=1\]\[\Rightarrow \]\[{{x}^{2}}+a=x\] \[\Rightarrow \] \[{{x}^{2}}-x=-\,a\] Now, \[\frac{{{x}^{2}}+x+a}{{{x}^{3}}-{{x}^{2}}}=\frac{x+1+\frac{a}{x}}{{{x}^{2}}-x}=\frac{2}{-\,a}\] |
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