A) \[\frac{5}{12}\]
B) \[\frac{1}{12}\]
C) \[\frac{1}{6}\]
D) \[\frac{25}{6}\]
Correct Answer: A
Solution :
\[\because \] \[x-y=2\] and \[xy=24\] \[\therefore \] \[{{(x+y)}^{2}}={{(x-y)}^{2}}+4xy\] \[={{(2)}^{2}}+4(24)=4+96=100\] \[\therefore \] \[x+y=10\] \[\frac{1}{x}+\frac{1}{y}=\frac{y+x}{xy}=\frac{10}{24}=\frac{5}{12}\]
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