A) 4
B) 6
C) 9
D) 3
Correct Answer: A
Solution :
[a] \[\frac{4x-\,3}{x}+\frac{4y-\,3}{y}+\frac{4z-\,3}{z}=0\] \[\Rightarrow \,\,\,\frac{4x}{x}-\,\frac{3}{x}+\frac{4y}{y}-\,\frac{3}{y}+\frac{4z}{z}-\,\frac{3}{z}=0\] \[\Rightarrow \,\,\,\frac{3}{x}+\frac{3}{y}+\frac{3}{z}=4+4+4=12\] \[\Rightarrow \,\,\,\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{12}{3}=4\]
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