A) \[xyz\]
B) \[x+y+z\]
C) 0
D) 1
Correct Answer: C
Solution :
Given, \[{{a}^{x}}={{b}^{y}}={{c}^{z}}=k\] (Let say) \[\Rightarrow \] \[a={{k}^{1/x}}\] \[b={{k}^{1/y}}\] An \[c={{k}^{1/z}}\] \[\therefore \] \[abc={{k}^{\frac{1}{x}+\frac{1}{y}+\frac{1}{z}}}\] \[\Rightarrow \] \[1={{k}^{\frac{1}{x}+\frac{1}{y}+\frac{1}{z}}}={{k}^{0}}\] (\[\because \] abc = 1, given) On comparing, \[\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0\] \[\Rightarrow \] \[xy+yz+zx=0\]
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