A) 1
B) 2
C) 0
D) -1
Correct Answer: A
Solution :
Given, \[\frac{\log x}{b-c}=\frac{\log y}{c-a}=\frac{\log z}{a-b}=k\](say) ?(i) \[\Rightarrow \]\[x={{e}^{k(b-c)}},y={{e}^{k(c-a)}},z={{e}^{k(a-b)}}\] Now, \[{{x}^{b+c}}.{{y}^{c+a}},{{z}^{a+b}}\] \[={{e}^{k(b-c)(b+c).}}{{e}^{k(c+a)(c-a)}}.{{e}^{k(a+b)(a-b)}}\] \[={{e}^{k({{b}^{2}}-{{c}^{2}})}}.{{e}^{k}}({{c}^{2}}-{{a}^{2}}).{{e}^{k}}({{a}^{2}}-{{b}^{2}})\] \[={{e}^{k({{b}^{2}}-{{c}^{2}}+{{c}^{2}}-{{a}^{2}}+{{a}^{2}}-{{b}^{2}})}}={{e}^{k.0}}=1\]
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