A) Always real and distinct
B) Always real and may be equal
C) May be imaginary
D) Always imaginary
Correct Answer: A
Solution :
[a] Let \[f\left( x \right)={{a}_{1}}\left( x-\alpha \right)\,\,\left( x-\beta \right)\] \[g\left( x \right)={{a}_{2}}\left( x-\beta \right)\,\left( x-\gamma \right)\] and \[h\left( x \right)={{a}_{3}}\left( x-\gamma \right)\,\left( x-\alpha \right)\] where \[{{a}_{1}},{{a}_{2}},{{a}_{3}}\]are positive. Let \[f\left( x \right)+g\left( x \right)+h\left( x \right)=F\left( x \right)\] \[\Rightarrow \,\,\,\,\,\,\,\,\,\,\,F(\alpha )={{a}_{2}}(\alpha -\beta )\,(\alpha -\gamma )\] \[F(\beta )={{a}_{3}}(\beta -\gamma )\,(\beta -\alpha )\] \[F(\gamma )={{a}_{1}}(\gamma -\alpha )(\gamma -\beta )\] \[\Rightarrow \,\,\,\,\,\,\,\,F(\alpha )F(\beta )\,F(\gamma )\] \[=-{{a}_{1}}{{a}_{2}}{{a}_{3}}{{(\alpha -\beta )}^{2}}{{(\beta -\gamma )}^{2}}{{(\gamma -\alpha )}^{2}}<0\] Hence, roots of \[F(x)=0\] are real and distinct.
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