A) one positive and one negative
B) both negative
C) both positive
D) both non-real complex
Correct Answer: C
Solution :
Let\[\alpha ,\,\,\beta \]are the roots of the equation\[(1-a){{x}^{2}}+3ax-1=0\], then\[\alpha +\beta =\frac{3a}{a-1}\],\[\alpha \beta =\frac{1}{a-1}\] As\[a>1\],\[\alpha +\beta >0\]and\[\alpha \beta >0\]and\[D=9{{a}^{2}}+4(1-a)\] \[=9\left( {{a}^{2}}-\frac{4}{9}a+\frac{4}{9} \right)=9\left\{ a\left( a-\frac{4}{9} \right)+\frac{4}{9} \right\}>0\], as\[a>1\] \[\therefore \]The equation has real and positive roots.
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