A) \[80\]
B) \[\frac{7}{2}\]
C) \[\frac{21}{2}\]
D) None of these
Correct Answer: C
Solution :
(c): \[2{{x}^{2}}-7kx+3\times {{k}^{3}}-1=0\] (Since \[{{e}^{3\,\,\log \,\,k}}={{e}^{\log \,\,{{k}^{3}}}}={{k}^{3}}\]) \[\therefore \alpha \beta =40=\frac{3{{k}^{3}}-1}{2}\] \[\Rightarrow 3{{k}^{3}}=81\Rightarrow k=+3\] \[\therefore \alpha +\beta =\frac{7k}{2}=\frac{7\times 3}{2}=\frac{21}{2}\]
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