A) \[\frac{2}{5}\]
B) \[\frac{2}{3}\]
C) \[\frac{2}{7}\]
D) \[\frac{3}{2}\]
Correct Answer: B
Solution :
| Let the zeroes of the polynomial |
| \[3{{x}^{2}}-8x+2k+1\] be \[\alpha ,\,\,\beta \] |
| \[\Rightarrow \,\,\,\alpha +\beta =\frac{8}{3}\] |
| We have, \[\alpha =7\beta \] |
| \[\Rightarrow \,\,\,\,\,7\beta +\beta =\frac{8}{3}\] |
| \[\Rightarrow \,\,\,\,\,\,\,\,\beta =\frac{1}{3}\] |
| \[\Rightarrow \,\,\,\,\,\,\,\,\,\alpha =\frac{7}{3}\] and \[\beta =\frac{1}{3}\] |
| \[\alpha \,.\,\beta =\frac{2k+1}{3}\] |
| \[2k=\frac{4}{3}\,\,\Rightarrow \,k=\frac{2}{3}\] |
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