A) \[\frac{1}{3}\]
B) \[-\frac{1}{3}\]
C) \[\frac{2}{3}\]
D) \[-\frac{2}{3}\]
Correct Answer: D
Solution :
| [d] Let \[f(x)=k{{x}^{2}}+2x+3k\] |
| If \[\alpha \] and \[\beta \] are the zeroes of \[f(x)\] |
| Then \[\alpha +\beta =-\frac{(2)}{k}\] |
| and \[\alpha \cdot \beta =\frac{3k}{k}=3\] |
| According to question, |
| \[\alpha +\beta =\alpha \beta \] |
| \[\Rightarrow \,\,\,\,\,\,\,\,\,\,\,-\frac{2}{k}=3\,\,\,\,\,\,\,\,\,\,\,\Rightarrow k=-\frac{2}{3}\] |
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