A) \[3\]
B) \[-3\]
C) \[\frac{1}{3}\]
D) \[-\frac{1}{3}\]
Correct Answer: A
Solution :
| [a] Let \[f(x)=3{{x}^{2}}+8x+k\] |
| According to question, \[\alpha \] and \[\frac{1}{\alpha }\]are the zeroes of the given polynomial \[f(x)\]. |
| \[\therefore \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\alpha \cdot \frac{1}{\alpha }=\frac{k}{3}\,\,\,\Rightarrow \frac{k}{3}=1\] |
| \[\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\,k=3\] |
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