A) 4
B) \[\frac{1}{4}\]
C) \[\frac{-1}{4}\]
D) 2
Correct Answer: B
Solution :
Given, \[\alpha\] and \[\frac{1}{\alpha }\] are the zeroes of quadratic polynomial \[2{{x}^{2}}-x+8k\]. |
Now, product of zeroes, |
\[\alpha \times \frac{1}{\alpha }=\frac{Constant\text{ }term}{Coefficient\,of\,{{x}^{2}}}\] |
\[\Rightarrow 1=\frac{8k}{2}\] |
\[\Rightarrow k=\frac{2}{8}=\frac{1}{4}\] |
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