A) \[\frac{1}{2}\]
B) \[-\frac{1}{2}\]
C) \[\pm \frac{1}{2}\]
D) \[\pm 1\]
Correct Answer: D
Solution :
[d] Let \[f(x)=a{{x}^{2}}+x+a\] |
Zeroes of the \[f(x)\] are equal |
\[\therefore \,\,\,\,\,\,\,\,\,\,\,\,\,\alpha \cdot \alpha =\frac{a}{a}\Rightarrow {{\alpha }^{2}}=1\] |
\[\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\alpha =\pm 1\] |
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