A) \[0\]
B) \[\frac{1}{2}\]
C) \[1\]
D) \[-1\]
Correct Answer: A
Solution :
Let \[p(x)={{x}^{2}}-1\] |
Now, sum of zeroes \[=-\frac{\text{Coefficient of x}}{\text{Coefficient of }{{\text{x}}^{2}}}\] |
\[\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\alpha +\beta =-\frac{0}{1}=0\] |
and product of zeroes \[=\frac{\text{Constant}\,\,\text{term}}{\text{Coefficient of }{{\text{x}}^{2}}}\] |
\[\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\alpha \cdot \beta =\frac{-1}{1}=-1\] |
\[\therefore \,\,\,\,\,\,\,\,\,\frac{1}{\alpha }+\frac{1}{\beta }=\frac{\alpha +\beta }{\alpha \beta }=\frac{0}{-1}=0\] |
So, option [a] is correct. |
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