A) \[-\,1\]
B) \[0\]
C) \[1\]
D) \[2\]
Correct Answer: A
Solution :
We have, \[f(x)={{x}^{2}}+\alpha {{x}^{2}}+\beta x+\gamma \] |
Roots of \[f(x)\]are eccentricity of parabola and rectangular hyperbola. |
Eccentricity of parabola = 1 |
Eccentricity of rectangular hyperbola \[=\sqrt{2}\] |
\[\therefore \]\[{{x}^{3}}+\alpha {{x}^{2}}+\beta x+\gamma =(x-1)(x-\sqrt{2})(x+\sqrt{2})\] |
\[\Rightarrow \]\[{{x}^{3}}+\alpha {{x}^{2}}+\beta x+\gamma \] \[=(x-1)({{x}^{2}}-2)={{x}^{3}}-{{x}^{2}}-2x+2\] |
\[\therefore \] \[\alpha =-1,\]\[\beta =-2,\]\[\gamma =2\] |
\[\therefore \] \[\alpha +\beta +\gamma =-1-2+2=-1\] |
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