A) \[2\]
B) \[4\]
C) \[8\]
D) \[6\]
Correct Answer: B
Solution :
\[{{x}^{2}}-x-a=0,\,\,D=1+4a=odd\] \[D\]must be perfect square of some odd integer. Let\[D={{(2\lambda +1)}^{2}}\Rightarrow 1+4a=1+4{{\lambda }^{2}}+4\lambda \] \[\Rightarrow \] \[a=\lambda (\lambda +1),\]as\[a\in [6,\,\,100]\] \[\Rightarrow \] \[a=6,\,\,12,\,\,20,\,\,30,\,\,42,\,\,56,\,\,72,\,\,90\] Thus, \['a'\] can attain 8 different values.You need to login to perform this action.
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