A) \[-1<a<0\]
B) \[0<a<1\]
C) \[a\ge 0\]
D) \[a>0\]or \[a<-1\]
Correct Answer: D
Solution :
(d): According to the given condition a lies between the roots. Let \[f(x)=2{{x}^{2}}-2\left( 2a+1 \right)x+a\left( a+1 \right)\] For a to lie between the roots, we must have \[D\ge 0\]and f (a) < 0. Now, \[D\ge 0\Rightarrow 4{{\left( 2a+1 \right)}^{2}}-8a\left( a+1 \right)\ge 0\] \[\Rightarrow 8\left[ {{a}^{2}}+a+\frac{1}{2} \right]\ge 0\], which is always true \[\therefore f(a)<0\Rightarrow 2{{a}^{2}}-2a\left( 2a+1 \right)+a\left( a+1 \right)<0\] \[\Rightarrow -{{a}^{2}}-a<0\Rightarrow {{a}^{2}}+a>0\] \[\Rightarrow a(a+1)>0\Rightarrow a>0\] or, \[a<-1\]
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