• # question_answer The values of a for which $2{{x}^{2}}-2\,(2a+1)\,\,x+a(a+1)=0$ may have one root less than a and other root greater than a are given by [UPSEAT 2001] A) $1>a>0$ B) $-1<a<0$ C) $a\ge 0$ D) $a>0\,\,\text{or }a<-1$

The given condition suggest that a lies between the roots. Let $f(x)=2{{x}^{2}}-2(2a+1)x+a(a+1)$ For ?a? to lie between the roots we must have Discriminant $\ge 0$ and$f(a)<0$. Now, Discriminant $\ge 0$ $\Rightarrow \,\,4{{(2a+1)}^{2}}-8a\,(a+1)\ge 0$ $\Rightarrow \,\,8({{a}^{2}}+a+1/2)\ge 0$ which is always true. Also $f(a)<0\Rightarrow 2{{a}^{2}}-2a(2a+1)+a(a+1)<0$ $\Rightarrow -{{a}^{2}}-a<0$$\Rightarrow {{a}^{2}}+a>0\Rightarrow a\,(1+a)>0$ $\Rightarrow \,\,a>0$ or $a<-1$.