Column I		Column II
(I)        (II)      (III)      (IV)	The real values of $a$ for which the quadratic equation $2x^2-(a^3+8a-1)x+a^2-4a=0$ possesses roots of opposite signs is given by If the equation $x^2+2(a+1)x+9a-5=0$ has only negative roots, then   The value of$a$ for which the inequality $x^2-2(4a-1)x+15a^2$$-2a-7>0$ is valid for all $x\in R$, is If $x\in R$, then $\frac{x^2+2x+a}{x^2+4x+3a}$ can take all real values, if	P)      Q)        R)       S)	$a\in (\frac{5}{9},1]\cup [6,\infty )$    $0<a<1$        $0<a<4$      $2<a<4$

         

(I) Roots of the given equation will be opposite sign, if product of roots $<0$.
 $\frac{a^2-4a}{2}<0\Rightarrow a^2-4a<0\Rightarrow 0<a<4$
(II) Equation has negative roots if discriminant  $\ge 0$
 $\alpha <0,\beta <0$
ie,  $\alpha +\beta <0andf\left(0\right)>0$
 $\therefore 4{(a+1)}^2-36a+20\ge 0$
 $\Rightarrow a^2-7a+6\ge 0\Rightarrow a\le 1ora\ge 6\left(i\right)$
 $\alpha +\beta <0\Rightarrow a+1>0\Rightarrow a>-1\left(ii\right)$ 
And  $f\left(0\right)>0$
 $9a-5>0\Rightarrow a>\frac{5}{9}\left(iii\right)$
From eq (i), (ii) and (iii), we get  $a\ge 6$
(III) Given,  $x^2-2(4a-1)x+15a^2-2a-7>0$
  discriminant <0 [  coeff. of $x^2>0$]
 $\Rightarrow 4{(4a-1)}^2-4(15a^2-2a-7)<0$
 $\Rightarrow a^2-6a+8<0\Rightarrow 2<a<4$
(IV) Let  $y=\frac{x^2+2x+a}{x^2+4a+3a}$
 $\Rightarrow x^2(y-1)+2(2y-1)x+a(3y-1)=0$
As,  $x\in R,D\ge 0$
 $\therefore 4{(2y-1)}^2-4(y-1)a(3y-1)\ge 0$
 $\Rightarrow (4-3a)y^2-4(1-a)y+1-a\ge 0$
 $\Rightarrow a<\frac{4}{3}anda(a-1)\le 0$
$0\le a\le 1\therefore 0<a<1$  (For a=0 or 1, y cannot all take real values)