Question

If both the roots of the equation $x^2-6ax+2-2a+9a^2=0$ exceed $3$, then

• A. $a>\frac{9}{11}$
• B. $a\ge \frac{11}{9}$
• C. $a>\frac{11}{9}$
• D. $a<\frac{11}{9}$

Answer 1

Answer: (C)

$a>\frac{11}{9}$

Both the roots are greater than a given number $k$ if the following three conditions are satisfied :
$D\ge 0$.....(1) ,
$-\frac{b}{2a}>k$.......(2) and
$kf\left(k\right)>0$......(3)
Now, here $k=3$ .
From (1), we have $a\ge 1$
From (2), we have $a>1$
From (3), we have $(9a-11)(a-1)>0$ ie $a<1$or $a>\frac{11}{9}$.
After taking the common region satisfied by all the above inequalities we get our answer as
$a>\frac{11}{9}$.